hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 1 hapter 03 Fundamentals of Organic hemistry Parts of Topics 10 and 20 from the IB HL hemistry urriculum What is an Organic ompound? Organic molecules contain carbon atoms. The carbon atoms are covalently bonded to other atoms, and various chains of carbon atoms can be found in most every molecule you will deal with in this chapter. As you might recall from your introductory class, carbon has four valence electrons, and therefore will make four bonds in accordance with the octet rule. All non-carbon-to-carbon bonds will be assumed to be carbon-hydrogen bonds, as hydrogen atoms are the most commonly found attached atom. Hydrogen has one valence electron, and will make one covalent bond. Use these facts to draw the structures of the following molecules in the space provided. H 4 2 H 6 3 H 8 4 H 10 4 H 10 (with different construction) 5 H 12 5 H 12 (with different construction) 5 H 12 (with different construction) 6 H 14
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 2 The carbon atom is capable of making single, double, and triple bonds, as well as bonding with oxygen, nitrogen, chlorine, or bromine. Oxygen has six valence electrons, and will make two covalent bonds. A single bond and a double bond are both possible for oxygen atoms. Nitrogen has five valence electrons, and will make three covalent bonds. Single, double, and triple bonds are all possibilities for nitrogen atoms. hlorine and bromine each have seven valence electrons, and will make one covalent bond. The presence of carbon-carbon double or triple bonds, as well as the presence of oxygen, nitrogen, chlorine, or bromine, is very significant and worth noting, as it will be extremely important throughout your study of organic chemistry. Be sure to look for patterns in the bonding when you encounter these situations. Use your best judgment to draw the following molecules in the spaces provided. 2 H 4 2 H 2 3 H 6 H 3 OH 2 H 5 OH 3 H 7 OH 3 H 7 OH (with different construction) H 2 O H 3 H 2 HO
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 3 H 3 H 2 OH 3 2 H 5 OOH 3 H 7 l 3 H 7 l (with different construction) H 3 OOH 3 2 H 5 NH 2 H 3 ONH 2 (H 3 ) 3 OH 2 H 5 N H 3 H 2 OH 2 H 3 H 3 H 2 Br H 3 (H 2 ) 3 OH 3
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 4 10.1 Introduction (4 Hours) Assessment Statement Obj Teacher s Notes 10.1.1 Describe the features of a homologous series. 2 Include the same general formula, neighboring members differing by H 2, similar chemical properties and gradation in physical properties. 10.1.2 Predict and explain the trends in boiling points of members of a homologous series. 3 10.1.3 Distinguish between empirical, molecular, and structural formulas. 2 A structural formula is one that shows unambiguously how the atoms are arranged together. A full structural formula (sometimes called a graphic formula or displayed formula) shows every atom and bond, for example, for hexane: A condensed structural formula can omit bonds between atoms and can show identical groups bracketed together, for example, for hexane: H 3 H 2 H 2 H 2 H 2 H 3 or H 3 (H 2 ) 4 H 3 10.1.4 Describe structural isomers as compounds with the same molecular formula but with different arrangements of atoms. 10.1.5 Deduce structural formulas for the isomers of the non-cyclic alkanes up to 6. 10.1.6 Apply IUPA rules for naming the isomers of the non- cyclic alkanes up to 6. 10.1.7 Deduce structural formulas for the isomers of the straight-chain alkenes up to 6. The use of R to represent an alkyl group and to represent the benzene ring can be used in condensed structural formulas. Although skeletal formulas are used for more complex structures in the hemistry data booklet, such formulas will not be accepted in examination answers. TOK: The use of the different formulas illustrates the value of different models with different depths of detail. 2 No distinctions need be made between different types of structural isomerism, such as chain and position isomerism and functional group isomerism. Knowledge of stereoisomerism is not required to the core 3 Include both straight-chain and branched-chain isomers 2 TOK: This could be discussed as an example of the use of the language of chemistry as a tool to classify and distinguish between different structures. 3 10.1.8 Apply IUPA rules for naming the isomers of the straight-chain alkenes up to 6. 2 The distinction between cis and trans isomers is not required. 10.1.9 Deduce structural formulas for compounds containing up to six carbon atoms with one of the following functional groups: alcohol, aldehyde, ketone, carboxylic acid, and halide. 3 ondensed structural formulas can use OH, HO, O, OOH, and F/l/Br/I. 10.1.10 Apply IUPA rules for naming compounds containing up to six carbon atoms with one of the following functional groups: alcohol, aldehyde, ketone, carboxylic acid, and halide. 2
