SQA Advanced Higher Chemistry Unit 3 Organic Chemistry

Transcription

1 SCHOLAR Study Guide SQA Advanced Higher Chemistry Unit 3 Organic Chemistry Peter Johnson Heriot-Watt University Brian T McKerchar Balerno High School Arthur A Sandison St Thomas of Aquin s High School Heriot-Watt University Edinburgh EH14 4AS, United Kingdom.

2 First published 2001 by Heriot-Watt University. This edition published in 2009 by Heriot-Watt University SCHOLAR. Copyright 2009 Heriot-Watt University. Members of the SCHOLAR Forum may reproduce this publication in whole or in part for educational purposes within their establishment providing that no profit accrues at any stage, Any other use of the materials is governed by the general copyright statement that follows. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, without written permission from the publisher. Heriot-Watt University accepts no responsibility or liability whatsoever with regard to the information contained in this study guide. Distributed by Heriot-Watt University. SCHOLAR Study Guide Unit 3: Advanced Higher Chemistry 1. Advanced Higher Chemistry ISBN Printed and bound in Great Britain by Graphic and Printing Services, Heriot-Watt University, Edinburgh.

3 Acknowledgements Thanks are due to the members of Heriot-Watt University s SCHOLAR team who planned and created these materials, and to the many colleagues who reviewed the content. We would like to acknowledge the assistance of the education authorities, colleges, teachers and students who contributed to the SCHOLAR programme and who evaluated these materials. Grateful acknowledgement is made for permission to use the following material in the SCHOLAR programme: The Scottish Qualifications Authority for permission to use Past Papers assessments. The Scottish Government for financial support. All brand names, product names, logos and related devices are used for identification purposes only and are trademarks, registered trademarks or service marks of their respective holders.

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5 i Contents 1 Introduction to Organic Chemistry Revision of naming of hydrocarbons The importance of organic chemistry Key concepts throughout the Unit Resources Hydrocarbons and Halogenoalkanes Hydrocarbons Halogenoalkanes Summary Resources End of Topic test Alcohols and Ethers Introduction Classification and nomenclature Physical properties Preparation of alcohols Reactions of alcohols Preparation of ethers Reactions of ethers Summary Resources End of Topic test Aldehydes, Ketones and Carboxylic Acids Introduction Physical properties Reactions of aldehydes and ketones Carboxylic acids Summary Resources End of Topic test Amines Introduction Naming and classification Physical properties Chemical properties Summary

6 ii CONTENTS 5.6 Functional groups: summary Resources End of Topic test Aromatics Introduction Benzene structure Benzene reactions Acidity of phenol and basicity of phenylamine (aniline) Summary Resources End of Topic test Stereoisomers Introduction Stereoisomerism Resources End of Topic test Elemental analysis and Mass spectrometry Introduction Elemental analysis Mass spectrometry High resolution mass spectrometry Example analysis of an unknown compound Summary Resources End of Topic test Infrared and Nuclear magnetic resonance spectroscopy and X-ray Introduction Infrared spectroscopy Nuclear magnetic resonance spectroscopy X-ray crystallography Summary Resources End of Topic test Medicines Introduction Aspirin development How a medicine functions Case studies Summary Resources End of Topic test End of Unit 3 Test (NAB) 209 Glossary 211

7 CONTENTS iii Further questions 216 Hints for activities 220 Answers to questions and activities Introduction to Organic Chemistry Hydrocarbons and Halogenoalkanes Alcohols and Ethers Aldehydes, Ketones and Carboxylic Acids Amines Aromatics Stereoisomers Elemental analysis and Mass spectrometry Infrared and Nuclear magnetic resonance spectroscopy and X-ray Medicines

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9 1 Topic 1 Introduction to Organic Chemistry Contents 1.1 Revision of naming of hydrocarbons The importance of organic chemistry Key concepts throughout the Unit Resources Prerequisite knowledge Before you begin this Unit, you should be able to understand the meanings of the following terms: homologous series; functional group; saturated; unsaturated; isomers; addition reaction; condensation reaction; hydrolysis; physical properties; chemical properties. Look up the glossary near the end of this booklet for the definition of any term about which you are unsure. Learning Objectives After studying this Unit, you should be able to: recognise and write equations for different types of reaction; describe, when appropriate, reaction mechanisms in terms of electron shifts; explain physical properties, such as melting point, boiling point and miscibility with water in terms of intermolecular forces.

10 2 TOPIC 1. INTRODUCTION TO ORGANIC CHEMISTRY 20 min ffl 1.1 Revision of naming of hydrocarbons Learning Objective ffi To revise the rules for the systematic naming of hydrocarbons You should also be familiar with the rules for the nomenclature of simple alkanes and branched alkanes. ffl Nomenclature of alkanes Learning Objective ffi To revise the rules for naming alkanes and practise the application of these rules The website shows an animation which runs through the rules to name an isomer of heptane, followed by a multiple choice test to give practice. This activity starts with a description of the rules for naming branched alkanes using an isomer of heptane as an example, followed by a multiple choice test to give practice. Unbranched alkanes are named by counting the number of carbon atoms in the chain, choosing the correct prefix (Table 1.1) and adding the name ending -ane. fi fl fi fl Table 1.1: Number of carbon atoms No. of C atoms Prefix No. of C atoms Prefix 1 meth- 6 hex- 2 eth- 7 hept- 3 prop- 8 oct- 4 but- 9 non- 5 pent- 10 dec- The naming of branched alkanes follows a set of rules so that each individual alkane has a unique name. This can be illustrated by using an isomer of heptane. Step 1 Identify the longest continuous carbon chain in the molecule (not necessarily a straight chain). This gives the name for the alkane as described above. In this case, the longest chain is 5 carbon atoms. So the basic name is pentane.

11 1.1. REVISION OF NAMING OF HYDROCARBONS 3 Step 2 Number the chain from the end nearest the first branch point. This keeps the numbers as low as possible. Step 3 Identify the branch points on the chain. Step 4 Identify the substituents. These will be alkyl groups - an alkane minus a hydrogen atom. In this case there are two methyl groups. Step 5 Now assemble the name. The substituents are written in front of the basic name, in alphabetical order if there is more than one type. The position of each substituent is shown by writing the number of the appropriate branch point in front of the name of the substituent. This gives the correct name: 2,3-dimethylpentane Note the comma between the numbers and the use of di- to show that there are two methyl groups attached to the chain. Now try the test. Name the following compounds from their structural formulae. Q1: a) 2,4-ethylpentane b) 2,4-methylpentane c) 2,4-dimethylpentane d) 2,4-dimethylheptane Q2: a) 2-ethyl-4-methylpentane

12 4 TOPIC 1. INTRODUCTION TO ORGANIC CHEMISTRY b) 3,5-dimethylhexane c) 4-ethyl-2-methylpentane d) 2,4-dimethylhexane Q3: a) 3-ethyl-4-methylpentane b) 3-ethyl-2-methylpentane c) 3-methyl-2-ethylpentane d) 2-ethyl-2-methylpentane Q4: a) 2-ethyl-2,3-dimethylbutane b) 2,3-dimethyl-3-ethylbutane c) 3,3,4-trimethylpentane d) 2,3,3-trimethylpentane See further questions on page 216. ffl 1.2 The importance of organic chemistry Learning Objective After studying this Topic, you should be able to: identify the type of bond fission occurring in a given reaction; classify molecules or ions as nucleophiles or electrophiles; ffi classify appropriate ions as carbanions or carbocations. Introduction The term organic was first used to describe those substances produced by living organisms. It was believed that such compounds could only be created in the presence of a special vital force found in living things and that they could not be made from substances obtained from non-living sources (inorganic compounds). However, the production of urea, which had previously been found in urine, by heating the inorganic compound, ammonium cyanate, was the beginning of the end for the vital force theory. NH 4 OCN NH 2 CONH 2 ammonium cyanate urea fi fl

13 1.2. THE IMPORTANCE OF ORGANIC CHEMISTRY 5 Despite this, the classification of compounds as organic or inorganic continues to this day. Since then it has become clear that the survival of living organisms depends on the individual properties of the huge variety of compounds of carbon and not on some vital force. Carbon makes up only 0.2% of the earth s crust and yet at least five million compounds have been identified, of which more than 95% are compounds of carbon. Why is the element carbon so special? The importance of carbon Visit the web version of this Topic to find an interactive diagram that shows some of the unique properties of carbon. On paper, try to write down as many reasons as you can to say why carbon is so special (i.e. what carbon can do that other elements cannot). 5 min The ability of carbon to form such a huge variety of compounds is due to important properties of the carbon atom itself. Organic chemistry is the chemistry of life itself and so is fundamental to biology and medicine.

14 6 TOPIC 1. INTRODUCTION TO ORGANIC CHEMISTRY The importance of organic chemistry 5 min On the website you can see another interactive diagram that can be used to reveal important areas where organic chemistry has had an impact. Organic chemistry has made, and continues to make, a huge contribution to modern day society. Potential solutions to many of the world s major problems may lie within the field of organic chemistry. 1.3 Key concepts throughout the Unit The study of such a huge variety of organic compounds is simplified by their classification into smaller groups. Any study of organic chemistry must involve consideration and explanation of the physical and chemical properties of these homologous series. Physical properties such as melting and boiling point and miscibility with water will be explained in terms of the intermolecular forces involved. Throughout the unit, you will encounter and be expected to identify a variety of different types of reaction, some of which will be new to you. These will include: Addition Condensation Hydrolysis Oxidation Reduction Substitution Elimination Acid / Base Possible mechanisms for some of these reactions will be discussed in terms of electron shifts. Bond breaking All chemical reactions involve the breaking and making of bonds. The products and the mechanism of a reaction will be strongly influenced by the way in which the bonds break. In covalent bonding, electrons are generally shared in pairs between two atoms, e.g. in the hydrogen bromide molecule (see Figure 1.1). or a general bond Figure 1.1: Covalent bond During bond breaking (also called bond fission), the electrons are redistributed between the two atoms. There are two ways in which this can occur. In homolytic fission, the two shared electrons separate equally, one going to each atom, as shown in Figure 1.2. Figure 1.2: Homolytic fission

15 1.3. KEY CONCEPTS THROUGHOUT THE UNIT 7 The curly half arrow () is used to represent the movement of a single electron. The single dot ( ) beside each atom represents the unpaired electron that has been retained by each atom from the shared pair. However, the atoms are highly reactive because unpaired electrons tend to attack other species. Highly reactive atoms or groups of atoms containing unpaired electrons are called free radicals. Free radicals are most likely to be formed when the bond being broken is essentially non-polar, i.e. the electrons are more or less equally shared. In heterolytic fission, both of the shared electrons go to only one of the two atoms producing ions, as shown in Figure 1.3. Figure 1.3: Heterolytic fission The full curly arrow (y) is used to represent the movement of a pair of electrons. Heterolytic fission is more likely when a bond is already polar. For example, the carbon to bromine bond in bromomethane is polar and can break heterolytically, the pair of electrons going to the more electronegative bromine atom (see Figure 1.4). Figure 1.4: C-Br bond fission Note that the H 3 C + ion (sometimes written as CH + 3 ) contains a carbon carrying a positive charge. The H 3 C + ion is an example of a carbocation (also called a carbonium ion). If the carbon atom is the more electronegative, heterolytic fission can lead to the formation of ions in which a carbon atom carries a negative charge (see Figure 1.5). Figure 1.5: Carbanion formation These negative ions are called carbanions. In general, both carbocations and carbanions are highly reactive and so very short-lived. Bond fission and carbocations Now try the online tests. These are designed to promote familiarity with the different modes of bond breaking. If you score less than 70% in the first test, you should revise this section and try the second test. Nucleophiles and electrophiles Nucleophile and electrophile are two other important terms which are essential to an 20 min

16 8 TOPIC 1. INTRODUCTION TO ORGANIC CHEMISTRY understanding of the mechanisms of organic reactions. Nucleophile means literally nucleus seeker. Since the nucleus is positively charged, nucleophiles are negatively charged ions, e.g. hydroxide ion, OH -, carbanions. Similarly, electrophile means literally electron seeker. Since electrons are negatively charged, electrophiles are positively charged ions, e.g. the hydrogen ion, H +, carbocations. With polar molecules, the situation is more complicated. A partially negative atom will act as a nucleophile because it is electron-rich, as in Figure 1.6. The nitrogen atom in an ammonia molecule with its lone pair of electrons is electron rich. Figure 1.6: Ammonia A partially positive atom will act as an electrophile because it is electron-deficient, as in Figure 1.7. The hydrogen end of a water molecule will be electron-deficient. Figure 1.7: Water So in a polar molecule, there will be both an electrophilic centre and a nucleophilic centre, as shown in Figure 1.8. Nucleophiles and electrophiles Figure 1.8: Polar molecule 20 min Now try the two online tests. These are designed to promote familiarity with the terms - nucleophile and electrophile. If you score less than 70% in the first test, you should revise this section and try the second test.

17 1.4. RESOURCES 9 ffl Summary of Key Points Learning Objective ffi To summarise the key points in the topic The summary is presented as a cloze test and takes the place of an end of Topic test. Copy and complete the summary using the words in the word bank. Summary - Key Points Organic Chemistry is the study of the compounds of.... The study of Organic Chemistry is essential to the understanding of the chemistry of.... Bond... can occur in two different ways - heterolytic and homolytic fission. Homolytic fission produces Heterolytic fission produces contain a carbon atom carrying a full positive charge.... contain a carbon atom carrying a full negative charge. Nucleophiles are... ions or... centres in molecules, which are attracted to... centres. Electrophiles are... ions or... centres in molecules, which are attracted to... centres. Word Bank fi fl 10 min life electron-deficient free radicals negative positive carbon electron-rich carbocations negative positive ions breaking carbanions 1.4 Resources Higher Still Support, Chemistry, Unit 3: Organic Chemistry, Learning and Teaching Scotland, ISBN Chemistry in Context: Hill and Holman, Nelson, ISBN Organic Chemistry: J. McMurry, Brooks/Cole Publishing, ISBN Chemistry: A. and P. Fullick, Heinemann, ISBN

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19 11 Topic 2 Hydrocarbons and Halogenoalkanes Contents 2.1 Hydrocarbons Bonding in alkanes Reactions of alkanes Bonding in alkenes Synthesis of alkenes Reactions of alkenes Halogenoalkanes Naming halogenoalkanes Reactions of monohalogenoalkanes Summary Resources End of Topic test Prerequisite knowledge Before you begin this Topic, you should be able to: use different ways to represent organic formulae (full structural formula, shortened structural formula, molecular formula); describe covalent and polar covalent bonding (Higher); describe types of bond breaking (Unit 3, Topic 1); recognise nucleophiles and electrophiles (Unit 3, Topic 1); recognise carbocations and carbanions (Unit 3, Topic 1). Learning Objectives After studying this Topic on Hydrocarbons, you should be able to: to describe the bonding in alkanes and alkenes in terms of hybridisation, ff bonds and ß bonds;

20 12 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES to describe the mechanism of the free radical substitution of alkanes by chlorine and bromine; to state two methods used for the preparation of alkenes in the laboratory; to predict the products of the addition reactions of alkenes with hydrogen, hydrogen halides, halogens and water; to explain the mechanisms involved for each of the above reactions. After studying this Topic on Halogenoalkanes, you should be able to: to name halogenoalkanes given the structural formula; to draw structural formulae given the correct name; to classify halogenoalkanes as primary, secondary or tertiary; to explain the mechanism of the S N 1 reaction and the S N 2 reaction; to predict the products of reaction of a monohalogenoalkane given the other reactant and conditions.

21 2.1. HYDROCARBONS Hydrocarbons Hydrocarbons are compounds containing the elements carbon and hydrogen only. Hydrocarbons can be subdivided into smaller subsets including alkanes, alkenes, alkynes and cycloalkanes amongst others (Figure 2.1). Figure 2.1: Hydrocarbon subsets Alkanes are saturated hydrocarbons which means that their molecules contain only single bonds. The simplest alkane is methane, CH 4. The structure of methane can be shown in various ways. Figure 2.2: Figure 2.3: live molecule - is available only on the web Figure 2.4: Methane 3D Figure 2.2 and Figure 2.3 show that there are four single covalent bonds but do not show the correct shape. Figure 2.4 is a two dimensional representation to show that the four bonds are tetrahedrally arranged around the carbon atom. For more complicated molecules, drawings like that in Figure 2.3 will be used but it must always be remembered that these do not show the true shape. Where shape is important, drawings like that in Figure 2.4 will be used, as well as rotating molecules on

22 14 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES the web. A structure for a more complicated alkane, 2,2-dimethylbutane, is shown in Figure 2.5. live molecule - is available only on the web Figure 2.5: Sometimes even the carbon and hydrogen symbols can be missed out. Figure 2.6 also represents 2,2- dimethylbutane. The end of a bond or an angle represents a carbon atom and its hydrogen atoms. ffl Bonding in alkanes Learning Objective ffi Figure 2.6: wire frame dimethylbutane To be able to describe the bonding in alkanes in terms of sp 3 hybridisation and ff bonds Covalent bonds are formed when atoms share electrons in pairs, usually one electron from each atom. This can be achieved by the overlap of half-filled atomic orbitals on each atom. A carbon atom has the electron arrangement shown in Figure 2.7. fi fl Figure 2.7 This shows that carbon atoms have two unpaired electrons and so might be expected to form only two covalent bonds. In fact, carbon atoms usually form four bonds. How can this be? A simple explanation might involve promotion of an electron from the 2s orbital to the empty 2p orbital, producing four unpaired electrons (Figure 2.8) which could then form four bonds with hydrogen. Figure 2.8 The energy required to promote the electron would be more than offset by the formation of two extra covalent bonds. However, the bonds might not be identical since one bond would involve the 2s orbital whereas the others would involve 2p orbitals. Spectroscopic

23 2.1. HYDROCARBONS 15 measurements show that all four bonds in methane are identical. A more satisfactory explanation involves the valence bond theory and hybridisation. The theory assumes that the 2s and three 2p orbitals combine during bonding to form four new identical hybrid orbitals (Figure 2.9). Figure 2.9 The shapes of atomic orbitals are derived from solutions to mathematical equations. By mathematically combining the four atomic orbitals, it is possible to predict the shape and orientation of the hybrid orbitals. The hybrid orbitals are known as sp 3 orbitals because they are formed by combining one s and three p orbitals. Figure 2.10: as orbital Figure 2.11: a p orbital Figure 2.12: asp 3 orbital The four sp 3 hybrid orbitals are arranged tetrahedrally, as would be predicted by simple electron pair repulsion theory (Unit 1, Topic 3). Compare the shape of the sp 3 orbital for carbon (Figure 2.12) with that of the p orbital (Figure 2.11). The hybrid orbital is more strongly directed in one direction. This provides better overlap when forming bonds and so produces stronger bonds. The sp 3 orbital consists of two lobes, one much smaller than the other. For simplicity, the smaller lobe is often omitted from diagrams. Overlap of a half-filled sp 3 hybrid orbital with a half-filled 1s orbital of a hydrogen atom forms a new molecular orbital within which the shared pair of electrons now move under the influence of both nuclei. A sigma(s) bond C H C H a sp 3 a 1s hybrid orbital orbital a molecular orbital Figure 2.13 The new molecular orbital (Figure 2.13 above) lies along the axis joining the two nuclei and is known as a sigma bond. A sigma bond (ff bond) is formed by end-on overlap of atomic orbitals.

24 16 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES Figure 2.14: Molecular orbitals in methane In methane, all four hybrid orbitals are used in forming ff bonds to hydrogen atoms (Figure 2.14). The same type of hybridisation occurs in all other alkanes. Each carbon atom uses the 2s and all three 2p orbitals in forming ff bonds to other carbon atoms and hydrogen atoms, e.g. in ethane. In Figure 2.15, three sigma bonds from carbon to hydrogen are already formed, leaving one half-filled hybrid orbital on each carbon atom. Figure 2.15: Ethane forming Overlap of the two hybrid orbitals forms a carbon to carbon ff bond. Figure 2.16: Ethane - molecular orbitals This model (Figure 2.16) is consistent with the fact that all the bond angles in alkanes are ffi (the tetrahedral bond angle).

25 2.1. HYDROCARBONS 17 ffl Reactions of alkanes Learning Objective To describe the mechanism of the free radical substitution of alkanes by chlorine and bromine ffi Alkanes are fairly unreactive. The only reactions of importance at normal temperatures are those with oxygen and halogens. Reaction with oxygen (combustion) is highly exothermic and so alkanes are widely used as fuels : fi fl The reaction of methane with chlorine does not occur in the dark but occurs explosively on exposure to sunlight to produce chloromethane and hydrogen chloride. A hydrogen atom has been replaced by chlorine on each methane molecule. A substitution reaction has taken place. Use the data booklet to find the bond enthalpies of the C-H bond and the Cl-Cl bond. Q1: C-H bond enthalpy (in kj mol -1 ) Q2: Cl-Cl bond enthalpy (in kj mol -1 ) Q3: Which bond will be easier to break? a) Cl-Cl b) C-H Q4: What type of bond fission is likely to occur? The following activity explains the mechanism of the reaction between methane and chlorine. Reaction between methane and chlorine This is a simulation (only available on the website) that illustrates the mechanism of the reaction between methane and chlorine. The chain reaction mechanism includes the following steps:- i) Initiation Light of the appropriate wavelength causes homolytic fission of the Cl-Cl bond, producing some chlorine atoms. (Figure 2.17) 15 min Figure 2.17: Initiation

26 18 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES ii) Propagation The chlorine atoms are free radicals and so are highly reactive. Each atom will combine with a hydrogen atom from a methane molecule, producing a new free radical. (Figure 2.18) Figure 2.18: Propagation step 1 The methyl radical is also highly reactive and, on colliding with a chlorine molecule, produces another chlorine radical. (Figure 2.19) Figure 2.19: Propagation step 2 The chlorine radical can then react with more methane as in Figure 2.18 and a self-sustaining cycle is set up. This is a free radical chain reaction. iii) Termination The chain reaction ends when two radicals collide and combine (Figure 2.20). Figure 2.20: Termination The concentration of radicals at any time is very small so these termination steps occur very infrequently.

27 2.1. HYDROCARBONS 19 Learning point Methane (and other alkanes) undergo substitution reactions with chlorine and bromine by a free radical chain reaction mechanism. All the processes are very rapid, hence the explosive nature of the reaction. A similar but slower reaction occurs when bromine is used instead of chlorine. With an excess of halogen, multiple substitution can occur, producing mixtures which must be further separated. With more complicated alkanes, containing many more hydrogen atoms, substitution can occur at a variety of different places in the molecule, giving complex mixtures, e.g: Figure 2.21 In Figure 2.21, replacing any one of the lighter shaded hydrogen atoms by a halogen atom would give rise to a different compound, i.e. there are nine different monohalogenoalkanes possible and many more dihalogenoalkanes. Free radical chain reaction Answer the following questions. Hydrogen and chlorine react together to form hydrogen chloride. statements describe aspects of the reaction mechanism. The following four 15 min 1. No reaction occurs in darkness. 2. If light of an appropriate wavelength is supplied, there is a violent reaction. 3. Thousands of HCl molecules are produced for each photon absorbed. 4. The presence of substances with unpaired electrons slows down the reaction. Answer the following questions: Q5: Why is light needed? Q6: What type of step is this? Q7: Write an equation for this step. Q8: Write the number of the statement which gives the best evidence that a chain reaction is involved.

28 20 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES Q9: Explain your answer to the previous question. Q10: Write equations for the propagation steps. Q11: Why should substances with unpaired electrons slow down the reaction? A Free Radical Chain Reaction is often initiated by light and involves an initiation step followed by propagation steps and finished by termination steps. ffl Bonding in alkenes Learning Objective To be able to describe the bonding in alkenes in terms of sp 2 hybridisation and ff bonds and ß bonds ffi Alkenes are unsaturated hydrocarbons whose molecules contain at least one carbon to carbon double bond. How is a double bond formed? Consider the bonding in an ethene molecule. Whatever model is used to describe the formation of the bond, it must be able to explain the observed facts. fi fl 1. The ethene molecule is flat and all bond angles are 120 ffi. 2. The carbon to carbon double bond is shorter than the single bond ( Table 2.1). 3. The C=C bond is intermediate in strength between a single and a triple bond but not twice as strong as a single bond (Table 2.1). 4. There is restricted rotation around the C=C bond but not around the C-C bond. There are two isomers of but-2-ene which differ only in the position of the methyl groups. In one, they are on the same side of the double bond, while in the other they are on opposite sides. (This is covered more fully in Topic 7, Stereoisomerism). Table 2.1: Carbon to carbon bonds C-C C=C C C Bond enthalpy / kj mol Bond length / nm These facts can again be explained by a model involving hybridisation. In this case, each carbon atom uses its 2s orbital and two of its 2p orbitals.

29 2.1. HYDROCARBONS 21 On mathematically mixing these, three identical hybrid orbitals are obtained. These lie in the same plane at 120 ffi to each other and at 90 ffi to the remaining p orbital (Figure 2.22). This is called sp 2 hybridisation and the new orbitals are called sp 2 hybrid orbitals. Figure 2.22: sp 2 hybridisation Two of the hybrid orbitals are used to form ff bonds to hydrogen atoms as with ethane (Figure 2.15). The remaining hybrid orbitals can overlap to form a carbon to carbon ff bond (Figure 2.23). Figure 2.23: C to C sigma bond forming As this bond forms, the unhybridised p orbitals are in the right position to overlap sideways on. This produces a new molecular orbital with lobes above and below the plane of the molecule.

