Organic chemistry II: A-level Chemistry AQA revision notes

3.3.7

Optical isomerism

Optical isomers are non-superimposable mirror images 3.3.7

Definitions

Chiral centre: a carbon atom bonded to four different atoms or groups, marked with an asterisk.
Enantiomers: a pair of stereoisomers that are non-superimposable mirror images of each other.

Notes

Butan-2-ol has a chiral centre bonded to H, OH, $CH_{3}$ and $CH_{2} CH_{3}$ , so it exists as two enantiomers.
Draw enantiomers in 3D: two bonds in the plane of the page, a solid wedge towards the viewer and a hashed wedge away, then reflect through a mirror plane.
Enantiomers have identical physical properties (melting point, boiling point, solubility) and identical reactions with achiral reagents; they differ in their effect on plane-polarised light and in their interactions with other chiral molecules such as enzymes.
Optical isomerism matters biologically because receptors and enzymes are themselves chiral. The two enantiomers of carvone smell different (spearmint and caraway), and the drug thalidomide had one enantiomer that was a sedative while the other caused birth defects; it also converts between forms in the body, so giving a single enantiomer would not have prevented the harm.
Common chiral molecules: butan-2-ol, 2-hydroxypropanoic acid (lactic acid), and every α-amino acid except glycine.
In 3D drawings, two of the four bonds should lie in the plane of the paper at roughly 109° apart, with one wedge and one hashed bond; the mirror image swaps the positions of any two groups.

Recognising a chiral centre 3.3.7

Key results

Butan-2-ol: C2 carries H, OH, $CH_{3}$ , $CH_{2} CH_{3}$ , so it is chiral.
Propan-2-ol: C2 carries two $CH_{3}$ groups, so it is achiral.

Method

Look only at $s p^{3}$ carbons: a C=C or C=O carbon cannot be a chiral centre.
For each candidate carbon, compare the whole groups attached, not just the first atom: $CH_{2} CH_{3}$ differs from $CH_{3}$ .
Discard any carbon carrying two identical groups, such as two H atoms or two methyl groups.

Notes

A molecule with one chiral centre always has a pair of enantiomers; molecules with more than one chiral centre can have more stereoisomers.
Worked example: 2-chlorobutane, $CH_{3} CHClCH_{2} CH_{3}$ , has one chiral centre (C2: H, Cl, $CH_{3}$ , $C_{2} H_{5}$ ); 1-chlorobutane has none. 2-hydroxypropanoic acid's C2 carries H, OH, $CH_{3}$ and COOH.
Ring carbons can be chiral too, if the two routes around the ring from that carbon are different; for A-level, check each substituted ring carbon as carefully as chain carbons.
A useful exam habit is to star every chiral centre on a skeletal formula before answering a question about isomers; molecules with $n$ chiral centres have up to $2^{n}$ stereoisomers.

Effect on plane-polarised light 3.3.7

Definitions

Optically active: rotates the plane of plane-polarised light.
Racemic mixture (racemate): an equimolar mixture of two enantiomers.

Notes

Under identical conditions, the two enantiomers rotate the plane by the same angle in opposite directions: (+) clockwise, (−) anticlockwise as viewed towards the light source.
A racemate shows no net rotation because the equal and opposite rotations cancel; it is optically inactive even though each molecule is chiral.
Comparisons need the same concentration, path length, temperature, wavelength and solvent. The sign of rotation is measured experimentally; it cannot be read from a structural drawing.
Plane-polarised light has its oscillations in a single plane, produced by passing ordinary light through a polariser. A polarimeter measures the angle through which a sample rotates this plane.
Labelling enantiomers (+) and (−) records the observed direction of rotation; it is not predictable from the structure and is independent of the R/S naming system, which is not required at A-level.
Optical activity is a test for an excess of one enantiomer: a racemic product shows no rotation even though it is made of chiral molecules, so measuring rotation tells a chemist whether a reaction gave a single enantiomer.

Formation of a racemic mixture 3.3.7

Notes

In ethanal the carbonyl carbon is trigonal planar, so the cyanide nucleophile is equally likely to attack from either face.
Attack from one face gives one enantiomer of 2-hydroxypropanenitrile and attack from the other gives its mirror image, in a 50:50 ratio.
The product is a racemate and is optically inactive. Methanal and propanone give no chiral product because the former carbonyl carbon ends up with two identical groups.
This matters in drug synthesis: often only one enantiomer is active, so the other must be separated or the synthesis designed to give a single enantiomer.
The same reasoning applies whenever a planar intermediate is attacked: in $S_{N} 1$ reactions of tertiary halogenoalkanes, the planar carbocation can be attacked from either face, so an optically active starting material gives a racemic product.
Separating enantiomers is difficult and expensive because they have identical physical properties. Pharmaceutical chemists instead use chiral catalysts or enzymes to make a single enantiomer, or start from naturally occurring chiral molecules.
Giving a racemate as a drug can mean half the dose is inactive, or worse, causes side effects; it can also require the doses and tests to be doubled.

Worked example

Propanal is treated with HCN (generated in situ from KCN and dilute sulfuric acid).

Explain why the product is optically inactive even though it contains a chiral centre.

Show worked solution

Nucleophilic addition of CN⁻ to the planar carbonyl carbon of propanal, $CH_{3} CH_{2} CHO$ , can occur with equal probability from either face of the flat C=O group, since nothing in the achiral starting material favours one face over the other.

This produces the two enantiomers of the hydroxynitrile product,

CH_{3} CH_{2} CH (OH) CN

in exactly equal amounts — a racemic mixture.

Although each individual enantiomer would rotate plane-polarised light, the equal and opposite rotations of the two enantiomers, present in equal amounts, cancel exactly, so the mixture as a whole shows no net optical activity.

3.3.8

Aldehydes and ketones

The carbonyl group 3.3.8

Definitions

Aldehyde: carbonyl carbon bonded to at least one H, giving –CHO at the end of a chain (named −al).
Ketone: carbonyl carbon bonded to two carbon groups (named −one).

Notes

The C=O bond is polar ( $C^{δ +} = O^{δ -}$ ) because oxygen is more electronegative, so the carbonyl carbon is attacked by nucleophiles.
The carbonyl carbon is trigonal planar (about 120°). This is why nucleophilic addition can produce a racemate.
Unlike alkenes, carbonyls undergo nucleophilic (not electrophilic) addition: the C=O π bond is polarised, while the C=C π bond is an electron-rich region.
Boiling points: propanal 48 °C and propanone 56 °C lie between butane (−1 °C, London forces only) and propan-1-ol (97 °C, hydrogen bonding), because carbonyl compounds have permanent dipole–dipole forces but no O–H to hydrogen bond with each other.
Small aldehydes and ketones dissolve in water because water molecules hydrogen bond to the carbonyl oxygen's lone pairs. Propanone is a common solvent because it mixes with both water and many organic compounds.
Naming: the aldehyde carbon is always C1 and needs no number (butanal); ketones need a locant from pentan-2-one onwards (propanone and butanone have no isomeric positions).

Oxidation and distinguishing tests 3.3.8

Key results

Aldehyde: $CH_{3} CHO + [O] \to CH_{3} COOH$ ; acidified dichromate turns from orange to green.
Tollens' reagent ( $[Ag(NH_{3})_{2}]^{+}$ ), warmed: aldehyde gives a silver mirror ( $Ag^{+} + e^{-} \to Ag$ ).
Fehling's solution (blue $Cu^{2 +}$ complex), warmed: aldehyde gives a brick-red precipitate of $Cu_{2} O$ .

Notes

The aldehyde is oxidised while it reduces the metal ion in the test reagent.
Ketones give no change with any of these mild oxidising agents because oxidation would require breaking a C–C bond.
Tollens' half-equation: $[Ag(NH_{3})_{2}]^{+} + e^{-} \to Ag + 2 NH_{3}$ . In alkaline solution the aldehyde is oxidised to its carboxylate ion, e.g. ethanoate.
Fehling's solution contains copper(II) ions complexed in alkaline solution; reduction to copper(I) gives the brick-red $Cu_{2} O$ precipitate. Benedict's solution in biology works the same way for reducing sugars, which contain aldehyde groups.
Practical safety: warm the tubes in a water bath, not over a flame, because the organic compounds are flammable; dispose of Tollens' reagent promptly, as explosive silver compounds can form on standing.
These tests make sense of the oxidation ladder: primary alcohol → aldehyde → carboxylic acid; secondary alcohol → ketone (stop).

Reduction with sodium borohydride 3.3.8

Key results

Aldehyde → primary alcohol: $CH_{3} CHO + 2 [H] \to CH_{3} CH_{2} OH$ .
Ketone → secondary alcohol: $CH_{3} COCH_{3} + 2 [H] \to CH_{3} CH(OH)CH_{3}$ .

