Organic Chemistry and Instrumental Analysis. Advanced Higher Chemistry

Transcription

1 Organic Chemistry and Instrumental Analysis Advanced Higher Chemistry

2 Starter State the definition of an orbital Name the sub-atomic particle involved in bonding Describe the formation of a covalent bond

3 Molecular orbitals When atoms approach each other, their separate sets of atomic orbitals merge to form a single set of molecular orbitals Copy and complete using your knowledge of orbitals Molecular orbitals have similar properties to atomic orbitals; they are populated by {electrons/protons} filling from the orbital with the {highest/lowest} energy AND they are considered full when they contain {two/one} electron(s)

4 Molecular orbitals The number of orbitals formed is always equal to the number of atomic orbitals involved For example, if there are two atomic orbitals involved, two molecular orbitals are formed Bonding molecular orbitals and anti bonding molecular orbitals are formed

5 Example: Hydrogen State the number of electrons found in a hydrogen atom Name the type of orbitals found in a hydrogen atom Write the electron configuration for a hydrogen atom

6 Example: Hydrogen When diatomic hydrogen is formed electrons are between the two hydrogen atoms because The two 1s atomic orbitals overlap to form a σ (sigma) molecular orbitals Sigma bond: a covalent bond formed by end on overlap of two atomic orbitals (type of orbital does not matter) lying along the axis of the plane (example)

7 Example: Hydrogen two 1s orbitals overlap σ molecular orbital

8 Example: Hydrogen The molecular orbital formed has a lower energy arrangement than when each atomic orbital is separate

9 Bonding A pi (π) bond is a covalent bond formed by the sideways overlap of two parallel atomic orbitals lying perpendicular to the axis of the bond. Pi bonds normally arise when atoms form multiple bonds

10 Bonding

11 Hybridisation Hybridisation is the process of mixing atomic orbitals within an atom to generate a set of new atomic orbitals called HYBRID ORBITALS Lets consider carbon; Write the electron arrangement for carbon Write the electron configuration for carbon Draw the orbital box diagram for carbon

12 Hybridisation From the orbital box diagram, you can identify two unpaired electrons Using this information, suggest how many covalent bonds carbon can make This suggests carbon can make two covalent bonds Is this consistent with the information we know from National 5? This is inconsistent as we know carbon can make four

13 Hybridisation It is hybridisation that makes this possible! In carbon, one 2s orbital mixes with three 2p orbitals to form 4 hybrid orbitals termed sp 3

14 Hybridisation

15 Hybridisation The 4 sp 3 hybrid orbitals can be described as degenerate State the definition of degenerate Identify the type and number of atomic orbitals in sp and sp 2 hybrid orbitals

16 Hybridisation In the compound Iodine heptafluoride (IF 7 ), the iodine atom uses sp 3 d 3 hybrid orbitals State the number and type of atomic orbitals that mix to form the set of sp 3 d 3 hybrid orbitals

17 Hybridisation Hybridisation can be used to explain the bonding in alkanes and alkenes Describe the difference between alkanes and alkenes

18 Alkanes In methane, all four hybrid orbitals are used to form σ bonds between the central carbon atom and hydrogen atoms

19 Alkanes In ethane, three hybrid orbitals are used to form σ bonds between the central carbon atom and hydrogen atoms However, the carbon-to-carbon single bond results from the overlap of sp 3 orbitals forming a σ bond

20 Alkanes Remember: σ bonds are formed by the end on overlap of two atomic orbitals As the carbon-to-carbon σ bond will lie along the line joining both atoms, they are free to rotate

21 Alkenes Hybridisation must also occur in alkenes In each carbon atom within the double bond, the 2s orbital mixes with two of the 2p orbitals to form 3 degenerate sp 2 hybrid orbitals This leaves one 2p orbital un-hybridised

22 Alkenes

23 Alkenes To minimise repulsion, sp 2 hybrid orbitals adopt a trigonal planar arrangement They are in the same plan at an angle of 120 o The 3 sp 2 hybrid orbitals of one carbon atom overlap to form two σ bonds with hydrogen atoms and one with the other carbon atom The un-hybridised 2p orbital forms a π bond (side on overlap)

24 Alkenes

25 Alkenes Summary of bonding and orbitals Bonding Type C - C C = C Bonding Orbitals Present 1σ 1σ and 1π 1σ and 2π

26 Activity Consider propene, Draw the structure of propene Identify the hybridisation present on each carbon atom Identify and label the bonding orbitals present