30 22 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES A bond formed by sideways overlap of two parallel atomic orbitals is called a pi bond ( ß bond). The carbon to carbon double bond involves the sharing of four electrons, one pair in a ff bond and the other pair in a ß bond. The two atoms are held more tightly together and so the bond is shorter and stronger than that in ethane. Sideways overlap is less effective than end-on overlap and so the ß bond is weaker than the ff bond (i.e. the C=C bond strength is less than twice the C C bond strength). Bonding in hydrocarbons 15 min This activity contains extension questions on bonding and hybridisation. Answer the following questions. It may help to draw Lewis dot structures for the molecule to identify the bonding and non-bonding pairs of electrons before answering the question. For the next three questions, consider the hydrocarbon ethyne, HC CH. Q12: What is the hybridisation of the carbon atoms required to form a C C bond? a) sp 3 b) sp 2 c) sp Q13: Which of the following statements is true about an ethyne molecule? a) There are three ß bonds and two ff bonds. b) There are two ß bonds and two ff bonds. c) There are two ß bonds and three ff bonds. d) There are three ß bonds and three ff bonds. Q14: What shape will the ethyne molecule be? ffl Synthesis of alkenes Learning Objective ffi To state two methods used for the preparation of alkenes in the laboratory Alkenes, particularly ethene and propene, are extremely important as feedstocks in the synthesis of a huge number of important chemicals. In industry, they are generally fi fl

31 2.1. HYDROCARBONS 23 obtained by the cracking of natural gas products or naphtha. They can be prepared in the laboratory by two main methods, both of which involve elimination reations. In an elimination reaction, a single reactant splits up to form two products, one of which is a small molecule like water. The other product will contain a multiple bond. Method 1 - Dehydration of an alcohol a) Using aluminium oxide as the catalyst. The alcohol vapour is passed over the hot aluminium oxide and the product gas is collected. Water is lost and an alkene is formed (Figure 2.24). Figure 2.24 This is only useful for volatile alcohols, e.g: C 2 H 5 OH! C 2 H 4 ethanol ethene b) Using a strong, non-volatile acid like sulphuric or phosphoric acid. Consider the dehydration of hexan-1-ol. Substance hexan-1-ol hex-1-ene H 2 SO 4 H 3 PO 4 b.pt. ( ffi C) The hexan-1-ol is warmed with the acid to about 80 ffi C. Q15: Which of the substances will boil at this temperature? Q16: From the information in the table, suggest how you could isolate the product from the reactant and catalyst. Method 2 - Elimination of hydrogen halide from monohalogenoalkanes A similar elimination reaction can occur with halogenoalkanes (often called haloalkanes or alkyl halides). In this case, potassium hydroxide in ethanol solution is used as a reactant. A molecule of the appropriate hydrogen halide is eliminated and an alkene is formed.

32 24 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES Figure 2.25 In all the examples so far, there has been only one possible product. However, in most cases, more than one alkene product is possible. Consequently, most reactions produce mixtures of products. Synthesis of alkenes 15 min Answer the following questions. Example Dehydration of butan-2-ol can give rise to two isomeric alkenes by loss of the hydroxyl group and a hydrogen atom from one of the adjacent carbon atoms. 1. Loss of a hydrogen atom from carbon atom number Loss of a hydrogen atom from carbon atom number 3. Q17: How many isomeric alkenes could be obtained by the dehydration of propan-2-ol? Q18: How many isomeric alkenes could be obtained from this compound by elimination of HCl?

33 2.1. HYDROCARBONS 25 Q19: How many isomeric alkenes could be obtained by the dehydration of this alcohol? Q20: How many isomeric alkenes could be obtained by the dehydration of this alcohol? See further questions on page 216. Mixtures of alkenes are normally obtained unless the starting material is a simple or very symmetrical molecule. At this point it would be a good idea to try the Prescribed Practical Activity on Preparation of Cyclohexene. PPA - Preparation of cyclohexene (Unit 3 PPA 1) Consult with your tutor to find out whether the PPA on preparing cyclohexene is to be completed at this stage. ffl Reactions of alkenes Learning Objective To be able to predict the products of the addition reactions of alkenes with hydrogen, hydrogen halides, halogens and water ffi To be able to explain the mechanisms involved for each of the above reactions Alkene molecules contain a C=C bond. Two pairs of electrons are shared, making the region between the two carbon atoms electron-rich. Consequently, alkenes can behave as nucleophiles. The carbon to carbon double bond is able to donate a pair of electrons to an electrophile. fi fl 120 min

34 26 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES Figure 2.26: Forming carbocation intermediate As the electrophile approaches the double bond, the pair of ß electrons move to form a bond between one of the carbon atoms and the electrophile, leaving the other carbon atom electron-deficient. This produces a carbocation intermediate (Figure 2.26). Figure 2.27: Attack of nucleophile This carbocation intermediate is then attacked by a nucleophile to form a saturated molecule (Figure 2.27). The overall reaction is addition. Since the initial attack is by an electrophile, the reaction is called Electrophilic Addition. This is the typical reaction of alkenes. A wide range of substances can be added across the double bond by a similar mechanism. The various products can then be used to synthesise other important organic chemicals. 1. Hydrogen halides Hydrogen chloride, hydrogen bromide and hydrogen iodide can all be added across the double bond to give the appropriate monohalogenoalkane. The mechanism involved in the reaction is just like the one discussed above ( Figure 2.26 and Figure 2.27). Consider the reaction of ethene with hydrogen chloride. The HCl molecule is polar.

35 2.1. HYDROCARBONS 27 Figure 2.28 The partially positive hydrogen atom is an electrophile. As the bond forms between the H atom and the C atom, the H-Cl bond breaks heterolytically. Figure 2.29 The negative chloride ion formed is a nucleophile which then attacks the positive carbon of the carbocation to form the product, chloroethane. Only one product is formed since it does not matter which carbon atom is attacked by the electrophile. What about the reaction of propene? This molecule is not symmetrical. The two carbon atoms of the double bond are not identical. So two different products are possible depending on which carbon is the subject of the electrophilic attack. Attack by HCl at the end carbon atom (carbon number 1) produces 2-chloropropane. Attack by HCl at the middle carbon atom (carbon number 2) produces 1-chloropropane. In fact, nearly all the product is 2-chloropropane. As a result of studying many similar reactions, a general rule, known as Markovnikov s Rule, was produced enabling us to predict the major product of this type of reaction. It can be stated: In the addition of HX to an alkene, the H atom attaches to the carbon atom with fewer alkyl groups attached to it and the X atom attaches to the carbon atom with more alkyl

36 28 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES groups attached to it. Markovnikov s Rule can be explained in terms of the relative stabilities of the carbocations involved (Figure 2.30). Figure 2.30: Relative stabilities of carbocations With propene, the two possible carbocations are: The H atom has been attached to carbon number 1. The H atom has been attached to carbon number 2. The major product is produced from the more stable carbocation intermediate, i.e. the one formed by attaching the H atom to the carbon with fewer alkyl groups attached to it. 2. Acid-catalysed addition of water Water can also be added to a C=C bond by a very similar mechanism to that described in the previous section. As before, using ethene as an example, the first step involves transfer of an H + ion from the acid catalyst to one of the carbon atoms of the double bond, forming the carbocation intermediate (in a similar way to that shown in Figure 2.28). The second step (Figure 2.31) involves nucleophilic attack by water on the electrondeficient carbon to form intemediate 2 shown in Figure Figure 2.31: Ethanol formation - stage 2 Intermediate 2 (Figure 2.31) then loses an H + ion to form the product, ethanol.

37 2.1. HYDROCARBONS 29 This is a general method for the conversion of alkenes into alcohols. Addition of water (hydration) to asymmetric alkenes can again give rise to mixtures of isomeric alcohols. Once more, Markovnikov s Rule can be used to predict the major product because this reaction involves a similar carbocation intermediate. 3. Addition reaction with halogens Electrophilic addition of halogens, like bromine and chlorine, also readily takes place, although there is strong evidence to suggest that a slightly different mechanism occurs. Consider the reaction of ethene with bromine. The bromine molecule is non-polar but, as the molecules approach, the electrons in the C=C bond repel those in the Br 2 molecule causing it to become polarised. Figure 2.32 As the molecules get closer, the Br-Br bond gradually breaks at the same time as the ß electrons of the double bond form a bond to the electrophilic bromine atom. There is strong evidence that the intermediate is not a carbocation but a cyclic bromonium ion (Figure 2.32). This ion is more stable because the positive charge is spread (delocalised) over three atoms. Figure 2.33 The bromide ion formed in step 1 (Figure 2.32 ), then attacks one of the carbon atoms in the ring that opens to form the product, 1,2-dibromoethane, as shown in Figure The addition of chlorine to alkenes is believed to occur by a similar mechanism.

38 30 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES 4. Catalytic addition of hydrogen A heterogeneous catalyst is used, usually a transition metal such as platinum, palladium or nickel. Adsorption of hydrogen onto the surface of the catalyst breaks up the molecules into atoms. The alkene also bonds to the surface using the electrons in the ß bond. Any hydrogen atoms which are close enough can then bond to the adsorbed alkene. The alkane which is formed has no ß electrons and cannot bond to the surface. As it departs, it leaves the surface free to accept another alkene molecule. Addition reactions 15 min Visit the website to complete the drag and drop exercise designed to give practice at predicting products and understanding mechanisms involved. ffl 2.2 Halogenoalkanes Learning Objective To name halogenoalkanes given the structural formula. To draw structural formulae given the correct name. To classify halogenoalkanes as primary, secondary or tertiary. To explain the mechanism of the S N 1 reaction and the S N 2 reaction. To predict the products of reaction of a monohalogenoalkane given the other reactant and conditions. ffi Halogenoalkanes are saturated organic compounds derived from alkanes by substituting one or more hydrogen atoms by halogen atoms. They are frequently referred to as alkyl halides or haloalkanes. fi fl

39 2.2. HALOGENOALKANES 31 Organic compounds containing halogens have a wide importance. Their uses include industrial solvents, anaesthetics, refrigerants and pesticides amongst many others. They also play an equally important role in the synthesis of other important organic chemicals Naming halogenoalkanes Halogenoalkanes are named according to the IUPAC rules (International Union of Pure and Applied Chemistry) which is a worldwide agreed system for naming chemicals. In Topic 1 of this Unit, you were reminded of the rules as they applied to alkanes. These rules are easily extended to cover halogenoalkanes. The presence of halogen substituents is shown by the appropriate prefix: prefix fluoro- chloro- bromo- iodo- Halogen fluorine chlorine bromine iodine Examples 1. Using the rules to name a compound Name the following compound: An animation on the web site shows the steps involved in naming this compound. Following these steps gives 2,3-dichloro-3-methylpentane. Note that the substituents are listed in alphabetical order. 2. Working out the structure from the name Draw the full structural formula for the following compound: 2-bromo-1-chloro-2-methylbutane

40 32 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES Identify the parent chain and number it. In this case - butane Add the substituents to the correct carbon atoms - Br to carbon 2, Cl to carbon 1 and the methyl to carbon Finally, add hydrogen atoms to all the remaining bonds Answer the following questions. In the first, you will be given a structural formula to name. In the second, you will be given the name and asked to draw the full structural formula. Further sets of questions can be used online for practice. Q21: Name this compound. Q22: Draw the full structural formula for 2-chloro-2,3-dimethylbutane. Monohalogenoalkanes can be classified as primary, secondary or tertiary depending on the environment of the halogen (Table 2.2, where X represents any halogen). Table 2.2 The halogen is attached to a carbon at the end of a chain. (This carbon has one alkyl group attached to it) The halogen is attached to a carbon in the middle of a chain. (This carbon has two alkyl groups attached to it) The halogen is attached to a carbon at a branch point in a chain. (This carbon has three alkyl groups attached to it)

41 2.2. HALOGENOALKANES 33 In each of the following questions, classify the halogenoalkane as primary, secondary or tertiary. Q23: What class of halogenoalkane is this? a) Primary b) Secondary c) Tertiary Q24: What class of halogenoalkane is this? a) Primary b) Secondary c) Tertiary Q25: What class of halogenoalkane is this? a) Primary b) Secondary c) Tertiary Q26: What class of halogenoalkane is this? a) Primary b) Secondary c) Tertiary

42 34 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES ffl Reactions of monohalogenoalkanes Learning Objective fi be able to explain the mechanism of the S N 1 reaction and the S N 2 reaction ffi be able to predict the products of reaction of a monohalogenoalkane given the other reactant and conditions. Halogenoalkanes undergo two main types of reaction: fl 1. Elimination reactions in which a hydrogen halide molecule is eliminated by reaction with a strong base (this reaction was discussed earlier as a method for synthesising alkenes, Figure 2.25). 2. Substitution reactions in which the halogen atom is replaced by a nucleophile. A nucleophile by definition is attracted to a positive centre and will be capable of accepting an H + ion. This means that good nucleophiles tend to be good bases. Eliminations reactions are favoured by the presence of a strong base. Consequently, elimination reactions and substitution reactions can occur simultaneously and compete with one another, frequently providing mixtures of products Elimination reactions In elimination reactions, a strong base, such as potassium hydroxide, is used in a solvent of relatively low polarity such as ethanol. An OH - ion attacks one of the hydrogen atoms of a carbon atom adjacent to the carbon atom carrying the halogen substituent. As the new O-H bond forms, the C-X bond breaks and a ß bond forms between the two carbon atoms. In effect, this elimination reaction is the opposite of the addition reaction in which HX is added to an alkene. As stated previously, if the halogenoalkane is not symmetrical, a mixture of products forms. Products of elimination reactions 10 min This is a problem solving activity that includes questions to give practice at predicting the products of elimination reactions of halogenoalkanes. For each of the following compounds, predict the product(s) that could be obtained by the elimination of a hydrogen halide molecule. You will need to draw the structural

43 2.2. HALOGENOALKANES 35 formula and name the products on paper before displaying the answer. Q27: What elimination product(s) will be produced from this compound? Q28: What elimination product(s) will be produced from this compound? Q29: What elimination product(s) will be produced from this compound? Q30: What elimination product(s) will be produced from this compound? See further questions on page Mechanisms of substitution reactions The other main type of reaction of monohalogenoalkanes involves substitution of the halogen atom by another atom or group. Consider this diagram of a bromoalkane: Q31: What type of bond is the C-Br bond? a) pure covalent

44 36 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES b) polar covalent with the carbon partially positive c) polar covalent with the carbon partially negative Q32: From your answer to the previous question, how will the carbon atom behave? a) as an electrophile b) as a nucleophile c) as a free radical Q33: What type of reactive species is liable to attack the carbon atom a) an electrophile b) a nucleophile c) a free radical Q34: How is the C-Br bond liable to break? a) homolytically b) heterolytically Substitution reactions can occur by either of two different mechanisms. A. S N 1 reaction In this mechanism, there are two steps. In the first, the C-Br bond breaks heterolytically to form a Br - ion and a carbocation intermediate. This step is slow. In the second step, a nucleophile (Nu - ) rapidly attacks the positive carbon atom to form the product. The overall rate of the reaction depends on the slow step (the rate determining step) which involves only one molecule. The reaction is said to have 1st order kinetics (see

45 2.2. HALOGENOALKANES 37 Advanced Higher, Unit 2, Topic 9). Consequently, this reaction mechanism is described as S N 1. B. S N 2 reaction In this mechanism, the nucleophile approaches the carbon atom from the opposite side to the halogen and a bond begins to form between the two atoms. At the same time, the bond between the carbon atom and the halogen begins to break. So there is only one step in this reaction and no carbocation intermediate. Instead, a transition state forms in which the new bond is half-formed and the old bond is halfbroken. The transition state can break down either to reform the reactants or to form the products. The overall rate of the reaction depends on the concentrations of both reactants. The reaction is said to show 2nd order kinetics (see Advanced Higher, Unit 2, Topic 9). So this reaction is described as S N 2.

46 38 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES Products of substitution reactions Using a variety of nucleophiles, a wide range of useful products can be obtained. Alcohols The reaction is carried out using an aqueous solution of an alkali such as sodium or potassium hydroxide. This provides a convenient route to specific alcohols (see Topic 3). Note that while the use of KOH in ethanol favours elimination ("E" forethanol - "E" for Elimination), the use of KOH in aqueous solution favours substitution. Ethers Alkoxides (containing the R-O - ion) are formed by the reaction of reactive metals with the appropriate dry alcohol (this reaction will be considered in Topic 3), e.g. sodium and ethanol. The nucleophilic ethoxide ion can react with a halogenoalkane to form an ether (also

47 2.2. HALOGENOALKANES 39 considered in more detail in Topic 3) Nitriles The use of potassium cyanide in ethanol provides a very useful route to compounds containing the nitrile functional group, -C N. This introduces another carbon atom and so increases the chain length by one. Note the change in name from chloroethane to propanenitrile. Nitriles can be readily hydrolysed to form carboxylic acids (see Topic 4). Amines In the first step of this reaction, the ammonia molecule acts as a nucleophile, using the lone pair on the nitrogen atom, and forms the positive methylammonium ion. Q35: What type of covalent bond is formed between the nitrogen atom and the carbon

48 40 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES atom? In the second step, an ammonia molecule (usually present in excess) acts as a base by removing an H + ion to form the product, methylamine - an example of an amine. Amines will be considered in more detail in Topic 5. Products of substitution reactions 15 min Questions to give practice in predicting the products of reactions of halogenoalkanes. All the questions in this activity refer to Figure Figure 2.34: Reaction Scheme Q36: What type of compound is compound 1? Q37: Write the name of reagent X. Q38: Draw the structural formula for the compound in box 2. Q39: What type of compound is compound 2? See further questions on page 217.

49 2.3. SUMMARY Summary The concept of hybridisation of atomic orbitals provides a useful model for bonding in organic compounds - sp 3 hybridisation for the tetrahedral arrangement of the bonds around a saturated carbon atom and sp 2 hybridisation for the formation of a C=C bond. The reaction between alkanes and bromine takes place by a free radical chain reaction, involving initiation, propagation and termination steps. Alkenes can be prepared either by dehydration of suitable alcohols or dehydrohalogenation of suitable monohalogenoalkanes using KOH in ethanol. Addition reactions of alkenes provide an important route to various types of organic molecule, including alkanes, mono- and dihalogenoalkanes, alcohols. The mechanisms of these reactions can be explained. Halogenoalkanes can be named systematically using IUPAC rules and can be classfied as primary, secondary or tertiary. Halogenoalkanes react by either elimination reactions or substitution reactions. Substitution reactions occur by eithers N 1 or S N 2 mechanisms and provide a synthetic route to many other types of organic compound, including alcohols, ethers, nitriles and amines. 2.4 Resources Higher Still Support: Advanced Higher Chemistry - Unit 3: Organic Chemistry, Learning and Teaching Scotland, ISBN Chemistry in Context: Hill and Holman, Nelson, ISBN Organic Chemistry: J. McMurry, Brooks/Cole Publishing, ISBN Chemistry: A. and P. Fullick, Heinemann, ISBN Website for representing molecules in different ways, for hybridisation and reaction mechanisms: End of Topic test An online assessment is provided to help you review this topic.

50 42 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES

51 43 Topic 3 Alcohols and Ethers Contents 3.1 Introduction Classification and nomenclature Physical properties Preparation of alcohols Reactions of alcohols Reaction with reactive metals Dehydration to form alkenes Forming esters Preparation of ethers Reactions of ethers Summary Resources End of Topic test Prerequisite knowledge Before you begin this Topic, you should be able to: use systematic nomenclature to name alkanols (Higher, Unit 2); classify alcohols as primary, secondary or tertiary (Higher, Unit 2); describe the nature and relative strengths of intermolecular forces (Higher, Unit 1); identify the different types of bond breaking (Topics 3.1 and 3.2); identify carbanions, carbocations, electrophiles and nucleophiles in reaction mechanisms (Topics 3.1 and 3.2); describe how alcohols and carboxylic acids can react reversibly to form esters. Learning Objectives After studying this Topic, you should be able to: use systematic nomenclature to name ethers;

52 44 TOPIC 3. ALCOHOLS AND ETHERS explain some physical properties of alcohols and ethers (such as solubility in water and boiling point) in terms of intermolecular forces; describe methods used to prepare alcohols and ethers; describe some reactions of alcohols and ethers.

53 3.1. INTRODUCTION Introduction The intoxicating effect of ethanol (an alcohol) has been known since ancient times. The anaesthetic properties of diethyl ether were discovered much more recently (in the 1840s). Both ethanol and diethyl ether are simple molecules containing carbon, hydrogen and oxygen atoms but arranged in different ways. Answer the following questions by drawing structural formulae on paper before revealing the answers. Q1: Draw full structural formulae for the two isomers of formula C 2 H 6 O and for the three isomers of formula C 3 H 8 O. Q2: Study the structures carefully and divide the five compounds into two groups according to similarities in structure. Q3: Look at the group of three compounds. What structural feature do they have in common? These three substances belong to the homologous series known as the alkanols which are a subset of the larger family of alcohols. A few examples of alcohols and their uses are shown in Figure 3.1. Figure 3.1: Some important alcohols Q4: Now look at the structure of the remaining two compounds. What structural feature do they have in common?

54 46 TOPIC 3. ALCOHOLS AND ETHERS Such compounds are known as ethers. Some examples of ethers and their uses are shown in Figure 3.2. Figure 3.2: Some important ethers These photographs show models of the most common alcohol and a simple ether. ethanol dimethyl ether ffl 3.2 Classification and nomenclature Learning Objective ffi To classify and name alcohols and ethers Alcohols At Higher level, the systematic naming of alkanols and their classification as primary, secondary or tertiary was introduced. The web version of this Topic has two animations which you can use for revision. Then answer the following questions. fi fl

55 3.2. CLASSIFICATION AND NOMENCLATURE 47 Q5: What class of alcohol is this? a) primary b) secondary c) tertiary Q6: What class of alcohol is this? a) primary b) secondary c) tertiary Q7: What class of alcohol is cholesterol (Figure 3.1)? a) primary b) secondary c) tertiary Q8: Draw the structural formulae for the two primary alcohols of formula, C 4 H 10 O. Give the correct name for both your structures. See further questions on page 217. Ethers Ethers are compounds containing the group R -O-R". ( R 1 and R 2 are often used instead of R -O-R".) Ethers can be classified as symmetrical - if R and R" are identical unsymmetrical - if R and R" are different. Ethers are named in a different way. In fact, there are two possible ways: The alkyl groups R and R" are named according to the number of carbon atoms and these names are written in front of the word ether. If the ether is symmetrical, the name is of the form - dialkyl ether. This method of naming is only useful for simple ethers. More complex ethers are considered as substituted alkanes.

56 48 TOPIC 3. ALCOHOLS AND ETHERS Consider the ether opposite: An animation on the website shows in detail how to name this compound. The smaller alkyl group and the oxygen atom are considered as a substituent. The other alkyl group is treated as a hydrocarbon and named according to the rules covered in Higher and summarised in Topic 1. Following this procedure gives: 2-methoxy-2-methylpropane. Note that the substituents are given in alphabetical order. Now, try the following questions. Q9: What is the alternative name for diethyl ether (shown in Figure 3.2)? Q10: What is the correct name for the ether shown opposite? Q11: What is the alternative name of ethyl methyl ether? Q12: What is the correct name for the ether shown opposite?

57 3.3. PHYSICAL PROPERTIES 49 ffl 3.3 Physical properties Learning Objective To explain some physical properties of alcohols and ethers (such as solubility in water and boiling point) in terms of intermolecular forces ffi The physical properties of organic compounds such as melting point, boiling point and solubility in water are dependent on the intermolecular forces involved. These forces are electrostatic in nature and involve attractions between molecules which contain dipoles. A dipole arises when there is an uneven distribution of charge in a molecule, such that one part has a partial positive charge and another part has an equal but opposite negative charge. Copy and complete the following table using the word bank. fi fl Word bank: Force Origin Example van der Waals Polar-polar attractions Hydrogen bonding between water molecules attraction between permanent dipoles between H-Cl molecules attractions involving HF, HO and HN bonds attraction between temporary dipoles between CH 4 molecules Q13: Which of the following shows the intermolecular forces in the correct order of strength? a) hydrogen bonding > polar-polar attraction > van der Waals forces b) van der Waals forces > polar-polar attraction > hydrogen bonding c) polar-polar attraction > van der Waals forces > hydrogen bonding d) hydrogen bonding > van der Waals forces > polar-polar attraction Boiling points of alcohols Here is a problem-solving exercise based on a graph of the boiling points of some organic compounds. The following graph (Figure 3.3) shows the boiling points for the first four members of the alkane (R-H), chloroalkane (R-Cl) and alkanol (R-OH) series. Study the graph carefully and answer the questions that follow on paper before revealing the answers. 15 min

58 50 TOPIC 3. ALCOHOLS AND ETHERS First consider the alkanes. Figure 3.3 Q14: What type of intermolecular force is involved when alkanes boil? a) van der Waals forces b) polar-polar attractions c) hydrogen bonding Q15: What happens to the boiling point as the number of carbon atoms increases? a) it decreases b) it increases Q16: Explain this trend. Now consider the chloroalkanes. Compare chloromethane (bp 249 K) with propane (bp 231 K). Q17: Why compare chloromethane with propane rather than methane? Q18: Explain why chloromethane has a higher boiling point than propane. Now consider the alkanols. Figure 3.4: Q19: For fairness, which alkane should be compared with ethanol (Figure 3.4)? Q20: For fairness, which chloroalkane should be compared with butanol? Q21: Using Figure 3.3, write a general statement about the boiling points of alkanols compared with alkanes and chloroalkanes.

