Question

Identify the product of the following reactions:
i) cyclopentanone \(+\) \(HO-NH_2 \xrightarrow{H^+}\) [A] (1)
ii) cyclohexanone \(+\) 2,4-dinitrophenylhydrazine \((H_2N-NH-C_6H_3(NO_2)_2) \rightarrow\) [B] (1)
iii) benzene-1,2-dicarboxylic acid (phthalic acid) \(+\) \(NH_3 \rightarrow\) [A] \(\xrightarrow[-H_2O]{\Delta}\) [B] (2)

Hint:

Ketone + \(NH_2OH\) gives an oxime; ketone + 2,4-DNP gives a 2,4-dinitrophenylhydrazone; phthalic acid + \(NH_3\) then heat gives phthalimide via the ammonium salt.

Answer 1

Concept: The carbonyl group of aldehydes and ketones reacts with ammonia derivatives \(H_2N-Z\) by nucleophilic addition followed by loss of water, giving \(>C=N-Z\) products; carboxylic acids form amides with ammonia.

i) Cyclopentanone + hydroxylamine (\(HO-NH_2\)), H+:
Hydroxylamine adds across the C=O and water is eliminated, giving the oxime.
Product [A] = cyclopentanone oxime, \(C_5H_8=N-OH\) (cyclopentanone oxime).
\[\text{cyclopentanone} + H_2N-OH \rightarrow \text{cyclopentanone oxime} + H_2O\]

ii) Cyclohexanone + 2,4-dinitrophenylhydrazine (2,4-DNP):
The reagent adds to C=O and loses water to give the hydrazone.
Product [B] = cyclohexanone 2,4-dinitrophenylhydrazone, \(C_6H_{10}=N-NH-C_6H_3(NO_2)_2\) (an orange-yellow solid, a test for the carbonyl group).

iii) Phthalic acid (benzene-1,2-dicarboxylic acid) + \(NH_3\), then heat:
Step 1: The two \(-COOH\) groups react with ammonia to form the ammonium salt.
Product [A] = ammonium phthalate (diammonium salt of phthalic acid).
Step 2: On strong heating this salt loses water (and \(NH_3\)) and cyclises through the ortho carboxyl groups.
Product [B] = phthalimide (benzene-1,2-dicarboximide), a five-membered cyclic imide fused to the benzene ring.
\[\boxed{[A]=\text{ammonium phthalate},\quad [B]=\text{phthalimide}}\]