27 Questions Analyse the structure of quinoline and determine the number of σ and π bonds present

28 Questions Which line in the table is correct for the following structure? Number of σ bonds Number of π bonds A 4 3 B 8 5 C 10 2 D 10 3

29 The Bonding Continuum Name the bonds which occur between atoms List the three types of bonding which occurs between atoms Describe how each type of bond is formed

30 The Bonding Continuum

31 Non-Polar Covalent Bonds Bonding electrons are shared equally between two atoms with the same electronegativities Both atoms have an equal attraction for the bonding electrons This results in the molecular orbital being completely symmetrical about the mid point between the two atoms

32 Polar Covalent Bonds Bonding electrons are unequally shared between two atoms with the different electronegativities The bonding electrons will be more strongly attracted to the atom with the {higher/lower} electronegativity This results in the formation of partial charges The atom with the higher electronegativity becomes partially {positive/negative} The atom with the lower electronegativity becomes partially {positive/negative}

33 Polar Covalent Bonds Due to the formation of charges, polar covalent bond have some character. The difference in the electronegativities between the atoms results in The molecular orbital being asymmetrical about the midpoint between the two atoms

34 Ionic Bonds Electrons are transferred from one atom to another resulting in the formation of ions There is a large difference in the electronegativity of the atom The molecular orbital is an extreme case of asymmetry with the bonding molecular orbitals being almost entirely located around just one atom.

35 Bonding Symmetrical molecular orbital formed by non-polar covalent bonds Asymmetrical molecular orbital formed by polar covalent bonds Extremely asymmetrical molecular orbital formed by ionic bonds

36 Molecular Orbitals Revision Describe the formation of a molecular orbital Copy and Complete The end-on overlap of two atomic orbitals results in the formation of a {σ / π } molecular orbitals. Anti-bonding orbitals are represented by σ*. The sideways overlap of two atomic orbitals results in the formation of a {σ / π } molecular orbitals. Anti-bonding orbitals are represented by π*. Electrons will fill bonding orbitals first as they have the energy. Under normal conditions, anti-bonding orbitals are.

37 Absorption of Visible Light by Organic Molecules Analyse the structures below and identify any similarities

38 Organic Molecules The presence of multiple double bonds can create a conjugated system This occurs when the single bonds alternate with double bonds in a structure A conjugated system allows electrons to become delocalised State the definition of delocalised

39 Organic Molecules Identify if the following structures contain any conjugation

40 Organic Molecules For an organic compound to be coloured it must contain a large degree of conjugation The energy of orbitals can be used to explain coloured organic compounds

41 Organic Molecules It is the energy difference between the HOMO and LUMO that determine the wavelength of the photon emitted HOMO the occupied molecular orbital with the highest energy LUMO the unoccupied molecular orbital with the lowest energy

42 Organic Molecules The conjugated system in an organic molecule can reduce the energy difference between the HOMO and LUMO This lowers the energy that will need to be absorbed to promote electrons This results in a {higher/lower} frequency and a {higher / lower} wavelength of the photons produced When the wavelength of light absorbed is in the visible region, the complementary colour can be observed

43 Chromophores The chromophore is the term given to the group of atoms within an organic molecule that are responsible for the absorption of light in the VISIBLE region The chromophore is responsible for coloured organic compounds Light is absorbed and the electrons in the chromophore gain energy and are promoted from the HOMO to the LUMO REMEMBER : The colour observed is complementary to that absorbed

44 Organic Chemistry and Instrumental Analysis Stereochemistry

45 Stereoisomerism State the definition of an isomer. Stereoisomers are molecules with the same molecular formula but different spatial arrangements of their atoms. There are two types of stereoisomers; geometric and optical

46 Geometric Isomerism Geometric isomerism usually occurs when there is a lack of rotation at a bond e.g. when a double bond is present REMEMBER: There is no rotation due to the π bond between the carbon atoms with a double bond Two geometric isomers can be labelled as cis or trans Lets look at but-2-ene as an example

47 Geometric Isomerism Here are two isomers of but-2-ene there is no rotation around the double bond Substituent groups on the same side of the molecule Substituent groups on different sides of the molecule

48 Geometric Isomerism The structure below is cis-but-2-enedioic acid. Name the function groups present Draw trans-but-2-enedioic acid

49 Geometric Isomerism Cis and trans isomers can react differently due to the positions of their function groups. e.g. cis-but-2-enedioic acid can form a ring structure whereas trans-but-2-enedioic acid cannot Geometric isomers can also have different melting points, boiling points and chemical properties.

