AQA A-Level Chemistry: Organic Chemistry Advanced

Benzene reacts with a mixture of concentrated nitric acid and concentrated sulfuric acid at a temperature below 50 °C to form nitrobenzene, C₆H₅NO₂.

Describe the mechanism of this reaction and explain why benzene undergoes substitution rather than addition. In your answer you should:

• show how the electrophile is generated from the two acids;
• describe how the electrophile reacts with the benzene ring, including the intermediate formed and how aromaticity is restored;
• explain, in terms of the bonding in benzene, why substitution is preferred to addition.

(6 marks)

A chemist prepares N-phenylethanamide (C₆H₅NHCOCH₃, Mᵣ = 135.0) from benzene (C₆H₆, Mᵣ = 78.0) by the three-step route below. One mole of benzene gives one mole of the final product when every step works perfectly.

Step	Conversion	Yield of step
1	benzene → nitrobenzene	80.0 %
2	nitrobenzene → phenylamine	75.0 %
3	phenylamine → N-phenylethanamide	60.0 %

(a) Calculate the overall percentage yield of N-phenylethanamide for the three-step route. (2 marks)

(b) Using your overall percentage yield, calculate the mass of benzene the chemist must start with to obtain 24.3 g of N-phenylethanamide. Show your working. (4 marks)

Four colourless liquids, W, X, Y and Z, are known to be (in some order) propanal (CH₃CH₂CHO), propanone (CH₃COCH₃), pentan-3-one (CH₃CH₂COCH₂CH₃) and butan-2-ol (CH₃CH(OH)CH₂CH₃). A student carries out four tests on each liquid. The results are shown below (✓ = positive result with the observation given; ✗ = no reaction).

Test (and positive observation)	W	X	Y	Z
2,4-DNPH — orange precipitate	✓	✓	✓	✗
Tollens' reagent — silver mirror	✓	✗	✗	✗
Warm with I₂ / NaOH(aq) — pale-yellow precipitate	✗	✓	✗	✓

Deduce the identity of each of W, X, Y and Z, giving a reason for each deduction from the test results. (5 marks)

Phenylamine (aniline), C₆H₅NH₂, is manufactured from benzene in two steps and is used to make dyes and drugs.

(a) Give the reagents and conditions for each of the two steps in the conversion of benzene → nitrobenzene → phenylamine, and name the intermediate. (3 marks)

(b) Phenylamine is a much weaker base than ethylamine, CH₃CH₂NH₂. Explain this difference. (2 marks)

A new polymer is made by reacting the two monomers below together. Both monomers are difunctional.

• Monomer 1: pentanedioic acid, HOOC–(CH₂)₃–COOH
• Monomer 2: butane-1,4-diamine, H₂N–(CH₂)₄–NH₂

(a) Name the type of polymer formed and the type of linkage in its chain. (2 marks)

(b) Write the structure of one repeat unit of the polymer (using condensed formulae such as –CO– and –NH–), and name the small molecule eliminated when the linkage forms. (2 marks)

Polymers can be classified as addition polymers or condensation polymers.

(a) State two ways in which a condensation polymer differs from an addition polymer. (2 marks)

(b) A nitrile, CH₃CH₂C≡N, can be converted into the amine CH₃CH₂CH₂NH₂. State the reagent(s) for this conversion. (1 mark)