Notes · ChemistryUK · A-Levels

Optical isomerism, carbonyls, carboxylic acids and aromatic chemistry (A-level)

This A-level-only topic develops the second half of organic chemistry. It covers optical isomerism and chirality, the nucleophilic-addition chemistry and tests of aldehydes and ketones, carboxylic acids and their derivatives (esters, acyl chlorides, anhydrides and amides), the delocalised structure and electrophilic substitution of benzene, and the basicity and reactions of amines.

5 sections~16 min reading time3 competenciesLevel Advanced 5

T·161616 / 18

Exam profile

AO1 · Recall the reactions, mechanisms and tests for carbonyls, carboxylic acids, arenes and aminesAO2 · Draw the nucleophilic-addition and electrophilic-substitution mechanisms and identify chiral centresAO3 · Explain the stability of benzene, the strength of bases and the outcomes of tests and reactions

Operators:stateexplainoutline the mechanismdeducepredictjustify

basic level

This entire topic is A-level only (A2); it is not assessed at AS.

higher level

The full A-Level treats optical isomerism, the nucleophilic addition of carbonyls, the reactions of carboxylic acids and their derivatives, aromatic electrophilic substitution and the basicity of amines, all with mechanisms.

Depth

Text

Contents · 5 sections

Optical isomerism, carbonyls, carboxylic acids and aromatic chemistry (A-level)

Optical isomerism and chirality#

AdvancedAQA 7405 3.3.9.1DfE GCE Chemistry - optical isomerism

A pair of enantiomers

Fig. 1A carbon bonded to four different groups is chiral; its mirror image cannot be superimposed on it (two enantiomers).

Optical isomerism is a form of stereoisomerism that arises when a molecule contains a chiral centre - a carbon atom bonded to four different groups. Because the four groups are all different, the molecule and its mirror image cannot be superimposed on one another, no matter how they are rotated, just as a left hand cannot be superimposed on a right hand. The two non-superimposable mirror images are called enantiomers (or optical isomers).

Enantiomers are identical in almost every physical and chemical property - the same melting point, boiling point and reactivity with ordinary reagents - because they contain the same bonds. They differ in only two ways. First, they rotate the plane of plane-polarised light in opposite directions (one clockwise, one anticlockwise) by the same amount; this optical activity gives the isomerism its name. Second, they can interact differently with other chiral molecules, which is why they often have very different effects in biological systems.

A racemic mixture (racemate) is an equal mixture of the two enantiomers. It is optically inactive, because the rotation caused by one enantiomer is exactly cancelled by the equal and opposite rotation of the other. Reactions that create a chiral centre from a flat (planar) intermediate, such as the nucleophilic addition of cyanide to a carbonyl, produce a racemic mixture, because the nucleophile can attack the planar carbonyl equally from either face.

Identifying a chiral centre is the key skill: look for a carbon bonded to four different groups. A molecule may have more than one chiral centre. The biological importance is considerable - many drugs are chiral, and often only one enantiomer has the desired effect while the other may be inactive or harmful, so the pharmaceutical industry works hard to make single enantiomers rather than racemates. When asked, mark the chiral carbon (often with an asterisk) and explain that its mirror image is non-superimposable.

Worked example

Recognising a chiral centre

Explain why butan-2-ol (CH3CH(OH)CH2CH3) shows optical isomerism but butan-1-ol does not.

Examine butan-2-ol
Carbon 2 is bonded to four different groups: H, OH, CH3 and CH2CH3, so it is a chiral centre.
Consequence
With four different groups, the molecule and its mirror image are non-superimposable, giving two enantiomers.
Examine butan-1-ol
In butan-1-ol no carbon is bonded to four different groups (C1 has two hydrogens), so there is no chiral centre and no optical isomerism.

Result: Butan-2-ol has a chiral carbon (four different groups) and shows optical isomerism; butan-1-ol has none.