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 5 Assessment Statement Obj Teacher s Notes 10.1.11 Identify the following groups when present in structural formulas: amino (NH 2 ), benzene ring, and esters (ROOR). 2 10.1.12 Identify primary, secondary, and tertiary carbon atoms in alcohols and halogenoalkanes. 2 The terms primary, secondary, and tertiary can also be applied to the molecules containing these carbon atoms. 10.1.13 Discuss the volatility and solubility in water of the compounds containing the functional groups listed in 10.1.9. 3 20.1 Introduction (1 Hour) Assessment Statement Obj Teacher s Notes 20.1.1 Deduce structural formulas for compounds containing up to six carbon atoms with one of the following functional groups: amine, amide, ester, and nitrile. 3 ondensed structural formulas can use NH 2, ONH 2, OO, and N for these groups. 20.1.2 Apply IUPA rules for naming compounds containing up to six carbon atoms with one of the following functional groups: amine, amide, ester, and nitrile. 2 Empirical and Molecular Formulas As you have learned in your study of general chemistry, the empirical formula of a chemical compound is the simplest whole-number ratio of elements in a compound. An empirical formula makes no reference to the shape, structure, or absolute number of atoms in a compound. Empirical formulas are the standard for most ionic compounds, such as calcium chloride, al 2, and for macromolecules like silica, SiO 2. The term empirical formula refers to the process of elemental analysis, a technique of analytical chemistry used to determine the relative percent composition of a pure chemical substance by element. Much like an empirical formula, a molecular formula contains a whole-number ratio of the elements in a compound. In contrast, the molecular formula does identify the absolute number of atoms in a compound. However, no shape or structure determination can be made based on a molecular formula. onsider the empirical formula H. From what you already know, there must be one hydrogen atom for every carbon atom. However, all three of the following molecules satisfy that requirement, and there is no discernable pattern in structure that can be discerned from the empirical formula. ethyne but-1-en-3-yne benzene
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 6 onsider the molecular formula 7 H 16. From what you know of covalent bonding, carbon atoms make four bonds, while hydrogen atoms make only one. There are many different ways to assemble carbons and hydrogens into 7 H 16, with many possible structures resulting. (Two such structures for 7 H 16 are found on the following page. As with empirical formulas, no determination of structure can be made from a molecular formula. H H H H H H H H H H H H H H H H heptane H H H H H H H H H H H H H H H 2,2-dimethylpentane H
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 7 Structural Formulas Organic compounds tend to be built according to a general scheme as follows: The carbon atoms form a skeleton that runs through the middle of the molecule. Hydrogen atoms and any other are attached to the carbon skeleton. One method of drawing organic molecules involves representing every single atom, with the bonds between them drawn as lines. This is typically referred to as a structural formula. The structural formula for 1-hexanol, or 6 H 13 OH, looks like this: Problem: Write the structural formula for 3 H 8. Organic molecules can also be represented in a more compact form, known as abbreviated structural formulas, also known as condensed structural formulas. This is done by omitting the bonds between atoms. The same compound, 1- hexanol, could therefore be represented as: Problem: Write the condensed structural formula for 3 H 8. Sometimes, it might be important to highlight the bonding that takes place between atoms in a molecule. In that case, a ball-and-stick model might be used: Problem: Draw a ball-and-stick model for 3 H 8.