Notes

Reagent: $NaBH_{4}$ in aqueous solution (or methanol), a source of the hydride ion $H^{-}$ as nucleophile.
Mechanism (nucleophilic addition): $H^{-}$ attacks the $δ +$ carbon and the π electrons move onto oxygen, forming an alkoxide ion; the alkoxide oxygen then takes a proton from water.
$NaBH_{4}$ does not reduce C=C bonds, because an electron-rich alkene is not attacked by a nucleophile.
Worked equation for butanone: $CH_{3} COCH_{2} CH_{3} + 2 [H] \to CH_{3} CH(OH)CH_{2} CH_{3}$ . The product, butan-2-ol, is chiral and forms as a racemic mixture, because hydride attacks the planar carbonyl from either face.
In the mechanism, draw the hydride ion with its lone pair ( $: H^{-}$ ), an arrow from it to the $δ +$ carbon, an arrow from the C=O π bond to oxygen, then an arrow from an oxygen lone pair of the alkoxide to an H of water (or $H^{+}$ ).
Reduction is the reverse of oxidation in this topic: aldehydes give primary alcohols and ketones secondary ones, which is useful in synthesis for moving between functional groups.

Nucleophilic addition: hydroxynitriles 3.3.8

Key results

$CH_{3} CHO + HCN \to CH_{3} CH(OH)CN$ (2-hydroxypropanenitrile); reagent KCN, then dilute acid.

Notes

Step 1: $CN^{-}$ attacks the carbonyl carbon through its carbon lone pair; the π pair moves onto oxygen. Step 2: the alkoxide is protonated by $H^{+}$ (or HCN).
KCN is used rather than HCN gas because HCN is a highly toxic, volatile gas; KCN is also toxic, and the reaction must be done in a fume cupboard.
The product has one more carbon than the carbonyl compound, so this extends the carbon chain.
With aldehydes (other than methanal) and unsymmetrical ketones the product is a racemic mixture, because the planar carbonyl is attacked equally from either face.
Hydroxynitriles are useful intermediates: hydrolysis with acid converts the nitrile into a carboxylic acid, so ethanal → 2-hydroxypropanenitrile → 2-hydroxypropanoic acid (lactic acid). Reduction converts the nitrile into an amine.
Naming counts the nitrile carbon as C1: $CH_{3} CH(OH)CN$ is 2-hydroxypropanenitrile, and the ketone product $(CH_{3})_{2} C(OH)CN$ is 2-hydroxy-2-methylpropanenitrile, which is not chiral.
The reaction is carried out at a slightly acidic pH so that there is enough $CN^{-}$ to attack and enough $H^{+}$ to protonate the alkoxide intermediate.

3.3.9

Carboxylic acids and derivatives

Carboxylic acids and their derivatives 3.3.9.1

Key results

Carboxylic acid –COOH (ethanoic acid); ester –COO– (methyl ethanoate); acyl chloride –COCl (ethanoyl chloride); acid anhydride –COOCO– (ethanoic anhydride); amide –CONH₂ (ethanamide).

Notes

Each derivative replaces the acid's OH with a different group on the same carbonyl carbon.
Esters are named alcohol part first: the alkyl group from the alcohol, then the -oate from the acid (methyl ethanoate = ethanoic acid + methanol).
Carboxylic acids have high boiling points because pairs of molecules form two hydrogen bonds to each other (dimers): ethanoic acid boils at 118 °C, above ethanol (78 °C) of similar $M_{r}$ .
Reactivity of the derivatives towards nucleophiles: acyl chlorides > acid anhydrides > esters ≈ carboxylic acids > amides. The better the leaving group, the faster the reaction.
Naming derivatives from the parent acid: ethanoic acid → ethanoyl chloride, ethanoic anhydride, ethanamide, and esters such as ethyl ethanoate.
Small carboxylic acids are soluble in water because they hydrogen bond with it; solubility falls with chain length, so long-chain fatty acids are insoluble.

Acidity, esterification and hydrolysis 3.3.9.1

Key results

Weak acid: $CH_{3} COOH + H_{2} O ⇌ CH_{3} COO^{-} + H_{3} O^{+}$ , only partially ionised.
Carbonate test: $2 CH_{3} COOH + Na_{2} CO_{3} \to 2 CH_{3} COONa + CO_{2} + H_{2} O$ (effervescence).
Esterification: acid + alcohol ⇌ ester + water, with a concentrated $H_{2} SO_{4}$ catalyst and heat.
Acid hydrolysis (dilute acid, heat) is reversible. Alkaline hydrolysis: $CH_{3} COOCH_{3} + NaOH \to CH_{3} COONa + CH_{3} OH$ goes to completion.

Notes

Alkaline hydrolysis is irreversible because the carboxylate ion cannot react with the alcohol; acidify afterwards to obtain the free acid.
Esters are used as solvents, plasticisers, perfumes and food flavourings.
Carboxylic acids react with metals (giving hydrogen), bases and carbonates like other weak acids: $CH_{3} COOH + NaOH \to CH_{3} COONa + H_{2} O$ . The carbonate test distinguishes them from phenols and alcohols.
Esterification is slow and reversible ( $K_{c} \approx 4$ for ethanol and ethanoic acid). Heating under reflux with the acid catalyst speeds it up; distilling off the volatile ester, or using an excess of one reactant, increases the yield.
Esters have fruity smells and lower boiling points than acids of similar $M_{r}$ because they cannot hydrogen bond to each other; this is why they are used in flavourings and perfumes.
Naming: the alcohol part first, then the acid part with -oate. $HCOOCH_{2} CH_{3}$ is ethyl methanoate; $CH_{3} CH_{2} COOCH_{3}$ is methyl propanoate.

Fats and oils: soap and biodiesel 3.3.9.1

Key results

Saponification: triglyceride + 3NaOH(aq), heated → 3 soap ( $RCOO^{-} Na^{+}$ ) + glycerol.
Biodiesel: triglyceride + 3CH₃OH, with a catalyst → 3 methyl esters ( $RCOOCH_{3}$ ) + glycerol.

Notes

Fats and oils are triesters of glycerol (propane-1,2,3-triol) and long-chain carboxylic acids (fatty acids); the three R groups need not be identical.
Soaps are sodium or potassium salts of long-chain carboxylic acids.
Biodiesel is a mixture of methyl esters of long-chain acids. Transesterification is reversible, so excess methanol pushes the equilibrium towards the esters.
Saturated fats (no C=C in the chains) pack closely and are solid at room temperature; unsaturated oils have cis C=C bonds that kink the chains, so weaker London forces and liquid oils result. Margarine is made by partially hydrogenating oils with hydrogen and a nickel catalyst.
Soaps clean because the ionic $-COO^{-} Na^{+}$ head is attracted to water while the long hydrocarbon tail dissolves in grease, so grease is lifted into micelles.
Biodiesel production uses a strong base catalyst such as KOH; the glycerol by-product is purified and sold, improving the overall economics of the process.
Biodiesel is described as renewable because the oils come from crops, but growing, harvesting and processing them uses energy and land, so it is not fully carbon-neutral.

Acylation with acyl chlorides and anhydrides 3.3.9.2

Key results

Water: $CH_{3} COCl + H_{2} O \to CH_{3} COOH + HCl$ .
Alcohol: $CH_{3} COCl + CH_{3} OH \to CH_{3} COOCH_{3} + HCl$ .
Ammonia: $CH_{3} COCl + 2 NH_{3} \to CH_{3} CONH_{2} + NH_{4} Cl$ .
Primary amine: $CH_{3} COCl + 2 CH_{3} NH_{2} \to CH_{3} CONHCH_{3} + CH_{3} NH_{3}^{+} Cl^{-}$ (N-methylethanamide).

Notes

Acyl chlorides react vigorously at room temperature, giving steamy fumes of HCl; with ammonia and amines the second molecule neutralises the HCl.
Acid anhydrides give the same acyl products but with a carboxylic acid (or its salt) as by-product.
Aspirin is made industrially from 2-hydroxybenzoic acid and ethanoic anhydride rather than ethanoyl chloride: the anhydride is cheaper, less corrosive, reacts less violently with water, and gives no HCl fumes.
Acid anhydrides react more slowly than acyl chlorides and usually need warming, which makes them easier to control on a large scale.
Aspirin forms when ethanoic anhydride acylates the phenolic –OH of 2-hydroxybenzoic acid, giving 2-ethanoyloxybenzoic acid; the by-product ethanoic acid is easily removed.
Acyl chlorides must be stored dry: moist air hydrolyses them to the carboxylic acid with steamy fumes of HCl, an observation used to recognise them.
In each reaction a hydrogen on O or N is replaced by an acyl group, $RCO-$ , which is why the process is called acylation.