Typical mistakes

Missing that all four groups on the carbon must be different for it to be chiral.
Saying enantiomers have different melting points or reactivity with ordinary reagents - they differ only in optical rotation and with other chiral species.

Active revision

Identify the chiral centre in 2-hydroxypropanoic acid ( $\text{CH}_3\text{CH(OH)COOH}$ ) and explain why it shows optical isomerism.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Chemistry 7405 specification (AQA)

Aldehydes and ketones: nucleophilic addition and tests#

AdvancedAQA 7405 3.3.10.1DfE GCE Chemistry - aldehydes and ketones

Nucleophilic addition of cyanide to a carbonyl

Fig. 2The cyanide nucleophile adds to the delta-positive carbonyl carbon; the pi electrons move onto oxygen to give an alkoxide, then protonation gives the hydroxynitrile.

Aldehydes and ketones both contain the carbonyl group, C=O, which is polar because oxygen is more electronegative than carbon, leaving the carbonyl carbon

\delta+

. This

\delta+

carbon is attacked by nucleophiles, so carbonyls undergo nucleophilic addition: the nucleophile adds to the carbon and the

\pi

electrons of the C=O move onto the oxygen, which is then protonated. The difference between the two classes is that an aldehyde has the carbonyl at the end of the chain (with a hydrogen attached) while a ketone has it within the chain (between two carbons).

Reduction adds hydrogen across the carbonyl. Using the reducing agent

\text{NaBH}_4

(which provides hydride ions,

\text{H}^-

, as the nucleophile), an aldehyde is reduced to a primary alcohol and a ketone to a secondary alcohol. The mechanism is nucleophilic addition of the hydride ion to the

\delta+

carbon, followed by protonation of the resulting alkoxide by water.

Addition of hydrogen cyanide (from

\text{KCN}

and acid) lengthens the carbon chain and introduces a nitrile group, giving a hydroxynitrile. The cyanide ion (the nucleophile) adds to the

\delta+

carbon and the oxygen is protonated. Because the carbonyl is planar, the cyanide can attack from either face with equal probability, so if the product has a chiral centre it is formed as a racemic mixture - a neat link to optical isomerism.

Aldehydes and ketones are distinguished by oxidation. Aldehydes are readily oxidised to carboxylic acids and so give a positive result with weak oxidising agents: Tollens' reagent (ammoniacal silver nitrate) gives a silver mirror, and Fehling's solution gives a brick-red precipitate of copper(I) oxide. Ketones are not oxidised by these reagents, so they give no reaction - the standard chemical way to tell an aldehyde from a ketone.

\text{CH}_3\text{CHO} + \text{HCN} \rightarrow \text{CH}_3\text{CH(OH)CN}

Addition of hydrogen cyanide

Cyanide adds to the carbonyl carbon and the oxygen is protonated, lengthening the chain and forming a hydroxynitrile (as a racemate).

Worked example

Why HCN addition gives a racemate

Explain why the reaction of hydrogen cyanide with ethanal produces a racemic mixture of 2-hydroxypropanenitrile.

The intermediate is planar
The carbonyl group of ethanal is planar, so the delta-positive carbon can be attacked from either face.
Two equally likely attacks
Cyanide attacks the two faces with equal probability, forming the two mirror-image products in equal amounts.
The product is chiral
The product carbon is bonded to four different groups (H, OH, CN, CH3), so the two products are enantiomers - a racemate.

Result: Attack on the planar carbonyl from either face with equal probability gives equal amounts of the two enantiomers, a racemic mixture.

Typical mistakes

Drawing the nucleophile attacking the oxygen instead of the delta-positive carbon.
Saying a ketone gives a silver mirror - only aldehydes are oxidised by Tollens' reagent.