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 8 The skeletal formula, or carbon framework, or line-segment formula of an organic compound is a shorthand representation of its molecular structure. Skeletal formulas are ubiquitous in organic chemistry because they show complicated structures clearly and they are quick and simple to draw. In a skeletal formula, none of the bonds connecting hydrogen to the carbon skeleton are shown. Every other bond is drawn as a line segment, and a carbon atom is located at the end of each line segment. If something other than carbon is attached to the skeleton, it is specified in the drawing. In the image below, the skeletal formula of 1-hexanol is shown: Because carbon makes a total of four bonds and hydrogens are omitted from skeletal formulas, the presence of hydrogens on the chain must be assumed. The formula for 1-hexanol is 6 H 13 OH. Look carefully at the structure above, and make sure you understand why there are a total of 13 hydrogens on 1-hexanol. Problem: Write the skeletal formula for 3 H 8. The simplest hydrocarbon is methane, containing only one carbon atom. Since each carbon atom can form four bonds, the formula of methane must be H 4. The bonding can be described as using sp 3 hybrid orbitals on carbon, giving the molecule a tetrahedral shape. This three-dimensional shape is not easy to show in two dimensions on a sheet of paper. One common representation is shown below as structure I. More often structure II is used because it is easier to draw, especially for more complex molecules. You should recall how to construct a Lewis dot structure from your study of inorganic chemistry. If a Lewis dot structure is required, it usually is drawn as in structure III. H H H H I Three-dimensional structure H H H H II Bond-line structure H.. H : :.. H H III Lewis structure To illustrate these differences, let us consider the two-carbon molecule known as ethane. With only single bonds (saturated), the compound has the molecular formula 2 H 6. It can be represented by structure IV (threedimensional) or, more commonly, by structure V (flat). H H H H H H IV Three-dimensional structure H H H H H H V Bond-line stucture (flat)
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 9 Structures IV and V above show all the bonds and indicate how each atom is bonded to others. In saturated hydrocarbons, each hydrogen is attached to a carbon by a single covalent bond. Each carbon must have four bonds, to other carbons or to hydrogens. If you are asked to write the structural formula of a compound, you should assume that a full structural formula is intended. The flat representation (V) usually will be adequate, however, it is sometimes prudent to draw the bonds at angles in order to highlight the sp 3 hybridized bonding. Problem: Draw the Lewis dot structure for 3 H 8. Problem: Draw the Three-dimensional structure for 3 H 8. Problem: Write the full structural formula for the compound with the condensed structural formula H 3 (H 2 ) 2 H 3. Problem: Write the structural formula for the hydrocarbon with five carbons in a continuous chain, containing only single carbon-carbon bonds; write its condensed structural formula. It is also possible to have a double or triple bond between the carbon atoms, giving an unsaturated compound. H H H VII H H H VIII An alkene Molecular formula: 2 H 4 ondensed structural formula: H 2 =H 2 or H 2 =H 2 An alkyne Molecular formula: 2 H 2 ; condensed structural formula: H H
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 10 Problem: Write the structural formula for H 3 H=HH 3. Problem: Write the empirical formulas for structures VII and VIII. Problem: A compound has an empirical formula of H 2. Write the molecular formula for a three-carbon alkene with that empirical formula. Write the structural formula. Problem: onsider the molecule at the right. Draw the structural formula for the molecule. Write the empirical and molecular formula for the compound. Structural Isomerism For a three carbon compound, we could write the following skeletons (for simplicity, the bonds to the hydrogen atoms are shown, but not the hydrogen atoms themselves): Since the arrangement of bonds about each carbon atom is tetrahedral, rather than planar, the two structures are identical. At this point it would be very helpful for you to make a model of the three-carbon compound (molecular formula 3 H 8 ). Notice that the three carbons are attached to each other in a line, with no branching of the chain. The drawings above simply show the same carbon skeleton from different viewpoints. One structure could be converted to another simply by rotating one part of the molecule about a carbon-carbon bond, relative to the rest of the molecule. You can see this best by using a model.
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 11 For the four-carbon molecule, we can simplify the representation of the carbon skeleton even further by showing only the carbon atoms and the bonds between them. In looking at this representation you must remember that each carbon will have other bonds in addition to those shown. All the other bonds are to hydrogen atoms, and each carbon has a total of four bonds. The two skeletal structures immediately above are different. If you use models, you can see that it is not possible to change one to the other simply by rotating one part of the molecule relative to the other. The only way to convert one to the other is by breaking a carbon-carbon bond. Notice also that the first structure has four carbons in a single chain. In the second structure, the longest continuous chain is three carbons, with the fourth carbon attached to the middle of the chain. We refer to the first structure as a straight chain, and the second as a branched chain. Since the structures are different, but the molecular formulas identical ( 4 H 10 ), this is an example of isomerism. This type of isomerism is known as structural isomerism. ontinuing our consideration of structural isomerism, we move to five-carbon chains. Problem: What is the molecular formula for a hydrocarbon containing a five-carbon chain, hydrogens, and single bonds throughout? Problem: onsider the following carbon skeletons. How many distinct molecules are there? Which molecules are identical to each other? (1) (2) (3) (4) (5) (6) Answer In the set of compounds above, (1) and (2) differ from each other, in that (1) is a straight chain of five carbons, and (2) is a branched chain with four carbons as the longest chain. Structure (3) is identical to (2). The branching is on the second carbon from the end of a four-carbon chain in each case. They appear different because one is written in the opposite direction from the other. Structure (4) is a different representation of structure (1), so they are identical. Since the longest carbon chain in structure (5) is three carbons, it is different from (1) and (2). Structure (6) is a different view of a four-carbon chain with a branch on the second carbon from the end, making it identical to structures (2) and (3). Thus we have found three different structures for five-carbon chains, and this is the maximum number of possibilities. Since the chemical formula is the same for each of the three different structures, these are structural isomers.
