Question

Identify 'B' in the following reaction
Ethyl cyanide \(\overset{\text{H}_2\text{O}\,}{\rightarrow }\text{A}\overset{\text{H}_2\text{O / dil HCl}\,}{\rightarrow }\text{B}+\text{NH}_3\)

• A. Ethanoic acid
• B. Ethanamide
• C. Propanoic acid
• D. Propanamide

Hint:

Nitrile partly hydrolyses to an amide (A). Further hydrolysis gives the acid (B) and ammonia.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Nitriles are hydrolysed in two stages. Partial hydrolysis with water gives an amide, and further hydrolysis with dilute acid gives a carboxylic acid with the loss of ammonia.

Step 2: Key Formula or Approach:
\[ \text{R-CN} \xrightarrow{\text{H}_2\text{O}} \text{R-CONH}_2 \xrightarrow{\text{H}_2\text{O/dil HCl}} \text{R-COOH} + \text{NH}_3 \]

Step 3: Detailed Explanation:
Ethyl cyanide is \(\text{CH}_3\text{CH}_2\text{CN}\), propanenitrile.
Step one gives A = \(\text{CH}_3\text{CH}_2\text{CONH}_2\), propanamide.
Step two hydrolyses the amide, which releases \(\text{NH}_3\) as stated in the equation, and forms B = \(\text{CH}_3\text{CH}_2\text{COOH}\), propanoic acid.
Ethanoic acid and ethanamide in (A) and (B) have two carbons, but ethyl cyanide has three carbons including the nitrile carbon. Propanamide in (D) is the intermediate A, not B.

Final Answer:
B is propanoic acid, option (C). \[ \boxed{\text{Propanoic acid (C)}} \]