Active revision

Give the mechanism for the reduction of propanone by $\text{NaBH}_4$ (as a source of $\text{H}^-$ ) and name the product.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Carboxylic acids and their derivatives#

AdvancedAQA 7405 3.3.11.1DfE GCE Chemistry - carboxylic acids and derivatives

The carboxylic-acid derivatives

Fig. 3The acid and its derivatives interconvert; the more reactive acyl chloride is used to make esters and amides.

Carboxylic acids contain the -COOH group and are weak acids: they partially ionise in water to give a carboxylate ion and

\text{H}^+

. They are acidic enough to react with reactive metals, bases and carbonates - and, distinctively, they release carbon dioxide from carbonates and hydrogencarbonates (the standard test for the group). Their acidity comes from the stability of the carboxylate ion, in which the negative charge is delocalised over the two oxygen atoms.

Carboxylic acids form esters by condensation with alcohols, catalysed by a little concentrated sulfuric acid:

\text{RCOOH} + \text{R'OH} \rightleftharpoons \text{RCOOR'} + \text{H}_2\text{O}

(esterification). Esters are sweet-smelling and are used as flavourings, perfumes and solvents. They are hydrolysed back to the acid (or its salt) and the alcohol by heating with acid or with alkali - alkaline hydrolysis (saponification) of the esters in fats produces soap.

The acyl chlorides (RCOCl) are the most reactive carboxylic-acid derivatives, made from the acid using a reagent such as

\text{SOCl}_2

. They react vigorously with nucleophiles - with water to give the carboxylic acid, with alcohols to give esters, and with ammonia or amines to give amides - releasing hydrogen chloride each time. Acid anhydrides ((RCO)

_2

O) react similarly but less violently, which is why ethanoic anhydride (rather than the corrosive, fuming ethanoyl chloride) is used to make aspirin industrially.

The reactivity order of the derivatives - acyl chloride

>

acid anhydride

>

ester/carboxylic acid

>

amide - guides synthesis: a more reactive derivative is used to make a less reactive one. When you predict a reaction, identify the nucleophile and the derivative, then give the product and the small molecule released (water, an alcohol, or hydrogen chloride). Amides (RCONH

_2

), containing the -CONH

_2

group, are made from acyl chlorides and ammonia and are the least reactive derivative.

\text{RCOOH} + \text{R'OH} \underset{\;conc.\ H_2SO_4\;}{\rightleftharpoons} \text{RCOOR'} + \text{H}_2\text{O}

Esterification

A carboxylic acid and an alcohol condense to an ester and water, catalysed by concentrated sulfuric acid.

Worked example

Making an ester two ways

Compare making ethyl ethanoate from (a) ethanoic acid and (b) ethanoyl chloride, each with ethanol.

From the acid
CH3COOH + C2H5OH -> CH3COOC2H5 + H2O, catalysed by concentrated sulfuric acid; this is a reversible, incomplete reaction.
From the acyl chloride
CH3COCl + C2H5OH -> CH3COOC2H5 + HCl; this is fast and essentially complete.
Compare
The acyl-chloride route gives a higher yield because it is not an equilibrium, but ethanoyl chloride is corrosive and releases HCl.

Result: Both give ethyl ethanoate; the acyl-chloride route is faster and complete, the acid route is a reversible equilibrium needing a catalyst.

Typical mistakes

Forgetting the concentrated sulfuric acid catalyst (and the equilibrium) in esterification.
Getting the reactivity order of the derivatives wrong (acyl chloride is the most reactive, amide the least).

Active revision

Write equations for the reactions of ethanoyl chloride with (a) water, (b) ethanol and (c) ammonia, naming each organic product.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Aromatic chemistry: benzene and electrophilic substitution#

AdvancedAQA 7405 3.3.12.1DfE GCE Chemistry - aromatic chemistry

The delocalised structure of benzene

Fig. 4Benzene is a regular hexagon with a delocalised ring of pi electrons (shown by aromatic bonds), not three separate double bonds.