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 12 Problem: Draw the structure of molecule (2) from the previous problem. Add the necessary hydrogen atoms to complete an accurate picture of what the molecule looks like. Problem: Draw the structure of molecule (5) from the previous problem. Add the necessary hydrogen atoms to complete an accurate picture of what the molecule looks like. Problem: How many isomers are possible for a carbon chain that contains six carbon atoms? Draw them (a carbon skeleton will suffice you do not have to include every hydrogen atom). As the number of carbon atoms increases, the number of possible isomers increases rapidly. For molecules containing only single bonds, the following table gives the maximum number of isomers for different numbers of carbon atoms. arbon Atoms Isomers 4 2 5 3 6 5 7 9 10 75 20 366 319
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 13 Homologous Series The straight-chain alkanes (compounds containing all single carbon-carbon bonds) all could be written with the following general condensed structural formula: H 3 (H 2 ) n H 3, where n = any number, including 0. A series of compounds showing this relationship, namely that members of the series differ only in the number of H 2 groups, is called a homologous series. Problem: Some of the following pairs of compounds are members of the same homologous series. Some are not. In the spaces at the right, identify the pair as homologous or not, and briefly explain your answer. (a) H 3 H 2 H 2 H 3, H 3 H(H 3 ) 2 (b) H 2 =HH 2 H 3, H 2 =HH 3 (c) lh 2 H 3, lh 2 H 2 H 2 H 2 l (d) H 3 H 2 H 2 H 3, H 3 H 2 H 2 H 2 H 2 H 3 (e) H 3 H 2 OOH, H 3 OOH (f) H 3 HOHH 2 H 3, H 3 OH Problem: Straight-chain alkanes have no double bonds and a general formula of n H 2n+2. Alkenes contain one double bond, and have a general formula of n H 2n. Alkynes contain a triple bond, and therefore have a general formula of n H 2n 2. Below, you will find 12 chemical formulas. Sort them into homologous series based on their formulas. 5 H 10 3 H 8 3 H 4 4 H 8 6 H 12 6 H 10 3 H 6 4 H 6 4 H 10 5 H 12 5 H 8 6 H 14 Alkanes: Alkenes: Alkynes:
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 14 Melting points and boiling points The boiling points and melting points of alkanes show a definite pattern, especially within a single homologous series. The following table gives the boiling points and melting points of the first 20 straight-chain alkanes. Alkane m.pt./ b.pt./ H 4-183 -162 2 H 6-172 -88 3 H 8-188 -42 4 H 10-138 0 5 H 12-130 36 6 H 14-95 69 7 H 16-91 98 8 H 18-57 126 9 H 20-54 151 10 H 22-30 174 11 H 24-26 196 12 H 26-10 216 13 H 28-5 235 14 H 30 6 254 15 H 32 10 271 16 H 34 18 287 17 H 36 22 302 18 H 38 28 317 19 H 40 32 331 20 H 42 36 344 The following figure shows the boiling points and melting points of the straight-chain alkanes plotted against number of carbon atoms. 400 300 200 Temperature/deg. 100 0-100 -200-300 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 b.pt/ m.pt./ Number of arbon Atoms
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 15 Answer the following questions before reading any further. Problem: Which compounds are gases at about 25? Problem: Which compounds are solids at about 25? Problem: What general trends do you notice for the boiling points and melting points? Problem: How could the general trend for boiling points be explained? Problem: What differences do you see in the two sets of data? Both boiling points and melting points increase with increasing size of the molecule. For boiling point, we are interested in the intermolecular forces in the liquid. Hydrocarbons are essentially non-polar because of elements of symmetry and the very low polarity of H bonds. Thus, intermolecular forces in the liquid are only London forces (van der Waals forces). Larger molecules have more surface area, so more attractive forces exist between molecules, requiring more energy to separate them, leading to higher boiling points. We can consider the boiling point to be an equilibrium between liquid and gaseous states. Thermodynamically, for vaporization, we can write G = H T S = 0. The result of this is that boiling temperature, T b = H v / S v. Thus, the enthalpies and entropies of vaporization determine the boiling temperature. For all hydrocarbons and most other organic compounds, S v has nearly the same value, so differences in boiling point are direct measures of differences in enthalpies of vaporization, which in turn depend on the strengths of intermolecular forces. Melting points follow the same general pattern, because intermolecular forces are weakened in going from the solid state to the liquid. The increase is not as regular as it is for boiling points because factors other than intermolecular forces play a role.