59 3.3. PHYSICAL PROPERTIES 51 Q22: Which is the most important type of intermolecular force being broken when alkanols are boiled (see Figure 3.4)? a) van der Waals forces b) polar-polar attractions c) hydrogen bonding See further questions on page 217. Alcohols exhibit hydrogen bonding and as a result have higher boiling points than other organic compounds of comparable relative formula mass. Boiling points of ethers Visit the web version of this Topic to find a drag and drop exercise as well as the following questions to investigate the boiling points of ethers. 10 min Figure 3.5 Figure 3.6 By considering the structures of the compounds in the table below and the probable intermolecular forces, answer the questions which follow. Ether bp / ffi C Alcohol bp / ffi C Alkane bp / ffi C CH 3 OCH 3-25 C 2 H 5 OH 78 C 3 H 8-45 CH 3 OC 2 H 5 11 C 3 H 7 OH 97 C 4 H 10-1 C 2 H 5 OC 2 H 5 35 C 4 H 9 OH 117 C 5 H Q23: Which family of compounds has the highest boiling points? a) ethers b) alcohols c) alkanes Q24: In which family will the intermolecular forces be strongest? a) ethers b) alcohols c) alkanes Q25: Consider the bonding in the ether functional group ( Figure 3.6) and compare with the structure of an alkane. Which of the following statements is false? a) Ethers are more polar b) Intermolecular forces will be stronger between ether molecules

60 52 TOPIC 3. ALCOHOLS AND ETHERS c) Alkanes will have higher boiling points d) Intermolecular forces will be weaker between alkane molecules. Q26: Consider the bonding in the ether functional group ( Figure 3.6) and the alcohol functional group (Figure 3.5). Which of the following statements is true? a) Hydrogen bonding occurs between ether molecules b) Ether molecules will be more polar than alcohol molecules c) Alcohols will tend to have higher boiling points d) Intermolecular forces will be stronger between ether molecules. See further questions on page 217. Due to the lack of hydrogen bonding, ethers have lower boiling points than the corresponding isomeric alkanols. Solubility in water Table 3.1 bp = -25 ffi C bp =78 ffi C bp = 100 ffi C From the information in Table 3.1, answer the following questions. Q27: Is hydrogen bonding possible between one water molecule and another? a) yes b) no Q28: Is hydrogen bonding possible between one ethanol molecule and another? a) yes b) no Q29: Is hydrogen bonding possible between one methoxymethane molecule and another? a) yes b) no Q30: Why is methoxymethane a gas at room temperature while the other two are liquids? See further questions on page 217. Table 3.2 shows solubility data in grams per 100 ml of water (g / 100 ml) for some ethers and alcohols.

61 3.4. PREPARATION OF ALCOHOLS 53 Ether Table 3.2: Solubility in water of ethers and alcohols Solubility (g / 100 ml) Alcohol Solubility (g / 100 ml) methoxymethane >10 ethanol >10 ethoxyethane 6.9 butan-1-ol ethoxybutane 0.21 hexan-1-ol butoxybutane < 0.1 octanol <0.1 Q31: Compare the solubilities of ethers in water with alcohols in water. Which of the following statements is true about the solubility of isomeric ethers and alcohols? a) The ether is more soluble. b) The alcohol is more soluble. c) The solubilities are roughly the same. Q32: Compare the solubilities of ethers in water with alcohols in water. What is the trend in solubility in both families as the chain length increases? In summary, ethers and alcohols of low molecular mass are soluble in water (miscible with water) due to their ability to form hydrogen bonds with water molecules. As the chain length increases, the solubility decreases. ffl 3.4 Preparation of alcohols Learning Objective ffi To describe methods used to prepare alcohols from alkenes or from halogenoalkanes For thousands of years, ethanol has been produced by fermentation. Nowadays, this process is still used on a much larger scale to produce alcoholic drinks, see Figure 3.7, and in some places such as Brazil, to produce fuel. fi fl

62 54 TOPIC 3. ALCOHOLS AND ETHERS Figure 3.7: Copper stills in the Benromach distillery and a wee dram There are two general methods used to prepare alcohols, both of which were described in Topic 3.2: Hydration of alkenes Nucleophilic substitution using halogenoalkanes Hydration of alkenes 10 min On the website, you can see a simulation of the mechanism for the addition of water to a double bond. Study the diagrams of the mechanism for addition of water to an alkene and then answer the questions which follow.

63 3.4. PREPARATION OF ALCOHOLS 55 Table 3.3: Hydration mechanism Q33: What role does the hydrogen atom of the acid catalyst play in the first stage of the reaction? a) free radical b) nucleophile c) electrophile Q34: Conversely, what role does the alkene play in the reaction? Q35: What type of carbocation intermediate is shown in Table 3.3? Q36: If the attack had taken place on the other carbon atom of the double bond, what type of carbocation would have formed? See further questions on page 218. Alcohols can be prepared from alkenes by the acid-catalysed addition of water. If the alkene is unsymmetrical, a mixture of products is obtained which requires further separation. Preparation from halogenoalkanes A series of questions to revise and reinforce the nucleophilic substitution of halogen atoms by a hydroxyl group. Study the diagrams and answer the questions which follow. 10 min

64 56 TOPIC 3. ALCOHOLS AND ETHERS Figure 3.8: Reaction mechanism 1 Figure 3.9: Reaction mechanism 2 Q37: How is the mechanism in Figure 3.8 normally described? Q38: How is the mechanism in Figure 3.9 normally described? Q39: Explain why halogenoalkanes are subject to nucleophilic attack. Q40: Write a structural formula and name the product formed when 2-bromopropane is reacted with aqueous KOH. See further questions on page 218. Alcohols can be prepared from halogenoalkanes by nucleophilic substitution using aqueous alkali. Both the alkene hydration method and the halogenoalkane substitution method can be

65 3.5. REACTIONS OF ALCOHOLS 57 used in the laboratory. Preparation from halogenoalkanes has the advantage in that generally only one product is formed, although care must be taken to choose conditions such that elimination does not occur (see Topic 3.2). In industry, alcohols (except for methanol) tend to be manufactured on a large scale by the acid-catalysed hydration of alkenes. Q41: Explain why methanol cannot be prepared by this method. ffl 3.5 Reactions of alcohols Learning Objective ffi To describe some reactions of alcohols Alcohols are amongst the most useful of organic compounds. They can be prepared from, and converted into, a wide variety of other compounds, some of which are shown in Figure fi fl ffl Reaction with reactive metals Learning Objective ffi Figure 3.10: Reactions of alcohols To describe the reaction of alcohols with sodium. Alcohols react with metals like sodium or potassium to produce hydrogen gas. Water also reacts with these metals to produce hydrogen gas. fi fl ethanol water There is an obvious similarity in structure between ethanol and water and so the similarity in reaction is not surprising.

66 58 TOPIC 3. ALCOHOLS AND ETHERS Q42: Write a balanced equation for the reaction of sodium with water. Q43: Apart from hydrogen, what other product is formed? Q44: Look closely at the equation for the reaction of sodium with water. Predict the other product formed when sodium reacts with ethanol. Draw its structure and suggest a name. In fact, this is a general reaction for alcohols. Figure 3.11: Formation of alkoxides The metal alkoxides are powerful bases as well as nucleophiles and are frequently used in organic synthesis. ffl Dehydration to form alkenes Learning Objective ffi To describe the dehydration of alcohols to give alkenes This reaction was discussed in Topic 3.2 under the heading Synthesis of alkenes. For simple alcohols, the alcohol vapour can be passed over hot aluminium oxide, for example: fi fl Figure 3.12: Dehydration of propan-2-ol A more general method involves the reaction of an alcohol with a strong non-volatile acid such as concentrated sulphuric acid or phosphoric acid. Q45: What class of reaction is this? a) substitution

67 3.5. REACTIONS OF ALCOHOLS 59 b) elimination c) neutralisation d) condensation Q46: If butan-2-ol is warmed with phosphoric acid, what product (or products) will be formed? You will not be asked to describe the mechanism of this type of reaction in detail but it is effectively the reverse of the mechanism described in Table 3.3 for the hydration of an alkene. ffl Forming esters Learning Objective ffi To describe the reaction of alcohols to form esters The reaction between alcohols and carboxylic acids to form esters was introduced at Higher level (Unit 2, World of Carbon). fi fl Figure 3.13: Esterification Q47: What class of reaction is shown in Figure 3.13? a) neutralisation b) elimination c) neutralisation d) condensation Q48: The reaction is frequently carried out in the presence of a small amount of concentrated sulphuric acid. Give two reasons for this. Q49: Name the ester formed in Figure Even in the presence of a catalytic amount of concentrated sulphuric acid, the reaction between alcohols and carboxylic acids is slow. Esters are normally prepared by an alternative two-stage process in which the carboxylic acid is first converted into an acid chloride by reaction with phosphorus pentachloride (PCl 5 ) or thionyl chloride (SOCl 2 ), see Figure 3.14.

68 60 TOPIC 3. ALCOHOLS AND ETHERS Figure 3.14: Making ester (stage 1) Q50: What type of reaction has taken place? a) neutralisation b) elimination c) substitution d) addition In stage 2, the acid chloride reacts with an alcohol. Acid chlorides are much more reactive than the parent acids and readily react with alcohols to form esters. Figure 3.15: Acid chloride to ester Q51: What type of reaction occurs in the second stage (Figure 3.15)? a) neutralisation b) addition c) elimination d) condensation ffl 3.6 Preparation of ethers Learning Objective ffi To describe the preparation of ethers from halogenoalkanes The standard method used to prepare ethers involves the nucleophilic substitution of a monohalogenoalkane using a metal alkoxide (formed as described in Figure 3.11). This reaction was briefly described in Topic 3.2. fi fl

69 3.7. REACTIONS OF ETHERS 61 Figure 3.16: Forming an ether This reaction (Figure 3.16) is a general one and can be used to prepare both symmetrical and unsymmetrical ethers. 3.7 Reactions of ethers Ether molecules are slightly polar (Table 3.1) but sufficiently non-polar to make them resistant to attack by most nucleophiles and electrophiles. They are also relatively resistant to reduction and, with some exceptions, oxidation. However they are highly flammable, particularly those with low molecular mass. In addition, on contact with air even for only a few days, ethers can form hydroperoxides and peroxides. Q52: The O-O bond in peroxides is fairly weak. What type of bond fission is likely to occur? As a result of this reactivity, these peroxides are extremely unstable and potentially explosive. Ethers are widely used as solvents because they can dissolve a large number of organic compounds. Because the ether molecule (Figure 3.6) is slightly polar, ethers readily dissolve non-polar molecules as well as molecules which are slightly polar. Ethoxyethane is widely used in solvent extractions since the extracted compounds are easily isolated by evaporation of the solvent. However, extreme care must be taken in

70 62 TOPIC 3. ALCOHOLS AND ETHERS using such ethers, particularly if the solvent is removed by distillation. No naked flames should be used anywhere near the solvent and the procedure should be carried out in a fume cupboard. To make sure that the solvent is free from peroxides, it can first be treated with a solution of an iron(ii) salt. 3.8 Summary Ethers (general formula - R -O-R") can be named according to IUPAC rules. Hydrogen bonding between alcohol molecules and lack of hydrogen bonding between ether molecules explains why alcohols (general formula - R-OH) have higher boiling points than isomeric ethers and other organic molecules of similar relative molecular mass. Hydrogen bonding between alcohol molecules and water, and between ether molecules and water, explains why ethers and alcohols of similar relative molecular mass have similar solubilities in water. Alcohols can be prepared from alkenes by hydration or from halogenoalkanes by nucleophilic substitution. In industry, alcohols (except methanol) are manufactured by the acid-catalysed hydration of alkenes. Alcohols react with some reactive metals to form alkoxides. Alcohols can be dehydrated to alkenes. Alcohols undergo condensation reactions slowly with carboxylic acids and more vigorously with acid chlorides to form esters. Ethers can be prepared by the reaction of halogenoalkanes with alkoxides. Ethers are highly flammable and on exposure to air may form potentially explosive peroxides. Ethers are mainly used as solvents because they are relatively inert chemically and will dissolve many organic compounds. 3.9 Resources Higher Still Support: Advanced Higher Chemistry - Unit 3: Organic Chemistry, Learning and Teaching Scotland, ISBN Chemistry in Context: Hill and Holman, Nelson, ISBN Organic Chemistry: J. McMurry, Brooks/Cole Publishing, ISBN Chemistry: A. and P. Fullick, Heinemann, ISBN Website

71 3.10. END OF TOPIC TEST End of Topic test An online assessment is provided to help you review this topic.

72 64 TOPIC 3. ALCOHOLS AND ETHERS

73 65 Topic 4 Aldehydes, Ketones and Carboxylic Acids Contents 4.1 Introduction Physical properties Solubility in water Reactions of aldehydes and ketones Oxidation Reduction Nucleophilic addition reactions Carboxylic acids Preparation of carboxylic acids Reactions of carboxylic acids Summary Resources End of Topic test Prerequisite knowledge Before you begin this Topic, you should be able to: recognise aldehydes, ketones, carboxylic acids and esters from their functional groups and give systematic names for simple examples; describe the oxidation of primary alcohols and secondary alcohols to aldehydes and ketones respectively and the further oxidation of aldehydes to carboxylic acids; describe oxidation as an increase in the oxygen to hydrogen ratio and reduction as a decrease in the oxygen to hydrogen ratio in an organic compound; describe how alcohols and acids can react reversibly to form esters. Learning Objectives After studying this Topic, you should be able to: explain the physical properties, such as boiling points and solubility in water, of aldehydes, ketones and carboxylic acids in terms of intermolecular forces;

74 66 TOPIC 4. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS distinguish between aldehydes and ketones using either Tollens reagent or Fehling s solution; describe some reactions of aldehydes, ketones and carboxylic acids; describe some methods of preparation of carboxylic acids.

75 4.1. INTRODUCTION 67 ffl 4.1 Introduction Learning Objective ffi To recognise functional groups and describe the bonding in the carbonyl group Molecules of aldehydes, ketones and carboxylic acids all contain a common feature, the carbonyl group- C=O. fi fl This is arguably the most important of functional groups. It is an essential part of many pharmaceuticals and most important biological molecules. The ability to identify correctly the various functional groups is crucial to a full understanding of organic chemistry. Carbonyl group The carbonyl group, C=O, appears as part of other functional groups. The formation of the carbon to oxygen double bond can be explained in a similar way to that of the carbon to carbon double bond (see Topic 3.2, section 1.3). Hybridisation of the same orbitals on the oxygen atom also occurs and overlap of the orbitals on both atoms gives the structure shown in Figure 4.1. Figure 4.1: Carbonyl bonding There is one crucial difference between the formation of the C=O bond and the formation of the C=C bond. Oxygen being more electronegative than carbon attracts the bonded electrons more strongly, particularly the ß electrons, making the C=O bond polar (Figure 4.2).

76 68 TOPIC 4. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS Figure 4.2 The physical properties and chemical properties of aldehydes, ketones and carboxylic acids are largely determined by the polarity of the C=O bond. Q1: Which orbitals in the carbon atom will be involved in hybridisation? a) 2s only b) 2s and one 2p orbital c) 2s and two 2p orbitals d) 2s and three 2p orbitals Q2: Will the carbon atom be an electrophilic centre or a nucleophilic centre? a) electrophilic b) nucleophilic Q3: Will the oxygen atom be an electrophilic centre or a nucleophilic centre? a) electrophilic b) nucleophilic Q4: What type of species will attack the carbon atom? Q5: What type of species will attack the oxygen atom? Q6: The oxygen atom is able to accept H + ions. In doing this, it is behaving like a..? ffl 4.2 Physical properties Learning Objective To explain the physical properties, such as boiling points and solubility in water, of aldehydes, ketones and carboxylic acids in terms of intermolecular forces ffi In the last Topic, the boiling points and solubility in water of alcohols and ethers were explained in some detail in terms of the intermolecular forces involved. The physical properties of aldehydes, ketones and carboxylic acids can be explained in a similar way. fi fl

77 4.2. PHYSICAL PROPERTIES 69 ffl Boiling points Learning Objective To be able to explain the trends in boiling points of aldehydes, ketones and carboxylic acids. ffi A data handling exercise based on graphs of the boiling points of compounds from different homologous series. The graph shows the boiling points for members of five different homologous series. Study it carefully and then answer the questions underneath. fi fl 15 min Q7: Which are the most important intermolecular forces being overcome when aldehydes and ketones boil? a) covalent bonds b) hydrogen bonds c) polar-polar attractions d) van der Waals forces Q8: Explain why aldehydes and ketones have higher boiling points than the alkanes of similar molecular mass. Q9: Which are the most important intermolecular forces being overcome when alcohols boil? a) covalent bonds b) hydrogen bonds c) polar-polar attractions d) van der Waals forces Q10: Explain why aldehydes and ketones have lower boiling points than the alcohols of similar molecular mass. See further questions on page 218. Figure 4.3

78 70 TOPIC 4. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS Figure 4.4 In pure carboxylic acids, the molecules pair up to form dimers ( Figure 4.4), in which each molecule forms two hydrogen bonds. This stronger bonding explains the higher boiling points of carboxylic acids. Aldehydes and ketones have higher boiling points than the corresponding alkanes because polar-polar attractions are stronger than van der Waals forces. They have lower boiling points than the corresponding alcohols because polar-polar attractions are weaker than hydrogen bonds. Both carboxylic acids and alcohols exhibit hydrogen bonding but this must be stronger in the acids since they have higher boiling points. ffl Solubility in water Learning Objective To explain the trends in solubility in water of aldehydes, ketones and carboxylic acids in terms of intermolecular forces and explain the acidity of carboxylic acids. ffi The solubility of an organic compound in water depends on the ability of the substance to form hydrogen bonds with water. fi fl

79 4.2. PHYSICAL PROPERTIES 71 Study the structures in the diagram above. Q11: Which of the structures shown will be able to form hydrogen bonds with water? a) all of them except propanone b) ethanoic acid only c) ethanoic acid and ethanol only d) all of them Q12: Which substances shown are likely to be soluble in water? a) all of them except propanone b) ethanoic acid only c) ethanoic acid and ethanol only d) all of them Q13: Use your knowledge to predict the effect of chain length on the solubility of aldehydes and ketones in water. Q14: Which of the following statements is false? a) Carboxylic acids are likely to be more soluble than the corresponding aldehydes and ketones. b) As the number of carbon atoms increases, the solubility of carboxylic acids decreases. c) Hexanoic acid is more soluble than ethanoic acid. d) Carboxylic acids of low molecular mass are very soluble in water. Acidity Although pure carboxylic acids exist as dimers ( Figure 4.4), such dimerisation does not occur in aqueous solution. Instead there is tendency for carboxylic acid molecules to dissociate according to the equation:

80 72 TOPIC 4. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS Figure 4.5 As a result, carboxylic acids are weak acids. Values for pk a can be used as a measure of the strength of an acid (see Unit 2, Topic 4). Substance pk a ethanoic acid 4.8 ethanol 18 water 15.7 Q15: Which is the strongest acid? a) ethanoic acid b) ethanol c) water Q16: Which is the weakest acid? a) ethanoic acid b) ethanol c) water With all three substances, the H + (aq) ion is produced by the heterolytic fission of an OH bond in a hydroxyl group. Why is the carboxyl group so acidic? The dissociation (Figure 4.5) is reversible. Any factor which pushes the equilibrium to the right will strengthen the acid. The ethanoate ion can be represented by two equivalent resonance structures: In fact, the electrons are spread (delocalised) over all three atoms of the group. Consideration of the bonding in the ethanoate ion helps to explain this ( Figure 4.6). The three atoms in the carboxylate group are sp 2 hybridised.

81 4.2. PHYSICAL PROPERTIES 73 Figure 4.6: Delocalisation in the ethanoate ion The unhybridised p orbitals are close enough to overlap to form a delocalised ß-system (Figure 4.6). Delocalisation of the charge makes the ion more stable and makes it less likely to accept a H + (aq) ion. Figure 4.7: Ethanol dissociation Figure 4.7 shows the similar dissociation of ethanol. Q17: Look at diagram (Figure 4.7). Can the alkoxide ion be stabilised in the same way? a) yes b) no Q18: In which group is the OH bond more likely to break? a) hydroxyl b) carboxyl Q19: Which will be more acidic? a) alkanols b) alkanoic acids Summary of solubility in water Aldehydes and ketones of relatively low molecular mass are miscible with water. Carboxylic acids of similar molecular mass are even more soluble. In both cases, as the chain length increases, the solubility decreases.

82 74 TOPIC 4. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS 4.3 Reactions of aldehydes and ketones ffl Oxidation Learning Objective ffi To distinguish between aldehydes and ketones using Tollens or Fehling s reagents The oxidation of aldehydes to carboxylic acids was introduced at Higher level (Unit 2, World of Carbon). This reaction can be used to distinguish between aldehydes and ketones. fi fl Figure 4.8 Aldehydes have a hydrogen atom attached to the carbonyl carbon atom ( Figure 4.8). This can be readily lost during oxidation. Ketones do not have such a hydrogen atom and are much more resistant to oxidation. Tollens reagent (named after its discoverer, Bernhard Tollens) contains complexed silver(i) ions, [Ag(NH 3 ) 2 ] +, which on reduction produce metallic silver. Figure 4.9: Reaction of an aldehyde with Tollens reagent The formation of a silver mirror (Figure 4.9) is often used as a test for the presence of aldehydes as opposed to ketones. Fehling s solution contains complexed copper(ii) ions and will also oxidise aldehydes but not ketones.

83 4.3. REACTIONS OF ALDEHYDES AND KETONES 75 Figure 4.10: Reaction of an aldehyde with Fehling s solution When reacted with an aldehyde, the copper(ii) ions are reduced to copper(i) and a reddish-brown precipitate of copper(i) oxide forms (Figure 4.10). Figure 4.11 The aldehyde is oxidised to an alkanoic acid (Figure 4.11). The symbol, [O], is often used in organic equations to show that a compound has been oxidised. ffl Reduction Learning Objective ffi To describe the reduction of aldehydes and ketones to alcohols Both aldehydes and ketones can be reduced to alcohols using reducing agents like lithium aluminium hydride, LiAlH 4, in ether. fi fl Figure 4.12: Reduction mechanism The AlH - 4 ion is able to provide a hydride ion, H -, which attacks the partially positive carbon atom of the carbonyl group to form the negative ion shown (Figure 4.12, stage 1).

84 76 TOPIC 4. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS Q20: What role does the hydride ion play in this step? a) acid b) base c) nucleophile d) electrophile When the ether solution is shaken with water, the negatively charged oxygen atom accepts a proton from water to form the alcohol product (Figure 4.12, stage 2). Q21: Which of the following describes the change in the arrangement of the bonds around the carbonyl carbon atom? a) from tetrahedral to trigonal planar b) from trigonal bipyramidal to tetrahedral c) from tetrahedral to tetrahedral d) from trigonal planar to tetrahedral Q22: If the starting material was an aldehyde, which class of alcohol would be produced? Q23: If the starting material was a ketone, which class of alcohol would be produced? Q24: Give the full systematic name of the product when the compound opposite is reduced using lithium aluminium hydride in ether: Q25: Give the full systematic name of the product when the compound opposite is reduced using lithium aluminium hydride in ether. ffl Nucleophilic addition reactions Learning Objective To describe the addition of reagents to the carbonyl group in aldehydes and ketones and the subsequent elimination of water where appropriate. ffi Aldehydes and ketones react with many reagents by nucleophilic addition reactions. It is easier to work out the various products if you first understand what goes on in the basic mechanism. fi fl

85 4.3. REACTIONS OF ALDEHYDES AND KETONES 77 In nucleophilic addition reactions, the mechanism is very similar to that described for the reduction using LiAlH 4. Overall, the reduction of an aldehyde or ketone corresponds to the addition of hydrogen across the carbonyl group (Figure 4.12). Figure 4.13: Stage 1 In the first stage (Figure 4.13), nucleophilic attack on the carbon atom of the carbonyl group is followed by electrophilic attack by a H + ion (protonation) of the oxygen atom to produce an alcohol. Note that the trigonal planar carbonyl compound is converted into a tetrahedral alcohol. In effect, H-Nu has added across the carbonyl group (where Nu is the nucleophile). If the nucleophilic atom has a hydrogen atom attached to it, the reaction can proceed further. This happens with some compounds related to ammonia, where one of the hydrogen atoms of the ammonia molecule has been replaced by another group (i.e. R-NH 2 ). Stage one of the reaction is as before (Figure 4.13). Figure 4.14: Stage 2 In the second stage (Figure 4.14), an acid-catalysed elimination of water occurs resulting in the formation of a carbon to nitrogen double bond.

86 78 TOPIC 4. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS Q26: Overall, what type of reaction has occurred? (i.e. Addition + elimination =?) Examples of nucleophiles that can react with aldehydes and ketones include those shown in Figure Figure 4.15: Hydrogen cyanide and hydrazines In the presence of cyanide ions, nucleophilic addition of hydrogen cyanide to an aldehyde or ketone produces a cyanohydrin (Figure 4.16). Figure 4.16: Cyanohydrin The cyanohydrin can be isolated because it does not react further by elimination of water as described in Figure The production of cyanohydrins is useful since hydrolysis can convert the cyanohydrin into a carboxylic acid with one more carbon atom than the starting material, e.g. On the other hand, hydrazine, NH 2 -NH 2 (Figure 4.15), is capable of nucleophilic addition followed by elimination of water. The product is known as a hydrazone (Figure 4.17).