Nucleophilic addition–elimination 3.3.9.2

Notes

Addition: a lone pair on the nucleophile (water, alcohol, ammonia or amine) attacks the $δ +$ carbonyl carbon; the C=O π pair moves onto oxygen, giving a tetrahedral intermediate.
Elimination: the oxygen lone pair reforms C=O and the C–Cl pair leaves as $Cl^{-}$ .
Proton transfer: a base removes $H^{+}$ from the positively charged O or N, giving the acid, ester or amide.
The carbonyl carbon is especially electrophilic in acyl chlorides because both O and Cl withdraw electron density, and $Cl^{-}$ is a good leaving group.
Mechanism checklist for full marks: the $δ +$ / $δ -$ on C=O (and C–Cl), a lone pair on the nucleophile with an arrow to the carbonyl carbon, an arrow from C=O to O, the tetrahedral intermediate with O⁻ and the positive nucleophile atom, an arrow from O⁻ back to reform C=O, an arrow from C–Cl to Cl, and finally loss of $H^{+}$ (removed by $Cl^{-}$ or a base).
The overall reaction is substitution (Cl replaced by OR, OH or NHR), but it happens in two steps, addition then elimination, which is why it is named this way.
Compare nucleophilic addition to aldehydes: there, no leaving group is attached to the carbonyl carbon, so the reaction stops after addition. Acyl chlorides have $Cl^{-}$ to lose, so elimination follows.

Addition–elimination with nitrogen nucleophiles 3.3.9.2

Notes

The mechanism is the same with ammonia or a primary amine: the nitrogen lone pair attacks the carbonyl carbon, C=O reforms as $Cl^{-}$ leaves, then a second molecule of ammonia or amine removes $H^{+}$ from nitrogen.
Ammonia gives a primary amide; a primary amine gives an N-substituted amide.
Because one equivalent of base is used to remove the proton, two moles of ammonia or amine are needed per mole of acyl chloride.
Worked example: ethanoyl chloride with ethylamine gives N-ethylethanamide, $CH_{3} CONHCH_{2} CH_{3}$ , and ethylammonium chloride. The N- in the name shows that the alkyl group is on nitrogen.
Nitrogen nucleophiles react faster than oxygen ones of comparable structure because nitrogen is less electronegative and its lone pair is more available.
This is the reaction used to make polyamides such as Kevlar from diacyl chlorides and diamines (3.3.12), and in biology the same type of bond (the peptide bond) is formed, though by a different mechanism.
The second molecule of amine is consumed as a salt, so the yield based on amine is at most 50% unless another base is added to remove the HCl.

3.3.10

Aromatic chemistry

Bonding in benzene 3.3.10.1

Notes

Benzene, $C_{6} H_{6}$ , is a planar hexagon with bond angles of 120°.
Each carbon forms three σ bonds and contributes one p electron; the p orbitals overlap sideways to form a ring of six delocalised π electrons above and below the plane.
All six C–C bonds are the same length, about 139 pm: between C–C (154 pm) and C=C (134 pm). There are no alternating single and double bonds, which is why it is drawn with a circle.
Kekulé (1865) proposed a ring of alternating single and double bonds. Three pieces of evidence rule it out: all C–C bonds are the same length (X-ray diffraction), benzene is much more stable than expected (hydrogenation enthalpy), and benzene does not decolourise bromine water as an alkene would.
Naming arenes: methylbenzene, chlorobenzene, nitrobenzene, phenylamine, benzoic acid, phenol; positions on the ring are numbered to give the lowest locants, e.g. 1,3-dinitrobenzene.
The $C_{6} H_{5}$ – group is called phenyl; compounds with a benzene ring are described as aromatic, those without as aliphatic.
Skeletal formulas of arenes should show the hexagon with a circle; if a Kekulé structure is drawn it must not be used to argue that benzene has separate double bonds.

Thermochemical evidence for delocalisation 3.3.10.1

Key results

Cyclohexene + $H_{2}$ → cyclohexane: $Δ H \approx - 118 kJ mol^{- 1}$ (one C=C).
Predicted for a localised triene: $3 \times (- 118) = - 354 kJ mol^{- 1}$ . Observed for benzene: $- 206 kJ mol^{- 1}$ .

Notes

Benzene releases about 148 kJ mol⁻¹ less energy than predicted, so it is about 148 kJ mol⁻¹ lower in enthalpy (more stable) than the hypothetical Kekulé structure.
This delocalisation stability explains why benzene resists addition reactions and instead undergoes substitution, which keeps the delocalised ring intact.
The 152 kJ mol⁻¹ figure sometimes quoted comes from slightly different data; what matters is that benzene's hydrogenation is about 150 kJ mol⁻¹ less exothermic than three isolated C=C bonds, a measure of its delocalisation (resonance) energy.
Draw the enthalpy diagram with the hypothetical cyclohexa-1,3,5-triene above benzene, both arrows ending at cyclohexane, and label the gap as the extra stability of benzene.
Cyclohexa-1,3-diene (−232 kJ mol⁻¹, close to 2 × −118) shows that two C=C bonds behave almost independently; only the complete ring of six p orbitals gives the large extra stability.

Nitration of benzene 3.3.10.2

Key results

Electrophile: $HNO_{3} + 2 H_{2} SO_{4} \to NO_{2}^{+} + H_{3} O^{+} + 2 HSO_{4}^{-}$ (often simplified to $HNO_{3} + H_{2} SO_{4} \to NO_{2}^{+} + HSO_{4}^{-} + H_{2} O$ ).
Overall: $C_{6} H_{6} + HNO_{3} \to C_{6} H_{5} NO_{2} + H_{2} O$ ; concentrated acids, about 50 °C.

Notes

Mechanism (electrophilic substitution): the delocalised electrons attack $NO_{2}^{+}$ , forming a positively charged intermediate in which delocalisation is partly lost; the C–H pair then returns to the ring, restoring the delocalised system, and $HSO_{4}^{-}$ takes the proton, regenerating $H_{2} SO_{4}$ .
Above about 55 °C further substitution gives dinitrobenzene.
Nitration is important in making explosives (TNT) and is the first step to aromatic amines used in dyes.
Mechanism checklist: an arrow from the ring's circle to the N of $NO_{2}^{+}$ ; the intermediate drawn as a horseshoe of delocalised electrons open towards the tetrahedral carbon, with + inside the horseshoe and the H and $NO_{2}$ on that carbon; an arrow from the C–H bond back into the ring.
Nitro compounds are used in explosives (TNT is 2-methyl-1,3,5-trinitrobenzene) and, after reduction to aromatic amines, in the manufacture of dyes and pharmaceuticals.
Sulfuric acid acts as a catalyst: it protonates nitric acid, which then loses water to form the nitronium ion, and is regenerated when $HSO_{4}^{-}$ removes $H^{+}$ from the intermediate.
The reaction is substitution rather than addition because restoring the delocalised ring releases energy; addition would leave a less stable, non-aromatic product.

Friedel–Crafts acylation 3.3.10.2

Key results

Electrophile: $CH_{3} COCl + AlCl_{3} \to CH_{3} CO^{+} + AlCl_{4}^{-}$ (acylium ion).
Overall: $C_{6} H_{6} + CH_{3} COCl \to C_{6} H_{5} COCH_{3} + HCl$ (phenylethanone).

Notes

$AlCl_{3}$ is a halogen carrier: it accepts a lone pair from chlorine, generating the acylium ion. Conditions: anhydrous $AlCl_{3}$ , warm, then aqueous work-up.
After attack and loss of $H^{+}$ , $AlCl_{4}^{-} + H^{+} \to AlCl_{3} + HCl$ regenerates the catalyst.
Acylation is a useful way to form a C–C bond to an aromatic ring and give an aromatic ketone.
Conditions must be dry because aluminium chloride reacts vigorously with water, which would destroy the catalyst and the acyl chloride.
Mechanism checklist: show the acylium ion $CH_{3} C^{+} = O$ , an arrow from the ring to the positive carbon, the horseshoe intermediate, and the C–H bond breaking back into the ring; then $H^{+} + AlCl_{4}^{-} \to AlCl_{3} + HCl$ .
Acylation is valuable because it forms a C–C bond to the ring; the ketone produced can then be reduced to an alcohol with $NaBH_{4}$ , opening routes to many aromatic compounds.
Other acyl chlorides give other aromatic ketones: propanoyl chloride gives 1-phenylpropan-1-one.

3.3.11

Amines

Primary, secondary and tertiary amines 3.3.11

Key results

Primary: one carbon group on N ( $RNH_{2}$ ). Secondary: two ( $R_{2} NH$ ). Tertiary: three ( $R_{3} N$ ). Quaternary ammonium ion: four ( $R_{4} N^{+}$ ).

Notes

The classification counts carbon groups on nitrogen, unlike alcohols where it counts carbons on the C–OH carbon.
The nitrogen lone pair makes amines both bases (they accept $H^{+}$ ) and nucleophiles (they donate the pair to an electron-deficient carbon).
Naming: methylamine $CH_{3} NH_{2}$ , dimethylamine $(CH_{3})_{2} NH$ , trimethylamine $(CH_{3})_{3} N$ ; phenylamine $C_{6} H_{5} NH_{2}$ . In more complex molecules the group is called amino, e.g. 2-aminopropanoic acid.
Small amines smell fishy and are soluble in water, because N–H groups and the N lone pair hydrogen bond with water.
Primary amines boil lower than alcohols of similar size (methylamine −6 °C, methanol 65 °C), because N–H···N hydrogen bonds are weaker than O–H···O ones; nitrogen is less electronegative than oxygen.
Tertiary amines have no N–H bond, so they cannot hydrogen bond to each other, although they can still accept hydrogen bonds from water.

