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

Step 1 (general pattern): Ketones undergo condensation with nitrogen nucleophiles of the type \(H_2N-G\): an unstable carbinolamine forms first and then dehydrates to a \(C=N-G\) linkage. Which product forms depends on \(G\).

Step 2 (reaction i): With \(G = OH\) (hydroxylamine) the C=N-OH function is an oxime. From cyclopentanone, [A] is cyclopentanone oxime.

Step 3 (reaction ii): With \(G = NH-C_6H_3(NO_2)_2\) (2,4-dinitrophenylhydrazine) the product is a 2,4-dinitrophenylhydrazone. From cyclohexanone, [B] is cyclohexanone 2,4-dinitrophenylhydrazone, the coloured derivative used to confirm a carbonyl compound.

Step 4 (reaction iii): Phthalic acid has two neighbouring \(-COOH\) groups. Ammonia first neutralises them to the ammonium salt [A] (ammonium phthalate). Heating drives off water so the two acyl groups close a ring with one nitrogen, yielding the cyclic imide phthalimide [B].
\(\boxed{A = \text{ammonium phthalate},\; B = \text{phthalimide}}\)