87 4.3. REACTIONS OF ALDEHYDES AND KETONES 79 Figure 4.17: Hydrazone formation Compounds based on hydrazine, in which one of the hydrogen atoms has been replaced by another group, react easily with aldehydes and ketones in exactly the same way to form crystalline products that are easily purified. These products are known as derivatives. The pure derivatives have sharp, distinctive melting points. A commonly used example is 2,4-dinitrophenylhydrazine (Figure 4.15). When this is used, the products are known as 2,4-dinitrophenylhydrazones. By comparing the melting point of a derivative with tables of the melting points of known derivatives, it is possible to identify the original aldehyde or ketone. Learning point Hydrazine and 2,4-dinitrophenylhydrazine react with aldehydes and ketones by nuceophilic addition to form hydrazones and 2,4-dinitrophenylhydrazones with distinctive melting points which can be used to identify unknown aldehydes and ketones. At this point it would be a good idea to try the Prescribed Practical Activity on Identification by Derivative Formation. PPA - Identification by Derivative Formation (Unit 3 PPA 2) Consult with your tutor to find out whether the PPA on Identification by Derivative Formation is to be completed at this stage. 120 min

88 80 TOPIC 4. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS Although aldehydes and ketones differ in their ease of oxidation, the rest of their chemistry is very similar, although in general aldehydes are more reactive than ketones. If both the alkyl groups in a ketone are large and bulky, they may hinder the nucleophile from attacking the carbon atom of the C=O group and so reduce the reactivity of the ketone. However, this does not explain why simple ketones, which allow relatively unhindered access to nucleophiles, are less reactive than the corresponding aldehydes. In Topic 3.2 (section 2.1.3), the relative stabilities of carbocations were discussed (Figure 4.18). Alkyl groups are able to push electrons towards a positively charged carbon atom and so stabilise the positive charge. The more alkyl groups, the greater the stabilisation. Figure 4.18: Relative stabilities of carbocations Q27: Which carbocation is most susceptible to attack by a nucleophile? Now look at the structures of aldehydes and ketones. Q28: Which structure is better able to stabilise the positive charge on the carbon atom? a) aldehyde b) ketone Q29: In which is the carbonyl atom more susceptible to attack by nucleophiles? a) aldehyde b) ketone

89 4.4. CARBOXYLIC ACIDS Carboxylic acids ffl Preparation of carboxylic acids Learning Objective ffi To describe some methods of preparation of carboxylic acids. Most of the normal methods used to prepare carboxylic acids have been discussed already and are summarised in Figure fi fl Figure 4.19: Preparation of acids Amides contain the functional group shown opposite and their preparation from carboxylic acids is considered in the next section. Q30: Name the class of compound X. Q31: What does the symbol, [O], represent? Q32: What general type of reaction occurs when esters, nitriles and amides are converted into carboxylic acids? Q33:

90 82 TOPIC 4. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS Which of the reactions in the grid above could not be used to prepare 2-methylpropanoic acid? a) A b) B c) C d) D See further questions on page 219. PPA - Preparation of benzoic acid by hydrolysis of ethyl benzoate (Unit 3 PPA3) 120 min Consult with your tutor to find out whether the PPA on Preparation of benzoic acid by hydrolysis of ethyl benzoate is to be completed at this stage. ffl Reactions of carboxylic acids Learning Objective ffi To describe some reactions of carboxylic acids. Some of the reactions of carboxylic acids are shown in Figure fi fl Figure 4.20: Reactions of carboxylic acids

91 4.4. CARBOXYLIC ACIDS 83 ffl Salt formation Learning Objective ffi To describe the reactions of carboxylic acids that result in the formation of salts. Carboxylic acids behave as typical acids by forming salts with alkalis, basic metal oxides, carbonates and some metals. On paper, write balanced equations, including states, for the reactions of aqueous ethanoic acid with the following substances: fi fl Q34: sodium hydroxide solution Q35: solid copper(ii) carbonate Q36: magnesium metal Q37: solid copper(ii) oxide ffl Amide formation Learning Objective ffi To describe the reaction of carboxylic acids with ammonia and amines to form amides. Ammonia and the related compounds, amines, are examples of bases that can react with carboxylic acids to form salts (see Topic 3.5), e.g: fi fl If these salts are heated strongly, further reaction occurs, producing amides, e.g:

92 84 TOPIC 4. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS Amides contain a functional group in which the carbon atom of a carbonyl group is bonded to nitrogen. There are three classes of amide. Figure 4.21: Classification of amides The functional group shown in Figure 4.21(b) was introduced at Higher level (Unit 2, World of Carbon, Polymers and Natural Products). It is formed when certain monomers join to form polymers, some natural and some synthetic. Q38: What important class of naturally occurring polymers contains this functional group? a) fats and oils b) carbohydrates c) proteins d) vitamins Q39: What alternative name is given to this linking group? Q40: Synthetic polymers containing this group are called polyamides. Give two examples of such polymers. ffl Ester formation Learning Objective ffi To describe the reaction of carboxylic acids with alcohols to form esters. Carboxylic acids react reversibly with alcohols to form esters. This reaction was first covered at Higher level (Unit 2, World of Carbon). It was briefly discussed in Topic 3.3 in the section on Reactions of alcohols. Since the reaction is reversible, it provides a fi fl

93 4.4. CARBOXYLIC ACIDS 85 convenient method for preparing carboxylic acids from esters (this Topic, section 4.1, Figure 4.19). The reaction is catalysed by acid. Although detailed knowledge of the reaction mechanism is not required, it is worth pointing out the similarity with other carbonyl reactions. Figure 4.22: Ester formation - mechanism The first stage involves nucleophilic addition to the carbonyl group (Figure 4.22). Protonation of the oxygen atom of the carbonyl group makes the carbon atom more susceptible to nucleophilic attack by the oxygen atom of the alcohol. In the second stage, water is eliminated to form the ester product, Q41: What other term can be used to describe such addition-elimination reactions? Q42: Name and draw the structural formula of the ester formed when propanoic acid and ethanol react. Q43: Name the ester formed when benzoic acid and methanol react. Q44: Name the ester formed when octanoic acid and propanol react. Q45: Which of the structures in the above grid does not represent methyl propanoate?

94 86 TOPIC 4. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS ffl Reduction Learning Objective To describe the reduction of carboxylic acids to primary alcohols with lithium aluminium hydride. ffi Earlier in this Topic (section 3.2), the reduction of aldehydes and ketones to alcohols using lithium aluminium hydride in ether was described (Figure 4.12). Carboxylic acids are more resistant to reduction than aldehydes and ketones and LiAlH 4 is one of the few reagents powerful enough to reduce carboxylic acids to primary alcohols. fi fl The mechanism for the reduction of acids is more complicated but the initial step is again likely to be nucleophilic attack by hydride ion, H -, on the partially positive carbon atom of the carbonyl group. Reactions of aldehydes, ketones and carboxylic acids 15 min This is a problem solving activity to test your knowledge of reactions of aldehydes, ketones and carboxylic acids. A hydrocarbon of formula, C 4 H 8, was reacted to form two isomeric compounds, P and Q, which were separated and further reacted as shown in the reaction scheme below (Figure 4.23). Using knowledge gained in this and the previous Topic, identify the unknown compounds in the scheme. Write your answers on paper and then confirm them by answering the questions which follow. If you are really stuck, there is a hint for this activity at the back of the book.

95 4.5. SUMMARY 87 Q46: Which is the original hydrocarbon? a) cyclobutane b) but-1-ene c) but-2-ene d) 2-methylpropene Q47: Compound P is a) primary b) secondary c) tertiary Q48: Which compound is a ketone? Q49: Name compound S. Q50: What type of compound is T? Q51: What is the formula for reagent Z? Q52: What is the formula for reagent Z? Figure Summary Aldehydes and ketones have higher boiling points than the corresponding alkanes and lower boiling points than the corresponding alcohols. These differences can

96 88 TOPIC 4. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS be explained in terms of the relative strengths of the intermolecular forces Pure carboxylic acids have relatively high boiling points due to the formation of dimers held together by hydrogen bonding. Aldehydes, ketones and carboxylic acids of low relative formula mass are miscible with water due to their ability to form hydrogen bonds with water molecules. The solubility in water decreases as chain length increases. Carboxylic acids are weak acids because they dissociate slightly in water. This is can be explained by the stability of the carboxylate anion caused by electron delocalisation. Aldehydes and ketones can be reduced to primary and secondary alcohols respectively using lithium aluminium hydride in ether, and undergo nucleophilic addition reactions with a variety of reagents. Nucleophilic addition with 2,4-dinitrophenylhydrazine followed by elimination of water (also described as condensation) produces 2,4-dinitrophenylhydrazones that have characteristic melting points useful in the identification of unknown aldehydes and ketones. Carboxylic acids can be prepared by the oxidation of primary alcohols or aldehydes and by the hydrolysis of esters, nitriles or amides. Carboxylic acids can react to form salts, esters and amides and can be reduced with lithium aluminium hydride to form primary alcohols. 4.6 Resources Higher Still Support: Advanced Higher Chemistry - Unit 3: Organic Chemistry, Learning and Teaching Scotland, ISBN Chemistry in Context: Hill and Holman, Nelson, ISBN Organic Chemistry: J. McMurry, Brooks/Cole Publishing, ISBN Chemistry: A. and P. Fullick, Heinemann, ISBN Website End of Topic test An online assessment is provided to help you review this topic.

97 89 Topic 5 Amines Contents 5.1 Introduction Naming and classification Physical properties Chemical properties Summary Functional groups: summary Resources End of Topic test Prerequisite knowledge Before you begin this topic, you should be able to: identify the functional group present in amines and relate this to ammonia and its properties; describe the formation of amides and polyamides from amines and carboxylic acids. (Higher Unit 2); relate the strength of nylon polymers to the hydrogen bonding between chains. (Higher Unit 2). Learning Objectives After studying this Topic, you should be able to: classify amines as primary, secondary or tertiary and name them according to IUPAC rules; explain why the boiling points of primary and secondary amines are higher than alkanes of comparable relative formula mass and explain why the lower amines are soluble in water; explain how amines can act as proton acceptors and describe some reactions resulting from their basic nature.

98 90 TOPIC 5. AMINES 5.1 Introduction The amines are a family of nitrogen-containing organic compounds derived from ammonia. Proteins are condensation polymers made up of many amino acids molecules contain the amine functional group and the polymers called polyamides are made from monomers containing amine groups. Nylon is one versatile polyamide that has important engineering uses related to its strength. This strength comes from the hydrogen bonding between the polymer chains. The monomer unit used to make nylon is 1,6- diaminohexane. The amines with low molecular mass like methylamine have a fishy smell. Amines are released by all decaying animal flesh and this macabre aspect of amine chemistry is reflected in the names of two diamines: Cadaverine (from cadaver-a dead body) NH 2 (CH 2 ) 5 NH 2. Putrescine (from putrefy - to decay with an odour) NH 2 (CH 2 ) 4 NH 2. The physical and chemical properties of the simpler members of the amines resemble that of ammonia, but are modified by the presence of alkyl groups. Most of the properties are due to the lone pair of electrons on the nitrogen atom. The illustration shows ammonia and a typical amine (trimethylamine). ffl 5.2 Naming and classification Learning Objective To be able to classify amines as primary, secondary or tertiary and name them according to IUPAC rules. ffi Amines are named according to IUPAC rules. The system is quite complex however, and amines can be named in several ways depending on their structure. Either the e from the parent alkane is removed and the suffix amine added, or the amine is added to the name of the substituent group (Figure 5.1). fi fl

99 5.2. NAMING AND CLASSIFICATION 91 Figure 5.1 The prefix di- or tri- is sometimes necessary if there are substituent groups with the same name, remembering to list the substituents alphabetically (Figure 5.2 (a)). If the amine group is a branch from the main chain, a number is used to describe its position and the prefix amino- is used. Although IUPAC rules suggest that 2-aminobutane (Figure 5.2(b)) should be named 1-methylpropylamine, texts like Salters Advanced Chemistry and A-Level Chemistry (Ramsden), use 2-aminobutane.To avoid confusion with textbooks, the name 2-aminobutane is preferred. The website listed in the resources section provides further detail. Amino groups rank low in seniority in compounds with multiple groups, so in these cases, the prefix amino is used, with the compound being classed as belonging to the family with the senior functional group. See Figure 5.2(c); in this case, an alcohol. Answer the following questions. Figure 5.2 Q1: Name this compound: a) methylamine b) dimethylamine c) diethylamine d) 2-butylamine Q2: Name this compound:

100 92 TOPIC 5. AMINES a) methylethylamine b) propylamine c) propan-2-amine d) 2-aminopropane Amines can be classified as primary (1 o ), secondary (2 o ) or tertiary (3 o ), depending on the number of carbon atoms connected to the nitrogen ( Figure 5.3). Figure 5.3: Primary, secondary and tertiary amines Amines are regarded as organic compounds based on ammonia. When one hydrogen is replaced by an alkyl or aryl (benzene related) group, the amine is primary. Two alkyl or aryl groups make the amine secondary, and three groups make it tertiary. The alkyl or aryl groups may be the same or different. The simplest aryl amine has one hydrogen replaced by benzene and is commonly named aniline. In this Topic, as in common with most text books, the name phenylamine (Figure 5.4) will also be used for this compound. In summary: Figure 5.4: Phenylamine

101 5.2. NAMING AND CLASSIFICATION 93 Class General Formula Functional group primary 1 o RNH 2 secondary 2 o RR NH tertiary 3 o RR R N ffl Classification and naming of amines Learning Objective To be able to classify amines as primary, secondary or tertiary and name them according to IUPAC rules ffi A drag and drop version of the following exercise is available on the web site summarising the naming and classification of amines Match up the structures (shown as (a), (b) and (c) in Figure 5.5), with the available spaces and then match names to all the structures. Lastly, classify the amines as primary (1 ffi ), secondary (2 ffi ), or tertiary (3 ffi ). fi fl 15 min

102 94 TOPIC 5. AMINES Figure 5.5: Amine exercise These questions will guide you through part of the exercise. Q3: Which of these structures is tertiary: a, b, or c? Q4: What name would you give to the tertiary structure you selected? a) ethylamine b) diethylmethylamine c) 2-aminopropane

103 5.3. PHYSICAL PROPERTIES 95 d) ethyldimethylamine Q5: Classify diethylamine. a) primary b) secondary c) tertiary Q6: What name and class would you give to the structure in the lower left box? a) ethyldimethylamine, tertiary b) methyldiethylamine, tertiary c) ethyldimethylamine, secondary d) methydiethylamine, secondary Amines are named according to IUPAC rules and can be classified as primary, secondary or tertiary, according to the number of carbon atoms connected to the nitrogen. ffl 5.3 Physical properties Learning Objective To explain why the boiling points of primary and secondary amines are higher than alkanes of comparable relative formula mass. ffi To explain why the lower amines are soluble in water. The physical properties of the amines such as melting point, boiling point and solubility are largely influenced by the presence of polar N-H bonds and their capability of forming hydrogen bonds with other amines or with water. Primary and secondary amines form hydrogen bonds readily (Figure 5.6) but tertiary amines have no N-H bonds and therefore cannot associate with each other. The only intermolecular forces present in a pure tertiary amine are van der Waals attractions. fi fl Figure 5.6

104 96 TOPIC 5. AMINES Examine the boiling point data (Table 5.1), comparable relative formula mass. comparing amines with alkanes of Table 5.1: Amines and alkanes Amine Boiling point Alkane of equivalent Boiling point /C o mass /C o (a) CH 3 NH (e) CH 3 CH 3-89 (b) (CH 3 ) 2 NH 7.5 (f) CH 3 CH 2 CH 3-42 (c) (CH 3 ) 3 N 3 (g) CH 3 CH(CH 3 )CH 3-12 (d) CH 3 CH 2 CH 2 NH 2 49 (h) CH 3 CH 2 CH 2 CH 3-1 Compare the isomeric amines (c) and (d). These have identical relative formula masses. Q7: Which amine has the higher boiling point? a) (c) b) (d) Q8: Which amine has the stronger intermolecular bonds? a) (c) b) (d) Q9: Which amine is tertiary? a) (c) b) (d) Q10: Which class of amine tends to have lower boiling points because the molecules cannot associate by hydrogen bonding? Q11: Compare amine (a) with alkane (e) which has approximately the same formula mass (Table 5.1). Which has the higher boiling point? a) (a) b) (e) Q12: Which has the stronger intermolecular bonds? a) (a) b) (e) Q13: Which associate with hydrogen bonds as well as van der Waals bonds? a) (a) b) (e) Learning Point Primary and secondary amines, but not tertiary amines, associate by hydrogen bonding. As a result, they have higher boiling points than isomeric tertiary amines and alkanes with comparable relative formula masses.

105 5.4. CHEMICAL PROPERTIES 97 Water is able to form hydrogen bonds with all amines, including tertiary amines. Because of this, amines with small alkyl groups are soluble. The larger the alkyl groups, the less soluble the amine, because large groups disrupt the hydrogen bonding with water (Figure 5.7). Figure 5.7: Hydrogen bonding in aqueous amines Learning Point Amine molecules can form hydrogen bonds with water molecules. This explains the appreciable solubility of the lower amines in water. ffl 5.4 Chemical properties Learning Objective To explain how amines can act as proton acceptors and describe some reactions resulting from their basic nature. ffi Basicity The chemical properties of the amines are similar to the properties of ammonia and are strongly influenced by the lone pair of electrons on the nitrogen atom. The lone pair of electrons gives amines the ability to behave as proton acceptors, i.e. as bases. The lone pair forms a dative covalent bond in which both electrons come from the same atom (the nitrogen). When an amine dissolves in water, an alkaline solution is formed as the reaction produces an ammonium ion and a hydroxide ion. In the example shown, methylamine dissolves to form the methylammonium ion and hydroxide (Figure 5.8). The equilibrium lies well to the left so methylamine is a weak base and the solution is alkaline. fi fl

106 98 TOPIC 5. AMINES Figure 5.8: Methylamine and water One of the most convenient ways of measuring the basicity of an amine is to look at the pk a value of the conjugate acid (Figure 5.9). The higher the pk a of the conjugate acid, the stronger the base. Look back to Topic 4.4 of Unit 2 if revision is needed. Answer the following questions: Figure 5.9 Table 5.2 Base Conjugate acid (ion) pk a ammonia ammonium 9.3 ethylamine ethylammonium 10.7 diethylamine diethylammonium 11.1 phenylamine phenylammonium 4.6 Q14: In general, are the alkylamines shown in the table (Table 5.2) weaker or stronger bases than ammonia? Q15: Is phenylamine a weaker or stronger base than ammonia? Q16: Which of these is the most likely ph of a phenylamine solution? a) 2 b) 5 c) 8 d) 14 Salt formation When a bottle of concentrated hydrochloric acid is opened near a bottle of concentrated

107 5.4. CHEMICAL PROPERTIES 99 methylamine, dense white fumes of methylammonium chloride are produced. The methylamine acts as a base, the hydrogen chloride as an acid and methylammonium chloride is the salt produced (Figure 5.10). Opening a bottle of concentrated methylamine and concentrated hydrochloric acid close to each other allows the fumes from each to mix in the air around the bottles. The reaction produces a white cloud of tiny crystals of methylammonium chloride. A neutralisation reaction has taken place. Figure 5.10: Methylamine and acid equation Answer these questions: Q17: The product of the reaction is a salt. What type of bonding is present in all salts? Q18: If the amine used was ethylamine, what name would the salt have? Amines react with other aqueous mineral acids and also with carboxylic acids to form salts (Figure 5.11). The reaction with a carboxylic acid is often the first stage in the production of amides. The ammonium salt of the carboxylic acid is heated to produce the amide. In this case (Figure 5.11), the methylammonium ethanoate will decompose on heating to form a secondary amide. ( see Topic 4.2 of Unit 3 on Amide formation).

108 100 TOPIC 5. AMINES Figure 5.11: Salt formation 5.5 Summary Amines can be considered to be derivatives of ammonia (NH 3 ), whose properties are closely associated with the lone pair of electrons on the nitrogen and the possibilities for hydrogen bonding. Amines can be classified as primary if there is one carbon containing group attached to the nitrogen, secondary if there are two groups and tertiary if there are three. The names of animes are governed by IUPAC rules, although in many cases the names used do not conform to these rules. Hydrogen bonding is possible between amines which have an N-H bond, i.e. primary and secondary amines. This leads to higher boiling points than those of isomeric tertiary amines (which have no N-H bonds) and higher boiling points than alkanes with comparable relative formula masses. The presence of the nitrogen lone pair on amines of all classes allows hydrogen bonding with water molecules. This is responsible for the solubility of the shorter chain amines. The lone pair of electrons can act as a proton acceptor (a base) and, in water, produces hydroxide ions that lead to amines behaving as weak alkalis. One of the reactions that amines undergo when acting as a weak base is to form salts with a range of mineral or carboxylic acids.

109 5.6. FUNCTIONAL GROUPS: SUMMARY 101 ffl 5.6 Functional groups: summary Learning Objective To recognise the characteristic chemical sections of an organic molecule that classify its chemical family ffi ffl Recognising functional groups Learning Objective ffi To give practice at recognising functional groups A version of this drag and drop exercise is available on the web site. The diagram contains all the classes of organic compound which have been introduced so far. Copy the diagram (Figure 5.12), or take a photocopy and draw structures for each of the functional groups in the appropriate boxes. Then answer the questions which follow. fi fl fi fl 5 min Figure 5.12 Look carefully at the structure of vanillin, which provides the flavour in vanilla.

110 102 TOPIC 5. AMINES Q19: Which functional group is labelled A? Q20: Which functional group is labelled B? Q21: Which functional group is labelled C? Look carefully at the structure of aspirin, a painkiller. Q22: Which functional group is labelled A? Q23: Which functional group is labelled B? Now, look carefully at the structure of testosterone, a male sex hormone. Q24: Which of the following statements is true for testosterone? a) It is a saturated alcohol b) It is an unsaturated aldehyde c) It is a saturated ketone d) It is an unsaturated ketone Q25: Which of these functional groups does not contain a carbonyl group? a) ester b) ether c) ketone d) aldehyde

111 5.7. RESOURCES Resources Chemical Storylines: Salters Advanced Chemistry, Heinemann ISBN Chemical Ideas: Salters Advanced Chemistry, Heinemann, ISBN Chemistry in Context: Hill and Holman, Nelson ISBN Chemistry: A. and P. Fullick, Heinemann, ISBN A-level chemistry: E.N.Ramsden, Stanley Thornes ISBN X General Chemistry: Ebbing, Houghton Mifflin, ISBN Higher Still Support: Advanced Higher Chemistry- Unit 3: Organic Chemistry. Learning and Teaching Scotland, ISBN Nomenclature in the IUPAC system can be investigated at: End of Topic test An online assessment is provided to help you review this topic.

112 104 TOPIC 5. AMINES

113 105 Topic 6 Aromatics Contents 6.1 Introduction Benzene structure Benzene reactions Halogenation (attack by a halogen) Nitration Sulphonation Alkylation Acidity of phenol and basicity of phenylamine (aniline) Summary Resources End of Topic test Prerequisite knowledge Before you begin this Topic, you should be able to: describe the bonding in ethene in terms of sigma and pi bonds (Topic 3.2); describe benzene as the simplest aromatic compound with distinctive structure and properties (Higher Unit 2); explain the stability of benzene and its resistance to addition reactions on the basis of delocalised electrons (Higher Unit 2); explain that aromatic compounds can be made by substituting the hydrogen atoms in benzene and that a benzene ring which has one hydrogen substituted by another group is known as the phenyl group ( -C 6 H 5 ) (Higher Unit 2); Explain that using benzene as a feedstock yields a range of consumer products (Higher Unit 2). Learning Objectives After studying this Topic, you should be able to: describe the stability of aromatic hydrocarbons like benzene in terms of sp 2 hydridisation, sigma and pi bonding and electron delocalisation;

114 106 TOPIC 6. AROMATICS explain how benzene resists addition reactions but undergoes electrophilic substitutions, giving examples of reactions and describing the mechanisms; describe how the delocalised electrons in the phenyl group explain the acidity of phenols and the basicity of aromatic amines compared with their aliphatic equivalents.

115 6.1. INTRODUCTION 107 ffl 6.1 Introduction Learning Objective ffi To revise some of the main points previously learned in Higher grade. The word aromatic was originally applied to hydrocarbons which had a sweet odour, but is now used to classify compounds with benzene-like structures (C 6 H 6 ) (Figure 6.1). Benzene has a distinctive structural formula and stable nature due to the delocalised electrons in the ring. Compounds that have benzene-like structures are quite different from aliphatic structures (Figure 6.1). The systematic name for the family of alkyl substituted aromatic hydrocarbons is the arenes. Removing one hydrogen leads to the formation of the phenyl group, C 6 H 5 -, present in many consumer products such as aspirin, polystyrene, dyestuffs, perfumes and fibres. fi fl Figure 6.1 Despite the fact that benzene is unsaturated, its reactivity is unlike that of the alkenes or alkynes. These unsaturated families undergo addition reactions (see Topic 3.2 section 1.5) whereas benzene resists addition reactions but undergoes electrophilic substitution reactions. ffl 6.2 Benzene structure Learning Objective To describe the bonding in benzene in terms of sp 2 hybridisation, sigma and pi bonds and electron delocalisation and to explain how this can account for its unexpected stability. ffi Although known to have a molecular formula C 6 H 6, the structure of benzene remained a source of controversy for many years until Friedrich Kekulé proposed a hexagonal ring with alternating double and single bonds. The two possible arrangements are identical and he proposed that the benzene ring oscillated between the two structures (Figure 6.2(b) and (c)). fi fl

116 108 TOPIC 6. AROMATICS Figure 6.2: The structure of benzene The delocalisation occurs because benzene is a flat symmetrical hexagon with the six carbons sp 2 hybridised (see Topic 3.2 section 1.3). The carbons are bonded to each other and to their hydrogen atoms by a sigma (ff) bond and each carbon has a p orbital perpendicular to the plane of the six membered ring. Each p orbital contains a single (ß) electron and overlaps sideways with both neighbouring orbitals to form a pi-bond. This results in a doughnut shaped region of negative charge above and below the plane of the ring. This is described as a delocalised ß- electron system because the electrons are evenly distributed. Each carbon to carbon bond is neither single nor double, and benzene is often shown as in Figure 6.2 (d). Structures Figure 6.2 (b) and (c) show extreme or localised arrangements with double and single bonds in definite regions. These are called resonance forms of benzene and Figure 6.2 (d) is a hybrid of these. Kekulé s ideas of the two structures therefore anticipated the idea of resonance that was introduced sixty years later. Reaction pathways in benzene chemistry are often shown using the localised arrangements because bond making and breaking are shown more clearly. A model of benzene gives some idea of the arrangement. (Table 6.1). Table 6.1 This model of benzene shows the delocalised ß- electron system above and below the hexagon of carbons.

