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How do benzene, carbonyl compounds and carboxylic acids react?

The structure and electrophilic substitution reactions of benzene, the reactions of aldehydes and ketones, and the reactions of carboxylic acids and their derivatives.

An Edexcel 9CH0 Topic 18 answer covering the structure and electrophilic substitution of benzene, the reactions of aldehydes and ketones, and carboxylic acids and their derivatives.

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What this topic is asking

Edexcel Topic 18 wants you to describe the delocalised bonding in benzene and its electrophilic substitution reactions (with mechanisms), and to explain the reactions of aldehydes, ketones and carboxylic acids and their derivatives, including the tests that distinguish them.

The answer

Benzene and electrophilic substitution

The evidence for delocalisation is that all six C-C\text{C-C} bonds are the same length (between a single and double bond), and the enthalpy of hydrogenation is far less exothermic than the Kekule (three localised double bonds) model predicts. The three key reactions are:

  • Nitration: concentrated HNO3\text{HNO}_3 with a concentrated H2SO4\text{H}_2\text{SO}_4 catalyst at 50 C50\ ^\circ\text{C}, electrophile NO2+\text{NO}_2^+.
  • Halogenation: Cl2\text{Cl}_2 or Br2\text{Br}_2 with a halogen-carrier catalyst (e.g. AlCl3\text{AlCl}_3, FeBr3\text{FeBr}_3).
  • Friedel-Crafts acylation: an acyl chloride with an AlCl3\text{AlCl}_3 catalyst, electrophile RCO+\text{RCO}^+.

The electrophilic substitution mechanism

The electrophile accepts a pair of electrons from the delocalised ring, forming an unstable intermediate in which the delocalisation is partly broken. The intermediate then loses an H+\text{H}^+, restoring the stable aromatic ring. Show a curly arrow from the ring to the electrophile, then a curly arrow from the C-H\text{C-H} bond back into the ring as the proton leaves.

Aldehydes and ketones

The C=O\text{C=O} group is polar (the carbon is δ+\delta+) and undergoes nucleophilic addition, for example with HCN\text{HCN} (in the presence of KCN\text{KCN}) to give a hydroxynitrile. Aldehydes are oxidised to carboxylic acids by Tollens' reagent (giving a silver mirror) and Fehling's solution (giving a brick-red precipitate of Cu2O\text{Cu}_2\text{O}); ketones are not oxidised by either reagent. This difference is the basis of the test to tell aldehydes from ketones. Both classes are reduced by NaBH4\text{NaBH}_4 to alcohols (aldehyde to primary alcohol, ketone to secondary alcohol).

Carboxylic acids and derivatives

Acyl chlorides are made from carboxylic acids with SOCl2\text{SOCl}_2 and react vigorously with water, alcohols and amines, releasing HCl\text{HCl}. They are the most useful synthetic route to esters and amides because the reactions go essentially to completion (unlike the reversible Fischer esterification).

Examples in context

Example 1. Aspirin synthesis. Aspirin is made by reacting salicylic acid (which has both a phenol OH\text{OH} and a COOH\text{COOH}) with ethanoic anhydride, an acid derivative. The reactive anhydride acylates the phenol OH\text{OH} to form an ester linkage, giving acetylsalicylic acid. This is a direct application of the reactivity order of acid derivatives: the anhydride is reactive enough to esterify the otherwise unreactive phenol, which a carboxylic acid alone could not do efficiently.

Example 2. Distinguishing glucose by Fehling's solution. Glucose contains an aldehyde group in its open-chain form, so it gives a brick-red Cu2O\text{Cu}_2\text{O} precipitate with Fehling's solution (a "reducing sugar"), whereas sucrose, which has no free aldehyde, does not. This is the same aldehyde-versus-ketone chemistry from Topic 18 applied to a biological molecule, and it underpins the classic clinical test once used to detect glucose in urine.

Try this

Q1. Explain why benzene undergoes substitution rather than addition. [2 marks]

  • Cue. The delocalised pi system gives extra stability; substitution preserves it, whereas addition would disrupt the stable ring.

Q2. Describe a chemical test to distinguish an aldehyde from a ketone. [2 marks]

  • Cue. Warm with Tollens' reagent; the aldehyde gives a silver mirror, the ketone gives no change.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20185 marksBenzene reacts with concentrated nitric acid in the presence of concentrated sulfuric acid at 50 C50\ ^\circ\text{C}. (a) Name the mechanism and the electrophile. (b) Write the equation for generating the electrophile. (c) Explain why benzene undergoes substitution rather than addition.
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Identify the nitration mechanism, generate the electrophile, and use delocalisation to justify substitution.

(a) Electrophilic substitution; the electrophile is the nitronium ion NO2+\text{NO}_2^+ (1).

(b) HNO3+2H2SO4NO2++2HSO4+H3O+\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+ (1). The sulfuric acid protonates the nitric acid, which then loses water to give NO2+\text{NO}_2^+ (1).

(c) Benzene has a stable delocalised pi system of six electrons (1). Substitution restores this stable ring, whereas addition would permanently break the delocalisation and lose that stability, so substitution is strongly preferred (1).

Edexcel 20214 marksA student has two unlabelled bottles, one containing propanal and one containing propanone. (a) Describe a chemical test using Fehling's solution that distinguishes them, including the observation for each. (b) Give the equation, using [O][\text{O}], for any reaction that occurs.
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Use the fact that only aldehydes are oxidised by Fehling's solution.

(a) Warm each compound with Fehling's solution (1). Propanal (an aldehyde) reduces the blue Cu2+\text{Cu}^{2+} to a brick-red precipitate of Cu2O\text{Cu}_2\text{O} (1). Propanone (a ketone) gives no change; the solution stays blue (1).

(b) For propanal: CH3CH2CHO+[O]CH3CH2COOH\text{CH}_3\text{CH}_2\text{CHO} + [\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{COOH} (1). Tollens' reagent (silver mirror) would be an equally valid alternative test.

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