When formaldehyde, dimethylamine, and acetophenone are mixed under acidic conditions, which compound is formed as the Mannich product?

hard MCQ 3 marks Class 12 · Chemistry · Amines

β‑Phenyl‑2‑dimethylamino‑propanone $(C H_{3})_{2} N ‑ C H_{2} ‑ CO ‑ C_{6} H_{5}$
N,N‑dimethyl‑2‑phenylacetaldimine $(C H_{3})_{2} N = CH ‑ C_{6} H_{5}$
N,N‑dimethyl‑2‑phenylethylamine $(C H_{3})_{2} N ‑ C H_{2} ‑ C H_{2} ‑ C_{6} H_{5}$
2‑Phenyl‑1,3‑dioxolane

Answer

β‑Phenyl‑2‑dimethylamino‑propanone

(C H_{3})_{2} N ‑ C H_{2} ‑ CO ‑ C_{6} H_{5}

The correct answer is β‑Phenyl‑2‑dimethylamino‑propanone $(C H_{3})_{2} N ‑ C H_{2} ‑ CO ‑ C_{6} H_{5}$ — Step 5: Compare with the given options; only option (a) matches the Mannich product.

Step-by-step solution

Step 1: Formaldehyde reacts with dimethylamine to give the iminium ion $(C H_{3})_{2} N = C H_{2}^{+}$ .
Step 2: The enol form of acetophenone (generated under acidic conditions) attacks the electrophilic carbon of the iminium ion.
Step 3: This carbon‑carbon bond formation yields a β‑aminocarbonyl compound – the classic Mannich product.
Step 4: The structure of the product is $(C H_{3})_{2} N ‑ C H_{2} ‑ CO ‑ C_{6} H_{5}$ , i.e., β‑dimethylamino‑propiophenone.
Step 5: Compare with the given options; only option (a) matches the Mannich product.

Common mistake: Mistaking the product for a Schiff base (option b) which arises from aldehyde + amine without the carbonyl compound.

Concept tested: Mannich reaction of amines

Published 7 September 2026 · How this solution was prepared and checked

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