117 6.2. BENZENE STRUCTURE 109 ffl Aromatic bonding Learning Objective To describe the bonding in benzene in terms of sp 2 hybridisation, sigma and pi bonding and electron delocalisation and explain how this can account for its unexpected stability ffi View the animation of the benzene structure on the website and then answer the questions that follow. These pictures (Figure 6.3) show the p orbitals (left) and the delocalised ß- electron system (right) in benzene. Use these pictures and the information given previously to answer the questions. fi fl 10 min Figure 6.3: Aromatic bonding Q1: What word describes the type of bond forming the hexagonal framework? Q2: What word describes the type of bond formed above and below the framework? Q3: What word describes the type of bond holding the hydrogens in place? Q4: How many electrons does each p-orbital contain? a) 1 b) 2 c) 3 d) 6 Q5: How many electrons are contained in total within the doughnut shaped regions? a) 1 b) 2 c) 3 d) 6

118 110 TOPIC 6. AROMATICS Q6: What type of hybridisation occurs in benzene? a) sp b) sp 2 c) sp 3 d) sp 4 The cyclic delocalisation of the electrons in benzene makes the molecule more stable than would be expected with a Kekulé type of structure. The term aromatic has come to mean any system that contains a ring of atoms stabilised by delocalised ß electrons. Bonding in benzene can be described in terms of sp 2 hybridisation, sigma and pi bonds and electron delocalisation. This can explain its unexpected stability. ffl 6.3 Benzene reactions Learning Objective To explain how benzene resists addition reactions but undergoes electrophilic substitutions, giving examples of reactions and describing the mechanisms. ffi Benzene tends to undergo electrophilic substitution reactions. The stable nature of the benzene ring system gives benzene the tendency to undergo reactions in which the aromatic character is preserved. The regions of high electron density above and below the hexagon attract reagents with a positive charge (electrophiles). The reaction which takes place is, in the first stage, similar to the electrophilic attack on ethene (see Topic 2 section 2.1.5), but differs in the second stage. The animation on the website shows the general mechanism for electrophilic substitutions on benzene with the reagent E + representing an electrophile with either a whole or a partial positive charge. This is shown in Figure 6.4. fi fl Figure 6.4: Electrophilic substitution The electrophile attacks one of the carbons atoms in the ring, forming an intermediate ion with a positive charge that can be stabilised by delocalisation around the ring. The intermediate ion then loses H + to regain aromatic character and restore the system. Four specific examples of electrophilic attack on benzene will now be considered. When writing reaction mechanisms for aromatic systems, it is often useful to represent them as Kekulé structures (the localised arrangements Figure 6.2 (b) and (c)); this will

119 6.3. BENZENE REACTIONS 111 be done in the four examples shown. It should always be remembered however that this is not a true picture of the benzene molecule. ffl Halogenation (attack by a halogen) Learning Objective To describe the mechanism involved in the electrophilic substitution of benzene by a halogen in the presence of a suitable catalyst. ffi Benzene undergoes electrophilic substitution with bromine or chlorine if a catalyst like iron(iii) bromide, iron(iii) chloride or aluminium(iii) chloride is present. The catalyst polarises the halogen molecule (here it is bromine) by accepting a pair of electrons from one atom and creating an electrophilic centre on the other atom (Figure 6.5). fi fl Figure 6.5: Bromine electrophile production This partially positive bromine atom attacks one of the carbons on the benzene and creates a carbocation which is stabilised by delocalisation on the ring. The intermediate then loses a hydrogen ion by heterolytic fission from the benzene, regaining the aromatic character and forming bromobenzene. The iron(iii) bromide is regenerated (Figure 6.6).The same mechanism applies if chlorine is used with a suitable catalyst. Figure 6.6: Electrophilic substitution of bromine on benzene Answer these questions concerning the similar reaction of chlorine with benzene in the presence of aluminium(iii) chloride.

120 112 TOPIC 6. AROMATICS Q7: What kind of reaction would take place? a) Electrophilic addition b) Nucleophilic substitution c) Electrophilic substitution d) Nucleophilic addition Q8: What is generally the best name for the product? a) Chlorophenyl b) Chlorobenzene c) Benzene chloride d) Phenylchloride Q9: What type of fission does the chlorine molecule undergo in this reaction? a) Nuclear b) Electrolytic c) Homolytic d) Heterolytic Q10: What type of organic ion is created as an intermediate in this reaction? ffl Nitration Learning Objective To describe the mechanism involved in the electrophilic substitution of benzene by a nitronium ion leading to the formation of nitrobenzene. ffi Benzene will react with nitric acid when a mixture of concentrated nitric acid and concentrated sulphuric acid (known as a nitrating mixture) is used and the temperature is kept below 55 ffi C (Figure 6.7). The nitrating mixture generates the nitronium ion, NO 2 +. fi fl Figure 6.7: Nitrating mixture The nitronium ion attacks the benzene ring and creates an intermediate ion that is stabilised by the same mechanism as in the halogen reactions. The ring then regains its aromatic stability by the removal of a hydrogen ion and the product molecule nitrobenzene is formed (Figure 6.8).

121 6.3. BENZENE REACTIONS 113 Figure 6.8: Electrophilic substitution of nitronium on benzene This is an important industrial reaction since nitrobenzene is used to produce aniline (phenylamine) (Figure 6.14), which is important in the dyestuff industries. If the temperature is allowed to rise above 55 ffi C, further substitution of the ring can take place and small amounts of the di- and tri- substituted compounds can result (Figure 6.9). Figure 6.9: Di- and trinitrobenzene structures Answer these questions. Q11: What type of reactant is the nitronium ion? a) carbanion b) nucleophile c) carbocation d) electrophile Q12: What type of intermediate is formed? a) free radical b) carbocation c) carbanion d) nitronium anion Q13: Describe how the intermediate is stabilised. Try writing a sentence before checking your answer.

122 114 TOPIC 6. AROMATICS Q14: What name would you suggest for structure (x)? (Figure 6.9). a) 1,1-dinitrobenzene b) 1,2-dinitrobenzene c) 1,3-dinitrobenzene d) 1,4-dinitrobenzene ffl Sulphonation Learning Objective To descibe the mechanism involved in the electrophilic substitution of benzene by sulphuric acid leading to the formation of benzenesulphonic acid. ffi Benzene will react with concentrated sulphuric acid if the reactants are heated together under reflux for several hours. If fuming sulphuric acid is used (sulphuric acid enriched with sulphur trioxide) under cold conditions -SO 3 H substitutes onto the ring. This suggests that the active species is sulphur trioxide. The sulphur atom in the sulphur trioxide molecule carries a partial positive charge and can attack the benzene ring. The mechanism is the same as that shown for nitration (Figure 6.10). fi fl Figure 6.10: Sulphonation of benzene Many detergents contain salts of alkylbenzenesulphonic acid that are produced from arene sulphonic acids. They are an important industrial chemicals. Q15: In this reaction is the sulphur trioxide acting as a nucleophile or a electrophile? Q16: Does it react with the benzene in an addition or a substitution reaction? ffl Alkylation Learning Objective To describe the mechanism involved in the electrophilic substitution of benzene by an alkyl group in the presence of a suitable catalyst. ffi The aluminium chloride catalyst mentioned in the halogenation reactions can be used to increase the polarisation of halogen containing organic molecules like halogenoalkanes. This allows electrophilic carbon atom to attack the benzene ring and builds up sidechains. The reaction is called a Friedel-Crafts reaction, after the scientists who discovered it. fi fl

123 6.3. BENZENE REACTIONS 115 Figure 6.11: Halogenoalkane and aluminium chloride The catalyst increases the polarity of the halogenoalkane producing the electrophilic centre that can attack the benzene ring (Figure 6.11). The carbocation formed is stabilised by delocalisation and the intermediate so formed regains its stability by loss of a hydrogen ion forming an alkylbenzene (Figure 6.12). Answer these questions. Figure 6.12: Alkylation of benzene Q17: What word describes the role of the aluminium(iii) chloride? Q18: What type of fission does the halogenoalkane undergo in this reaction? a) nuclear b) homolytic c) heterolytic d) electrolytic Q19: Name the chloroalkane necessary to produce ethylbenzene in a Friedel-Crafts reaction. ffl Benzene reactions summary Learning Objective To give examples of electrophilic substitution reactions on a benzene system, name the reactants, products and reagents used and describe the mechanism. ffi Visit the online version of this Topic to try the drag and drop exercise summarising the main electrophilic substitution reactions on the benzene ring system Answer these questions about Figure 6.13 showing a summary of the main electrophilic substitution reactions on the benzene ring system. fi fl 10 min

124 116 TOPIC 6. AROMATICS Figure 6.13: Benzene electrophilic substitution Q20: What reagent would be used in reaction 1? a) Cl 2 /AlCl 3 b) RCl/AlCl 3 c) Br 2 /FeBr 3 d) RCl/H 2 SO 4 Q21: What reagent would be used in reaction 2? a) Cl 2 /AlCl 3 b) RCl/AlCl 3 c) Br 2 /FeBr 3 d) RCl/H 2 SO 4 Q22: What name would you give to the product in reaction 3? (carried out at room temperature).

125 6.4. ACIDITY OF PHENOL AND BASICITY OF PHENYLAMINE (ANILINE) 117 Q23: What name would you give to the product in reaction 4? Q24: What name would you give to the product in reaction 6 if the reagent used in 5 was iron(iii) bromide and bromine? Benzene undergoes electrophilic substitution reactions. These include chlorination and bromination to produce chlorobenzene and bromobenzene respectively, nitration to produce nitrobenzene, sulphonation to produce benzenesulphonic acid and alkylation to produce alkylbenzenes. ffl 6.4 Acidity of phenol and basicity of phenylamine (aniline) Learning Objective To describe the effect that delocalisation of charge over the benzene ring has on the acidity of phenols and the basicity of aromatic amines compared with their aliphatic equivalents. ffi Phenol The delocalised electrons present in a phenyl ring have a strong influence on the properties of functional groups attached to the aromatic system. Phenol and phenylamine (also called aniline) (Figure 6.14) are two compounds which show modified pk a values when compared to aliphatic molecules with the same functional groups. fi fl Figure 6.14 Table 6.2 shows a comparison of phenol and ethanol in terms of the pk a value and the reaction with sodium. In each case the sodium is reacting with hydrogen ions to produce hydrogen gas. The lower the pk a of an acid, the stronger the acid. (See Unit 2 Topic 4 section 4.3 if revision is needed). Table 6.2 Alcohol pk a Reaction with sodium ethanol 18.0 phenol 9.9 slow release of hydrogen fast release of hydrogen

126 118 TOPIC 6. AROMATICS (Note: the phenol reacting with sodium is dissolved in a little solvent, as phenol is normally solid.) Q25: Which reacts most rapidly with sodium? a) ethanol b) phenol Q26: Which contains the highest concentration of hydrogen ions? a) ethanol b) phenol Q27: Does the pk a value given back up your answer? a) yes b) no Q28: The pk a value for water is Which of these is the strongest acid? a) ethanol b) phenol c) water Phenol is acting as a stronger acid than ethanol because ionisation ( Figure 6.15) produces the phenoxide ion. The negative charge on the oxygen can be delocalised over the aromatic ß system. Figure 6.15: Phenol dissociation The resonance structures that are possible for the phenoxide ion stabilise it and make it less likely to accept a hydrogen ion (Figure 6.16). The equilibrium position for the phenol dissociation therefore lies more to the right.

127 6.4. ACIDITY OF PHENOL AND BASICITY OF PHENYLAMINE (ANILINE) 119 Figure 6.16 Compared with ethanol, which would have to carry the negative charge without delocalisation when it ionised, phenol is therefore a stronger acid than the aliphatic alcohols. Although weak, phenol can undergo neutralisation reactions and reacts with active metals in the same way as other acids to give a salt and hydrogen gas. Phenylamine (aniline) Phenylamine (also called aniline) is an aromatic amine and, like other amines, is basic. It is able to accept a hydrogen ion and form the phenylammonium ion (Figure 6.17). Figure 6.17: Phenylamine acting as a base The ability of phenylamine to act as a base, and a comparison with other amines and ammonia, is illustrated in Table 6.3 which shows values of pk a for the conjugate acids. The lower the value of pk a of the conjugate acid, the weaker the base. (See Unit 2 Topic 4 section 4.4 if revision is needed). Table 6.3 Base Conjugate acid (ion) pk a ammonia ammonium 9.3 ethylamine ethylammonium 10.7 diethylamine diethylammonium 11.1 phenylamine phenylammonium 2.7

128 120 TOPIC 6. AROMATICS Q29: Which conjugate acid ion has the highest value of pk a? a) ammonium b) ethylammonium c) diethylammonium d) phenylammonium Q30: Which conjugate acid ion has the lowest value of pk a? a) ammonium b) ethylammonium c) diethylammonium d) phenylammonium Q31: Which base is the strongest? a) ammonia b) ethylamine c) diethylamine d) phenylamine Q32: Which base is the weakest? a) ammonia b) ethylamine c) diethylamine d) phenylamine Phenylamine(aniline) is considerably less basic than any of the other substances in the table. This is because the lone pair of electrons on the nitrogen atom is delocalised around the aromatic ring. This lone pair is therefore less available to act as a base and combine with a hydrogen ion. The delocalisation can be represented by these resonance structures (Figure 6.18). Figure 6.18

129 6.5. SUMMARY 121 The equilibrium shown in (Figure 6.17) therefore lies to the left. 6.5 Summary The unexpectedly stable structure of benzene can be described in a variety of ways: 1. The Kekulé description involving two different hexagonal forms with double bonds. 2. A description involving sp 2 hybridisation resulting in sigma and pi bonding and delocalisation of the p orbital electrons into a pi bond system above and below the hexagon of carbon atoms. All of the descriptions of the bonding in benzene can be useful in understanding the stability of benzene and the reaction mechanisms involved in electrophilic substitution reactions. Electrophilic substitution in benzene involves the attack of a positive or partly positive species on the delocalised electron cloud and, in the first instance, the formation of monosubstituted benzene compounds. The presence of delocalised electrons creates stabilised carbocation intermediates in the phenyl group during electrophilic substitution. The presence of delocalised electrons in the phenyl group can also be used to explain: 1. The stronger acidic nature of phenol compared with aliphatic alcohols. 2. The weaker basic nature of phenylamine(aniline) compared with other aliphatic amines. 6.6 Resources Higher Still Support: Advanced Higher Chemistry - Unit 3: Organic Chemistry, Learning and Teaching Scotland, ISBN Chemistry in Context: Hill and Holman, Nelson, ISBN Organic Chemistry: J. McMurray, Brooks/Cole Publishing, ISBN Chemistry: A. and P. Fullick, Heinemann, ISBN Chemical Ideas: Salters advanced chemistry, Heinemann, ISBN Website for representing molecules in different ways, for hybridisation and reaction mechanisms:

130 122 TOPIC 6. AROMATICS Website for IUPAC naming of organic compounds: End of Topic test An online assessment is provided to help you review this topic.

131 123 Topic 7 Stereoisomers Contents 7.1 Introduction Stereoisomerism Geometric isomerism Optical isomerism Summary Resources End of Topic test Prerequisite knowledge Before you begin this Topic, you should be able to: define isomers as organic compounds which have the same molecular formula but different structural formulae (Standard Grade). Learning Objectives After studying this Topic, you should be able to: state that in stereoisomers, the same atoms are bonded together in the same order but have a different arrangement in space making them non-superimposable; describe geometric isomerism (cis-trans isomerism) in terms of lack of free rotation around a bond; describe some differences in physical and chemical properties between geometric isomers; explain how optical isomerism occurs in compounds with four different groups arranged around a carbon atom and that this usually leads to chiral molecules which are non-superimposable mirror images of each other; state that, apart from optical activity, optical isomers have identical physical and chemical properties except when in a chiral environment; state that in biological systems only one of a pair of optical isomer is usually present.

132 124 TOPIC 7. STEREOISOMERS ffl 7.1 Introduction Learning Objective ffi To revise knowledge of isomers from S.G. and Higher and from earlier Topics. Isomerism arises whenever there is more than one way to organise a given number of atoms. It is very common in organic chemistry. Note that when talking about isomers we are always referring to two or more molecules. One molecule is not an isomer by itself but can be an isomer of one or more other molecules. Look at the molecules in Figure 7.1. The use of molecular models, in addition to the Study Guide and the Scholar website, is strongly recommended at all stages of this Topic. A B fi fl Figure 7.1: Isomeric molecules Q1: To what class of compounds does A belong? Q2: To what class of compounds does B belong? These two substances are isomers. They both have the same molecular formula, C 2 H 6 O. However, the atoms in each are connected in a different order, C-O-C in one and C-C-O in the other. Isomers like this are known as structural isomers In this case, the two isomers belong to different homologous series but often the structural difference is more subtle (Figure 7.2). Figure 7.2: Structural isomers Q3: What is the structural difference between these two isomers?

133 7.2. STEREOISOMERISM 125 Learning Point Structural isomers have different physical and chemical properties and can even belong to different homologous series. ffl 7.2 Stereoisomerism Learning Objective To state that in stereoisomers, the same atoms are bonded together in the same order but have a different arrangement in space making them non-superimposable. ffi In stereoisomerism, the molecules have the same molecular formula and the same structural formula (the atoms are connected in the same order in each molecule). However, in each molecule, the atoms have a different three dimensional arrangement in space which makes them non-superimposable. This means that no matter how you twist and turn the molecules, one isomer cannot fit exactly on top of the other. There are two main types of stereoisomerism as shown in Figure 7.3. fi fl Figure 7.3: Isomerism In this Topic, we will concentrate on stereoisomerism in organic compounds, although this phenomenon also occurs in other compounds such as transition metal complexes.

134 126 TOPIC 7. STEREOISOMERS ffl Geometric isomerism Learning Objective To describe geometric isomerism (cis-trans isomerism) in terms of lack of free rotation around a bond. To describe some differences in physical and chemical properties between geometric isomers. ffi Geometric isomerism can arise when there is lack of free rotation around a bond, often a C=C bond. Consider 1,2-dichloroethene. There are two geometric isomers (Figure 7.4). fi fl Figure 7.4: Dichloroethene isomers They exist because rotation about the C=C bond is very difficult. In Topic 2, the C=C bond was described as consisting of a ff bond and a ß bond (Figure 7.5).

135 7.2. STEREOISOMERISM 127 Figure 7.5: Carbon to carbon double bond Rotation is restricted because this would involve breaking the ß bond. The two geometric isomers are not superimposable. Geometric isomers caused by restricted rotation around a bond are distinguished by using the terms - cis and trans : cis - both groups are on the same side of the double bond, trans - the groups are on opposite sides of the double bond ( trans means across). Molecule (a) in Figure 7.4 is trans-1,2-dichloroethene, while molecule (b) in Figure 7.4 is cis-1,2-dichloroethene. A similar situation occurs with alkenes containing 4 or more carbon atoms, e.g. Figure 7.6 Figure 7.7 3D molecules Experiment with the live 3D molecules available on the website. Q4: Name the molecule in Figure min Q5: Name the molecule in Figure 7.7.

136 128 TOPIC 7. STEREOISOMERS Geometric isomerism can also arise in disubstituted cycloalkanes, e.g. Figure 7.8 Q6: Name the molecule in Figure 7.8(a). Q7: Name the molecule in Figure 7.8(b). Physical properties Look at the information in the table and answer the questions that follow. 15 min Isomer mp ( ffi C) bp ( ffi C) cis-but-2-ene trans-but-2-ene cis-1,2-dichloroethene trans-1,2-dichloroethene cis-pent-2-ene trans-pent-2-ene Q8: Look at the melting points. In general, which isomer has the higher melting point? a) cis b) trans Q9: Which isomer has the stronger intermolecular forces? a) cis b) trans The differences in melting point for each pair of isomers can be explained in terms of the shapes of the molecules. This determines how closely the molecules can pack together.

137 7.2. STEREOISOMERISM 129 Q10: What effect will the packing have on the strength of the intermolecular forces? a) The closer the molecules pack, the weaker the van der Waals forces b) The closer the molecules pack, the stronger the van der Waals forces c) The closer the molecules pack, the weaker the repulsive forces d) The further apart the molecules, the stronger the intermolecular forces Q11: In general, molecules of which isomer can be more closely packed? a) cis b) trans The boiling points of the isomers of 1,2-dichloroethene are significantly different. Consider the types of bond within these molecules. Q12: Which of these correctly describes the types of bond present? a) pure covalent only b) polar covalent only c) pure covalent and polar covalent d) pure covalent and ionic Q13: By looking at the shape, and considering the bonding within the molecules, suggest an explanation why the cis-isomer has a higher boiling point than the trans-isomer. Geometric isomers often display differences in physical properties such as melting point and boiling point. Sometimes geometric isomers can show differences in chemical properties although this is much less common than for structural isomers. A simple example involves the geometric isomers of butenedioic acid (Figure 7.9). Figure 7.9: Isomers of butenedioic acid

138 130 TOPIC 7. STEREOISOMERS Cis-butenedioic acid readily eliminates water on heating. The product is known as a cyclic anhydride (Figure 7.10). 60 min Figure 7.10: Dehydration of cis-butenedioic acid On the other hand, trans-butenedioic acid cannot eliminate water under the same conditions since the two carboxyl groups are on opposite sides of the double bond (Figure 7.9). ffl Case Study - Fats and edible oils Learning Objective To study examples of relevance of geometric isomerism in consumer products such as the health issues associated with cis and trans fatty acids ffi This case study illustrates the differences between fats and edible oils, including the differences in physical and chemical properties between cis and trans isomers of unsaturated fatty acids. The biological importance of cis-fatty acids is also discussed. PLEASE NOTE: Detailed knowledge of the names, their structures and how they work does not have to be memorised. Read through the case study. If you are given it as a homework exercise, follow the instructions about preparation. If you are only interested in gathering information in order to complete the Fats and oils exercise, you may skim through looking for the necessary information. Preparation If you are making a presentation to a group then you will need to prepare: fi fl a hand-out information sheet. audio visual illustrations (blackboard or OHP or data projection). a short script for yourself. Include within your presentation: the similarities in structure between fats and edible oils the differences between fats and edible oils

139 7.2. STEREOISOMERISM 131 differences in behaviour between saturated and unsaturated fatty acids the biological importance of unsaturated fats Fats and oils Fats, also called lipids, are the third main class of food type needed in the human diet, the others being proteins and carbohydrates. Oils are simply fats that are liquid at room temperature. In foods derived from animals, the main sources of fat are dairy produce and meat although most foods contain some fat. Some of the richer vegetable sources of dietary fat are nuts and seeds, soyabeans, olives and peanuts. Fats are an important component of our diet and at least a minimum intake is required. However, many health problems, particularly in the western world, are associated with an excessive intake of fat. The main functions of fat in the body are as an energy reserve and for insulation. Fats can be burned to release energy when we need it and are not getting enough from the carbohydrates in our diet. Fatty tissue around internal organs help to protect them from trauma and temperature change by providing padding and insulation. They also have other uses. They are important in transporting other nutrients such as the vitamins A, D, E and K which are not water-soluble. Fats also form an essential part of the cell membrane. Finally, they are also a source of essential fatty acids. Structure of Fats Triglycerides are by far the most common type of fat and make up about 95% of the lipids in food and in our bodies. All triglycerides have a similar structure (Figure 7.11). They are esters formed from the condensation of fatty acids with glycerol (propane-1,2,3-triol). Fatty acids are long, straight chain carboxylic acids which contain an even number of carbon atoms, typically between 12 and 20 atoms. Figure 7.11: Structure of a saturated fat The three fatty acids present in a triglyceride are generally different, giving rise to a large number of possible combinations. Fats can be classified as saturated or unsaturated according to the nature of the fatty acids present. The fatty acids in turn can be classified as saturated, monounsaturated or polyunsaturated depending on the presence and number of C=C bonds (Table 7.1).

