Q.An aromatic compound 'A' on treatment with aqueous ammonia and heating forms compound 'B' which on heating with $Br_2$ and KOH forms a compound 'C' of molecular formula $C_6H_7N$. Write the structures and IUPAC names of compounds A, B and C.

Concept: Formation of an amide from a carboxylic acid, followed by the Hofmann bromamide degradation.

Step 1 – Identify compound C from its molecular formula

$C_6H_7N$ matches aniline, $C_6H_5NH_2$ (6 carbons, 1 nitrogen, degree of unsaturation 4 = one benzene ring).

Step 2 – Work backwards from C to B

C (aniline) comes from B via $\ce{Br2/KOH}$ — the Hofmann bromamide degradation, which converts a primary amide to an amine with one fewer carbon. So B must be the corresponding amide: benzamide, $C_6H_5CONH_2$.

Step 3 – Work backwards to A …

The Concept: Nucleophilic Substitution Reactions in Aromatic Systems

The problem hinges on two key transformations:

1. Conversion of a carboxylic acid to an amide — here, an aromatic carboxylic acid reacts with aqueous ammonia under heat. This is a straightforward nucleophilic acyl substitution: ammonia attacks the carbonyl carbon, displacing the hydroxyl group.

2. The Hoffmann bromamide degradation — an amide treated with bromine and a strong base (KOH) loses the carbonyl carbon (it ends up as carbonate, $K_2CO_3$, in the alkaline medium), and the nitrogen atom ends up bonded to the alkyl/aryl group that was originally attached to the carbonyl. The net result: $R-CONH_2 \rightarrow R-NH_2$, with the loss of one carbon atom.

The final compound C has molecular formula $C_6H_7N$. That formula is characteristic of aniline ($C_6H_5NH_2$). Counting: benzene ring ($C_6H_5$) plus $NH_2$ gives $C_6H_7N$ — exactly right.

Since C comes from B via Hoffmann degradation, B must be the amide of the same aromatic carboxylic acid. That means B is benzamide, $C_6H_5CONH_2$.

And since B comes from A by reaction with aqueous ammonia, A must be the corresponding carboxylic acid: benzoic acid, $C_6H_5COOH$.

Let's verify the molecular formula logic:

• Benzoic acid ($C_7H_6O_2$) + $NH_3$ → benzamide ($C_7H_7NO$) + $H_2O$
• Benzamide + $Br_2$ + 4 KOH → aniline ($C_6H_7N$) + $K_2CO_3$ + 2 KBr + 2 $H_2O$

The carbon count drops from 7 to 6 — the carbonyl carbon leaves as carbonate ($K_2CO_3$) in the Hoffmann rearrangement. Everything fits.

Step-by-Step Solution

1. Identify compound C from its molecular formula

The formula $C_6H_7N$ has six carbons and one nitrogen. For an aromatic compound, the simplest possibility is aniline — a benzene ring with an amino group. The degree of unsaturation: for $C_6H_7N$, using the formula $DU = \frac{2C + 2 + N - H}{2} = \frac{12 + 2 + 1 - 7}{2} = 4$. Four degrees of unsaturation match a benzene ring (which has four $\pi$ bonds/rings). So C is aniline.

2. Work backwards from C to B …

Here is the clear, step-by-step solution using the Retrosynthetic Analysis method (working backwards from the given molecular formula).

Step 1: Identify Compound C from its molecular formula

The molecular formula of C is $C_6H_7N$.

• Degree of Unsaturation (DU):

• Interpretation: 4 degrees of unsaturation strongly suggests an aromatic ring.
• Functional group: a benzene ring plus one nitrogen with formula $C_6H_7N$ points to aniline ($C_6H_5NH_2$).

Therefore, Compound C is Aniline.

Step 2: Work backwards from C to B (Hoffmann Bromamide Degradation)

The reaction converting B to C is:

• Name the reaction: $Br_2$ with KOH acting on a compound to give an amine is the Hoffmann bromamide degradation.
• What it does: it converts a primary amide ($R\text{-}CONH_2$) into a primary amine ($R\text{-}NH_2$) with one less carbon atom.
• Applying it here: since C is aniline ($C_6H_5NH_2$), the amide B must have been benzamide ($C_6H_5CONH_2$).

Therefore, Compound B is Benzamide.

Step 3: Work backwards from B to A (acid + aqueous ammonia, heat)

The reaction converting A to B is:

• Read the reagents carefully: "aqueous ammonia and heating" is the classic two-stage conversion of a carboxylic acid to its amide — the acid first neutralises ammonia to give the ammonium salt, and strong heating then dehydrates the salt to the amide: $C_6H_5COOH + NH_3 \rightarrow C_6H_5COO^-NH_4^+ \xrightarrow{\Delta} C_6H_5CONH_2 + H_2O$ …

Here are the mistakes students actually make on this A/B/C identification problem, and how to avoid each. The correct chain, for reference:

✗ Mistake 1: Taking A to be benzoyl chloride

• Why students do it: benzoyl chloride does give benzamide with ammonia, so it looks like a valid A.
• Why it misses the clue: an acid chloride reacts with ammonia vigorously without any heating — the question's specific wording, "aqueous ammonia and heating", describes the carboxylic-acid route, where heat is essential to dehydrate the intermediate ammonium salt ($C_6H_5COO^-NH_4^+$) to the amide. The standard answer for this sequence is benzoic acid.
• ✓ Correct: A = benzoic acid; $C_6H_5COOH + NH_3 \rightarrow C_6H_5COO^-NH_4^+ \xrightarrow{\Delta} C_6H_5CONH_2 + H_2O$.

✗ Mistake 2: Reading "aqueous ammonia + heat" as ammonolysis of an aryl halide (chlorobenzene → aniline)

• Why it's wrong: the carbon–chlorine bond of chlorobenzene does not undergo nucleophilic substitution with aqueous ammonia on heating — aryl halides need extremely harsh, specialised conditions for amination, far beyond anything this question implies. Worse, this reading leaves the second step unexplainable: aniline with $Br_2/KOH$ is not a Hoffmann degradation (there is no amide), so no compound of formula $C_6H_7N$ with one fewer carbon could result.
• ✓ Correct: work backwards from the $Br_2/KOH$ step instead — it only makes sense if B is a primary amide.

✗ Mistake 3: Treating $Br_2$ + KOH as ring bromination

• Why it's wrong: count the atoms. C ($C_6H_7N$) contains no bromine and has one carbon fewer than benzamide ($C_7H_7NO$). Ring bromination would add Br and keep all seven carbons. The reagent pair $Br_2 + KOH$ acting on a primary amide is the Hoffmann bromamide degradation — the carbonyl carbon leaves as carbonate ($K_2CO_3$), shortening the skeleton by one carbon. …