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 16 ompare the following data for isomers of pentane. ompound m.pt./ b.pt./ H 3 (H 2 ) 3 H 3 130 36 H 3 H 2 H(H 3 ) 2 160 28 (H 3 ) 4 17 10 In general, chain-branching causes a slight decrease in boiling point, compared to the straight-chain compound with the same number of carbons. This is because the molecule becomes more compact, with a smaller surface area, reducing the intermolecular attractions. Melting points are often lowered for the same reason, except for highly symmetric molecules, such as (H 3 ) 4, which may have unexpectedly high melting points. The enthalpy of fusion, or melting, H f, is always much less than H v for the same compound, because intermolecular forces are weakened during melting rather than completely broken, as during boiling. In addition, different compounds have different values of S f, depending on their structures. For a highly symmetric molecule, S is small for the transition solid to liquid, since a symmetric molecule, like a ball, can assume many different orientations in the solid. As S f deceases, T m increases. Paralleling the thermodynamics math of boiling temperature seen above, the melting temperature, T m = H f / S f. Solubilities and densities Other important physical properties of the alkanes are their low solubilities in water and their densities. The alkanes have such low solubility in water that we consider them insoluble. This is because dissolving would require that strong intermolecular forces between water molecules (due to hydrogen bonding) be broken and replaced by the much weaker van der Waals forces (London forces) between water molecules and non-polar alkane molecules. The net result of this process is energetically highly unfavorable, so it does not occur to any significant extent. All alkanes, regardless of whether they are gas, liquid, or solid, are less dense than water. Most liquid hydrocarbons have a density of about 0.7 g cm -3. Practical consequences of this property include the fact that hydrocarbons such as oil float on water. Oil fires cannot be put out with water, because the burning oil floats on top of water and may be spread more widely. Names of straight-chain alkanes The following table gives the molecular formula and the condensed structural formula of each of the first eight straight-chain alkanes, as well as its name according to the IUPA system. Molecular formula ondensed structural formula Name H 4 H 4 Methane 2 H 6 H 3 H 3 Ethane 3 H 8 H 3 H 2 H 3 Propane 4 H 10 H 3 H 2 H 2 H 3 Butane 5 H 12 H 3 H 2 H 2 H 2 H 3 Pentane 6 H 14 H 3 H 2 H 2 H 2 H 2 H 3 Hexane 7 H 16 H 3 H 2 H 2 H 2 H 2 H 2 H 3 Heptane 8 H 18 H 3 H 2 H 2 H 2 H 2 H 2 H 2 H 3 Octane
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 17 Note the following points: 1. All the names end in ane, indicating that the compound is an alkane, the name for a saturated hydrocarbon. 2. The prefixes pent-, hex-, hept- and oct-, are the Greek prefixes for the numbers 5-8. Higher alkanes are also named with Greek prefixes. The table gives the molecular formula for each of the first eight alkanes. These formulas follow a pattern. The general formula for any alkane is n H 2n+2. This is the largest proportion of H to that is possible for any organic compound. Problem: Write the molecular formula for alkanes with the following number of carbons: (a) 10 (b) 18 (c) 56 Problem: Which of the following formulas could not correspond to an alkane? (a) 4 H 8, (b) 9 H 18, (c) 14 H 30, (d) 10 H 8 Since the different isomers for an alkane with a specific number of carbons all have the same molecular formula, the general formula of n H 2n+2 applies to all alkanes, straight-chain and branched-chain.