Base strength 3.3.11.2

Key results

$RNH_{2} + H_{2} O ⇌ RNH_{3}^{+} + OH^{-}$ .
Base strength: primary aliphatic amines > ammonia > aromatic amines (e.g. methylamine > ammonia > phenylamine).

Notes

Alkyl groups are electron-releasing, increasing electron density on nitrogen, so the lone pair is more available to accept a proton.
In phenylamine the lone pair is partly delocalised into the benzene ring, so it is less available to accept a proton.
Amines react with acids to form salts: $CH_{3} NH_{2} + HCl \to CH_{3} NH_{3}^{+} Cl^{-}$ (methylammonium chloride). The salts are ionic, soluble and odourless; adding NaOH regenerates the free amine.
Base strength depends on how available the nitrogen lone pair is to accept a proton: anything that pushes electron density onto N strengthens the base, anything that pulls it away (such as delocalisation into a benzene ring) weakens it.
Explanations should name the effect explicitly: 'the alkyl group is electron-releasing (positive inductive effect)' and 'the lone pair is delocalised into the ring', then link to lone-pair availability.
Many drugs are amines and are sold as their hydrochloride salts, which are more water-soluble and stable than the free bases.

Preparation of amines 3.3.11.1

Key results

From a halogenoalkane: $CH_{3} CH_{2} Br + 2 NH_{3} \to CH_{3} CH_{2} NH_{2} + NH_{4} Br$ (excess ethanolic ammonia, heat under pressure).
Nitrile reduction: $CH_{3} CN + 4 [H] \to CH_{3} CH_{2} NH_{2}$ ( $LiAlH_{4}$ in dry ether, or $H_{2}$ with a nickel catalyst).
Aromatic: nitrobenzene + 6[H] → phenylamine + 2H₂O (tin and concentrated HCl, reflux, then NaOH).

Notes

The halogenoalkane route gives a mixture because the product amine can react further; the nitrile route gives only the primary amine.
In the tin/HCl reduction the amine forms as its ammonium salt; NaOH releases the free amine. Aromatic amines are used to make dyes.
The nitrile route is used when a pure primary amine is needed, because the nitrile can be reduced only to the primary amine. It also adds a carbon to the chain: bromoethane → propanenitrile → propylamine.
Catalytic hydrogenation of nitriles (hydrogen with a nickel catalyst) is preferred industrially to $LiAlH_{4}$ , which is expensive and reacts violently with water.
The reduction of nitrobenzene is a key step in making dyes: phenylamine is converted into diazonium salts and then coupled to give brightly coloured azo dyes (beyond the specification, but explains its importance).
In the tin and acid reduction the amine leaves the reaction as phenylammonium ion, $C_{6} H_{5} NH_{3}^{+}$ ; NaOH removes the proton and the free amine is separated by steam distillation.

Nucleophilic substitution with halogenoalkanes 3.3.11.3

Notes

Step 1: the nitrogen lone pair attacks the $δ +$ carbon and bromide leaves, giving an alkylammonium ion. Step 2: a second ammonia or amine molecule removes $H^{+}$ .
Each product still has a lone pair on nitrogen, so substitution continues: ammonia → primary → secondary → tertiary amine → quaternary ammonium salt.
The quaternary ion forms directly in one step because the tertiary amine's nitrogen has no H to lose. Excess ammonia favours the primary amine; excess halogenoalkane favours the quaternary salt.
Mechanism for the first step uses the nitrogen lone pair and the $δ +$ carbon exactly as for ammonia, then a second amine removes a proton; with each substitution the product amine is a slightly stronger nucleophile, which is why mixtures form so readily.
Worked sequence: ammonia with excess bromomethane gives $CH_{3} NH_{2}$ , then $(CH_{3})_{2} NH$ , $(CH_{3})_{3} N$ and finally tetramethylammonium bromide, $(CH_{3})_{4} N^{+} Br^{-}$ .
Products can be separated by fractional distillation, but the poor selectivity is why other routes (nitrile reduction, acylation then reduction) are preferred for single amines.

Acylation: forming amides 3.3.11.3

Key results

$CH_{3} COCl + 2 CH_{3} NH_{2} \to CH_{3} CONHCH_{3} + CH_{3} NH_{3}^{+} Cl^{-}$ .
$(CH_{3} CO)_{2} O + 2 CH_{3} NH_{2} \to CH_{3} CONHCH_{3} + CH_{3} NH_{3}^{+} CH_{3} COO^{-}$ .

Notes

Ammonia and primary amines react with acyl chlorides and anhydrides by nucleophilic addition–elimination (see 3.3.9) to give amides.
The second molecule of amine acts as a base and neutralises the acidic by-product.
Amides are much weaker bases than amines, because the nitrogen lone pair is delocalised onto the adjacent C=O group; they do not react with dilute acids to form salts.
Amides can be hydrolysed back to a carboxylic acid and an amine by heating with acid (giving the ammonium salt) or alkali (giving the carboxylate salt), the same chemistry used to break polyamides and proteins.
Paracetamol is made by acylating 4-aminophenol with ethanoic anhydride: the more nucleophilic N–H, not the O–H, reacts.
Use the N- prefix in names for groups on nitrogen: $CH_{3} CONHCH_{3}$ is N-methylethanamide, an isomer of propanamide.

Quaternary ammonium salts as surfactants 3.3.11.3

Notes

Quaternary ammonium salts with at least one long alkyl chain, such as dodecyltrimethylammonium bromide, are cationic surfactants.
The long hydrocarbon tail is hydrophobic and the $N^{+}$ head is hydrophilic; the head is attracted to negatively charged surfaces such as hair and fabric, so they are used in fabric softeners and hair conditioners.
The quaternary nitrogen has four C–N bonds and no lone pair, so it is permanently charged and is neither a base nor a nucleophile.
Quaternary ammonium salts are made by reacting a tertiary amine with excess halogenoalkane, the final stage of the substitution sequence.
They are cationic surfactants, unlike soaps (anionic, $RCOO^{-}$ ). The positive heads adsorb onto negatively charged surfaces such as wet fabric and hair, leaving a soft hydrophobic layer that reduces static.
Some quaternary ammonium salts are also disinfectants, disrupting bacterial cell membranes; they are used in antiseptic sprays and mouthwashes.
Their permanent charge means they are soluble in water in their salt form at any pH, unlike amines, whose charge depends on pH.

3.3.12

Polymers

Condensation polymerisation 3.3.12.1

Definitions

Condensation polymerisation: monomers with two reactive functional groups join, eliminating a small molecule (usually water or HCl) at each new link.

Notes

Polyesters contain ester links, –COO–; polyamides contain amide links, –CONH–.
Contrast with addition polymers, where C=C opens and all monomer atoms are retained.
Each monomer must have two functional groups (or two of the same group on both ends of each of two monomers) so the chain can keep growing in both directions.
To find the monomers from a polymer, break the ester or amide linkage and add back the atoms of water: the C=O side gets –OH (or –Cl for an acyl chloride monomer) and the O or N side gets –H.
Typical pairs: diol + dicarboxylic acid (or diacyl chloride) → polyester; diamine + dicarboxylic acid (or diacyl chloride) → polyamide; a hydroxycarboxylic acid or amino acid can polymerise on its own.
Using a diacyl chloride rather than a dicarboxylic acid makes the reaction faster and complete at room temperature, at the cost of producing HCl.

Polyester: Terylene (PET) 3.3.12.1

Key results

Ethane-1,2-diol + benzene-1,4-dicarboxylic acid → poly(ethylene terephthalate) + water.
Repeating unit: $- OCH_{2} CH_{2} O - CO - C_{6} H_{4} - CO -$ .

Notes

Used for clothing fibres and plastic drinks bottles.
To draw a repeating unit, join one of each monomer by the new link and remove the atoms lost in the small molecule (H from one end, OH from the other), leaving open bonds through the brackets.
PET fibres are strong and resist stretching and creasing, which makes them ideal for clothing (often blended with cotton) and for lightweight, shatter-resistant bottles.
Draw the repeating unit with the ester linkage in the main chain: $- O-CH_{2} -CH_{2} -O-C(=O)-C_{6} H_{4} -C(=O)-$ , with continuation bonds through both brackets and $n$ outside.
PET can be recycled mechanically (melted and remoulded, e.g. into fleece fabric) or chemically, by hydrolysis or methanolysis of the ester links back to monomers that can be repolymerised.
The planar benzene rings let chains pack closely, contributing to the polymer's strength and its relatively high melting point (about 250 °C).

