Edexcel A-Level Chemistry: Organic Chemistry II

Propanoyl chloride, CH₃CH₂COCl, reacts vigorously and irreversibly with propan-1-ol, CH₃CH₂CH₂OH, at room temperature to form the ester propyl propanoate together with misty fumes of a gas.

Describe and explain the mechanism of this reaction. In your answer you should:

• name the type of mechanism and identify the nucleophile;
• explain why the carbonyl carbon of the acyl chloride is attacked;
• describe the addition step, the tetrahedral intermediate and the elimination step, naming the species lost;
• give the organic product and identify the misty fumes.

(6 marks)

A chemist prepares N-ethylbutanamide (CH₃CH₂CH₂CONHCH₂CH₃, Mᵣ = 115.0) from butan-1-ol (CH₃CH₂CH₂CH₂OH, Mᵣ = 74.0) by the three-step route below. The route preserves the four-carbon chain, so one mole of butan-1-ol can give one mole of product when every step works perfectly.

Step	Conversion	Reagents and conditions	Yield of step
1	butan-1-ol → butanoic acid	acidified K₂Cr₂O₇, heat under reflux	70.0 %
2	butanoic acid → butanoyl chloride	SOCl₂, room temperature	90.0 %
3	butanoyl chloride → N-ethylbutanamide	excess ethylamine (CH₃CH₂NH₂)	80.0 %

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

(b) Using your overall percentage yield, calculate the mass of butan-1-ol the chemist must start with to obtain 11.5 g of N-ethylbutanamide. Show your working. (4 marks)

The amino acid 2-aminopropanoic acid (alanine), CH₃CH(NH₂)COOH, has two ionisable groups: the carboxyl group (pKₐ ≈ 2.3) and the protonated amino group (pKₐ ≈ 9.7). Its isoelectric point is at pH 6.0.

(a) Draw, using condensed formulae, the structure of the zwitterion of alanine that predominates at its isoelectric point, and explain why this is the dominant form there. (2 marks)

(b) Deduce the predominant structure of alanine in (i) a strongly acidic solution at pH 1, and (ii) a strongly alkaline solution at pH 13. Give a reason for each. (3 marks)

A fragment of a candidate local-anaesthetic drug, labelled G, has the structure:

H₂N–C₆H₄–COOCH₂CH₃ (ethyl 4-aminobenzoate)

It contains an aromatic primary amine group and an ester group on a benzene ring.

(a) G is treated with ethanoyl chloride, CH₃COCl, at room temperature. Identify which functional group of G reacts, name the organic product formed at that group, and state the type of mechanism. (3 marks)

(b) A separate sample of G is heated under reflux with aqueous sodium hydroxide. Identify the functional group that reacts and name the two organic products formed. (2 marks)

A new packaging polymer is made by reacting the two difunctional monomers below.

• Monomer 1: hexanedioic acid, HOOC–(CH₂)₄–COOH
• Monomer 2: propane-1,3-diol, HO–(CH₂)₃–OH

(a) Name the type of polymer formed and the type of linkage in its chain, and name the small molecule eliminated. (2 marks)

(b) Write the structure of one repeat unit of the polymer, using condensed formulae such as –CO–O– and –(CH₂)ₙ–, showing a continuation bond at each end. (2 marks)

Three colourless liquids are an aldehyde (propanal, CH₃CH₂CHO), a ketone (propanone, CH₃COCH₃) and a carboxylic acid (ethanoic acid, CH₃COOH).

(a) Name one reagent that, in a single test, distinguishes the aldehyde from the ketone, and state the observation with each. (2 marks)

(b) State a simple test that identifies the carboxylic acid, giving the reagent and the observation. (1 mark)