50 Geometric Isomerism Examples Trans fats are mentioned as a health risk The cis formation of combretastatin is known for anticancer activity

51 Geometric Isomerism 1. Identify which isomer has the highest melting point 2. Identify which isomer has the strongest intermolecular forces 3. Identify which isomer can most tightly pack together

52 Geometric Isomerism Geometric isomerism does not only occur when a molecule contains a double bond Saturated ring structures can also have geometric isomers Name the family of saturated ring structures

53 Geometric Isomerism The structure below is 1,2-dichlorocyclopropane Identify if this structure is the cis or trans isomer Draw the other isomer of this structure

54 Optical Isomerism Optical isomers are asymmetric molecules that are nonsuperimposable mirror images of each other Chiral molecules are non-superimposable mirror images of each other A chiral molecule is one in which a central carbon is surrounded by four different atoms

55 Optical Isomerism Chiral example Mirror images but do not super impose These molecules are OPITCAL ISOMERS (or enantiomers)

56 Optical Isomerism Lactic acid is shown below: Explain why lactic acid can exhibit optical isomerism Draw the optical isomer of lactic acid

57 Optical Isomerism Optical isomers (enantiomers) have identical physical properties except they rotate polarised light in a opposite but equal way. They can be describe as OPTICALLY ACTIVE. When an equal number of moles (equimolar) of both isomers exist, the mixture is optically inactive This mixture is known as a RACEMIC mixture.

58 Optical Isomerism Many organic compounds found in nature can be described as chiral In living organisms, only one of the optical isomers are present, resulting in optical activity For example, amino acids, proteins, enzymes and sugars.

59 Organic Chemistry and Instrumental Analysis Synthesis

60 Bond Fission In any chemical reaction we know there must be bond making and bond breaking. A covalent bond (shared pair of electrons) can break in two different ways: Homolytic fission Heterolytic fission

61 Homolytic Fission Homolytic fission involves the breaking of bonds When the bond is broken one electron from the shared pair goes to each atom A B A. + B. Two free radicals are formed in this breaking process: the unpaired electron is shown by the.

62 Homolytic Fission The free radicals formed at neutral over all as the number of protons equals the number of electrons in the atom However, they are highly reactive due to their unpaired electron causing instability

63 Homolytic Fission Name the three stages of free radical formation Each stage can be shown using arrows to represent the movement of electrons Single headed arrows represent the movement of one electron Double headed arrows represent the movement of two electrons

64 Heterolytic Fission Heterolytic fission involves the breaking of bonds When the bond is broken the pair of electrons from the bond go to one atom A B A + + B - Ions (charged particles) are formed in this breaking process Heterolytic fission is more likely when the bond is polar

65 Heterolytic Fission If these ions go on to react with ions of the opposite charge the pair of electrons can become shared A polar bond is formed B - + C + B C

66 Heterolytic Fission When this occurs the negative ion acts as an electron donor and the positive ion acts as an electron acceptor Nucleophiles are electron donors (nucleus lovers) Electrophiles are electron acceptors (electron lovers) In the previous example label the nucleophile and the electrophile

67 Examples Nucleophiles Electrophiles Br - H + OH - NO + 2 NH 3 SO 3

68 Organic Chemistry and Instrumental Analysis Synthesis Haloalkanes

69 Haloalkanes Haloalkanes are substituted alkanes One or more hydrogen atoms have been replaced with a halogen atom They are named in a similar way to branched alkanes

70 Naming haloalkanes For example: 1,2-dichloroethane 2-bromo-2-chloro-1,1,1-trifluoroethane

71 Naming haloalkanes Monohaloalkanes can be classified as primary, secondary and tertiary PRIMARY SECONDARY Tertiary two hydrogen atoms on the carbon bonded to the halogen atom one hydrogen atoms on the carbon bonded to the halogen atom no hydrogen atoms on the carbon bonded to the halogen atom Draw an example of each haloalkane

72 Reactions of Haloalkanes Monohaloalkanes can undergo nucleophilic substitution reactions 1. With alkalis to form alcohols 2. With alcoholic alkoxides to form ethers 3. With ethanoic cyanide to form nitriles which can be hydrolysed to form carboxylic acids

73 Reactions of Haloalkanes 1. With alkalis to form alcohols The polar bond in the molecule results in the carbon atom having a partially positive charge This carbon atom can act as an electrophile (electron acceptor) This can result in this carbon atom being attacked by a nucleophile (electron donator)

74 Reactions of Haloalkanes 1. With alkalis to form alcohols The oxygen in the hydroxide ion can donate a lone pair of electrons to the partially positive carbon atom creating a new covalent bond The halogen is thrown out as a halide ion The halogen has been substituted

