Q.(a) Distinguish, with the help of a chemical test, between ethylamine and dimethylamine. [1 mark]
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Start your 14-day free trial to unlock the full solution →(a) The carbylamine test distinguishes 1 degree amines (positive) from 2 degree amines (negative). (b) Gabriel synthesis needs an SN2-reactive alkyl halide, which aryl halides are not, so it cannot make aromatic primary amines like aniline. (c) Hofmann bromamide degradation converts an amide to an amine with one fewer carbon (A); aniline's ring is so activated that bromine water brominates all three free ortho/para positions at once (B).
(a) Distinguishing ethylamine (1 degree amine) from dimethylamine (2 degree amine) - Carbylamine test:
Warm each compound with chloroform and alcoholic KOH.
- Ethylamine (a primary amine) reacts to give ethyl isocyanide, which has an intensely unpleasant/offensive smell: C2H5NH2 + CHCl3 + 3KOH -> C2H5NC + 3KCl + 3H2O (positive test).
- Dimethylamine (a secondary amine) does NOT have an N-H available in the way required and gives NO reaction / no foul smell (negative test). This clean positive/negative distinction identifies which sample is the primary amine.
(b) Why aniline cannot be prepared by the Gabriel phthalimide synthesis:
The Gabriel synthesis requires an SN2 reaction between potassium phthalimide and an ALKYL halide (R-X, where the carbon bearing the leaving group is sp3 and accessible to nucleophilic attack). Aryl halides (like chlorobenzene/bromobenzene, needed to eventually give aniline) do NOT undergo nucleophilic substitution under these conditions - the C-X bond in an aryl halide has partial double-bond character (due to resonance donation of the halogen lone pair into the ring) and the sp2 carbon is not accessible for backside attack, so no substitution occurs. Hence aniline (an aromatic primary amine) cannot be synthesized by this route; the Gabriel method only works for aliphatic (alkyl) primary amines.
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