Question

An organic compound (P) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with Br₂ and KOH forms compound (R) having molecular formula C₆H₇N. Names of P, Q and R respectively are.

• A. Phenylethanoic acid, phenylethanamide, benzamine
• B. Benzoic acid, 4-methylbenzamide, 4-methylaniline
• C. Benzoic acid, benzamide, aniline
• D. Toluic acid, methylbenzamide, 2-methylaniline

Hint:

Recognizing named reactions is a superpower in organic chemistry. Hofmann Bromamide Degradation (Amide → Amine with one less carbon) is very frequently tested. When a molecular formula is given for an aromatic compound, try to fit it to common structures like benzene, aniline, phenol, etc.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:

The question involves a two-step reaction sequence and asks us to identify the compounds P, Q, and R. The most efficient approach is to start from the final product whose molecular formula is given, and then work backward using the named reactions.

Step 2: Detailed Explanation:

Analysis of the final product R (C₆H₇N):

The molecular formula C₆H₇N corresponds to an aromatic amine. Aniline, with the structure C₆H₅NH₂, has the exact molecular formula C₆H₇N.

Therefore, R is Aniline.

Analysis of the reaction Q → R:

The given reagents are bromine (Br₂) and potassium hydroxide (KOH) with heating. This reaction is characteristic of the Hofmann bromamide degradation.

In this reaction, a primary amide (R–CONH₂) is converted into a primary amine (R–NH₂) with the loss of one carbon atom.

Since the product R is aniline (C₆H₅NH₂), the corresponding amide must be benzamide (C₆H₅CONH₂).

The reaction can be written as:
\( C_6H_5CONH_2 + Br_2 + 4KOH \rightarrow C_6H_5NH_2 + K_2CO_3 + 2KBr + 2H_2O \)

Thus, Q is Benzamide.

Analysis of the reaction P → Q:

Compound Q (benzamide) is formed by treating compound P with aqueous ammonia under heating.

Carboxylic acids react with ammonia to form ammonium salts, which upon heating undergo dehydration to give amides.

Therefore, compound P must be the corresponding carboxylic acid, i.e., benzoic acid (C₆H₅COOH).

The reaction is:
\( C_6H_5COOH + NH_3 \rightarrow [C_6H_5COO^-NH_4^+] \xrightarrow{\Delta} C_6H_5CONH_2 + H_2O \)

Hence, P is Benzoic Acid.

Step 3: Final Answer:

The compounds are:
P: Benzoic acid
Q: Benzamide
R: Aniline

This set correctly corresponds to option (C).