75 Reactions of Haloalkanes 2. With alcoholic alkoxides to form ethers The polar bond in the molecule results in the carbon atom having a partially positive charge This carbon atom can act as an electrophile (electron acceptor) This can result in this carbon atom being attacked by a nucleophile (electron donator)

76 Reactions of Haloalkanes 2. With alcoholic alkoxides to form ethers The oxygen in the methoxide ion can donate a pair of electrons to the partially positive carbon atom creating a new covalent bond The halogen is thrown out as a halide ion The halogen has been substituted

77 Reactions of Haloalkanes 3. With ethanoic cyanide to form nitriles which can be hydrolysed to form carboxylic acids This is reaction occurs in 2 steps Step 1 Step 2 involves the formation of the nitrile (Nucleophilic substitution) involves the conversion of a nitrile to a carboxylic acid (acid hydrolysis)

78 Reactions of Haloalkanes All these reactions must be heated under reflux (see your Researching Chemistry notes)

79 Reactions of Haloalkanes Monohaloalkanes can undergo elimination reactions. These reactions involve heating under reflux with ethanolic potassium (or sodium) hydroxide ethanolic potassium (or sodium) hydroxide a solution of potassium (or sodium) hydroxide in ethanol An alkene is always formed

80 Activity a. Identify the type of reaction b. Draw the organic products formed 2-chloro-2-methylbutane 1. reacting with aqueous potassium hydroxide 2. reacting with ethanolic potassium hydroxide

81 Reaction Mechanisms A monohaloalkane will undergo nucleophilic substitution by one of two different mechanisms; S N 1 or S N 2

82 Reaction Mechanisms Remember first order means only one substance is involved in the slow rate determining step

83 Reaction Mechanisms These reactions require two steps 1. Slow rate determining step in which an intermediate product is formed termed a carbocation 2. Fast second step in which the final product is formed The overall reaction involves heterolytic fission and nucleophilic attack

84 Reaction Mechanisms The reaction of 2-bromo-2-methylpropane with hydroxide ions (nucleophile) The rate equation for this reaction was determine and can be written as follows: rate = k[(ch 3 ) 3 CBr]

85 Reaction Mechanisms Remember second order means that one molecule of each reactant is involved in the rate determining step

86 Reaction Mechanisms These reactions require one step A transition state is formed which often adopts a trigonal bipyramidal shape with partial bonds in the same plane The overall reaction involves nucleophilic attack

87 Reaction Mechanisms The reaction of bromoethane with hydroxide ions (nucleophile) The rate equation for this reaction was determine and can be written as follows: rate = k[ch 3 CH 2 Br][OH - ]

88 S N 1 or S N 2 When a haloalkane undergoes nucleophilic substitution it is important to identify which reaction mechanism is taking place Factors that must be considered include 1. whether the carbocation intermediate is primary, secondary or tertiary (stability) 2. The size of the alkyl groups on the haloalkane (steric hinderance)

89 S N 1 or S N 2 Alkyl groups have a positive inductive effect This means they are electron donating; they push electrons onto the positively charged carbon and increase stability Therefore, tertiary cations are more stable than primary cations Tertiary most likely to undergo S N 1 mechanism

90 S N 1 or S N 2 The size of the alkyl groups is known as a steric effect In S N 2 mechanisms, the nucleophile attacks from the side opposite to the halogen atom Larger alkyl groups can limit access to the partially positive carbon atom, therefore, tertiary haloalkanes very unlikely to undergo S N 2 reactions Primary haloalkanes are the most likely to undergo S N 2 reactions

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94 Organic Chemistry and Instrumental Analysis Synthesis Preparing Alcohols

95 Alcohols Properties of Alcohols (Higher knowledge) - Alcohols have hydrogen bonded to a highly electronegative atom creating a polar bond - This results in hydrogen bonding being present between molecules (intermolecular) - This is the strongest type of intermolecular bonding, increasing the energy to break - This increases the boiling point of alcohols

96 Alcohols Properties of Alcohols (Higher knowledge) - Smaller alcohols (methanol, ethanol, propan-1-ol) are soluble in water as they are polar (hydrophilic) - Like dissolves like - Larger alcohols (heptan-1-ol) become more non-polar due to the increase in the number of non-polar bonds so they are insoluble in water (hydrophobic)

97 Alcohols Properties of Alcohols (Higher knowledge) Alcohols can be primary, secondary or tertiary Draw and Name one of each type of alcohol