Benzene,

\text{C}_6\text{H}_6

, is a planar, regular hexagon of six carbon atoms, each bonded to one hydrogen. Each carbon uses three of its electrons in

\sigma

bonds (to two carbons and one hydrogen), leaving one electron in a p-orbital perpendicular to the ring. These six p-orbitals overlap sideways all the way round to form a single delocalised

\pi

system - a ring of electron density above and below the plane - rather than three separate double bonds. This is why benzene is drawn as a hexagon with an inner circle.

The evidence for delocalisation is thermochemical and structural. All six carbon-carbon bonds are the same length, intermediate between a single and a double bond, whereas the Kekule model of alternating single and double bonds predicts two different lengths. And the enthalpy of hydrogenation of benzene is much less exothermic than expected: hydrogenating one double bond (in cyclohexene) releases about

120\ \text{kJ mol}^{-1}

, so three would be expected to release about

360\ \text{kJ mol}^{-1}

, but benzene releases only about

208\ \text{kJ mol}^{-1}

. Benzene is therefore about

150\ \text{kJ mol}^{-1}

more stable than the Kekule model predicts - the delocalisation (resonance) stabilisation.

Because its delocalised

\pi

system is a region of high electron density, benzene is attacked by electrophiles; but unlike an alkene it undergoes substitution rather than addition, so that the stable delocalised ring is preserved. In electrophilic substitution, the electrophile bonds to the ring to form an unstable intermediate, then a hydrogen ion is lost to restore the aromatic ring. Nitration uses a mixture of concentrated nitric and sulfuric acids to generate the nitronium ion

\text{NO}_2^+

as the electrophile, substituting a hydrogen to give nitrobenzene.

Friedel-Crafts acylation introduces an acyl group (

\text{RCO-}

) onto the ring using an acyl chloride and an aluminium chloride catalyst, which generates the acylium electrophile

\text{RCO}^+

; this is a key way of forming a new carbon-carbon bond to build up aromatic molecules. In every case the mechanism is the same: generate the electrophile, the ring's delocalised electrons attack it to form the intermediate, then lose

\text{H}^+

to regenerate the aromatic system. Being able to outline this mechanism and explain why substitution (not addition) occurs is central to aromatic chemistry.

\text{C}_6\text{H}_6 + \text{HNO}_3 \xrightarrow{\;conc.\ H_2SO_4\;} \text{C}_6\text{H}_5\text{NO}_2 + \text{H}_2\text{O}

Nitration of benzene

Concentrated nitric and sulfuric acids generate NO2+, which substitutes a hydrogen to give nitrobenzene.

Enthalpy of hydrogenation evidence

Fig. 5Benzene releases far less energy on hydrogenation than the Kekule model predicts; the difference (about 150 kJ/mol) is the delocalisation stabilisation.

Worked example

Delocalisation stabilisation from hydrogenation data

The enthalpy of hydrogenation of cyclohexene is -120 kJ mol^-1 and that of benzene is -208 kJ mol^-1. Estimate the delocalisation (resonance) stabilisation of benzene.

Expected value
If benzene had three separate C=C double bonds, its hydrogenation would release 3 x 120 = 360 kJ mol^-1.
Actual value
Benzene actually releases only 208 kJ mol^-1 on hydrogenation.
Difference
The stabilisation is 360 - 208 = 152 kJ mol^-1: benzene is more stable than the Kekule model predicts.
$360 - 208 = 152\ \text{kJ mol}^{-1}$

Result: Benzene is about 152 kJ mol^-1 more stable than three isolated double bonds would be - its delocalisation stabilisation.

Typical mistakes

Drawing benzene with three localised double bonds (the Kekule model) rather than the delocalised ring.
Saying benzene undergoes addition like an alkene - it substitutes, to preserve the stable delocalised ring.

Active revision

Outline the mechanism for the nitration of benzene, including how the electrophile is generated and how the aromatic ring is restored.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Amines: base strength and reactions#

AdvancedAQA 7405 3.3.13.1DfE GCE Chemistry - amines

Base strength of amines

Fig. 6The more available the nitrogen lone pair, the stronger the base: ethylamine > ammonia > phenylamine.