Polyamide: nylon 6,6 3.3.12.1

Key results

Hexane-1,6-diamine + hexanedioic acid → nylon 6,6 + water.
Repeating unit: $- NH(CH_{2})_{6} NH - CO(CH_{2})_{4} CO -$ .

Notes

"6,6" records six carbons in each monomer.
A diacyl chloride can replace the dicarboxylic acid; then HCl is eliminated instead of water.
Used for clothing, ropes and carpets.
Hydrogen bonds between N–H and C=O groups on adjacent chains make nylon strong and give it a high melting point (about 265 °C), which is why it can be spun into fibres from the melt.
Nylon was originally developed as a synthetic replacement for silk, itself a natural polyamide (protein).
Worked repeating unit check: hexane-1,6-diamine contributes $- NH(CH_{2})_{6} NH -$ and hexanedioic acid $- CO(CH_{2})_{4} CO -$ ; together they make one repeating unit, with one water lost at each link (two per repeating unit when counted across the chain).
Polyamides absorb a little water by hydrogen bonding, which slightly changes their dimensions and strength.

Polyamide: Kevlar 3.3.12.1

Key results

Benzene-1,4-diamine + benzene-1,4-dicarbonyl dichloride → Kevlar + HCl (or benzene-1,4-dicarboxylic acid with water eliminated).

Notes

The rigid, linear aromatic chains pack closely and form many hydrogen bonds between N–H and C=O groups on neighbouring chains.
This gives very high tensile strength for its mass: used in bulletproof vests, helmets and protective clothing.
Weight for weight, Kevlar is about five times stronger than steel. Its rigid, rod-like chains line up in parallel in fibres, held by a regular network of hydrogen bonds, with further attraction between stacked benzene rings.
Kevlar is made from benzene-1,4-diamine and benzene-1,4-dicarbonyl dichloride; using the acyl chloride gives HCl as the small molecule and allows reaction at low temperature.
The 1,4-substitution on each ring keeps the chain straight; using 1,3-substituted monomers gives a kinked chain (as in Nomex), which is less strong but more heat-resistant.
Uses: body armour, helmets, cut-resistant gloves, ropes and composite materials for sports equipment and aircraft.

Amino acids form polyamides 3.3.12.1

Notes

Each amino acid has both an $- NH_{2}$ and a $- COOH$ group, so a single monomer can polymerise by condensation.
The amide link formed between amino acids is called a peptide bond; the polymers are polypeptides and proteins.
Proteins are polyamides built from about 20 different amino acids, each differing only in its side chain R. The sequence of side chains, not the backbone, gives each protein its properties.
Synthetic polyamides such as nylon are made from two monomers in alternation, whereas a protein's sequence is irregular and specified genetically.
Like other polyamides, proteins are hydrolysed back to their monomers by heating with concentrated acid; this is the first step in analysing a protein's composition (3.3.13).
The peptide bond is planar because the nitrogen lone pair is delocalised into the C=O group, which restricts rotation and helps proteins fold into regular shapes.

Forces between polymer chains 3.3.12.1

Notes

Polyalkenes: London forces only.
Polyesters: London forces plus permanent dipole–dipole forces between polar C=O and C–O groups.
Polyamides: also hydrogen bonds between N–H on one chain and C=O on another. These stronger interchain forces make polyamides strong fibres with relatively high melting points.
The strength of interchain forces explains properties directly: polyalkenes are soft and melt at relatively low temperatures; polyesters are stronger; polyamides such as nylon and Kevlar are the strongest fibres.
Chain arrangement matters as much as force type: long, unbranched, regular chains pack closely in crystalline regions, maximising contact, while branching or bulky side groups keep chains apart and weaken attraction.
Heating a thermoplastic gives the chains energy to overcome these intermolecular forces and slide past each other, so it softens and can be remoulded; no covalent bonds in the chains are broken.
In exam answers, name the force, say between which groups on which chains it acts, and link its strength to the property asked about.

Hydrolysis and biodegradability 3.3.12.2

Key results

Polyester + $OH^{-}$ → carboxylate + diol. Polyamide + $OH^{-}$ → carboxylate + amine.
Acid hydrolysis of a polyamide gives the dicarboxylic acid and the ammonium salt of the amine.

Notes

The polar C=O carbons in ester and amide links are attacked by nucleophiles, so condensation polymers can be hydrolysed and are potentially biodegradable.
Polyalkenes have only non-polar C–C and C–H bonds in the backbone, so they are chemically inert and non-biodegradable.
Hydrolysable does not mean fast: PET and nylon can persist for years in the environment.
Worked example: alkaline hydrolysis of nylon 6,6 with NaOH(aq) gives hexane-1,6-diamine and sodium hexanedioate; acid hydrolysis gives hexanedioic acid and the hexane-1,6-diammonium salt.
Biodegradable polymers designed for packaging, such as poly(lactic acid) made from renewable starch, contain ester links that are hydrolysed by water and microbial enzymes, though often only in industrial composting conditions.
Polyalkenes may break into smaller pieces through photo-oxidation, but this produces microplastics rather than complete breakdown into carbon dioxide and water.

Reuse, recycling and disposal 3.3.12.2

Key results

Landfill: accepts mixed waste, but polymers persist and land is used up.
Incineration: energy can be recovered, but it releases $CO_{2}$ and needs pollution controls (e.g. HCl from PVC).
Recycling: conserves crude oil and reduces landfill, but sorting, cleaning and transport cost energy, and quality can fall. Some condensation polymers can be chemically recycled by hydrolysis to monomers.

Notes

Reuse extends a product's life and is often the lowest-impact option where practical.
A full comparison considers raw materials, energy, emissions and waste at every stage, a life-cycle assessment. Recycling usually wins for single, clean polymer types; incineration with energy recovery may be preferable for mixed, contaminated waste.
Mechanical recycling degrades polymers slightly with each cycle (chains shorten), so recycled material is often used in lower-grade products such as fleece or garden furniture.
Chemical recycling of condensation polymers by hydrolysis recovers monomers of the original purity, a key advantage over polyalkenes, which can only be cracked to a mixture of hydrocarbons.
Landfill wastes the energy and carbon stored in the polymer and takes up land for centuries; it is the least sustainable option.

3.3.13

Amino acids, proteins and DNA

Amino acids: charge depends on pH 3.3.13.1

Definitions

Zwitterion: an ion with both a positive and a negative charge and no overall charge, $H_{3} N^{+} CHRCOO^{-}$ .
Isoelectric point: the pH at which the amino acid exists mainly as the zwitterion.

Key results

Acidic solution: $H_{3} N^{+} CHRCOOH$ (charge +1). Alkaline solution: $H_{2} NCHRCOO^{-}$ (charge −1).

Notes

Amino acids are amphoteric: the carboxyl group donates $H^{+}$ and the amino group accepts it.
Amino acids have high melting points and dissolve in water because of strong ionic attractions between zwitterions.
All α-amino acids except glycine (R = H) have a chiral α-carbon.
Worked example: alanine (2-aminopropanoic acid). In acid it is $CH_{3} CH(NH_{3}^{+}) COOH$ ; at its isoelectric point $CH_{3} CH(NH_{3}^{+}) COO^{-}$ ; in alkali $CH_{3} CH(NH_{2}) COO^{-}$ .
Amino acids with an extra acidic or basic group in the side chain (e.g. aspartic acid, lysine) have more possible ionic forms; the extra group is protonated or deprotonated depending on pH in the same way.
Because the zwitterion is the main form in the solid, amino acids are crystalline solids that decompose rather than melt cleanly, and they are much less soluble in non-polar solvents than their $M_{r}$ suggests.
Naturally occurring α-amino acids are almost all a single enantiomer, a direct consequence of being made by stereospecific enzymes.

Peptide bonds 3.3.13.2

Key results

Glycine + alanine → Gly–Ala + $H_{2} O$ ; the new –CONH– link is a peptide bond.

Notes

Two different amino acids can form two different dipeptides (Gly–Ala and Ala–Gly), depending on which amino group reacts with which carboxyl group.
Hydrolysis by heating with concentrated aqueous HCl breaks the peptide bonds; the amino acids form as their protonated (cationic) forms in the acid.
Worked example: glycine with alanine can give two dipeptides, Gly–Ala ( $H_{2} NCH_{2} CONHCH(CH_{3}) COOH$ ) and Ala–Gly ( $H_{2} NCH(CH_{3}) CONHCH_{2} COOH$ ). Three different amino acids can make six tripeptides, each used once.
Laboratory hydrolysis of a protein uses 6 mol dm⁻³ hydrochloric acid heated under reflux for many hours; in the body, enzymes (proteases) hydrolyse peptide bonds rapidly at 37 °C.
After acid hydrolysis, the amino acids are present as their protonated forms, $H_{3} N^{+} CHRCOOH$ ; neutralising the solution is needed before chromatography.
A chain of many amino acids is a polypeptide; one or more polypeptides folded into a functional shape is a protein.