98 Preparation of Alcohols Alcohols can be prepared from 1. monohaloalkanes by heating under reflux with aqueous potassium or sodium hydroxide (nucleophilic substitution) 2. alkenes by reacting with water using sulfuric acid as a catalysts (acid-catalysed addition or hydration) 3. aldehydes/ketones by reacting with lithium aluminium hydride dissolved in ether (reduction)

99 Reactions of Alcohols Alcohols can react with 1. Reactive metals to form alkoxides (displacement/redox) 2. Aluminium oxide/ concentrated sulfuric acid/ concentrated phosphoric acid to form alkenes (dehydration/elimination) 3. Carboxylic acids/ acid chlorides to form esters using sulfuric acid as catalyst (condensation)

100 Reactions of Alcohols Alcohols can react with 1. Reactive metals to form alkoxides (displacement/redox) Alcohols are similar to water so react similarly

101 Reactions of Alcohols

102 Reactions of Alcohols The alkoxides formed are powerful bases (neutralise acids) and nucleophiles (electron donators)

103 Reactions of Alcohols 2. Aluminium oxide/ concentrated sulfuric acid/ concentrated phosphoric acid to form alkenes (dehydration/elimination)

104 Reactions of Alcohols 3. Carboxylic acids/ acid chlorides to form esters using sulfuric acid as catalyst (condensation) REMEMBER: alcohol + acid ester + water OR alcohol + acid chloride ester + hydrochloric acid REMEMBER: These are condensation reactions (produce a small molecule e.g. H 2 O or HCl)

105 Organic Chemistry and Instrumental Analysis Synthesis Ethers

106 Ethers Ethers can be synthesised from haloalkanes and alkoxides. Ethers have the general structure R-O-R where R and R represent alkyl groups e.g. CH 3 Ethers can be describe as being an alkane with an alkoxy group attached (R -O) Examples: methoxy (CH 3 -O), ethoxy (CH 3 CH 2 -O)

107 Naming Ethers Ethers are named following the IUPAC rules Ethers are named by 1. Identify the longest chain of carbon atoms and name after the alkane containing the same number of carbon atoms 2. Identify the carbon on which the alkoxy group is attached 3. Name the alkoxy group based on the number of carbon atoms in the chain

108 Properties of Ethers Ethers have lower boiling points than those organic molecules with same molecular masses e.g. ethoxyethane has a boiling point of 34.5 o C compared to butan-1-ol that has a boiling point of 119 o C But why???

109 Properties of Ethers Alcohols have a hydroxyl group which can result in hydrogen bonding between its molecules Hydrogen bonding is the strongest of the intermolecular forces requiring more energy to break which results in an increased boiling point Ethers do not have any opportunity for hydrogen bonding between its molecules resulting in the lower boiling point

110 Properties of Ethers Ethers are soluble in water The oxygen results in two polar bonds being formed and can then hydrogen bond to water However, like with alcohols, as the chain length increases the solubility decreases as the molecule becomes more non-polar

111 Properties of Ethers

112 Uses of Ethers Can dissolve a wide variety of polar organic compounds Relatively unreactive: not polar enough to be attacked by a nucleophile/ an electrophile Resistant to oxidation and reduction Extremely volatile (easily evaporated) and highly flammable

113 Organic Chemistry and Instrumental Analysis Synthesis Alkenes

114 Preparation of Alkenes REMEMBER: Alkenes are unsaturated molecules Alkenes can be prepared by 1. The dehydration of alcohols using aluminium oxide 2. Base-induced elimination of hydrogen sulfides from monohaloalkanes

115 Preparation of Alkenes Alkenes can be prepared by 1. The dehydration of alcohols using aluminium oxide These are termed ELIMINATION REACTIONS Elimination reactions are reactions in which a small molecule is eliminated from a larger molecule

116 Preparation of Alkenes 2. Base-induced elimination of hydrogen sulfides from monohaloalkanes This reaction required heating a monohaloalkane under reflux with ethanolic potassium (sodium) hydroxide ethanolic potassium (sodium) hydroxide potassium (sodium) hydroxide dissolved in ethanol

117 Preparation of Alkenes 2. Base-induced elimination of hydrogen sulfides from monohaloalkanes Lone pair of electrons on oxygen are donated to create an O-H covalent bond This results in the C-H bond breaking and a second C-C bond forming The carbon atom now has 5 bonds so undergoes heterolytic fission throwing out the bromine

