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A-Level Carboxylic acids, acyl chlorides and esters

What the A-Level syllabus expects for Carboxylic acids, acyl chlorides and esters, and how to practise it.

What the syllabus expects

How it's examined

Questions on this topic most often ask you to explain, suggest. About 5% of the past-paper style questions in Rae's bank for this subject sit in this topic.

Worked examples

Example 1 (3 marks)

Draw the structural formulae of the organic products formed when propanoyl chloride reacts with tryptophan (trp).

Show the worked answer

Tryptophan is (indol-3-yl)-CH2-CH(NH2)-COOH. Propanoyl chloride (CH3CH2COCl) is an acyl chloride: it acylates the primary amine -NH2 to give an amide, releasing HCl. The -COOH group and (under these conditions) the weakly nucleophilic indole N-H do not react. The organic product is the N-acyl amino acid.

Example 2 (4 marks)

Acetylcholine, [CH3COOCH2CH2N(CH3)3]+, is broken down by the enzyme acetylcholinesterase. At the active site a serine residue (treated as a surface-attached -OH) and a histidine residue (treated as a surface-attached imidazole ring) act as the binding points for acetylcholine. Choline, [(CH3)3NCH2CH2OH]+, is one of the products. Propose a mechanism for the breakdown of acetylcholine that shows how choline is formed.

Show the worked answer

The reaction is nucleophilic acyl substitution (transesterification) of the ester carbonyl of acetylcholine, CH3-CO-O-CH2CH2N(CH3)3+. 1. The histidine imidazole nitrogen acts as a base and removes the H from the serine -OH, making the serine oxygen a strong nucleophile. 2. The serine oxygen lone pair attacks the carbonyl carbon of the acetyl group (curly arrow O to C); the C=O pi electrons move onto the oxygen, giving a negatively charged tetrahedral intermediate. 3. The tetrahedral intermediate collapses: the C=O reforms and the C-O bond to the choline oxygen breaks (curly arrow), expelling the choline alkoxide -O-CH2CH2N(CH3)3+. 4. This alkoxide is protonated (by the protonated histidine) to give choline, [(CH3)3NCH2CH2OH]+, leaving the acetyl group esterified onto the serine (acetyl-enzyme).

Example 3 (2 marks)

Propose one procedure that could be performed during the preparation, and explain why it works, to eliminate salicylic acid that is contaminating the aspirin crystals.

Show the worked answer

Purify the crude aspirin by recrystallisation. Dissolve the impure aspirin crystals in the minimum volume of hot solvent (e.g. hot ethanol or a hot ethanol/water mixture) so that everything just dissolves, then allow the solution to cool slowly and filter off the crystals that form. This works because salicylic acid is present only as a small amount of impurity: on cooling, the aspirin (the major component) becomes saturated and crystallises out, while the small amount of more soluble salicylic acid remains dissolved in the cold mother liquor and is removed with the filtrate.

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More A-Level H2 Chemistry topics

The make-up of the atom and how its electrons are arranged · How atoms bond and how that governs a substance's behaviour · Ideal gases and working with gas mixtures · Competing definitions of acids and bases · Trends in the elements across a period and down a group · The mole and reacting-quantity calculations · all of A-Level H2 Chemistry