Protein structure 3.3.13.2

Key results

Primary: the sequence of amino acids, held by peptide bonds.
Secondary: α-helix or β-pleated sheet, held by hydrogen bonds between backbone C=O and N–H groups.
Tertiary: the overall 3D fold, held by side-chain interactions including hydrogen bonds, ionic attractions and disulfide (S–S) bridges between cysteine residues.

Notes

Heat or extreme pH break hydrogen bonds and ionic interactions, changing the 3D shape; covalent peptide and disulfide bonds survive mild conditions.
In an α-helix each C=O hydrogen bonds to the N–H four residues further along the same chain; in a β-pleated sheet hydrogen bonds link N–H and C=O groups of neighbouring strands lying side by side.
Tertiary structure is held by hydrogen bonds between polar side chains, ionic attractions between $-NH_{3}^{+}$ and $-COO^{-}$ side groups, disulfide bridges (covalent S–S between cysteine residues) and hydrophobic interactions between non-polar side chains clustered away from water.
Changing pH alters the charges on side chains, breaking ionic interactions; heating breaks hydrogen bonds. Either can denature a protein by changing its 3D shape, even though the primary sequence is unchanged.
Disulfide bridges make some proteins very resilient: keratin in hair is rich in cysteine, and perming works by breaking and re-forming these bridges.

Identifying amino acids by thin-layer chromatography 3.3.13.2

Key results

Example: spot at 20.8 mm, solvent front at 32.0 mm gives $R_{f} = 0.65$ .

Method

Hydrolyse the protein to free amino acids.
Spot the hydrolysate and reference amino acids on a TLC plate and develop in a solvent.
Amino acids are colourless: spray with ninhydrin (or use UV light) to locate the spots.
Measure $R_{f} = \frac{distance moved by spot}{distance moved by solvent front}$ and compare with standards run under the same conditions.

Notes

Ninhydrin reacts with amino groups to give purple (or yellow, for proline) spots once the plate is warmed; wear gloves, because it also stains skin.
Run known amino acids alongside the hydrolysed sample on the same plate, so that the solvent, temperature and plate are identical; compare $R_{f}$ values and spot positions directly.
If two amino acids have similar $R_{f}$ values in one solvent, two-dimensional chromatography (running the plate a second time at 90° in a different solvent) can separate them.
$R_{f}$ values are always between 0 and 1 and have no units; quote them to two decimal places from careful measurements to the centre of each spot.

Enzymes 3.3.13.3

Notes

Enzymes are protein catalysts. Substrates bind in an active site with a complementary shape and complementary interactions.
The active site is stereospecific: it often binds only one enantiomer of a chiral substrate.
Inhibitors are drugs that bind to the active site and block the substrate. Computer modelling of the active site is used to design such drugs.
Enzymes lower activation energy by binding the substrate in an orientation and environment that stabilises the transition state; like all catalysts they do not change $Δ H$ or the position of equilibrium.
Because the active site is chiral, an enzyme usually acts on only one enantiomer of a substrate. This stereospecificity is why many drugs are now made as single enantiomers.
Drug design: the 3D structure of the active site (from X-ray crystallography) is used in computer modelling to design molecules that fit it, often mimicking the substrate (competitive inhibitors); candidate molecules are then synthesised and tested.
A competitive inhibitor's effect can be overcome by raising substrate concentration; a non-competitive inhibitor, binding elsewhere, cannot be outcompeted in this way.

DNA: a polymer of nucleotides 3.3.13.4

Key results

Nucleotide: a phosphate group, 2-deoxyribose and one base (adenine, thymine, guanine or cytosine).
Base pairing: A–T (two hydrogen bonds) and G–C (three hydrogen bonds).

Notes

The sugar–phosphate backbone is joined by covalent phosphodiester links between the 3′ carbon of one sugar and the 5′ carbon of the next.
Two antiparallel strands are held together by hydrogen bonds between complementary bases and twist into a double helix.
Pairing is specific because each base pair's hydrogen-bond donors and acceptors line up only with its partner.
The structures of the nucleotide components are given in the data booklet; questions ask you to link them (e.g. show the condensation that forms the phosphodiester bond) rather than recall them.
Each base pair has one larger purine (A or G) and one smaller pyrimidine (T or C), so the double helix has a constant width. G–C pairs, with three hydrogen bonds, hold the strands together more strongly than A–T pairs.
Hydrogen bonds are individually weak, so the strands can be separated during replication and transcription, while the covalent sugar–phosphate backbone keeps the sequence intact.
The negative charges on the phosphate groups make DNA a polyanion, which helps it dissolve in water and bind positively charged proteins and metal complexes such as cisplatin's aqua forms.

Cisplatin: an anticancer drug 3.3.13.5

Notes

Cisplatin is $c i s$ - $[PtCl_{2} (NH_{3})_{2}]$ , a square-planar Pt(II) complex with the two chloride ligands adjacent.
In cells a chloride ligand is replaced by water; Pt then bonds to a nitrogen atom on guanine. Both chlorides are replaced, forming a cross-link between adjacent guanines on one DNA strand.
The distorted DNA cannot be copied, so cell division (replication) is prevented in rapidly dividing cancer cells.
It also damages healthy cells (e.g. hair loss, kidney damage), so doses are limited: a balance between benefit and adverse effects.
Ligand replacement happens mainly inside cells because the chloride ion concentration there is much lower than in blood plasma; outside cells, high $[Cl^{-}]$ keeps the drug in its unreactive dichloro form.
Transplatin, the trans isomer, is ineffective: its two leaving groups point in opposite directions, so it cannot bind two neighbouring guanines on the same strand.
Side effects arise because cisplatin also binds DNA in healthy cells, especially rapidly dividing ones such as hair follicles and the gut lining; it is also toxic to the kidneys. Doses are kept as low as effective and given in short courses.
This is a direct application of three topics: square planar complexes and cis–trans isomerism (3.2.5), ligand substitution, and DNA structure.

3.3.14

Organic synthesis

Choosing reactions and conditions 3.3.14

Key results

Ethene → ethanol: steam, $H_{3} PO_{4}$ catalyst, 300 °C, high pressure. Ethanol → ethene: concentrated $H_{2} SO_{4}$ , about 170 °C.
Bromoethane + NaOH(aq), reflux → ethanol (substitution); + KOH in ethanol, heat → ethene (elimination); + excess ethanolic $NH_{3}$ , sealed tube → ethylamine.

Notes

The same reagent can give different products under different conditions: always state solvent and temperature as well as reagent.
Summary of halogenoalkane routes: aqueous NaOH, reflux → alcohol; ethanolic KOH, heat → alkene; excess ethanolic $NH_{3}$ , heat in sealed tube → primary amine; ethanolic KCN, reflux → nitrile.
Summary of alkene routes: HBr → bromoalkane; $Br_{2}$ → dibromoalkane; steam with $H_{3} PO_{4}$ → alcohol; concentrated $H_{2} SO_{4}$ then water → alcohol; polymerisation → polyalkene.
Learn each reaction as a triple: reagent, conditions and type of reaction. Exam synthesis questions award marks separately for each, and 'heat' without 'under reflux' or the solvent is often not enough.
Whenever two reactions compete (substitution versus elimination), state the conditions that favour the one you want.

Controlling oxidation and reduction 3.3.14

Key results

Ethanol → ethanal: $K_{2} Cr_{2} O_{7} / H_{2} SO_{4}$ , limited oxidant, warm and distil.
Ethanol → ethanoic acid: excess oxidant, heat under reflux.
Propan-2-ol → propanone: acidified dichromate, reflux. Reverse reductions: $NaBH_{4}$ (aq).

Notes

$NaBH_{4}$ reduces aldehydes and ketones back to alcohols but does not reduce carboxylic acids.
To identify which product formed, use the tests from 3.3.5: the aldehyde gives a silver mirror; the acid fizzes with sodium hydrogencarbonate; the ketone gives neither.
Reduction of carboxylic acids to alcohols needs a stronger reducing agent such as $LiAlH_{4}$ , not $NaBH_{4}$ ; this is beyond the specification but explains why $NaBH_{4}$ is described as selective.
Because oxidation and reduction interconvert alcohols and carbonyl compounds, a synthesis can use them to move a functional group's oxidation level up or down as needed.

Making esters and amides 3.3.14

Key results

Acid + alcohol, concentrated $H_{2} SO_{4}$ , reflux ⇌ ester + water.
Acyl chloride + alcohol, room temperature → ester + HCl (fast and not reversible).
Acyl chloride + excess $NH_{3}$ → amide + $NH_{4} Cl$ .

