Question

Observe the following reaction sequence:
\text{(P) }\xrightarrow{\text{NH}_3, \Delta} \text{(Q)} \xrightarrow{\text{Br}_2, \text{KOH}} \text{(R)}
Which of the following is the correct structure for P, Q, and R?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

In Hofmann degradation, an amide is converted into a primary amine with one less carbon atom.

Answer 1

The Correct Option is A

Let's analyze the given reaction sequence and determine the structures for P, Q, and R.

The reaction sequence is:

\(P \xrightarrow{\text{NH}_3, \Delta} Q \xrightarrow{\text{Br}_2, \text{KOH}} R\)

Step 1: Identify Compound P

The reaction starts with P, which undergoes a reaction with \(\text{NH}_3\) and heat.

This step commonly involves converting a carboxylic acid (benzoic acid) to an amide. Therefore, P is likely benzoic acid: \(\text{C}_6\text{H}_5\text{COOH}\).

Step 2: Formation of Compound Q

When benzoic acid reacts with ammonia and heat, it forms benzamide:

\(\text{C}_6\text{H}_5\text{COOH} + \text{NH}_3 \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{CONH}_2\)

So, Q is benzamide.

Step 3: Formation of Compound R

The Hofmann Bromamide reaction (using \(\text{Br}_2, \text{KOH}\)) converts an amide to a primary amine.

Thus, applying this to benzamide, we obtain aniline as the final product:

\(\text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow{\text{Br}_2, \text{KOH}} \text{C}_6\text{H}_5\text{NH}_2\)

Therefore, R is aniline.

Conclusion

Based on the reactions, the structures of P, Q, and R are:

1. P = Benzoic acid
2. Q = Benzamide
3. R = Aniline

The correct option that matches these structures is Option 1.