118 Reactions of Alkenes Alkenes can undergo many different electrophilic addition reactions 1. Catalytic addition of hydrogen to form alkanes (hydrogenation) 2. Addition of halogens to form dihaloalkanes (halogenation) 3. Addition of hydrogen halides using Markovnikov s rule to form monohaloalkanes (hydrohalogenation) 4. Acid-catalysed addition of water Markovnikov s rule to form alcohols (hydration)

119 Markovnikov s Rule When a hydrogen halide or water is added to an unsymmetrical alkene, two products can be formed Markovnikov states that when H-X or H-OH is added to an unsymmetrical alkene, the major product is the one in which the H atom becomes attached to the carbon atom with the greatest starting number of attached hydrogen atoms Let s analyses the addition of hydrogen chloride to but-1-ene

120 Reactions of Alkenes Alkenes can undergo many different electrophilic addition reactions 1. Catalytic addition of hydrogen to form alkanes (hydrogenation) Don t need to know mechanism

121 Reactions of Alkenes 2. Addition of halogens to form dihaloalkanes (halogenation) This is a two-step process Step 1 Step 2- the bromine molecule acts as an electrophile and attacks the double bond of the alkene (nucleophile) the bromine ion acts as a nucleophile and attacks the intermediate cyclic ion (electrophile)

122 Reactions of Alkenes 3. Addition of hydrogen halides using Markovnikov s rule to form monohaloalkanes (hydrohalogenation) This is a two-step process Step 1 Step 2 The H-X molecule acts as the electrophile and attacks the double bond in the alkene forming an intermediate carbocation (positive carbon ) The bromine ion acts as a nucleophile and attacks the carbocation to form the haloalkane

123 Reactions of Alkenes 4. Acid-catalysed addition of water Markovnikov s rule to form alcohols (hydration) This is a three step process Step 1 Step 2 Step 3 The hydrogen ion of the acid is an electrophile and attacks the double bond in propene forming a carbocation Carbocation undergoes nucleophilic attack by a water molecule forming a protonated alcohol Protonated alcohols are strong acids and readily lose protons to produce alcohol

124 Organic Chemistry and Instrumental Analysis Synthesis Carboxylic Acids

125 Preparation of Carboxylic Acids Carboxylic acids can be prepared by 1. Oxidising a primary alcohol or an aldehyde 2. Hydrolysing nitriles 3. Hydrolysing amides 4. Hydrolysing esters

126 Preparation of Carboxylic Acids Carboxylic acids can be prepared by 1. Oxidising a primary alcohol or an aldehyde This required an oxidising agent State the definition of an oxidising agent Name three oxidising agents

127 Heating with Copper(II) oxide Heating under reflux with acidified potassium dichromate Reacting with acidified potassium permanganate

128 Preparation of Carboxylic Acids Carboxylic acids can be prepared by 2. Hydrolysing nitriles Heating a nitrile under reflux with aqueous acid (H 2 O/H + ) The hydrogen ions of the acid catalyse the reaction

129 Preparation of Carboxylic Acids Carboxylic acids can be prepared by 3. Hydrolysing amides Heating an amide under reflux with aqueous acid or aqueous alkali (H 2 O/H + or H 2 O/OH - ) The hydrogen ions/ hydroxide ions act as to catalyse the reaction

130 Preparation of Carboxylic Acids Carboxylic acids can be prepared by 4. Hydrolysing esters Heating an amide under reflux with aqueous acid or aqueous alkali (H 2 O/H + or H 2 O/OH - ) The hydrogen ions/ hydroxide ions act as to catalyse the reaction

131 Preparation of Carboxylic Acids Carboxylic acids can be prepared by 4. Hydrolysing esters When an alkali is used the equilibrium position is shifted to the right (products) The alkali can also react with the carboxylic acid

132 Reactions of Carboxylic Acids Carboxylic acids can be used in the same way as a common acid such as hydrochloric acid They can react with 1. Metals to form salts and hydrogen gas 2. Carbonates to form salts, carbon dioxide and water 3. Alkalis to form salts and water

133 Organic Chemistry and Instrumental Analysis Synthesis Amines

134 Amines Amines are found in medicines, plastics, cosmetics and other useful substances. The are organic derivatives of ammonium (NH 3 ) The function group is the NH 2 Examples: CH 3 -NH 2 CH 3 CH 2 NH 2

135 Classifying Amines Amines can be primary, secondary or tertiary. PRIMARY: The nitrogen atom has 2 hydrogen atoms attached SECONDARY: The nitrogen atom has 1 hydrogen atom attached TERTIARY: The nitrogen atom has no hydrogen atoms attached

136 Naming Amines The carbon chains on the amine are named like branches e.g. methyl, ethyl, propyl etc. Prefixes used to indicate when there are more than one of the dame type of branch e.g. dimethyl or trimethyl

137 Properties of Amines Primary and secondary have much higher boiling points than the alkanes with similar formula masses. Example methylamine: mass= 31 bp= -7.5 o C ethane: mass= 30 bp= -89 o C Using your knowledge of chemistry, explain this difference. Include why this not true for tertiary amines in your answer.

