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A-Level Amines, amides and amino acids

What the A-Level syllabus expects for Amines, amides and amino acids, and how to practise it.

What the syllabus expects

How it's examined

Questions on this topic most often ask you to compare, name, outline. About 3% of the past-paper style questions in Rae's bank for this subject sit in this topic.

Worked examples

Example 1 (3 marks)

Compare the relative basicities of tryptophan and lysine, giving your prediction along with an explanation.

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Lysine has a side chain -(CH2)4-NH2 ending in a primary aliphatic amine. The amino nitrogen has a lone pair that is not delocalised, and the alkyl chain is electron-donating, making the lone pair readily available to accept a proton. Tryptophan's side chain contains an indole ring; the nitrogen lone pair of the indole is part of the aromatic pi system (delocalised into the ring to preserve aromaticity), so it is far less available for protonation. Therefore lysine is more basic than tryptophan.

Example 2 (3 marks)

Propose a synthetic route of three stages that converts acetanilide (C6H5NHCOCH3) into sulfapyridine (H2N-C6H4-SO2NH-(2-pyridyl)), beginning from acetanilide together with chlorosulfuric acid (HOSO2Cl) held at 80 C. For every stage, name the intermediate produced along with the reagents and reaction conditions used.

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Stage 1: Chlorosulfonation. React acetanilide (C6H5NHCOCH3) with chlorosulfuric acid, HOSO2Cl, at 80 C. The -SO2Cl group is introduced para to the acetamido group (which is o/p-directing). Intermediate: 4-acetamidobenzenesulfonyl chloride, CH3CONH-C6H4-SO2Cl. Stage 2: Sulfonamide formation. React this sulfonyl chloride with 2-aminopyridine (2-pyridylamine), usually with a base such as pyridine, at room temperature. The -SO2Cl reacts with the amine to form the sulfonamide. Intermediate: 4-acetamido-N-(2-pyridyl)benzenesulfonamide, CH3CONH-C6H4-SO2NH-(2-pyridyl). Stage 3: Hydrolysis of the amide (removal of the acetyl protecting group). Reflux with dilute aqueous acid (e.g. dilute HCl) - or dilute NaOH - then neutralise. This hydrolyses the CH3CONH- amide to the free -NH2 group. Product: sulfapyridine, H2N-C6H4-SO2NH-(2-pyridyl).

Example 3 (2 marks)

In nicotine, a pyridine ring (with ring nitrogen labelled N-1) is joined to an N-methylpyrrolidine ring (with ring nitrogen labelled N-2). The pyridine nitrogen N-1 has a pKb of 10.88. Compare how basic the nitrogen atom in pyrrole is relative to N-1 in nicotine, and give reasons for your comparison.

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The nitrogen in pyrrole is much less basic than N-1 (the pyridine-type nitrogen) in nicotine. In pyrrole, the nitrogen's lone pair is delocalised into the aromatic ring: it forms part of the 6 pi-electron aromatic system. Because this lone pair is tied up in maintaining aromaticity, it is not available to accept a proton, so pyrrole is a very weak base (essentially non-basic). In nicotine, N-1 is a pyridine-type nitrogen: its lone pair sits in an sp2 orbital in the plane of the ring and is NOT part of the aromatic pi system, so it is available to accept a proton (pKb 10.88). Therefore pyrrole's nitrogen is less basic than N-1 in nicotine.

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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