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 18 Naming of alkanes In the table on the previous page, we saw the names of the first eight straight-chain alkanes. We also have seen that alkanes containing more than three carbons can have branched chains. In fact, the number of branched-chain alkanes is much greater than the number of straight-chain compounds. We will now see how to name these branched-chain alkanes. The procedure is simple and logical and based on the name (already known) for the longest continuous chain in the compound. Let us proceed in steps to develop the basic principles of this system. 1. Write the carbon skeleton for the compound to be named, showing all carbon atoms and the bonds between them. Examples: H 3 H(H 3 ) 2 (H 3 H 2 ) 2 HH 3 2. Find the longest continuous carbon chain and give it the appropriate name as shown in the table on page 1. Examples: propane pentane Note that the longest chain may not be on one line. octane 3. Find the points where the chain branches. The branches, or substituents, are called alkyl groups. (from alkane + yl). (a) The names of the substituents are derived from the alkane with the same number of carbons. Thus H 3 - = methyl, H 3 H 2 - = ethyl, and so on in the same way that the general name, alkyl group, was derived from alkane. (b) The substituent name is placed before the name of the longest chain, without a space between the two parts of the name.
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 19 Examples: methylpropane methylpentane ethylhexane (c) If there is more than one substituent of the same type, use the prefixes di-, tri-, etc. Example: trimethylpentane (d) Two or more different substituents should be listed in alphabetical order ethylmethylpentane 4. To indicate the position of the substituent, where there is more than one possibility, the carbons in the longest continuous chain are numbered. The numbering can be started from either end of the chain. The direction of numbering should be chosen so that when the two possible directions are compared, the number should be the lowest at the first point where the numbers differ. The numbers are separated from the rest of the name by hyphens. 3-methylpentane (Both directions give the same result.) 3-ethylhexane (preferred to 4-ethylhexane)
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 20 2,2,3-trimethylbutane (Numbers have to be given for multiple substituents, even when they are identical.) methylbutane (No number is necessary since no other position is possible.) Problem: Name each of the following compounds. (a) (H 3 ) 2 HH 2 H 3 (b) H 3 H(H 3 )H(H 3 ) 2 (c) H 3 H(H 3 )H 2 H 2 H 2 H 3 (d) H 3 H(H 3 )H(H 2 H 3 )H 2 H 3
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 21 Problem: Draw the structural formula for each of the following compounds. (a) 2-methylpentane (b) 4-ethyloctane (c) 2,3-dimethylhexane (d) 3-methylheptane Names of straight-chain alkenes The presence of double bonds is indicated by changing the ending -ane for an alkane to -ene for alkenes. H 3 H=H 2 propene If the multiple bond can be in more than one position in the molecule, number the position of the first carbon in the multiple bond, starting from the end of the chain which gives the lowest number. H 3 H 2 H 2 H=H 2 pent-1-ene (or 1-pentene) Problem: Write the structural formula of one other straight-chain pentene, and give its name.
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 22 Problem: Name each of the following compounds. Draw structural formulas first. (a) H 3 H=HH 3 (b) H 3 H=HH 2 H 3 Problem: Draw the structural formula for each of the following compounds. (a) 2-pentene (b) 1-butene
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 23 The importance of functional groups All of our discussion so far has been concerned with hydrocarbons, compounds containing only carbon and hydrogen. Many organic compounds contain other elements, especially oxygen, nitrogen and occasionally, halogen. The presence of these other atoms changes both the chemical and physical properties of the compound relative to the hydrocarbon with the same carbon skeleton. These atoms, or groups of atoms, that determine the characteristic chemical and physical properties of a type of organic compound are called functional groups. The carbon-carbon double bond, even though it contains only carbon atoms and perhaps hydrogen atoms, is also considered a functional group because alkenes undergo many reactions that are not possible for alkanes. As an example of how a functional group can affect properties, let us compare some properties of ethane (H 3 H 3 ) and ethanol (H 3 H 2 OH). ethane gas (boiling point, -88 ) insoluble in water burns readily does not react with most chemical oxidizing agents does not react with acids forms ethene at very high temperatures ethanol liquid (boiling point, 78 ) soluble in water in all proportions burns readily reacts readily with chemical oxidizing agents reacts with acids forms ethene readily The two compounds are closely related structurally; ethanol is like ethane except for the replacement of an H by an OH. Except for combustion, a characteristic of almost all organic compounds, the physical and chemical properties of the two compounds are very different. H 8 H 17 3 Another example of a compound in which a functional group makes a difference is cholesterol, a compound H 3 much in the news. holesterol is needed by the body, but too much cholesterol is thought to be one of the factors contributing to the risk of heart disease. The cholesterol structure of cholesterol looks very complex, because HO it has four rings attached to each other, but you can probably see that it contains two familiar functional groups: a carbon-carbon double bond and an OH group. hemists have found that the chemistry of cholesterol is very similar to that of other compounds containing OH or =. Since functional groups behave similarly, whether in simple or complex molecules, our further study of the physical and chemical properties of organic compounds will be organized according to the type of functional group present. Before dealing with the properties of functional groups, however, we need to become familiar with the structures, names and ways of representing some of the most common functional groups.