138 Properties of Amines All amines are soluble in water. Explain why this is true. Water can hydrogen bond with all amines (like dissolves like) Solubility decreases as the number of carbon atoms increases as the molecule becomes non-polar overall

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142 Properties of Amines Amines are weak bases so there is not a full dissociation of ions resulting in an equilibrium being reached between reactants and products The non bonded pair of electrons on the nitrogen atom can accept a hydrogen ion (proton) from water producing an alkylammonium ion and hydroxide ions (left from water) The increased concentration of hydroxide ions, increases the ph resulting an alkaline solution

143 Properties of Amines As amines are bases, they can be used to neutralise acids such as HCl, H 2 SO 4, or HNO 3. Like all neutralisation reactions, a salt is produced. They can react with carboxylic acids as shown in the previous section.

144 Organic Chemistry and Instrumental Analysis Synthesis Benzene

145 Aromatics Many every day products contain aromatic compounds For example, trichlorophenol in antiseptics or toluene in permanent markers

146 Aromatics The simplest aromatic structure is named benzene Benzene has the molecular formula C 6 H 6 Benzene is unsaturated (it contains double bonds)

147 Structure of Benzene Structural formula Kekule Structures

148 Aromatics The Kekule structures did not explain all the features of benzene Benzene does not decolourise bromine water which you would expect - benzene must resist addition reactions - benzene must be more stable

149 Aromatics Benzene has a flat structure with a hexagonal shape - all bonds are of the same length and strength Each carbon atom is sp 2 hybridised - each carbon makes 3 sigma bonds 1. 2 x carbon to carbon 2. 1 x hydrogen - each carbon atom has one electron in the p-orbital - all the p orbitals overlap forming one pi molecular orbital

150 Aromatics

151 Aromatics 12 sigma bonds One pi molecular orbital The electrons in the pi orbital are shared with all carbon atoms delocalised

152 Benzene Benzene can undergo substitution where one of the hydrogen atoms is replaced The cyclic structure is then called a phenyl group (C 6 H 5 )

153 Reactions of Benzene Benzenes can undergo 1. Alkylation using a haloalkane and metal halide 2. Nitration using concentration sulfuric and nitric acid 3. Sulfonation using concentrated sulphuric acid 4. Halogenation using a halogen or metal halide

154 Organic Chemistry and Instrumental Analysis Elemental Microanalysis

155 Elemental Microanalysis Used to determine the masses of C, H, O, S and N in an organic sample This information can be used to determine the empirical formula of a substance The substance must be combusted for this analysis

156 Empirical Formulae The empirical formula can be determined from the elemental masses

157 Organic Chemistry and Instrumental Analysis Mass Spectrometry

158 Mass Spectrometry Mass spectrometry can be used to determine the accurate molecular mass and structural features of organic compounds Mass spectrometers do three things 1. Vaporise a very small sample of a compounds (10-4 g) 2. Ionise the vaporised molecules by bombarding the sample with electrons (forms positive ions) 3. Separates and analyses the ions according to their mass/charge ratio producing the mass spectrum

159 Mass Spectrometry Positive ions with a lower mass/ charge ratio are deflected more than those with a higher mass/charge ratio The peak on a mass spectrum with the highest mass/charge ratio is the GFM of the compound and arises due to the molecular ion (the molecule with one positive charge)

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161 Look at the mass spectra of benzoic acid and methyl benzoate and identify the ions responsible for the major peaks in each case.

162 Look at the mass spectra of benzoic acid and methyl benzoate and identify the ions responsible for the major peaks in each case.

163 Mass Spectrometry From a mass spectrum and empirical formula, it is possible to determine the molecular formula of a sample.