Notes

Using an acyl chloride gives a higher yield of ester than direct esterification because the reaction is not an equilibrium.
Esters can also be made from acid anhydrides and alcohols, giving the carboxylic acid as by-product; this is preferred industrially for the same reasons as for aspirin.
Amides from primary amines and acyl chlorides are N-substituted amides; the amine must be in excess (or another base added) to neutralise HCl.
Ester hydrolysis in alkali, followed by acidification, is a reliable way to obtain a pure carboxylic acid and alcohol from an ester.

Extending the carbon chain 3.3.14

Key results

Halogenoalkane + KCN in ethanol, reflux → nitrile; e.g. bromoethane → propanenitrile.
Nitrile → amine with $LiAlH_{4}$ in dry ether (propanenitrile → propylamine); or → carboxylic acid by heating with dilute acid.
Aldehyde or ketone + KCN, then dilute acid → hydroxynitrile.

Notes

Both cyanide reactions form a new C–C bond, adding one carbon to the chain.
Friedel–Crafts acylation also forms a new C–C bond, attaching a carbon chain to a benzene ring; it is the aromatic counterpart of the cyanide reactions.
Plan by counting carbons: if the target has one more carbon than the starting material, look for a step that introduces $CN^{-}$ ; if it has the same number, no C–C bond-forming step is needed.
Worked route: ethanol → ethene (concentrated acid, heat) → bromoethane (HBr) → propanenitrile (KCN) → propanoic acid (acid hydrolysis), adding one carbon overall.
Cyanide is highly toxic, so in a risk assessment state the hazard, use small quantities, work in a fume cupboard, and avoid acidifying cyanide solutions (which releases HCN gas).

Aromatic routes 3.3.14

Key results

Benzene → nitrobenzene: concentrated $HNO_{3}$ / $H_{2} SO_{4}$ , 50–55 °C. Nitrobenzene → phenylamine: Sn/concentrated HCl, reflux, then NaOH(aq).
Benzene → phenylethanone: $CH_{3} COCl$ , anhydrous $AlCl_{3}$ , warm.

Notes

Each route uses substitution so the delocalised ring is retained.
Aromatic routes rely on electrophilic substitution to introduce a group, then on standard reactions of that group: a nitro group is reduced to an amine; a ketone is reduced to a secondary alcohol; an amine is acylated to an amide.
Worked route to paracetamol-type amides: benzene → nitrobenzene → phenylamine → N-phenylethanamide (with ethanoyl chloride or ethanoic anhydride).
Conditions for nitration must be controlled (below about 55 °C) to avoid multiple substitution, which would give dinitro products.
Arenes are unreactive towards nucleophiles and do not undergo the substitution reactions of halogenoalkanes, so the order of steps matters when planning routes.

A worked multi-step synthesis 3.3.14

Method

Target ethyl propanoate: the ester link splits it into propanoic acid (3 C) and ethanol (2 C).
Ethene + HBr → bromoethane (2 C).
Bromoethane + KCN in ethanol, reflux → propanenitrile (3 C).
Propanenitrile + dilute HCl, reflux → propanoic acid.
Propanoic acid + ethanol, concentrated $H_{2} SO_{4}$ , reflux → ethyl propanoate.

Notes

Plan backwards from the target (identify links and count carbons), then check forwards that every reagent and condition is correct.
A second worked route: target ethyl ethanoate from ethene only. Ethene + steam ( $H_{3} PO_{4}$ ) → ethanol; split the ethanol, oxidising part with excess acidified dichromate under reflux to ethanoic acid; then react the acid with the rest of the ethanol (concentrated $H_{2} SO_{4}$ , reflux).
Where an intermediate has the right carbon skeleton but the wrong functional group, look for a one-step interconversion before adding steps; most A-level routes need two to four steps.
When asked for 'reagents and conditions', give names or formulae of reagents (not just 'acid') and conditions such as reflux, temperature, solvent or catalyst; state the intermediate structures as well.
Check each step's type of reaction against the functional groups present: a reagent that would also attack another group in the molecule may give unwanted products.

Designing an efficient synthesis 3.3.14

Key results

Atom economy = $\frac{M _{r} ( desired product )}{\sum M _{r} ( reactants )} \times 100%$ .
$C_{2} H_{4} + H_{2} O \to C_{2} H_{5} OH$ : 100%. $C_{2} H_{5} Br + NaOH \to C_{2} H_{5} OH + NaBr$ : $46/149 \approx 31%$ .
Four steps at 80% each: overall yield $0.8 0^{4} \approx 41%$ .

Notes

Efficient syntheses use few steps, high atom economy, safer reagents and little or no solvent.
Yield and atom economy measure different things: a reaction can have high yield but produce much waste.
Green chemistry principles also include using renewable feedstocks, catalysts instead of stoichiometric reagents, safer solvents (water, supercritical $CO_{2}$ ), and designing products that degrade after use.
Worked example: if a three-step synthesis has yields of 90%, 75% and 60%, the overall yield is $0.90 \times 0.75 \times 0.60 = 40.5%$ . The lowest-yield step is the best target for improvement.
Purification adds losses: recrystallisation, distillation and chromatography each leave some product behind, which is one reason shorter syntheses are preferred.
Pharmaceutical routes also aim to produce the required single enantiomer, avoiding wasteful separation of a racemate.

3.3.15

Nuclear magnetic resonance spectroscopy

Sample, reference and chemical shift 3.3.15

Notes

In a strong magnetic field, $^{1} H$ and $^{13} C$ nuclei absorb radio-frequency energy; the exact frequency depends on the local electron environment.
Chemical shift, δ in ppm, is measured relative to tetramethylsilane (TMS), $Si(CH_{3})_{4}$ , set at δ = 0. TMS gives a single sharp peak (12 equivalent H, 4 equivalent C), is inert, non-toxic and volatile, so is easily removed.
Samples are dissolved in deuterated solvents such as $CDCl_{3}$ , or in $CCl_{4}$ , which contain no $^{1} H$ to give a solvent signal.
Electronegative atoms nearby withdraw electron density, deshielding the nucleus and increasing δ.
Approximate $^{1} H$ chemical shifts (use the data booklet for exact ranges): R–CH₃ 0.7–1.2; H–C–C=O 2.1–2.6; H–C–O 3.3–4.3; aromatic C–H 6.5–8.0; aldehyde CHO 9.0–10.0; carboxylic acid COOH 10–12 ppm. Alcohol O–H and amine N–H vary widely (about 1–5 ppm).
Nuclei behave like tiny magnets: in the applied field they can align with it or against it, and radio-frequency energy flips them between the two states. The energy needed depends on how much the nucleus is shielded by its own electrons.
Chemical shift is used rather than frequency because it does not depend on the strength of the instrument's magnet, so spectra from different machines can be compared directly.
MRI scanners in hospitals use the same principle, detecting hydrogen nuclei mainly in water and fat to image soft tissues without ionising radiation.

Carbon-13 NMR 3.3.15

Key results

Ethyl ethanoate: four carbon environments, at about δ 171 (C=O), 60 (O–CH₂), 21 (CH₃–C=O) and 14 (CH₃).
Propanone: three carbons but only two environments, because the methyl carbons are equivalent by symmetry.

Notes

The number of peaks equals the number of carbon environments; peak heights do not count carbons.
Use the data booklet ranges: C=O in esters and acids 160–185 ppm, aldehydes and ketones 190–220 ppm.
Approximate $^{13} C$ shifts (use the data booklet for exact ranges): C–C 5–40; C–Cl or C–Br 10–70; C–N 25–60; C–O 50–90; C=C and aromatic carbons 90–150; ester and acid C=O 160–185; aldehyde and ketone C=O 190–220 ppm.
Worked counts: butanone has 4 environments; benzene has 1; methylbenzene has 5 (the methyl carbon, the ring carbon bearing it, two equivalent ortho, two equivalent meta and one para); propan-2-ol has 2.
$^{13} C$ makes up only about 1.1% of carbon, so signals are weak and spectra are recorded with many scans. Two $^{13} C$ atoms are rarely next to each other, so C–C splitting is not seen.
Use symmetry to count environments: draw the molecule and look for a mirror plane or rotation that maps carbons onto each other.

Proton NMR: shift, integration and splitting 3.3.15

Key results

Ethyl ethanoate: δ ≈ 4.1 (2H, quartet, –OCH₂–); δ ≈ 2.1 (3H, singlet, CH₃C=O); δ ≈ 1.3 (3H, triplet, –CH₂CH₃).

Notes

Number of signals = number of proton environments.
The integration trace gives the relative number of protons in each environment (here 2:3:3).
Splitting tells you about neighbours; chemical shift tells you about the environment (here the O–CH₂ is deshielded by the adjacent oxygen).
The integration trace rises in steps; the ratio of step heights gives the ratio of protons, which may need scaling to the molecular formula (e.g. 1 : 1.5 : 1.5 means 2 : 3 : 3).
O–H protons can be identified by shaking the sample with $D_{2} O$ : deuterium exchanges with the O–H hydrogen, and that peak disappears from the spectrum.
When interpreting a spectrum, work through each signal stating its shift (environment), area (number of H) and splitting (number of neighbouring H), then assemble fragments into a structure consistent with the molecular formula.
Aldehyde protons (9–10 ppm) and carboxylic acid protons (10–12 ppm) are distinctive and quickly identify these functional groups.