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 24 Structures and names of functional groups In Table 1, the names of the main functional groups are given in the first column. The common name of the type of compound in which each occurs appears in column 2. Systematic names sometimes used within the IUPA system are given in parentheses below the common name. The structural formulas in column 3 introduce a new symbol which simplifies the writing of organic structures. R stands for any alkane in which a hydrogen has been replaced by another atom or group of atoms. Recall from hapter 2 that such a group is called an alkyl group. The simplest possible group is that derived from methane. The alkyl group is thus H 3 -, or methyl. Thus ROH can represent any alkane in which a hydrogen atom has been replaced by an OH group. Notice that the structure of cholesterol above contains two methyl groups and a 8 H 17 group. This also is an alkyl group, in this case the octyl group. Representing compounds in this way is consistent with the observation that the properties of the compound depend much more on the nature of the functional group than on the hydrocarbon portion of the molecule. olumn 3 of the table is especially important. You should be able to recognize and write functional groups in both full and condensed forms. Ethyl ethanoate, for example, could be written in any of the following forms: H 3 OOH 2 H 3, H 3 H 2 OOH 3, H 3 OH 2 H 3, H 3 OH O 3 H 3 O 2 H 2 H 3 O In the fourth column are given systematic names and formulas of specific compounds, as well as common names in parentheses.
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 25 Table 1: Functional Groups Name of group Type of compound Structural formulas Example hydroxyl alcohol ROH ethanol (H 3 H 2 OH) (alkanol) (ethyl alcohol) carbonyl aldehyde (alkanal) RHO R O H methanal (HHO) (formaldehyde) carboxyl ketone (alkanone) carboxylic acid (alkanoic acid) ROR' R O R' ROOH R O OH propanone (H 3 OH 3 ) (acetone) ethanoic acid (H 3 OOH) (acetic acid) halogen halogenoalkane RX Bromomethane (H 3 Br) acyloxy carboxylic acid ester (alkanoic acid ester) ROOR' R'OOR R O OR' amino amine RNH 2 RNHR', etc. amido (only HL) amide RONH 2 R O NH 2 ethyl ethanoate (H 3 OOH 2 H 3 ) (ethyl acetate) Ethylamine (H 3 H 2 NH 2 ) ethanamide (H 3 ONH 2 ) (acetamide) nitrilo (only HL) nitrile RN R N Ethanenitrile (H 3 N) R represents an alkyl group; R', if appearing in the same structure as R, represents a second alkyl group. It may be the same or different from the first. Notice that for amines, two different structures have been written. These differ in the number of carbon-containing groups attached to the nitrogen. The number of groups can be 1, 2, or 3, and they can be any combination of alkyl or aryl (aromatic) groups.
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 26 Naming compounds containing functional groups To name simple compounds with a single functional group, only a slight extension of the rules for naming hydrocarbons is needed. Rule 1: For a compound with a straight chain of carbons, name it as for an alkane, but replace the final e with an ending to indicate the functional group. alcohol alkane alkanol e.g., ethanol (H 3 H 2 OH) aldehyde alkane alkanal e.g., propanal (H 3 H 2 HO) ketone alkane alkanone e.g., propanone (H 3 OH 3 ) carboxylic acid alkane alkanoic acid e.g., ethanoic acid (H 3 OOH) Note: For aldehydes, carboxylic acids and amides, the functional groups can only occur at the ends of carbon chains, so they do not need numbers to indicate their positions. For alcohols and ketones, the groups may occur at different places in the chain, so numbers may be necessary to distinguish different isomers. For alcohols, this would apply to all compounds except methanol and ethanol. The number usually is placed immediately before the suffix indicating the functional group. Numbering starts from the end which gives the lowest number for the functional group. Examples: H 3 H 2 HOHH 3 = butan-2-ol H 3 H 2 H 2 OH 2 H 3 = hexan-3-one (not hexan-4-one) In many textbooks in the U.S., these compounds would be named 2-butanol and 3-hexanone, respectively. Problem: How many different isomers of pentanol are possible? Draw skeletal formulas for them. Problem: What is the shortest straight-chain ketone which requires a number to indicate the position of the carbonyl group?