164 Organic Chemistry and Instrumental Analysis Infrared Spectrometry

165 Infrared Spectroscopy Infrared spectroscopy uses infrared electromagnetic radiation (see Unit 1) When infrared is absorbed by organic compounds, the energy results in the bonds vibrating The energy is not sufficient enough to break the bonds

166 Infrared Spectroscopy There are different types of vibration that can occur 1. Stretching changing distance between atoms in molecule 2. Bending changing angle between atoms in molecule

167 Infrared Spectroscopy

168 Infrared Spectroscopy The energy absorbed is dependent on the type of atoms that are involved in the bond and the strength of the bond General rules include 1. Atoms with low mass combined with stiff bonds absorb radiation of a lower wavelength, higher frequency and higher energy 2. Atoms with high mass combined with loose bonds absorb radiation of a higher wavelength, lower frequency and lower energy

169 Infrared Spectroscopy Infrared spectroscopy can be used with samples in any state (solid, liquid or gas) 1. IR radiation is passed through the sample 2. Sample absorbs radiation of different wavelengths causing bond vibrations 3. Transmitted radiation passes to detector 4. Intensity at different wavelengths measured 5. Results shown as an IR spectrum

170 Infrared Spectroscopy In the spectrum produced, each peak represents absorption Absorption is represent by a wavenumber (cm -1 ) REMEMBER = wavelength is equal to 1/λ Each peak represents a type of bond in the structure Infrared data is given in the databook

171 Question The following spectra are of ethanoic acid and ethanoic anhydride. Determine which spectrum is due to which compound.

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174 Infrared Spectroscopy Given the spectrum of an unknown sample, infrared spectroscopy can be used to suggest suitable structural formula

175 Organic Chemistry and Instrumental Analysis Proton Nuclear Magnetic Resonance Spectroscopy

176 Low Resolution 1 H NMR Using the information from the databook, predict where we would find each peak in propanoic acid

177 Low Resolution 1 H NMR You need to be able to Identify the different proton environments Determine the chemical shift values for different proton environment Create low resolution NMR spectra based on these values

178 High Resolution 1 H NMR 1 H NMR spectroscopy is an analytical technique which can provide information on the different environments of hydrogen atoms (protons) in an organic molecule This requires a higher frequency radio wave than low resolution 1 H NMR spectroscopy The spectrum produced will contain more detail

179 High Resolution 1 H NMR The interaction between hydrogen atoms on neighbouring carbon atoms can result in splitting of the peak as the hydrogen atoms in one environment will experience slightly different magnetic fields The peak produced is split into multiple peaks called multiplets The number of hydrogen atoms on the neighbouring carbon atom determines the number of peaks

180 High Resolution 1 H NMR Name of multiplet Number of hydrogen atoms on neighbouring carbons, n Numbers of peaks in spectrum (n+1) Doublet 1 2 Triplet 2 3 Quartet 3 4

181 High Resolution 1 H NMR Example ethyl ethanoate Draw the full structural formula for ethyl ethanoate Identify the different proton environments Determine the chemical shift values for different proton environment in a low resolution spectrum In a high resolution spectrum we can predict how many peaks will form the multiplet

182 High Resolution 1 H NMR Example ethyl ethanoate No neighbouring protons n=0 n+1 = = 1 Remain as a single peak 3 neighbouring protons n=3 n+1 = = 4 Observe four peaks 2 neighbouring protons n=2 n+1 = = 3 Observe three peaks

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184 Organic Chemistry and Instrumental Analysis Pharmaceutical Chemistry

185 Pharmaceutical Chemistry Drugs are substances which alter biochemical processes in the body Drugs can be used as medicines if they have beneficial effects Medicines usually contain the drug mixed with other ingredients

186 Pharmaceutical Chemistry Example: Paracetamol Paracetamol is considered to have benefits when taking at the correct dosage. However, can cause liver failure if the stated dosage is exceeded. Paracetamol has a range of key features Identify and name the main structural features of paracetamol

187 Pharmaceutical Chemistry There are many plant based drugs 1. Opium from poppies used in morphine and codeine 2. Salicin from willow bark which is now synthesised to make aspirin (acetylsalicyclic acid)

188 Drug Classification Drugs can be classified as agonists or antagonists dependent on the response they trigger when binding to a receptor site Agonists produce a similar response to the body s natural active compound Antagonists prevent the action of the body s natural active compound but produce no response themselves

189 How Drugs Work Drugs work by binding to receptors Receptors are usually protein molecules on the surface of cells which can interact with small biologically active molecules Catalytic receptors are enzymes that catalyse biological chemical reactions

190 How Drugs Work The structural fragment of a drug molecule which confers pharmacological activity upon it normally consists of different functional groups correctly orientated with respect to each other The size and shape of the drug molecule must be able to fit a binding site The drug can interact and bind to the receptor through correctly positioned functional groups

191 How Drugs Work

192 How Drugs Work These interactions are intermolecular forces London dispersion forces Permanent dipole- permanent dipole interactions Hydrogen bonds Describe how each of these intermolecular forces are formed