140 132 TOPIC 7. STEREOISOMERS Stearic Palmitic Oleic Linoleic Linolenic Arachidonic Physical properties Table 7.1: Examples of fatty acids Acid Formula Type mp ( C) C 17H 35COOH saturated C 15H 31COOH C 17H 33COOH C 17H 31COOH C 17H 29COOH C 19H 31COOH saturated monounsaturated polyunsaturated polyunsaturated polyunsaturated Saturated fats have higher melting points than unsaturated fats. This can be explained in terms of differences in the packing of the molecules in the solid state. The long hydrocarbon chains in saturated fat molecules can be packed quite closely together which maximises the van der Waals forces (Figure 7.11). o Figure 7.12: Structure of a cis-unsaturated fat Unsaturated fats (oils) contain one or more C=C bond. Almost invariably the C=C bonds have a cis configuration. This introduces a kink into the hydrocarbon chain since there is restricted rotation around this bond. This makes close packing much more difficult (Figure 7.12). The intermolecular forces are weaker and so unsaturated fats have lower melting points than saturated fats. Figure 7.13: Structure of a trans-unsaturated fat

141 7.2. STEREOISOMERISM 133 It is worth pointing out that triglycerides made from trans unsaturated fatty acids can pack together much more closely (Figure 7.13) and so have higher melting points than their cis equivalents. Essential fatty acids About 40 different fatty acids occur naturally. Table 7.1 shows some of the most common. The most important essential fatty acids are linoleic, linolenic and arachidonic acids, collectively known as vitamin F (Figure 7.14). These are all polyunsaturated fatty acids that are needed by the body. However, they cannot be synthesised from other compounds in the human body, although if there is sufficient linoleic acid it can be converted to arachidonic acid. They must be obtained from vegetable oils in the diet. (The term essential is also used for those amino acids which must be obtained in the diet). Linoleic acid is found in most foods while linolenic acid is common in oily fish. Figure 7.14: Essential fatty acids These fatty acids are important for normal growth, especially of the blood vessels and nerves, and to keep the skin and other tissues youthful and supple through their lubricating properties. They are also essential precursors in the biosynthesis of a group of compounds known as prostaglandins. These are hormone-like substances present in small amounts in all body tissues and fluids. They are all C 20 carboxylic acids that contain a five-membered ring with two long side chains. Figure 7.15: Prostaglandin The basic structural similarity between arachidonic acid and the prostaglandin shown in Figure 7.15 is obvious. One step in the conversion involves the formation of a five

142 134 TOPIC 7. STEREOISOMERS membered ring between carbons 8 and 12 in the fatty acid chain. If the configuration of either of the two central C=C bonds in arachidonic acid was trans, such a transformation would be impossible. Prostaglandins have an extraordinarily wide range of biological effects, including controlling blood pressure and controlling inflammation. They are also involved in blood clotting, kidney function and the reproductive system and are the subject of a great deal of research to produce new drugs. Diet Many health problems are associated with an excessive intake of fat in the diet. Levels of fat intake are strongly linked to body weight and therefore obesity. Fats provide about 42% of the calories in the average American diet. A diet in which about 25% or less of total calories is derived from fat would be healthier and help to reduce blood cholesterol levels and risk of blood and heart disease. The type of fat in the diet is also important. Increasing the proportion of unsaturates and polyunsaturates in the diet has been shown to produce a significant reduction in blood cholesterol levels. Current suggestions for a healthy diet include: reduce the total amount of fat in the diet specifically reduce the saturated fat intake by reducing consumption of red meats and dairy produce reduce intake of hydrogenated fat found in cooking oils and margarines raise the levels of polyunsaturated fats found in vegetable oils such as corn oil and sunflower oil as well as in most seeds and nuts eat more fish, particularly oily fish like herring and mackerel rather than red meat and of course, keep blood pressure under control, don t smoke and exercise regularly. Summary exercise Either copy the table below or photocopy it and then complete it. The words and phrases in the wordbank will help. Some may be used more than once or not at all. Alternatively visit the website for a drag and drop version of this exercise.

143 7.2. STEREOISOMERISM 135 ffl Optical isomerism. Learning Objective To explain how optical isomerism can occur in compounds with four different groups arranged around a carbon atom. To explain that this leads to chiral molecules that are non-superimposable mirror images of each other. ffi In the same way as your left hand and your right-hand are mirror images of each other, many chemical compounds can exist in two mirror image forms. Right and left-hands cannot be superimposed on top of each other so that all the fingers coincide and are therefore not identical. A right hand glove does not fit a left hand and is said to be chiral. The hands and the compounds have no centre of symmetry, plane of symmetry or axis of symmetry. Chirality arises from a lack of symmetry. Lack of symmetry is called asymmetry. Look at this picture (Figure 7.16) and answer the questions. fi fl

144 136 TOPIC 7. STEREOISOMERS Figure 7.16 The lady is holding one hand in front of the mirror and one hand behind. Q14: Which hand is the lady holding in front of the mirror? a) left b) right Q15: Which hand appears to be in the mirror? Q16: Which word describes the property exhibited by these two hands? a) identical b) symmetrical c) asymmetrical Q17: Which of these objects could be described as chiral? a) golf ball b) pencil c) shoe d) methane molecule Carbon atoms and symmetry. Optical isomerism occurs in other compounds such as transition metal complexes. In this Topic however, we are concentrating on carbon based compounds. The four single bonds around a carbon atom are arranged tetrahedrally. If there are four different atoms or groups attached to the carbon, it is asymmetric. It can exist in two isomeric forms. If you have access to molecular models or a kit from which to build models, it is recommended that you build models of the two isomeric forms shown in the first picture (Picture 1). Use a different coloured sphere for each of the four attached groups. This will help in understanding why they are not identical.

145 7.2. STEREOISOMERISM 137 Picture 1 shows the two isomers side by side and an attempt to line them up identically is shown in Picture 4. The arrow shows one of the points of mismatch. It is not possible to superimpose one on the other. This exercise is developed in the activity Optical isomers of alanine in the next section of this Topic. The tetrahedral carbon atom with four different groups is asymmetric. Molecules containing one or more asymmetric carbon atoms are usually (but not always) also asymmetric. Asymmetric atoms or molecules are described as chiral with the carbon atom being called the chiral centre. They differ from each other only in that they are mirror images of each other and exhibit optical activity. Such isomers are called enantiomers or optical isomers.the crystals of enantiomers are mirror images of each other. A mixture containing equal amounts of each enantiomer is known as a racemic mixture. Learning Points Optical isomerism can occur in compounds with four different groups arranged around a carbon atom. This leads to chiral molecules that are non-superimposable mirror images of each other. Optical isomers are stereoisomers since the atoms are connected in the same order but the arrangement of the atoms in space is different Biological systems. ffl Learning Objective To explain how optical isomers are non-superimposable mirror images of each other and that in biological systems only one optical isomer of each organic compound is usually present. ffi Optical isomerism is immensely important in biological systems. In most biological systems only one optical isomer of each organic compound is usually present. For example, when the amino acid alanine is synthesised in the laboratory, a mixture of the two possible isomers (a racemic mixture) is produced (Figure 7.17). When alanine is isolated from living cells, only one of the two forms is seen. fi fl

146 138 TOPIC 7. STEREOISOMERS 10 min ffl Optical isomers of alanine Learning Objective ffi Figure 7.17 To be able to explain how optical isomers are non-superimposable mirror images of each other. This activity involves 3-dimensional representations of amino acids with text and questions. It will help you understand why only one optical isomer of each organic compound is usually present in biological systems. Visit the Scholar website to find an interactive activity in which you can manipulate the structure of alanine. This illustration (Figure 7.18) shows one of the enantiomers of alanine, drawn in a different way from Figure fi fl The carbon atom surrounded by four different groups is the chiral centre. As well as the hydrogen atom it has an amine group, a carboxyl group and a methyl group. Figure 7.18 Q18: Compare Figure 7.18 with the structures shown in Figure Which structure is it? a) Structure 1 b) Structure 2 If you have access to molecular models, you should build the molecule on the left of Picture 1 by starting with a carbon and using different coloured spheres for the four groups. (The suggested colours would be red for carboxyl, blue for amine, yellow

147 7.2. STEREOISOMERISM 139 for methyl and white for hydrogen.) Try to build the other enantiomer as the exercise progresses. Picture 1 shows two isomers side by side and when one of them is taken away and replaced with a mirror, the two images in Picture 2 look the same as in Picture 1. In Picture 3, the isomer that was removed is placed behind the mirror, showing that it is identical to the image. Picture 4 points out how the two isomers are non-superimposable because even with the two substituent groups on the left of each isomer lined up, the two arrowed groups do not match. Answer the following questions. Q19: The molecules shown in Picture 4 are a) superimposable b) non-superimposable Q20: The two molecules are said to be a) symmetric b) asymmetric

148 140 TOPIC 7. STEREOISOMERS Q21: The two molecules can be said to be a) identical b) chiral Optical isomers can occur in compounds with four different groups arranged around a carbon atom and this leads to chiral molecules which are non-superimposable mirror images of each other. In biological systems only one of a pair of optical isomer of each organic compound is usually present. The proteins in our bodies are built up using only one of the enantiomeric forms of amino acids. Only the isomer of alanine with structure 2 (Figure 7.19) occurs naturally in organisms such as humans. Since amino acids are the monomers in proteins such as enzymes, enzymes themselves will be chiral. The active site of an enzyme will only be able to operate on one type of optical isomer. An interesting example of this is the action of penicillin, which functions by preventing the formation of peptide links involving the enantiomer of alanine present in the cell walls of bacteria. This enantiomer is not found in humans. Penicillin can therefore attack and kill bacteria but not harm the human host, as this enantiomer is not present in human cells. Figure 7.19: Alanine reflections Polarised light ffl Learning Objective To explain that enantiomers have identical physical and chemical properties, except when in a chiral environment. ffi To describe the effect enantiomers have on plane polarised light. Enantiomers behave in the same way in ordinary test tube reactions. Physical properties like melting, density and solubility are also identical. But enantiomers behave differently when in the presence of other chiral molecules. For example, enantiomers can taste differently due to the chiral nature of the taste bud receptors in the tongue. One enantiomeric form of amino acids all taste sweet, the other isomer is sometimes bitter. Chemicals with different enantiomeric forms can smell different. One isomer of limonene (Figure 7.20) smells of oranges, the other of lemons. fi fl

149 7.2. STEREOISOMERISM 141 Figure 7.20: Limonene enantiomers Enantiomers have an opposite effect on plane polarised light. This is why they are said to be optically active. An instrument called a polarimeter can detect this effect. ffl The polarimeter Learning Objective To be able to explain how a polarimeter can be used to determine the effect an optically active isomer has on plane polarised light. ffi Go to the website to see the animations that illustrate how a polarimeter works and the effect optical isomers have on plane polarised light Look at the diagrams and answer the questions which follow. fi fl 10 min Figure 7.21: Polarised light The waves of normal light vibrate in all directions at right angles to the direction of travel. Some substances will allow light vibrating in only a single plane to pass through them, producing plane polarised light. See Figure Polaroid sunglasses use this property to reduce glare. Figure 7.22: Polarimeter A compound which, when inserted into a beam of plane polarised light can rotate it so that the light vibrates in a different plane, is said to be optically active. Enantiomers have this property, since solutions of one kind rotate the plane polarised light in a clockwise

150 142 TOPIC 7. STEREOISOMERS direction (measured as +x ffi ), and the other enantiomer rotates plane polarised light by exactly the same amount in the opposite direction (Figure 7.22) (measured as -x ffi ). A mixture containing equal amounts of each enantiomer (a racemic mixture) is optically inactive as the rotations cancel each other. Q22: The function of the polariser in a polarimeter is to: a) rotate normal light b) produce normal light c) rotate plane polarised light d) produce plane polarised light Q23: One enantiomer of a compound is found to give a reading of ffi when inserted into a polarimeter. What reading would you expect the same amount of the other enantiomer to produce in the polarimeter? a) 0 ffi b) ffi c) ffi d) +27 ffi Q24: What reading would you expect a mixture of equal amounts of the two enantiomers to produce? a) 0 ffi b) +27 ffi c) -27 ffi d) 360 ffi Optical isomers (enantiomers) have identical physical and chemical properties, except when in a chiral environment. However, they have an opposite effect on plane polarised light. Mixtures containing equal amounts of both optical isomers are optically inactive. Many medicines are produced as a mixture of enantiomers, only one of which is pharmacologically active, as it can prove costly to separate the isomers. One of the enantiomers of salbutamol used in the treatment of asthma, is 68 times more effective than the other (see Unit 3, Topic10, section 4). Great care has to be taken when using drugs with enantiomeric forms as this has led to tragedy in the past, specifically in the case of thalidomide (Figure 7.23). A mixture of the isomers was used to treat nausea during pregnancy, and one enantiomer, which was thought to be inactive, turned out to cause damage to the unborn child. Many handicapped babies were born before the drug was recognised as being responsible. Screening of pharmaceuticals has to be very thorough. Regulations were tightened significantly after the thalidomide tragedy to ensure that both enantiomeric forms of chiral drugs are tested.

151 7.2. STEREOISOMERISM 143 Figure 7.23: Thalidomide Mechanisms for substitution reactions In Unit 3, Topic 2, the S N 2 mechanism for nucleophilic substitution reactions of halogenoalkanes was described (Figure 7.24). The strongest evidence for this mechanism comes from the use of chiral compounds. Figure 7.24: SN2 mechanism If this mechanism occurs with a chiral compound, then the product will also be chiral but with the configuration inverted (a bit like an umbrella being blown inside out by a strong wind). Figure 7.25: SN1 mechanism, step1 Figure 7.25 shows the mechanism for the S N 1 reaction which occurs in two steps. Figure 7.25 shows the first step in the S N 1 mechanism. In the second step, a nucleophile will attack the planar carbocation to form the product.

152 144 TOPIC 7. STEREOISOMERS Q25: If the original halogenoalkane in this S N 1 reaction is chiral, what will the product be? a) chiral with the same configuration as the original b) chiral with inverted configuration c) a racemic mixture d) a non-chiral molecule Q26: Explain your answer to the previous question Summary Structural isomers are molecules with the same molecular formula but different structural formula (in each molecule, the atoms are connected in a different order). In stereoisomers, the same atoms are bonded together in the same order but have a different arrangement in space making them non-superimposable. Geometric isomers are a type of stereoisomer that arise due to lack of free rotation around a bond, often a C=C bond. Geometric isomers are labelled cis and trans, according to whether the substituents are on the same side or opposite sides of the bond. They normally display differences in some physical properties and sometimes chemical properties. Optical isomers are a type of stereoisomer with four different groups or atoms around a chiral centre. Such a centre is asymmetric (has no symmetry) and gives rise to a pair of isomers which are non superimposable mirror images of each other. These optical isomers are also called enantiomers. Optical isomers have identical physical and chemical properties except when in the presence of other chiral molecules. They also have an opposite effect on plane polarised light, rotating it by exactly the same amount but in opposite directions. Mixtures containing equal amounts of each enantiomer (a racemic mixture) are optically inactive as the rotations cancel each other. In biological systems only one optical isomer of each organic compound is usually present. 7.3 Resources Higher Still Support: Advanced Higher Chemistry - Unit 3: Organic Chemistry, Learning and Teaching Scotland, ISBN Chemistry in Context: Hill and Holman, Nelson, ISBN Organic Chemistry: J. McMurray, Brooks/Cole Publishing, ISBN Chemistry: A. and P. Fullick, Heinemann, ISBN Chemical Ideas: Salters advanced chemistry, Heinemann, ISBN

153 7.4. END OF TOPIC TEST 145 For the case study on Fats and edible oils : Fats and Oils: Unilever Educational booklet: Advanced Series, Unilever Website for animations of stereochemistry:: Website for IUPAC naming of organic compounds: End of Topic test An online assessment is provided to help you review this topic.

154 146 TOPIC 7. STEREOISOMERS

155 147 Topic 8 Elemental analysis and Mass spectrometry Contents 8.1 Introduction Elemental analysis Mass spectrometry Interpreting mass spectra Fragmentation patterns High resolution mass spectrometry Example analysis of an unknown compound Summary Resources End of Topic test Prerequisite knowledge Before you begin this Topic, you should be able to: manipulate empirical and molecular formulae (Higher); describe stereoisomerism in organic compounds (Topic 7). Learning Objectives After studying this Topic, you should be able to: calculate empirical formulae from elemental analysis data; explain the operation of a mass spectrometer, and be able to interpret simple mass spectra.

156 148 TOPIC 8. ELEMENTAL ANALYSIS AND MASS SPECTROMETRY 8.1 Introduction Organic compounds can show a very large degree of structural complexity. Remember, it took nearly a century to work out the basic chemical structure of DNA, from its isolation in 1869 by Friedreich Miescher to the double-helical structure of Watson and Crick in The methods available to these early chemists have now been refined, and the development of new instrumental techniques and powerful computers means that the working out of complex organic structures is almost routine today. Each of the methods described here, and in Topic 3.9, provides specific information about an organic compound, with these individual features combining ultimately to provide a full picture of the structure. ffl 8.2 Elemental analysis Learning Objective To determine % composition from combustion analysis data, and to work out empirical formulae. ffi From a chemical point of view, the most fundamental data about a compound are: which elements are present and how much of each are in the compound? This information leads to the simplest empirical formula for the material. The apparatus The percentage of carbon, hydrogen, nitrogen and sulphur in a compound is found by burning a known weight of the material in an oxygen atmosphere and measuring the weights of carbon dioxide, water, nitrogen and sulphur dioxide that are produced. Other elements known to be present (e.g. halogens) have to be measured by other methods. The equipment used is shown in a video Modern Chemical Techniques produced by the Royal Society of Chemistry. The output from the analysis gives the % composition for C, H, N and S. fi fl Q1: Combustion analysis cannot give the % oxygen in a compound. Can you suggest a reason why? The calculation Knowing the percentage composition of a compound can lead us to what is really required - the chemical formula. To convert from a percentage composition by mass to the ratio of the atoms in a substance, the percentages are divided by the appropriate atomic mass to give a ratio of moles, which are then converted to whole number ratios of atoms.

157 8.2. ELEMENTAL ANALYSIS 149 Example : Elemental analysis Analysis of an alcohol shows it to contain 37.5% carbon and 12.5% hydrogen. What is the formula? The % carbon + % hydrogen add up to only 50%. The question says that we are dealing with an alcohol (i.e. contains C,H and O), so we can assume that the remaining 50% of the compound is oxygen. Divide by atomic mass Divide to give whole numbers Element Carbon Hydrogen Oxygen % by weight / /1 50.0/16 Ratio of atoms (=3.125) (=12.50) (=3.125) Whole number ratio 3.125/3.125= /3.125= /3.125= 1 The formula for this compound is CH 4 O. Try the following problem. Q2: A liquid is found to contain 52.17% carbon and 13.04% hydrogen, with the remainder being oxygen. What is its formula? Q3: The answer to Q2 gives no information about the structure of this compound. From your knowledge of isomers, suggest two possible structures for this substance. Not only does the formula above give no indication of structure, but it only indicates the ratio of atoms in the compound. It is an empirical formula. Q4: A compound of C,H and O contains 40.0% carbon and 6.667% hydrogen. What is the empirical formula? The empirical formula CH 2 O would be obtained for methanal (molecular formula HCHO), but would also be the result from the analysis of ethanoic acid, CH 3 COOH (C 2 H 4 O 2 ), glyceraldehyde (C 3 H 9 O 3, HOCH 2 CHOHCHO ), glucose (C 6 H 12 O 6 ) or many other compounds. Obviously, some further information is required to enable a molecular formula to be found. Q5: A white solid is found to contain % carbon, % hydrogen and % nitrogen. What is its empirical formula? Q6: The compound in the previous question was found to have a molecular mass of 108. What is the molecular formula? Q7: An antibiotic contains C,H,N,S and O. Combustion of g of the compound in excess oxygen yielded g CO 2, g H 2 O, g SO 2 and g N 2. What is its percentage composition?

158 150 TOPIC 8. ELEMENTAL ANALYSIS AND MASS SPECTROMETRY Q8: What is the antibiotic s empirical formula? See further questions on page 219. ffl 8.3 Mass spectrometry Learning Objective To describe the various parts of a simple mass spectrometer, and understand their functions. ffi Mass spectrometry is a very powerful technique for obtaining information about molecular masses and the structure of molecules. It requires only a very small amount of sample. The instrument (Figure 8.1) is divided into four main sections: fi fl 1. The inlet system. This is maintained at a high temperature, so that any sample introduced will be vaporised rapidly. The interior of a mass spectrometer must be maintained at high vacuum to minimise collisions, so the injector and the rest of the instrument are connected to vacuum pumps. 2. The ion source. Some of the vaporised sample enters the ionisation chamber where it is bombarded with electrons. Provided the energy of the collision is greater than the molecular ionisation energy, some positive ions are produced from molecules in the sample material. Some of these molecular ions (sometimes called parent ions) will contain sufficient energy to break bonds, producing a range of fragments, some of which may also be positive ions (daughter ions). The parent ions and ion fragments from the source are accelerated and focussed into a thin, fast moving ion beam by passing through a series of focusing slits. 3. The analyser. This stream of ions is then passed through a powerful magnetic field, where the ions experience forces that cause them to adopt curved trajectories. Ions with the same charge (say, 1+) will experience the same force perpendicular to their motion, but the actual path of the ion will depend on its mass. Heavier ions will be deflected less than lighter ones. 4. The detector. Ions with the same mass/charge (m/z) ratio will have the same trajectory and can be counted by a detector. As the magnetic field strength is altered ions with different m/z ratios will enter the detector so that a graph of abundance (the ion count) against m/z values can be constructed. This is the mass spectrum of the sample compound(s).

159 8.3. MASS SPECTROMETRY 151 Figure 8.1: Diagram of mass spectrometer You can try a simple experiment at home. Get someone to roll golf balls and table tennis balls (heavy and light ions!) across a table. As the balls pass you, blow gently across their path. The two types of ball are about the same size, will experience about the same deflecting force from your breath, but will have different amounts of deflection owing to their different masses. ffl Interpreting mass spectra Learning Objective To understand how molecular ions and fragments are produced in a mass spectrometer, and to observe the general features of mass spectra. ffi When organic molecules are bombarded with energetic electrons, two basic processes can occur: An electron can be captured to form a negatively charged radical ion, M+e -! M.- Or an electron can be removed to form a positively charged radical ion, M+e -! M.+ +2e - fi fl

160 152 TOPIC 8. ELEMENTAL ANALYSIS AND MASS SPECTROMETRY (It is conventional in mass spectrometry to call a molecule M) This latter process is most common (80% of cases), so that positive ion mass spectrometry is mostly employed. The positive radical ion has one electron removed from a pair in a filled orbital (giving it an odd number of electrons). For simplicity at the level of Advanced Higher this radical electron (. ) is omitted from the representation of these ions. Molecular ions are, therefore, called M +. When a molecular ion fragments, it can form other ions (with either an odd or even number of electrons) and neutral molecules. The molecular ion and associated fragment ions will be measured in the mass spectrometer and the resulting trace is called the mass spectrum. The neutral molecules, and any negatively charged ions will not be registered. A typical mass spectrum (Figure 8.2) is shown for benzoic acid. It consists of a series of vertical lines, whose height represents the relative abundance of ions of a particular mass/charge (m/z) ratio. The most abundant peak is called the base peak and is assigned an abundance of 100%. Other abundances are given smaller percentages, relative to the base peak. The peak with the highest m/z value is often the molecular ion, but in cases where this is particularly unstable, its abundance might be very small, or undetectable. Figure 8.2: Mass spectrum of benzoic acid Q9: In the spectrum of benzoic acid (Figure 8.2), what is the m/z value of the base peak? Q10: What is the m/z value of the molecular ion?

161 8.3. MASS SPECTROMETRY 153 ffl Fragmentation patterns Learning Objective To work out some structural features of compounds from analysis of their mass spectra. ffi Molecular ions, formed by the collision of a molecule with energetic electrons, can be either stable enough to remain intact during their progress through the mass spectrometer, or will have sufficient energy imparted by the collision to break specific bonds and produce a pattern of fragment ions at the detector. The mass spectrum (Figure 8.3) of an aromatic hydrocarbon naphthalene, shown in Figure 8.4, has a stable molecular ion, C 10 H 8 +, giving a very large response at m/z 128, with only minor fragmentation peaks elsewhere. In this case, it is very clear that the base peak is the molecular ion and the compound has a molecular mass of 128. The molecular ion of naphthalene is so stable that a peak of m/z 64 is seen. This is a doubly charged ion, M 2+, with m = 128 and z = 2, so that m/z = 64. This type of fragmentation pattern is fairly typical of aromatic compounds, and is further evidence of the stability of this group of compounds (see Topic 6). fi fl Figure 8.3: Mass spectrum of naphthalene Figure 8.4 In contrast, the mass spectrum of 2-methylpentane (C 6 H 14 ), Figure 8.6, has only a small abundance of the molecular ion at m/z 86, but many fragment ions. This pattern of fragment ions reflects the fact that the lifetime of the molecular ion is so short

162 154 TOPIC 8. ELEMENTAL ANALYSIS AND MASS SPECTROMETRY (< 10-6 s) that fragmentation occurs in the source. 2-methylpentane has several bonds which have broken to give fragments with lifetimes > 10-5 s, long enough to have reached the detector. The bonds which break in the molecular ion will always be the weakest, and the fragment abundances reflect their stability (see Topic 3.1) so that observation of the fragmentation pattern can give information about the structure of the molecule. Figure 8.5: 2-methylpentane fragments In Figure 8.5 above, the wavy lines represent broken bonds. The numbers on each side of the line show the masses of the fragments produced when that bond breaks. 15 min ffl Figure 8.6: Mass spectrum of 2-methylpentane The fragments of 2-methylpentane Learning Objective To show that breaking specific bonds in a molecule leads to fragments detected in a mass spectrum. ffi Visit the online version of this Topic, where you are able to click on the fragment and see the corresponding mass spectrum position. Use Figure 8.6 to answer the following questions. fi fl Q11: From the fragmentation pattern in Figure 8.6, do you think that C - C or C - H bonds are weaker? Q12: Does this comparison of bond strengths agree with the bond enthalpy data on page 9 of the data booklet?

163 8.3. MASS SPECTROMETRY 155 Q13: The m/z 43 ions are most abundant in Figure 8.6. Which bond in 2-methylpentane is the weakest? The breaking of specific bonds in a molecule leads to the various ion abundance peaks in a mass spectrum. Fragmentation patterns are useful in two ways: If the energy of the electron beam is set to a standard value, the fragmentation pattern produced by a particular compound is very reproducible. The mass spectra of many thousands of compounds are now available in digitised format, so that a computer can search the data base of known mass spectra to identify an unknown sample whose mass spectrum has been obtained using electron-impact ionisation. A study of a mass spectrum can give information about the structure of the compound. ffl The fragments of 2-ethoxybutane Learning Objective ffi To show that the mass spectrum is dependent on the molecular structure. Figure 8.7 shows the six largest peaks in the mass spectrum of 2-ethoxybutane. fi fl 10 min Figure 8.7: Simplified mass spectrum of 2-ethoxybutane The molecular ion of 2-ethoxybutane can fragment as shown in Figure 8.8 to produce

164 156 TOPIC 8. ELEMENTAL ANALYSIS AND MASS SPECTROMETRY the ions in the mass spectrum. Can you identify which fragments of the structure are responsible for each of the lines on the mass spectrum? Figure 8.8: Fragments of 2-ethoxybutane The fragment m/z values are produced from the various bonds in the molecule breaking. The relative abundances of the fragment ions reflects the stability of the bonds breaking to produce them and the stability of the ions themselves. The molecular ion (M + ) of 2-ethoxybutane has m/z 102. The ion of m/z 87 is 15 m/z units less than M +, written (M - 15) +. This is a common difference, often due to the loss ofach 3 - group. Differences of 15 between m/z values infer the presence of a methyl group in the compound. Table 8.1 below shows some common mass differences and the groups they suggest. Table 8.1: Fragment masses Mass difference Suggested group 15 CH 3 17 OH 28 C=O or C 2 H 4 29 C 2 H 5 31 CH 3 O 45 COOH 77 C 6 H 5 Example : Structure from fragmentation pattern Compounds A and B both have the same molecular formula, C 3 H 6 O, but their molecular structures are different. The mass spectra of A and B are shown below.

