Question

Compound 'P' undergoes the following sequence of reactions : (i) NH₃ (ii) $\Delta$ $\rightarrow$ Q (i) KOH, Br₂ (ii) CHCl₃, KOH (alc), $\Delta$ $\rightarrow$ NC-CH₃. 'P' is : 

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

The Hofmann Bromamide reaction always removes the carbonyl carbon from the amide. If your final amine has $n$ carbons, your starting amide (and acid) must have $n+1$ carbons.

Answer 1

The Correct Option is D

Let's analyze the given reaction sequence and determine the structure of compound 'P'.

1. Compound 'P' first reacts with \(NH_3\), and then under the influence of heat \((\Delta)\) forms compound 'Q'. This indicates that 'P' undergoes partial conversion with ammonia, likely forming an amide.
2. Next, 'Q' is treated with \(KOH\) and \(Br_2\), followed by \(CHCl_3\) and alcoholic \(KOH\) with heat. This sequence suggests a Hoffman degradation, which produces a primary amine with a reduced carbon chain (loss of one carbon from the amide carbon).
3. The final product is NC-CH₃. This implies that the amide initially involved in the reaction must have had a methylene adjacent to the carbonyl group, which may have originally been a carboxylic acid.

Based on this reasoning, the initial structure of 'P' should be a carboxylic acid. Examining the given options:

1. Option 1: An amide - not likely since it cannot form an amide with ammonia.
2. Option 2: An aldehyde - does not fit the reaction sequence.
3. Option 3: A ketone - unsuitable for forming an amide.
4. Option 4: A carboxylic acid - fits the criteria.

Therefore, the correct answer is Option 4.
Thus, the compound 'P' is a carboxylic acid, which successfully undergoes a series of reactions to eventually form an alkyl cyanide.