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 27 Exceptions to rule 1: (a) Esters are structurally related to carboxylic acids, with an alkyl group replacing the acidic hydrogen. They are named by placing the alkyl group first, then naming the acidic portion by replacing the e of the alkane with the same number of carbon atoms by oate. Acid O H 3 O H ethanoic acid Ester H 3 O O H 2 H 3 ethyl ethanoate (b) amines are named as substituted amines, regardless of the number of alkyl groups attached to nitrogen. H 3 H 2 NH 2 = ethylamine H 3 H 2 NHH 3 = ethylmethylamine (alphabetic order of alkyl groups) (c) halogenoalkanes are named as substituted alkanes H 3 HBrH 3 = 2-bromopropane Rule 2: Branching in the main carbon chain is indicated by numbering the position of the alkyl substituent(s). Numbering starts from the carbon of a functional group if it is at the end of a chain. If the functional group is not at the end of the main chain, number from the end that will give the functional group the lowest possible number. (H 3 ) 2 HH 2 OOH = 3-methylbutanoic acid H 3 HOHH(H 3 ) 2 = 3-methylbutan-2-ol Problem: Name the following compounds, and identify any functional group or groups present. (a) H 3 H 2 HOHH 3 (e) (H 3 ) 2 HHO (b) 6 H 5 OH 3 (f) 6 H 5 OOH 3 (c) H 3 HIH 3 (g) H 3 HOHOOH (d) H 3 H 2 OO (h) 6 H 5 NH 2
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 28 Problem: Write the formula for each of the following, showing clearly the functional group present. (a) propanoic acid (d) ethylmethylamine (b) pentan-2-ol (e) pentanal (c) butanone (f) 3-bromohexane Problem: Draw abridged structures (no hydrogens needed) of the following. Name each of the following compounds (a) (H 3 H 2 ) 2 HHO (b) H 3 OH(H 3 ) 2 (c) H 3 HOHH 2 H 2 H 3 (d) H 3 H 2 H 2 OOH 3
hempocalypse Now! hapter 03 Fundamentals of Organic hemistry Page 29 Isomerism due to functional groups We know that isomers refer to compounds with the same molecular formula, but different structures. We have encountered these already in hydrocarbons, with different arrangements of the carbon atoms, and in compounds containing functional groups, when a substituent such as a halogen atom or hydroxyl group can be attached to different positions in the chain. In some cases, we have compounds with different functional groups but the same formula. For example, the following two compounds, an acid and an ester, are isomers, both with the molecular formula 2 H 4 O 2 H 3 =O OH H=O OH 3 H 3 OOH, HOOH 3 ethanoic acid methyl methanoate Two other functional group isomers are propanal and propanone. Problem: Write the structural formulas of propanal and propanone. Draw their structural formulas. Problem: Which of the following are isomers of butanoic acid, H 3 H 2 H 2 OOH? (a) (H 3 ) 2 HOOH (b) HOH 2 H=HH 2 OH (d) H 3 H=HH 2 OH (e) H 3 H 2 H 2 HO (c) H 3 OOH 2 H 3 Because the functional groups are different, such isomers can be expected to have quite different physical and chemical properties from each other. Problem: For each of the following pairs, indicate whether or not the two compounds are isomers of each other. (a) H 3 H 2 H 2 OH 3 and H 3 H 2 OH 2 H 3 (b) H 3 HOHH 2 H 3 and H 3 H 2 HOHH 3 (c) H 3 H 2 H 2 H 2 l and (H 3 ) 2 HH 2 l (d) H 3 H 2 (H 3 ) 2 H 2 OH and H 3 HOHH(H 3 ) 2