165 8.3. MASS SPECTROMETRY 157 A1. For compound A, which group of atoms could be lost when the ion of m/z 43 forms from the ion of m/z 58? A2. Suggest a formula for the ion of m/z 43 in the spectrum of compound A. A3. Identify A.

166 158 TOPIC 8. ELEMENTAL ANALYSIS AND MASS SPECTROMETRY B1. For compound B, what atom or groups of atoms could be lost when i) the molecular ion changes into the ion of m/z 57? ii) the ion of m/z 57 changes into the ion of m/z 29? B2. Suggest formulae for the ions of m/z 28,29, and 57 in the spectrum of compound B. B3. Identify B. A1. The mass difference from 58 to 43 is 15. Reference to Table 8.1 indicates that a methyl group is likely to be responsible for this. A2. The ion of m/z 43 is probably due to loss of a methyl group from the molecule, so that a fragment of formula C 2 H 3 O would form the m/z 43 ion. A likely structure for this is CH 3 CO. A3. A is propanone, CH 3 COCH 3, which has major fragments of CH 3 CO and CH 3. B1. i) A loss of 1 mass unit from 58 to 57 can only be due to loss of a hydrogen atom. The molecule must have a single, loose H. ii) The mass difference from 57 to 29 is 28, probably due to loss of C=O. (See Table 8.1) B2. The ion of m/z 28 is the C=O + fragment; addition of 1 mass unit ( an H) to this would form acho + (aldehyde) as the m/z 29 peak. The m/z 57 is loss of hydrogen from the molecule, leaving a fragment of formula C 3 H 5 O +. We know this contains a C=O group, so is likely to be C 2 H 5 CO. B3. B is propanal, CH 3 CH 2 CHO, which loses the aldehyde H, then the CO. Here is a series of questions for you to try for yourself. The questions concern three of the four isomers of butanol. Q14: Draw the structures for the four isomers of butanol. The mass spectra X, Y and Z below are for three isomers of butanol, excluding 2-methylpropan-1-ol. There is one example of a primary alcohol, a secondary alcohol and a tertiary one, but not necessarily in that order.

167 8.3. MASS SPECTROMETRY 159 Relative Abundance Mass spectrum Z m/z 56 74(M+) Given that alcohols tend to fragment at the C-C bond adjacent to the C-OH group (i.e. R-CH 2 OH would produce R + and CH 2 OH + fragments), can you explain, with reasons, which spectrum is for which isomer? Q15: What is the m/z value for the molecular (M + ) ion of all three butanols? Answer with one integral value.

168 160 TOPIC 8. ELEMENTAL ANALYSIS AND MASS SPECTROMETRY If you would like to try to link the mass spectra with the appropriate molecular structures, have a go now, by working out the main fragments lost from the M + ion. Otherwise, the remaining questions will guide you through to an answer, with the conclusions summarised at the end. Q16: Spectrum X has a base peak at m/z 59. Using Table 8.1, what fragment has been lost from the molecular ion to produce this m/z 59 ion? Q17: Using the fragmentation information, which of the three butanols would be likely to produce a CH 3 fragment most readily? Q18: Closer examination of spectrum Y shows a base peak at m/z 45. What could be the structure of this ion? Q19: When the C-C bond adjacent to the the C-OH bond breaks in butan-1-ol what are the structures and m/z values of the fragments? Q20: In which spectrum are these most prominent? Q21: Summarise, with reasons, which spectrum applies to which structure. ffl 8.4 High resolution mass spectrometry Learning Objective To use very accurate molecular and fragment masses to distinguish between different formulae. ffi Some high resolution mass spectrometers provide an accuracy of 1 in 10 6 atomic mass units, so that very accurate masses can be obtained for the peaks in the mass spectrum. If you were asked for the molecular masses of, say, ethane (C 2 H 6 ) and methanal (HCHO), the data booklet information would give This is a 1 decimal place figure using an average of the masses of the isotopes normally present in these compounds on the Earth. However, a mass spectrometer would separate the individual molecules containing different isotopes; 12 C 1 H 2 16 O would appear as a different peak from 13 C 1 H 2 16 O, one mass unit higher. Indeed, the first use of mass spectrometry was to measure the masses of isotopes of elements. In addition, a high resolution mass spectrometer would be able to distinguish between the two compounds above because their accurate masses are not the same. Q22: Given that the atomic mass for 12 C is H is O is What is the accurate molecular mass of ethane ( 12 C 2 1 H 6 )? fi fl

169 8.5. EXAMPLE ANALYSIS OF AN UNKNOWN COMPOUND 161 Q23: What is the accurate molecular mass of methanal ( 12 C 1 H 2 16 O)? So, without further analysis, an accurate mass for these compounds would identify one against the other. This ability can be extended to give molecular formulae for unknown samples, the accurate mass being unique for a particular combination of atoms. This, of course, also applies to the fragments in a full mass spectrum, so that mass spectrometry with an accurate mass capability is a powerful technique for determining molecular structure. 8.5 Example analysis of an unknown compound A summary series of questions Here is a series of questions to enable you to see how the techniques described in this Topic can lead to a compound s structure. The questions concern analysis of a colourless, pleasant-smelling liquid, containing only carbon, hydrogen and oxygen. (Compound L) Q24: When g of the liquid is completely burned in an excess of oxygen, g of CO 2 and g of H 2 O are produced. What is the empirical formula? Q25: The mass spectrum of the compound is shown below (Figure 8.9). Figure 8.9: Mass spectrum of L What is the m/z value of the highest m/z ion? Q26: Does this ion s mass agree with the empirical formula; in other words, is this mass a simple multiple of the empirical formula?

170 162 TOPIC 8. ELEMENTAL ANALYSIS AND MASS SPECTROMETRY Q27: Therefore, what is the molecular formula for the compound? Q28: What is the m/z value of the base peak? Table 8.1 might help answer the next few questions Q29: From the mass difference from the molecular ion to base peak, what fragment might have been lost? Q30: What is a likely structure for the ion of m/z 77? Q31: What is a possible fragment composition lost when the m/z 105 ion forms the m/z 77 ion? Q32: These three fragments (m/z ) fortunately add to the molecular mass (136). So what is a likely structure for the compound? Q33: Name compound L. In this example, the evidence from mass spectrometry is good enough to establish the structure of the compound. However, this is not always the case and Topic 9 will describe other analytical techniques that are employed to determine a compound s structure. 8.6 Summary Elemental microanalysis, by combustion in oxygen, can be used to determine the masses of C, H, S and N in a known weight of an organic compound. Other elements are determined by other methods. The % composition information can then be used to find the empirical formula. Mass spectrometry can be used to determine accurate molecular masses and to indicate structural features of an organic compound. A mass spectrometer consists of: an injector, often at a high temperature to vaporise the sample; an ion source, where the molecules of sample are bombarded with electrons causing the formation of positive molecular ions. Depending on the stability of these ions, and the energy imparted by the collision, bonds in these molecular ions may break in a characteristic fashion to produce fragment ions The parent molecular ion and ion fragments are then accelerated into a fine, fast moving beam of ions; an analyser, consisting of a magnetic field which causes ions to be deflected in a curved trajectory dependent on their mass/charge ratio; a detector, which counts the ions of particular m/z value, to enable a mass spectrum of abundance against m/z to be constructed. Investigation of the mass spectrum can yield information about the molecular mass (if the parent, M + ion is present), and about the structure of the sample, from the distribution of the fragment ions.

171 8.7. RESOURCES 163 Where very accurate masses are available from the mass spectrometer, these alone can often provide chemical formulae for the parent and fragment ions. 8.7 Resources Chemical Ideas: Salters Advanced Chemistry, Heinemann, ISBN General Chemistry: Ebbing, Houghton Mifflin, ISBN Higher Still Support: Advanced Higher Chemistry - Unit 3: Organic Chemistry, Learning and Teaching Scotland, ISBN Organic Spectroscopy: Kemp, Macmillan, ISBN Mass Spectrometry Barker Analytical Chemistry by Open Learning series, John Wiley, ISBN Interpreting Organic Spectra Whittaker The Royal Society of Chemistry, ISBN Video: Modern Chemical Techniques, Royal Society of Chemistry, and associated book, ISBN End of Topic test An online assessment is provided to help you review this topic.

172 164 TOPIC 8. ELEMENTAL ANALYSIS AND MASS SPECTROMETRY

173 165 Topic 9 Infrared and Nuclear magnetic resonance spectroscopy and X-ray crystallography Contents 9.1 Introduction Infrared spectroscopy The infrared spectrometer Interpreting IR spectra Functional group identification Nuclear magnetic resonance spectroscopy The NMR spectrometer Interpreting NMR spectra Magnetic resonance image (MRI) scanning X-ray crystallography The diffraction of X-rays by crystals Electron density maps Summary Resources End of Topic test Prerequisite knowledge Before starting this Topic, you should be able to: manipulate empirical and molecular formulae (Higher); describe stereoisomerism in organic compounds (Topic 7). Learning Objectives By the end of this Topic, you should be able to: explain how infrared spectra are obtained by the interaction of radiation with molecules, and be able to interpret spectra in terms of functional groups;

174 166 TOPIC 9. INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY AND X-RAY CRYSTALLOGRAPHY state how nuclear magnetic resonance spectra are obtained and be able to use simple spectra to define the numbers and environments of hydrogen atoms in organic compounds; explain the use of X-ray crystallography to produce electron-density maps of organic molecules.

175 9.1. INTRODUCTION Introduction The methods described in the previous Topic to determine molecular formulae and provide structural information about organic compounds are extended in this Topic to include infrared and nuclear magnetic resonance spectroscopy and X-ray crystallography, which are used to identify functional groups in organic compounds, to investigate the different types of hydrogen atom environments in organic compounds, and to reveal the 3D structure, respectively. 9.2 Infrared spectroscopy When you sit in front of a fire, which gives out infrared (IR) radiation, you feel warm. This is because the IR radiation absorbed by molecules in your skin is of a suitable frequency to increase their vibrational energy. The greater a molecule s vibrational energy, the hotter it is. This property is used in "night-vision" equipment, which visualises changes in temperature as shown in Figure 9.1- a man holding a match. Figure 9.1: An infrared image of a man holding a match Infrared photography is used to show cloud cover in meteorology. The meteorological office web site, has a satellite picture of Europe updated every six hours, see ( Vibrational energy levels are quantised (see Unit 1, Topic 2.2), and measuring the particular IR frequencies that are absorbed by different molecules forms the basis of infrared spectroscopy. Units of electromagnetic radiation. In Unit 1 Topic 1.2, you saw a diagram giving the wavelength and frequency of the entire electromagnetic spectrum. The range of the infrared frequencies used for spectroscopy lies between to Hz. This corresponds to wavelengths bewteen 2.5 and 15 μm wavelength. In fact many chemists use wavenumber to describe the energies of IR radiation.

176 168 5 min TOPIC 9. INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY AND X-RAY CRYSTALLOGRAPHY ffl The units of IR spectroscopy Learning Objective ffi To revise the interrelations between frequency, wavelength and wavenumber. Put the correct numbers from the bank below into the table to show the comparison of frequency, wavelength and wavenumber over the range used in IR spectroscopy. fi fl Frequency (Hz) 1.2 x x10 13 Wavelength (μm) Wavenumber (cm ) x x10 13 In infrared spectra, the wavelength of the infrared radiation is expressed in wavenumbers, the reciprocal of length in centimetres (cm -1 ). How radiation and matter (molecules) interact. Simple diatomic molecules, for example halogens, have only one bond, which can vibrate only by stretching and compressing. The two halogen atoms can pull apart and then push together. The frequency of this vibration is determined by the mass of the two atoms and the strength of the bond. If you think of this system as two spheres attached by a spring, then the frequency of the natural, resonant vibration will decrease as the spheres increase in mass, and increase as the spring gets stronger. Intermediate Slower (heavier spheres) Faster (stronger spring) Similar principles apply to molecular vibrations. Table 9.1 shows the bond enthalpies for hydrogen halides. Table 9.1: Bond enthalpies for hydrogen halides Compound Bond Enthalpy/kJ mol -1 HCl 432 HBr 366 HI 298

177 9.2. INFRARED SPECTROSCOPY 169 Q1: Which molecule has the strongest bond? a) HCl b) HBr c) HI Q2: Which molecule has the largest molecular mass? a) HCl b) HBr c) HI Q3: Which molecule will have the largest vibration frequency? a) HCl b) HBr c) HI These diatomic molecules have a single, specific frequency at which they vibrate. When the whole range of IR frequencies is passed through a cell containing, say, HCl, only that frequency at which this molecule vibrates will be absorbed, leaving most of the radiation to pass through intact. Some of the HCl molecules will then vibrate at a higher quantum state. These molecules return to the lower state by giving up the energy in collisions which alters their velocity. Table 9.2 below gives the frequency of vibration for hydrogen halides and the IR absorption in wavenumbers. Table 9.2: Frequency and wavenumber for hydrogen halide molecules Compound Vibration frequency/hz IR absorption/cm -1 HCl 8.7 x HBr 7.7 x HI 6.7 x (When you see animations of molecular vibrations, remember that they are very considerably slowed down. The actual bond stretching frequency is around times per second!) Most molecules contain more than two atoms and this gives rise to modes of vibration other than simple stretching of bonds. Consider a water molecule. Each of the O-H bonds will have a characteristic stretching frequency, but these can interact and stretch either symmetrically or asymmetrically, giving rise to two slightly different IR absorption frequencies for the O-H bonds in a water molecule. See Figure 9.2

178 170 TOPIC 9. INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY AND X-RAY CRYSTALLOGRAPHY Symmetric stretch Asymmetric stretch Figure 9.2: Stretching modes in water molecules In addition to the stretching mode, molecules can also have bending modes. The hydrogen atoms in a methylene - (CH 2 group) can move in the same plane (either in a scissor motion, or a rocking motion), or can waggle in and out of the plane (again, either twisting or wagging). These are shown in Figure 9.3. All these vibration modes will have different frequencies. Scissor bend Twisting deformation Rocking bend Wagging deformation Figure 9.3: Bending modes Animations of these stretching and bending modes can be seen on the website. In general, it takes more energy to stretch the molecule than to bend it. Q4: Two wavenumbers applicable to the C-H bond are given below, one for stretching and one for bending. Think about which requires most energy, and therefore which will have the highest frequency. Which wavenumber applies to the C-H stretch? a) 1460 b) 2930 Molecules can vibrate in all of these fashions, and most of these will have characteristic infrared absorbance wavenumbers.

179 9.2. INFRARED SPECTROSCOPY 171 LEARNING POINTS Molecules vibrate with a set of specific frequencies dependent on the types of the bonds and the groups or atoms at the ends of these bonds. The energies associated with these vibrations are quantised. Infrared radiation causes parts or all of a molecule to vibrate in a higher vibrational state The infrared spectrometer Figure 9.4 shows a diagram of an infrared spectrometer. The source of infrared radiation has its beam split into two. One part of the beam passes through the sample; the other through a reference cell, which might contain the solvent, for example. The monochromator grating scans the wavelengths prior to a detector which compares the intensity of the two beams. The amplified signal is plotted as % transmission or absorbance. Figure 9.4: Infrared spectrometer There is a description of an IR spectrometer on the video "Modern Chemical Techniques" from the Royal Society of Chemistry, which describes the preparation of samples and the running of the spectrometer. Q5: In IR spectroscopy, discs of sodium chloride or potassium bromide ("salt flats") are often used to hold a thin film of liquid samples in the IR beam. What property must NaCl or KCl have that allows them to be used in this way? Figure 9.5 shows an IR spectrum for liquid paraffin (a mixture of alkanes).

180 172 TOPIC 9. INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY AND X-RAY CRYSTALLOGRAPHY Figure 9.5: IR spectrum of alkane mixture This spectrum shows the simplest IR spectrum, for an organic compound consisting of only two types of bonds ( C - C and C - H). Many other organic compounds will have this carbon backbone with the addition of other functional groups. In an earlier question you were given 2930 cm -1 as the C-H stretch wavenumber, and 1460 cm -1 as C-H bend. The complete IR spectrum for liquid paraffin (Figure 9.5) also has absorbance at 2880 cm -1, the asymmetric stretch, and a further bending mode at 1380 cm Interpreting IR spectra The IR spectrum for propanone is shown (Figure 9.6), together with its structure (Figure 9.7).

181 9.2. INFRARED SPECTROSCOPY 173 Figure 9.6: IR spectrum of propanone Figure 9.7: Structure of propanone The spectrum contains some wavenumbers from vibrations of the C - H bonds, but there is another strong absorbance due to the carbonyl group. Q6: What is its wavenumber? Give your answer to the nearest 100 cm -1. The C=O stretch absorbance is characteristic of ketones and ocurrs always around this wavenumber. The carbonyl group also occurs in other organic compounds, such as aldehydes, carboxylic acids and esters. The C=O stretch frequency changes slightly when the group is not directly connected to alkyl or aryl groups and will be found with other wavenumbers characteristic of that group. Another common group in organic compounds is hydroxyl, found in alcohols and

182 174 TOPIC 9. INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY AND X-RAY CRYSTALLOGRAPHY phenols. Figure 9.8 shows the IR trace of ethanol. Figure 9.8: IR spectrum of ethanol Q7: What is the wavenumber of the O -H stretch? Give your answer to the nearest 100 cm -1, for the maximum absorption wavenumber. The infrared correlation table Because these infrared absorbance wavenumbers are common for a particular group in any compound, tables of values have been built up by observation of a large number of compounds. Page 13 of the Data Booklet gives these data. You will notice that there is a range of values for a particular group, because the value is slightly altered by the surrounding groups. Look at the data booklet, particularly the different C=O stretch wavenumbers in the functional groups containing the carbonyl group. You will be using this table extensively in the next section Functional group identification The main use of IR spectra, which can be obtained quickly and cheaply, is to identify the presence of functional groups and the carbon backbone type in unknown organic compounds. Use the correlation table on page 13 of the data booklet to answer the following

183 9.2. INFRARED SPECTROSCOPY 175 questions regarding the IR spectra labelled X, Y and Z.

184 176 TOPIC 9. INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY AND X-RAY CRYSTALLOGRAPHY Q8: Which compound contains an alcohol group? a) X b) Y c) Z Q9: Which compound is not aromatic? a) X b) Y c) Z Q10: Which class of compound is present in X? Q11: Which of the following could be Z? a) Benzyl alcohol (C 6 H 5 CH 2 OH) b) Butan-1-ol (C 4 H 9 OH) c) Benzonitrile (C 6 H 5 CN) d) Acetonitrile (CH 3 CN) The "fingerprint" region of an IR spectrum. The region of an IR spectrum from 4000 to 1400 cm -1 contains many absorbance wavenumbers for specific bond types. Below 1400 cm -1 IR spectra typically have a number of absorbances not assigned to a particular bond. In fact, these relatively low energy vibrations are due to complex vibrations which are unique to that molecule.

185 9.2. INFRARED SPECTROSCOPY 177 Comparison of this region for an unknown material with a set of standards run under identical conditions will allow identification. Q12: The IR spectrum for an analgesic (pain killer) is shown, together with standard spectra for aspirin, paracetamol, and phenacetin (Taken from the Pharmaceutical Codex). By comparing the spectra, can you tell which analgesic it is? Analgesic

186 178 TOPIC 9. INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY AND X-RAY CRYSTALLOGRAPHY a) Aspirin b) Paracetamol c) Phenacetin 9.3 Nuclear magnetic resonance spectroscopy Some atomic nuclei, such as those for 1 H and 13 C have a nuclear spin, rather like electrons (Unit 1, Topic 2.5). When an electrically charged object moves (for example a nucleus spins), it will create a magnetic field around it. When these nuclei are in normal conditions the spins and associated magnetic fields will be randomly distributed. However, if these nuclei are put into a strong magnetic field the nuclei (behaving like a compass needle in the Earth s magnetic field) will line up along the field. This can happen in only two ways because of quantum restrictions: some will align with their tiny magnetic fields in the same direction as the large field; others will align in an opposed direction. This change in condition is illustrated in Table 9.3. Table 9.3: Nuclei alignment without and with a magnetic field. These two states have slightly different energies, with the opposed state (dark spheres) being higher. If electromagnetic radiation of a suitable frequency is applied to nuclei in a strong magnetic field some will absorb energy and flip to the higher energy quantum state. This radiation must have the same frequency as the rotating nuclei, so the two are in resonance; hence the term nuclear magnetic resonance. (Much of the NMR data is obtained for 1 H, protons, so that the term proton magnetic resonance is sometimes

187 9.3. NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY 179 used.) The energy required for this transition is in the radio frequency region of the spectrum, typically from 60 MHz to 1000 MHz. Very strong magnetic fields are required for NMR, so that superconducting electromagnets cooled in liquid nitrogen are used. As the nuclei return to the lower state (relax) they emit radiation which can be detected The NMR spectrometer The equipment used to obtain low resolution NMR spectra is shown in Figure 9.9. Figure 9.9: Diagram of an NMR spectrometer The sample is dissolved in CDCl 3 or CD 3 COCD 3, "deuterated" solvents with hydrogens replaced by deuterium (D or 2 H). Q13: Why do you think deuterated solvents are used? a) They are denser than normal solvents so will not fly out of the sample tube so easily. b) They contain no 1 H which would swamp the NMR spectrum of the sample with its signal. c) They can be obtained in a purer state than normal solvents.

188 180 TOPIC 9. INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY AND X-RAY CRYSTALLOGRAPHY Absolute values are difficult to obtain, so the ppm values are obtained by reference to a standard arbitrarily assigned the ffi value 0. The standard is tetramethylsilane (TMS), see Figure 9.10, which has an NMR signal well separated from those found in most organic molecules. Figure 9.10: Structure of tetramethylsilane (TMS) Q14: Look at the structure of TMS (Figure 9.10). How many different chemical environments are there for the 12 H atoms? a) 1 b) 3 c) 4 d) 12 The Royal Society of Chemistry video "Modern Chemical Techniques" shows the operation of an NMR spectrometer. If the external magnetic field was the only influence on the nuclei, you would be able to detect the different response frequencies from, say, 1 H and 13 C, but since all organic compounds contain carbon and hydrogen, this would not be much use. However, there is another factor. All nuclei in atoms are surrounded by electrons, which, because they have their own spin, will shield the nuclei from the total effect of the magnetic field. The effective field experienced by any given nucleus is, therefore, modified by the extent to which it is surrounded by electrons. Look at the structure of a methanol molecule (Figure 9.11). Figure 9.11: Methanol Q15: Around which hydrogens (methyl or hydroxyl) are the electron clouds denser? (Hint: think about electronegativity) Q16: So which hydrogens (methyl or hydroxyl) will feel the effects of the external field more? The proton NMR spectrum of methanol (Figure 9.12) shows two different signals for the two different hydrogens in the different magnetic and chemical environments.

189 9.3. NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY 181 Methanol TMS chemical shift / ppm 0 Figure 9.12: NMR spectrum of methanol The two signals are said to have different chemical shifts. In this case the hydrogen atom in the -OH group is shifted more than those in -CH 3. Observe the relative size of the two signals from CH 3 and OH Q17: How many hydrogen atoms are in the "methyl" environment? Q18: How many hydrogen atoms are in the "hydroxyl" environment? Q19: What would you estimate is the ratio of peak areas for methyl H and hydroxyl H signals? a) 1to3 b) 1to1 c) 3to1 Since it is difficult to measure areas, NMR spectra often have an integration signal, generated electronically, overlaid on the NMR signal. The height of this line is proportional to the area under the NMR response. (See Figure 9.13) LEARNING POINT The areas under the signals in an NMR spectrum are in the ratio of hydrogen atoms in that part of the molecule. The chemical shift (ffi) values of protons ( 1 H) in different chemical groups in a molecule have approximately constant values, with respect to tetramethylsilane ffi = 0. Because the different signals from protons ( 1 H) in a compound reflect their chemical environment, proton nuclear magnetic resonance is one of the most useful instrumental techniques for organic chemists.