Amines are derivatives of ammonia in which one or more hydrogens are replaced by carbon groups: primary (

\text{RNH}_2

), secondary (

\text{R}_2\text{NH}

) and tertiary (

\text{R}_3\text{N}

). They act as bases because the nitrogen atom has a lone pair of electrons that can accept a proton, forming an alkylammonium ion (for example

\text{CH}_3\text{NH}_2 + \text{H}^+ \rightarrow \text{CH}_3\text{NH}_3^+

). The more available the lone pair, the stronger the base.

Base strength therefore depends on the electron density on the nitrogen lone pair. Aliphatic amines such as ethylamine are stronger bases than ammonia, because the alkyl group is electron-releasing (it pushes electron density onto the nitrogen), making the lone pair more available to accept a proton. Aromatic amines such as phenylamine are weaker bases than ammonia, because the nitrogen lone pair is partly delocalised into the benzene ring, so it is less available to bond to a proton.

Amines are also nucleophiles, because of the same lone pair. They react with halogenoalkanes by nucleophilic substitution to form further amines (ammonia gives a primary amine, which can react again), and with acyl chlorides to form amides. The lone pair attacks the electron-poor carbon, and the reaction can continue to substitute the amine further, so a mixture of products is often obtained.

Amines are prepared by two main routes: the reaction of a halogenoalkane with excess ammonia (nucleophilic substitution, giving a primary amine plus further-substituted amines), and the reduction of a nitrile (for example with a nickel catalyst and hydrogen, or with

\text{LiAlH}_4

), which gives a primary amine with one more carbon. Aromatic amines such as phenylamine are made by reducing the corresponding nitro compound (nitrobenzene) with tin and hydrochloric acid. When comparing base strengths, always argue from the availability of the nitrogen lone pair.

\text{CH}_3\text{NH}_2 + \text{H}^+ \rightarrow \text{CH}_3\text{NH}_3^{+}

An amine acting as a base

The nitrogen lone pair accepts a proton to form an alkylammonium ion.

Worked example

Comparing base strengths

Place ethylamine, ammonia and phenylamine in order of increasing base strength and explain the order.

Phenylamine (weakest)
The nitrogen lone pair is delocalised into the benzene ring, so it is least available to accept a proton.
Ammonia (intermediate)
The lone pair is freely available, with no ring to delocalise it and no strongly electron-releasing group.
Ethylamine (strongest)
The electron-releasing ethyl group increases the electron density on nitrogen, making the lone pair most available.

Result: Increasing base strength: phenylamine < ammonia < ethylamine, set by the availability of the nitrogen lone pair.

Typical mistakes

Saying phenylamine is a stronger base than ammonia - its lone pair is delocalised into the ring, making it weaker.
Forgetting that reacting a halogenoalkane with ammonia gives a mixture of further-substituted amines.

Active revision

Explain why ethylamine is a stronger base than ammonia, but phenylamine is a weaker base than ammonia.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Chemistry 7405 specification

Optical isomerism, carbonyls, carboxylic acids and aromatic chemistry (A-level)

5 sections~16 min reading time3 competenciesLevel Advanced 5

Optical isomerism, carbonyls, carboxylic acids and aromatic chemistry (A-level)

Key points

Recognising a chiral centre

Key points

Why HCN addition gives a racemate

Key points

Making an ester two ways

Key points

Delocalisation stabilisation from hydrogenation data

Key points

Comparing base strengths

Sources

Optical isomerism, carbonyls, carboxylic acids and aromatic chemistry (A-level)

Key points

Recognising a chiral centre

Key points

Why HCN addition gives a racemate

Key points

Making an ester two ways

Key points

Delocalisation stabilisation from hydrogenation data

Key points

Comparing base strengths

Sources