The n + 1 rule 3.3.15

Key results

n equivalent H on adjacent carbons split a signal into n + 1 peaks: singlet (n = 0), doublet (1), triplet (2), quartet (3).
Line intensities: 1:1, 1:2:1, 1:3:3:1.

Notes

An ethyl group (–CH₂CH₃) shows a characteristic quartet and triplet.
Equivalent protons do not split each other. O–H and N–H protons usually appear as broad singlets because they exchange rapidly.
Worked examples: ethanal $CH_{3} CHO$ gives a 3H doublet (next to one H) and a 1H quartet (next to three H). A propan-2-yl group, $(CH_{3})_{2} CH -$ , gives a 6H doublet and a 1H septet (6 + 1 = 7 lines).
Splitting is mutual: if signal A is split by B's protons, B is split by A's. A quartet of 2H paired with a triplet of 3H is the signature of an ethyl group.
Protons on the same carbon, or on equivalent carbons, do not split each other; neither (at A-level) do protons more than three bonds apart, so a $CH_{3}$ attached to C=O with no H on the neighbouring atom gives a singlet.
The intensities follow Pascal's triangle: 1 : 2 : 1, 1 : 3 : 3 : 1 and so on, which helps recognise multiplets whose outer lines are small.

Distinguishing isomers 3.3.15

Notes

Ethyl ethanoate and methyl propanoate are both $C_{4} H_{8} O_{2}$ with four $^{13} C$ and three $^{1} H$ environments, so peak counts alone cannot tell them apart.
Ethyl ethanoate has an O–CH₂ quartet at δ ≈ 4.1 (2H); methyl propanoate has an O–CH₃ singlet at δ ≈ 3.7 (3H). In $^{13} C$ , O–CH₂ is at about 60 ppm and O–CH₃ at about 52 ppm.
Combine NMR with the molecular formula from mass spectrometry and functional groups from IR.
Butanal and butanone ( $C_{4} H_{8} O$ ): butanal shows an aldehyde H at 9–10 ppm (a triplet) and four $^{1} H$ environments; butanone shows no peak above 3 ppm and three $^{1} H$ environments, including a 3H singlet near 2.1 ppm.
Propan-1-ol and propan-2-ol: propan-1-ol has four $^{1} H$ environments and three $^{13} C$ ; propan-2-ol has three $^{1} H$ environments (including a 6H doublet) and only two $^{13} C$ .
A strategy for structure questions: molecular formula (MS), functional groups (IR, chemical tests), carbon environments ( $^{13} C$ NMR), then proton environments and neighbours ( $^{1} H$ NMR), checking that the final structure fits every piece of data.

Worked examples

Worked example

Compound X has molecular formula C₄H₈O₂ and shows four ¹H NMR signals: δ 1.3 (3H, triplet), δ 2.1 (3H, singlet), δ 4.1 (2H, quartet), and no exchangeable proton.

Deduce the structure of X, justifying your answer from the data.

Show worked solution

The absence of any broad, exchangeable (D₂O-removable) signal rules out an alcohol or carboxylic acid, despite the O₂ in the formula, and points instead to an ester.

The δ 2.1 singlet (3H) has no neighbouring hydrogens, so it is a CH₃ group directly attached to a carbonyl carbon, C(=O)CH₃.

The δ 1.3 triplet (3H) and δ 4.1 quartet (2H) are a classic coupled pair: the CH₃ (triplet, split by 2 neighbouring H) and the CH₂ (quartet, split by 3 neighbouring H) of an ethyl group, with the CH₂ shifted downfield to δ 4.1 because it is attached directly to the ester oxygen, O–CH₂.

Putting the fragments together, CH₃CO–O–CH₂CH₃, gives ethyl ethanoate,

M_{r} = 12 (4) + 1 (8) + 16 (2) = 48 + 8 + 32 = 88

consistent with C₄H₈O₂.

Worked example

A carbonyl compound with molecular formula C₄H₈O gives a mass spectrum with a molecular ion peak at m/z = 72 and a prominent fragment peak at m/z = 43.

Suggest a structure for the compound and explain the origin of the fragment peak.

Show worked solution

$M_{r} = 72$ is consistent with C₄H₈O (:

12 \times 4 + 1 \times 8 + 16 = 72

The fragment at m/z = 43 corresponds to a loss of $72 - 43 = 29$ from the molecular ion, consistent with loss of a C₂H₅• (ethyl) radical.

A structure consistent with both pieces of evidence is butanone, $CH_{3} COCH_{2} CH_{3}$ : fragmentation next to the carbonyl group (the bond that breaks most readily in a ketone's mass spectrum) can lose the ethyl radical, leaving the stabilised acylium ion $CH_{3} CO^{+}$ at m/z = 43.

3.3.16

Chromatography

Thin-layer chromatography 3.3.16

Definitions

Chromatography separates components by their different balance between a mobile phase and a stationary phase.
$R_{f} = \frac{distance travelled by spot}{distance travelled by solvent front}$ .

Key results

Example: solvent front 8.0 cm; spot A 4.8 cm gives $R_{f} = 0.60$ ; spot B 2.4 cm gives $R_{f} = 0.30$ .

Method

Draw the origin in pencil (ink would separate) and spot the sample; let it dry.
Stand the plate in solvent below the origin, with a lid to saturate the air with vapour.
Mark the solvent front immediately; locate colourless spots with UV light or a stain.

Notes

The stationary phase is silica or alumina on a plate; the mobile phase is the solvent. A component held more strongly by the stationary phase has a lower $R_{f}$ .
$R_{f}$ values depend on conditions, so compare with standards on the same plate.
The solvent (mobile phase) is chosen to give $R_{f}$ values well spread between 0.2 and 0.8; a more polar solvent carries polar compounds further up a silica plate.
TLC is used to monitor reactions: spot the starting material, the reaction mixture and a co-spot side by side; the reaction is complete when the starting material's spot has disappeared.
Purity check: a pure compound gives one spot; extra spots show impurities, although two compounds with the same $R_{f}$ in that solvent would still appear as one.
Separation depends on the balance of adsorption onto the polar silica surface (stronger for polar compounds) and solubility in the mobile phase.

Column chromatography 3.3.16

Notes

A column is packed with silica or alumina (stationary phase); the solvent (eluent) flows down through it.
Components that interact less strongly with the stationary phase spend more time in the mobile phase and elute first; fractions are collected separately.
Unlike TLC, it can purify and collect useful quantities of each component.
The column is packed with a slurry of silica in the solvent to avoid air bubbles and cracks, which would let components run through unevenly; the sample is added as a concentrated band at the top.
A gradient of solvents, starting non-polar and becoming more polar, can wash off weakly held components first and then the strongly held ones.
Collected fractions are checked by TLC; those containing the same single compound are combined and the solvent evaporated to recover the purified product.
Column chromatography is the preparative version of TLC: the same stationary phase and solvent system, scaled up to separate grams of material.

Gas chromatography 3.3.16

Notes

For volatile components: the sample is vaporised and carried by an unreactive carrier gas (e.g. helium or nitrogen) through a long column in an oven.
The stationary phase is a high-boiling liquid on an inert solid support (or a solid).
Components separate according to their volatility and how strongly they interact with the stationary phase.
Raising the oven temperature reduces retention times because components spend more time in the gas phase; temperature programmes (ramping the temperature during a run) help separate mixtures with a wide range of boiling points.
Detectors respond to the components as they leave the column, giving a peak for each; the area under each peak is proportional to the amount of that component.
Uses: testing athletes' urine for drugs, measuring blood alcohol, detecting pesticide residues in food, analysing perfumes and identifying accelerants in fire investigations.
Gas chromatography needs only a tiny sample, which is why it suits forensic and environmental analysis where material is limited.

Retention time and GC–MS 3.3.16

Definitions

Retention time: the time from injection to the peak maximum for a component.

Notes

Components are identified by comparing retention times with standards under identical conditions (column, temperature, gas flow). A match supports but does not prove identity.
Relative peak areas indicate the relative amounts of the components.
GC–MS feeds each separated component into a mass spectrometer; the mass spectrum is compared with a database for a much more certain identification. Used in forensics, drug testing and environmental analysis.
Worked example: a mixture gives peaks with areas 1200, 3000 and 1800 units; the percentage composition is 20%, 50% and 30% (assuming the detector responds equally to each component).
Two different compounds can have the same retention time on one column, which is why GC–MS is far more conclusive: the mass spectrum gives a fingerprint for each separated component.
Limitations: compounds must be volatile and thermally stable; very similar compounds may not separate; and reference spectra or standards are needed for identification.
The same separate-then-identify approach is used with liquid chromatography coupled to mass spectrometry (LC–MS) for involatile molecules such as proteins and many drugs.