190 182 TOPIC 9. INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY AND X-RAY CRYSTALLOGRAPHY Interpreting NMR spectra Determining structure using NMR The NMR spectrum of a hydrocarbon is shown in Figure Hydrocarbon W TMS chemical shift / ppm 0 Figure 9.13: NMR spectrum of hydrocarbon W Q20: From the spectrum, how many different environments are there for hydrogen atoms? Q21: What is the ratio of areas for these peaks? Q22: From the correlation table (page 15 of the data booklet), can you identify the type of hydrogen in the larger peak? a) RCH 3 b) RCH 2 R c) ArCH 3 d) ArH Q23: What type of hydrogens are in the smaller peak? a) RCH 3 b) RCH 2 R c) ArCH 3 d) ArH You might find it useful to summarise the information in a table. Table 9.4: NMR table Peak 1 Peak 2 ffi (ppm) Type of H ArCH 3 ArH Number of H atoms 3 5 Group CH 3 C 6 H 5

191 9.3. NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY 183 Q24: What is the hydrocarbon? Q25: The proton NMR spectrum for a substance U, of formula C 2 H 5 OBr is shown in Figure Compound U TMS chemical shift / ppm 0 Figure 9.14: NMR spectrum for U. On paper, summarise the information in a table like the one above (Table 9.4), then name the compound. An explanation for the answer will be given in the display answer. The next three questions refer to N,N - diethylphenylamine (C 10 H 15 N), shown in Figure Figure 9.15: N,N - diethylphenylamine Q26: How many different types of H environment are there in the molecule? a) 1 b) 2 c) 3 d) 5

192 184 TOPIC 9. INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY AND X-RAY CRYSTALLOGRAPHY Q27: What is the ratio of H atoms in these groups? a) 2:1 b) 4:5:6 c) 1:2:3 d) 3:2:5 Q28: Can you predict the chemical shift (ffi) ranges for this compound? Q29: Sketch the NMR spectrum for N,N-diethylphenylamine Magnetic resonance image (MRI) scanning Our bodies consist largely of water, which exists in a large number of different environments in each tissue. Just as hydrogen atoms in molecules are shielded to different extents depending on the surrounding atoms, the protons in water in tissues will experience slight differences in a strong external magnetic field, and so will absorb slightly different radiofrequency radiation. The part of the body being investigated is moved into a strong magnetic field. By using computers to process the absorption data, a series of images of the different water molecule environments, is built up into a picture of the body s tissues. It is assumed that MRI scanning is harmless to health, unlike other imaging processes (such as CAT scans) which involve low doses of ionising X-rays. MRI scanning is particularly good for brain tissue where there is a large amount of fatty lipids which provides a different environment for water compared with the other tissues. The MRI image opposite shows fluid collection in the region that separates the brain from the skull. This is a blood clot which applies pressure to the brain and is very dangerous, perhaps even fatal.

193 9.4. X-RAY CRYSTALLOGRAPHY X-ray crystallography Introduction The previous sections have described methods for investigating the ways in which the atoms in a molecule are connected chemically to form the molecule. (Which functional groups are present, and what different types of hydrogens are there?) This section describes how X-ray crystallography can be used to determine the actual positions of atoms in molecules The diffraction of X-rays by crystals When monochromatic light passes through a series of evenly-spaced slits on a grating, the transmitted light produces a series of interference lines on a screen. Crystals of pure chemical compounds consist of very regularly repeating patterns of atoms, spaced about 100 pm apart. (You can check this approximation by looking at covalent and ionic radii in the data booklet.) If you then look at Topic 1.2 in Unit 1 you will see that X-rays have wavelengths about m. similar to the interatomic distances. (100 x m=10-10 m) By using X-rays, which are diffracted by atoms in much the same way as light is by a diffraction grating, a pattern of diffracted spots is obtained when radiation passes through a crystal of material. (See Figure 9.16). Figure 9.16: X-ray diffraction pattern Because X-rays are electromagnetic, it is the electrons in the atoms or ions in the crystal which are responsible for the scattering. The differing electron densities of atoms are useful in assigning the diffraction spots to different scattering species. The next questions are about sodium and potassium chlorides, which both crystallise in a face-centred cubic lattice. Q30: A sodium chloride crystal consists of an array of sodium and chloride ions. How many electrons are there in each sodium ion? Q31: How many electrons are there in each chloride ion?

194 186 TOPIC 9. INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY AND X-RAY CRYSTALLOGRAPHY Q32: Which ion would you expect to scatter X-rays to the greater extent? a) Sodium b) Chloride Q33: Work out how many electrons each potassium ion has. What do you notice about the numbers of electrons in potassium and chloride ions? a) More in potassium b) More in chloride c) Both the same You would expect the intensities of spots from planes of potassium and chloride ions to be similar, unlike sodium chloride with different intensities. These differences can be exploited to help identify the positions giving rise to the reflections, by deliberately inserting a heavy atom into a crystal and noting the more intense reflections produced from the modified crystal. Q34: In organic compounds which commonly-found atom would you expect to scatter X-rays least? Electron density maps The photograph of the diffraction pattern above (Figure 9.16) would require a very considerable input of time to measure the angles of the different spots and calculate the position of atoms in the crystal. Nowadays, the process is automated. The crystal is mounted on a frame, which can be rotated in all directions under the control of a computer. At any particular position, an array of detectors sends information about the scattered X-rays to the computer. The calculation capability of modern computers allows these data to be manipulated to provide an electron density map of the atoms in the molecules forming the crystal. The main points of the process are now illustrated. The X-ray diffraction pattern from a crystal of urea:

195 9.4. X-RAY CRYSTALLOGRAPHY 187 The electron density map derived from the previous pattern: Q35: In the compound urea, CO(NH 2 ) 2, which atom has the most electrons? a) carbon b) hydrogen c) nitrogen d) oxygen Q36: Which atom will show the highest electron density pattern? a) carbon b) hydrogen c) nitrogen d) oxygen Look at the electron density map and identify this atom. Q37: One of the types of atoms below typically shows a very weak response on account of its small number of electrons. Which element is this? a) carbon b) hydrogen c) nitrogen d) oxygen Identification of carbon and nitrogen by similar means allows the complete 3D structure of urea to be determined.

196 188 TOPIC 9. INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY AND X-RAY CRYSTALLOGRAPHY Arrangement of atoms in urea For urea this might not seem a great task, but the technique finds invaluable application for larger molecules, particularly, biomolecules. Learning point Electron density maps are produced from the positions and intensities of the "spots" in a diffraction pattern, produced when a crystal of an organic compound is exposed to X-rays of a single wavelength. From the electron density map the precise location of each atom in the molecule can be determined and since heavier atoms have more electrons than lighter ones each atom in a molecule can be identified. Since a hydrogen atom has a low electron density, it is not easily detected by X-rays. 9.5 Summary All chemical bonds vibrate with quantum energy states that depend on the mass of the atoms and the strength of the bonds. These vibrations, for molecules with more than two atoms, fall into groups of stretching and bending modes, whose energies are in the infrared (IR) range. These are usually expressed in wavenumbers. The IR wavenumbers involved are from 4000 to 400 cm -1. Infrared radiation is passed through a sample of the organic compound and then to a detector which measures the intensity of the transmitted radiation at different wavelengths. From correlation tables the different wavenumber absorptions can be assigned to characteristic groups. For example the carbonyl (C=O) group absorbs near 1700 cm -1. Additionally the "fingerprint" region below 1400 cm -1, which contains vibrations from the whole molecule, can be used for identification purposes. The nuclei of some atoms, in particular 1 H, have a spin which means that they behave as tiny magnets and will align with or against an externally applied magnetic field. Those aligned with the field have a lower energy than those

197 9.6. RESOURCES 189 opposed to it. Absorption of energy in the radiofrequency range "flips" the nuclei from low to high state. 1 H atoms in organic molecules have different energies for the low and high states of nuclear spin which are dependent on their different electron densities. A proton NMR spectrum is generated by spinning a sample in the field of a strong magnet and measuring the emission of radiofrequency radiation. The sample to be analysed is normally dissolved in a solvent containing no 1 H atoms. e.g. CDCl 3. Deuterium ( 2 H) has an NMR frequency a long way from 1 H. The low resolution NMR spectrum consists of peaks whose chemical shift (ffi, in ppm) from the signal of tetramethylsilane (set at zero) reflects the chemical group containing the protons and whose areas are proportional to the number of protons in that environment. Magnetic resonance imaging in medicine uses the same technique to visualise the different environments of protons in body tissues, in order to diagnose disease or injury. X-rays are of wavelength comparable to interatomic distances in molecules (around m). When X-rays pass through the regular 3D array of atoms in a crystal, a symmetrical arrangement of diffraction spots is produced, which can be used to determine the precise 3D structure. 9.6 Resources Chemical Ideas: Salters Advanced Chemistry, Heinemann, ISBN General Chemistry: Ebbing, Houghton Mifflin, ISBN Higher Still Support: Advanced Higher Chemistry - Unit 3: Organic Chemistry, Learning and Teaching Scotland, ISBN Organic Spectroscopy: Kemp, Macmillan, ISBN Interpreting Organic Spectra: Whittaker The Royal Society of Chemistry, ISBN Video: Modern Chemical Techniques, Royal Society of Chemistry, and associated book, ISBN End of Topic test An online assessment is provided to help you review this topic.

198 190 TOPIC 9. INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY AND X-RAY CRYSTALLOGRAPHY

199 191 Topic 10 Medicines Contents 10.1 Introduction Aspirin development How a medicine functions Enzyme function The pharmacophore Case studies Summary Resources End of Topic test Prerequisite knowledge Before you begin this Topic, you should be able to: describe enzyme function in relation to the molecular shapes of proteins (Higher Unit 2); use systematic names, full and shortened structural formulae, describe functional groups and reaction types and mechanisms for all the families of organic compounds preceeding this Topic in Advanced Higher. Learning Objectives After studying this Topic, you should be able to: state that medicines are beneficial drugs containing pharmacologically active compounds; describe the historic development of aspirin as an example of a medicine developed from synthesised derivatives of natural compounds found in plant extracts.; describe the interaction of biologically active molecules with receptor sites on cells and enzymes; explain the terms pharmacophore, agonist and antagonist and describe the functions they perform.

200 192 TOPIC 10. MEDICINES 10.1 Introduction Almost without realising it, many of us probably owe the quality of our lives to the ready availability of a wide range of medicines. The pharmaceutical industries are amongst Britain s largest companies and invest huge sums of money in research and development of medicinal substances. Any substance which alters the normal biochemical processes in the body is known as a drug. This includes a wide range of chemicals from alcohol and caffeine to more medicinal substances like aspirin and antibiotics. Those drugs which have a beneficial effect on health are called medicines. Many common pharmacy products contain aspirin. Some products have active drugs removed. Some have them added! ffl 10.2 Aspirin development Learning Objective To use aspirin as an example of a medicine from a plant source from which compounds and derivatives have been identified and synthesised. ffi Traditional medicines from plant and animal sources can provide strong clues to lead in chemicals for the pharmaceutical industry. The pharmacologically active compounds in such plant and animal extracts can be identified and used as medicines. Sometimes these compounds and their derivatives can be synthesised and modifications made to improve efficiency and reduce side effects. Morphine, for example, was isolated as the active principle from the extract of the juice of the opium poppy. It is a powerful euphoric and analgesic (pain killer). Many synthetic medicines related to morphine have been made, including codeine and methadone. Aspirin is a common medicine that has been developed from natural sources. The Greek physician Hippocrates recommended the use of willow bark for pain relief during childbirth and a boiled vinegar extract of willow leaves was recommended for pain relief in Roman times. fi fl

201 10.2. ASPIRIN DEVELOPMENT 193 In 1829, the compound salicin (Figure 10.1) was isolated as the active principle in willow showing the analgesic properties. Hydrolysis of the salicin yielded salicyl alcohol (Figure 10.1), which is metabolised in the body to give salicylic acid (Figure 10.1). This proved to be useful in treating pain and reducing the fever in rheumatic fever (antipyretic) but unfortunately was also found to cause gastric bleeding. The most effective derivative found was acetylsalicylic acid (Figure 10.1), better known as aspirin. The sodium and calcium salts of acetylsalicylic acid are soluble. It has become clear that aspirin is of benefit, not only as an analgesic and antipyretic but also in the reduction of blood clotting and as such it is involved in treatment and prevention of heart disease and stroke. (Remember, however, that as with most medicines, there are side effects and it should be used with care). Figure 10.1: Aspirin related compounds Q1: Which of the chemicals shown is an alcohol? a) A b) B c) C d) D

202 194 TOPIC 10. MEDICINES Q2: What other product is formed when A is hydrolysed to B? a) water b) alcohol c) aspirin d) glucose Q3: What type of reaction takes place when B forms C? a) esterification b) reduction c) hydrolysis d) oxidation Q4: What type of reaction takes place when C forms D? a) esterification b) reduction c) hydrolysis d) oxidation PPA - Preparation of aspirin (UNIT 3 PPA 4) 120 min Consult with your tutor to find out whether the PPA on Preparation of aspirin is to be completed at this stage. PPA - Aspirin determination (UNIT 3 PPA 5) 120 min Consult with your tutor to find out whether the PPA on Aspirin determination is to be completed at this stage. ffl 10.3 How a medicine functions Learning Objective To describe how medicines work by binding to receptor sites and functioning as agonist or antagonist molecules. ffi A living organism is a complicated system with a multitude of chemical reactions within it. Why, then, do medicines have such a specific effect on some parts of the body and where do they interact with the chemical systems? Many of today s medicines interact with receptors, which are usually protein molecules that are either on the surface of cells or are enzymes that catalyse chemical reactions (catalytic receptors). Receptors normally interact with a specific small biologically active molecule which can bind reversibly with the site. The forces involved are electrostatic attractions, varying in strength from van der Waals bonding to hydrogen bonding or even ionic bonding. Generally, the functional groups on both have to be in the correct position to interact and bind the medicine to the receptor. fi fl

203 10.3. HOW A MEDICINE FUNCTIONS 195 ffl Receptor function Learning Objective ffi To describe how medicines work by binding to receptor sites. Visit the web version of this Topic to access an animation and a drag and drop exercise that illustrate the way receptors function. Play the animation a few times. It illustrates how a receptor initiates activity in a cell. In this example, the cell being activated is a muscle cell. If the correct active molecule binds, it activates the cell to trigger a biological response and then reversibly leaves the site unchanged (Figure 10.2). fi fl 5 min Figure 10.2: Receptor function The biological response here is contraction of a muscle cell. A useful analogy might be a car engine and its ignition mechanism. The ignition key represents the biologically active molecule, the ignition lock is the receptor and the engine is the cell. The process of starting the engine is the same as initiating the biological response. Many of today s medicines interact with receptors, which are usually protein molecules that are either on the surface of cells or are enzymes that catalyse chemical reactions (catalytic receptors). Receptors normally interact with a specific small biologically active molecule which can bind reversibly with the site. Effective medicines can work by binding to the receptor site and either mimicking the response of the active molecule or blocking the effect of the active molecule. If the medicine mimics the natural active molecule, it will stimulate the same response and activate the biological response. In this case, the medicine is classified as an agonist. An agonist therefore is a drug that binds to a receptor, triggering or increasing a particular activity in that cell (Figure 10.3). This class of medicine is useful in situations where there is a shortage of the natural active molecule. For example salbutamol, is an agonist which mimics adrenaline and switches on receptors which lead to dilation (widening) of the airways. It can be used to treat asthma attacks. The agonist molecule is like a good copy of the original ignition key in the analogy used previously. It fits the lock, switches the ignition and starts the car (it activates the cell)( Figure 10.3).

204 196 TOPIC 10. MEDICINES Figure 10.3: Agonist function If the medicine binds onto the receptor site and does not switch it on, it prevents the action of the body s natural active molecule and is classed as an antagonist. An antagonist therefore is a drug which binds to a receptor without stimulating cell activity and prevents any other substances from occupying that receptor ( Figure 10.4 ). This class of medicine is useful if there is a surplus of natural messengers or where one wants to block a particular message. For example, propranolol is an antagonist which blocks the receptors in the heart that are stimulated by adrenaline. It is called a fi-blocker and is used to relieve high blood pressure. Figure 10.4: Antagonist function The antagonist molecule is like a badly copied key. It fits into the lock but will not turn and is therefore unable to start the engine (unable to activate the cell). Antagonists also sometimes stick in place, usually due to stronger bonding, preventing any other active molecule from binding. ffl Enzyme function Learning Objective ffi To describe how the catalytic receptor site in enzymes catalyses chemical reactions. Enzymes are large protein molecules which act as catalysts in biological processes. The protein in enzymes is normally folded into a three dimensional structure, part of which is a pocket known as the active site or the catalytic receptor. Since the catalytic receptor is a specific shape, it can only accommodate certain substrate molecules. Once in position, the substrate is chemically changed and the products formed. The products could be smaller (a cleavage reaction) (see Table 10.1) or larger (a synthesis reaction). fi fl

205 10.3. HOW A MEDICINE FUNCTIONS 197 Table 10.1: Enzyme action Medicines can act on enzyme active sites in similar agonistic or antagonistic ways, for example acting as a plug or inhibitor to switch off specific enzymes. For example, some smart drugs are designed to improve memory in Alzheimer s disease by blocking enzyme breakdown of a chemical called acetylcholine, necessary for neurotransmission. ffl The pharmacophore Learning Objective To identify the pharmacophore in a group of medicines and explain how it functions in conferring pharmacological activity. ffi The development of new medicines very often starts with a lead in clue from a plant or animal source. Research then seeks to find related compounds with fewer side effects or greater activity. The first stage in this is isolation of the minimum structural feature that gives pharmacological activity; this is called the pharmacophore. The shape of the pharmacophore complements that of the receptor site, with the functional groups on both structures correctly positioned to interact and bind them together. Once identified, chemists can design and synthesise related compounds, looking towards improving the performance of the medicine. The pharmacophore can often be identified by comparing the structures of medicines with similar pharmacological activity. The highly addictive analgesic morphine and some of its close modifications can provide an example. (Figure 10.5) fi fl Figure 10.5: Opiate structures In each structure, the lines are carbon to carbon bonds with a carbon at each corner and most hydrogens omitted.

206 198 TOPIC 10. MEDICINES 10 min The pharmacophore part of these structures is identified in the associated website activity. ffl Morphine pharmacophore Learning Objective To identify the pharmacophore in a group of medicines by comparison of their structures. ffi Visit the online activity which shows the slight structural changes made by chemists to modify the properties of morphine.the pharmacophore can then be identified. fi fl The pharmacophore can be identified by comparing the structures of medicines with similar pharmacological activity. The analgesic medicines shown here ( Figure 10.6) have a common pharmacophore. They are highly addictive and are legally controlled. Figure 10.6: Morphine derivatives and the pharmacophore. When the shape, dimensions and functional groups present have been identified, the pharmacophore can be cropped, added to and manipulated by chemists to produce compounds which are still analgesic but less addictive (Figure 10.7). The pharmacophore s shape and dimensions complement the receptor, usually through functional groups that are positioned by the shape and size of the structures to allow interaction and binding of the medicine to the receptor.

207 10.4. CASE STUDIES 199 Figure 10.7: Modified opiates For example, etorphine (a synthetic opiate), (Figure 10.7) is almost 100 times as potent as morphine and is used in veterinary medicine to immobilize large animals. ffl 10.4 Case studies Learning Objective To describe a common medicine which performs as either an agonist or antagonist in its treatment of a medical condition. ffi The following four case studies show how medicines function in different common applications. Detailed knowledge does not have to be memorised. Your tutor may wish you to focus on one of the four and give a short presentation to the rest of your group. If you are simply required to be aware of the case studies, read them through, paying particular attention to the names of drugs, mode of action and medical conditions they are effective against. You will then be able to tackle the final activity. If you are making a presentation to a group, then you will need to prepare: a hand-out information sheet; audio visual illustrations (blackboard or OHP or data projection); a short script for yourself. Include within your presentation: the medical conditions involved; fi fl

208 200 TOPIC 10. MEDICINES formulae, equations, diagrams; mode of action; side effects. 30 min Read each of the case studies. You may skim them looking for the necessary information to carry out the Medicines in use activity at the end of the Topic, where you will be asked to match up the medical problems with the correct medicine and its mode of action. ffl Case Study 1 Salbutamol: Learning Objective ffi To be aware of the structure and function of salbutamol as a treatment for asthma. A case study illustrating the development of salbutamol from pharmacologically active compounds from natural sources. PLEASE NOTE: Detailed knowledge of the medicines, their structure and how they work does not have to be memorised. Read each of the case studies. If you are presenting one as a homework exercise, chose one particular case study to focus on and follow the instructions about preparation. If you are only interested in gathering information in order to complete the Medicines in use activity, you may skim them looking for the necessary information which is italicised, and match up the medical problems with the correct medicine and its mode of action. For over 5000 years, the Chinese plant Ma Huang was used to treatasthma and bronchitis. It was boiled in water and the solutions drunk. The chemical responsible for the biological activity is ephedrine (Figure 10.8). Pure ephedrine has cardiovascular side effects. fi fl Figure 10.8: Natural bronchodilators Bronchodilators like ephedrine widen the airways in the lungs and relieve the breathlessness and wheezing of an asthma attack. One of the natural substances that perform this function is adrenaline (Figure 10.8), which is a hormone circulating in the blood supply. Using adrenaline to treat asthma also increases heart rate and raises blood pressure. Research has shown that there is a variety of different receptors on which adrenaline can act, each producing a different effect. These have been classed as alpha (ff) and beta (fi) type receptors, The main effects of adrenaline are shown in Table 10.2.

209 10.4. CASE STUDIES 201 Table 10.2: Effects of adrenaline Receptor Effect Outcome for asthmatics ff-activated increased blood pressure unhelpful fi 1 -activated increased heart rate unhelpful fi 2 -activated dilation of bronchi helpful Instead of adrenaline, which activated all these receptors and was quickly broken down in the body, what was required was a fi 2 specific agonist which lasted longer and would only activate the dilation effect. One of the first was isoprenaline (Figure 10.9) which has a bulky group attached to the nitrogen. This seemed to favour fi 2 activation. Figure 10.9 Adding an even bulkier group to the nitrogen, and replacing the 3-hydroxyl group on the ring with a hydroxymethyl group, yielded an even more effective medicine, salbutamol (Figure 10.9). The hydroxymethyl group slowed down the breakdown of the drug in the body so that it lasted about 4 hours. Interestingly, salbutamol has two enantiomeric forms (see Topic 3.7), one of which is 68 times more effective than the other (can you think why?). Salbutamol is a market leader that is delivered by inhalation from an aerosol spray (Figure 10.10). Figure Recent work to extend the time between doses has yielded a drug which is more fat soluble, and therefore takes longer to be metabolised. It has a duration of 12 hours.

210 202 TOPIC 10. MEDICINES 30 min Figure Salbutamol is an effective medicine for quick relief in an asthma attack. (Figure 10.11) provides longer spells of relief. ffl Case Study 2 Propranolol: Learning Objective ffi Salmeterol To be aware of the structure and function of propranolol as a treatment for asthma. A case study illustrating the development of propranolol from pharmacologically active compounds from natural sources. PLEASE NOTE: Detailed knowledge of the medicines, their structure and how they work does not have to be memorised. As before, read each of the case studies. If you are presenting one as a homework exercise, chose one particular case study to focus on and follow the instructions about preparation. If you are only interested in gathering information in order to complete the Medicines in use activity you may skim them looking for the necessary information which is italicised, and match up the medical problems with the correct medicine and its mode of action. The heart pumps blood through the arteries and veins of our cardiovascular system. Permanently high blood pressure (hypertension) is caused by increased output from the heart or from increased resistance to flow in the arteries or both. Hypertension can lead to heart attack and strokes. Another heart condition called angina results from restrictions in blood flow through the arteries that supply the heart muscles. This can lead to heart attack if untreated. Table 10.2 shows that adrenaline activates the fi 1 type receptors and also increases blood pressure. If an antagonist for the fi 1 receptors could be found then it could be used to block the receptor and allows the heart rate and blood pressure to fall. The search started with isoprenaline (Figure 10.12) which although an agonist, had shown specific activity to fi receptors rather than ff receptors. A large number of modifications were made to try to find an antagonist from isoprenaline, one of the first being pronethalol. fi fl

211 10.4. CASE STUDIES 203 Figure Further research led to the discovery of propranolol (Figure 10.13) which is a pure antagonist and has become the benchmark against which all fi blockers are rated. Interestingly, only one of the enantiomeric forms (the S form) is the active ingredient. (Can you suggest why?) Figure 10.13: Typical beta blockers A wide variety of these fi blockers are now used by people with heart disorders to keep the heart calm. ffl Case Study 3 Sulphonamides and penicillins: Learning Objective To be aware of the structure and function of sulphonamides and penicillins as treatments for bacterial diseases. ffi A case study illustrating the development of sulphonamides and penicillins from pharmacologically active compounds from natural sources. PLEASE NOTE: Detailed knowledge of the medicines, their structure and how they work does not have to be memorised. As before, read each of the case studies. If you are presenting one as a homework exercise, chose one particular case study to focus on and follow the instructions about preparation. If you are only interested in gathering information in order to complete the Medicines in use activity you may skim them looking for the necessary information which is italicised, and match up the medical problems with the correct medicine and its mode of action. In 1928, Alexander Fleming observed that a plate on which staphylococci bacteria were being cultured at St Mary s Hospital, London had been contaminated with a mould which fi fl 30 min

212 204 TOPIC 10. MEDICINES seemed to be killing the bacteria close to it. Although he recognised this, he failed to isolate the substance responsible for the effect. The active compound penicillin was eventually isolated by Chain and Florey in Figure 10.14: Penicillin and penicillium mould A wide variety of penicillins was soon produced. The basic structure of penicillin is shown in Figure 10.15, the structures differing only in the nature of the group R. Figure All penicillins are bactericidal (kill bacteria). During replication bacteria, need to make new cell wall material by crosslinking strands of polymer chains to give a strong netlike structure. The penicillins block the active site of the enzyme responsible for this crosslinking and thus act as antagonists, preventing further cell wall construction. Since mammal cells do not have cell walls, this enzyme is absent and the penicillin doesn t affect them. Antibiotics can therefore be used against bacterial infections like pneumonia, meningitis and septicaemia. It should be noted however that the increased use of antibiotics over the last 75 years has led to the development of resistance in some types of bacteria. Use them sparingly and always complete the course! Before the penicillin antibiotics were widely available, a dyestuff called prontosil (Figure 10.16) had been found to be an effective bacteriostatic agent, stopping the further growth of bacteria.