Q.A colourless substance 'A' ($C_6H_7N$) is sparingly soluble in water and gives a water soluble compound 'B' on treating with mineral acid. On reacting with $CHCl_3$ and alcoholic potash 'A' produces an obnoxious smell due to the formation of compound 'C'. Reaction of 'A' with benzenesulphonyl chloride gives compound 'D' which is soluble in alkali. With $NaNO_2$ and HCl, 'A' forms compound 'E' which reacts with phenol in alkaline medium to give an orange dye 'F'. Identify compounds 'A' to 'F'.

Concept: Oxidation Reactions — The key is recognising that compound A is an aromatic primary amine (aniline, $C_6H_5NH_2$), which fits the molecular formula $C_6H_7N$ and explains all the reactions.

Reasoning steps:

1. A ($C_6H_7N$) is sparingly soluble in water but dissolves in mineral acid → basic nature, likely an aromatic amine. With $CHCl_3$ + alcoholic KOH, it gives an obnoxious smell (carbylamine reaction) → confirms a primary amine, forming isocyanide C ($C_6H_5NC$).

2. A reacts with benzenesulphonyl chloride to give D soluble in alkali → Hinsberg test positive for a primary amine; D is $C_6H_5NHSO_2C_6H_5$ (sulphonamide with a free H, hence alkali-soluble). …

Concept and Intuition

The molecular formula C₆H₇N immediately suggests an aromatic amine — specifically aniline (C₆H₅NH₂), because six carbons form the benzene ring, five hydrogens on the ring plus two on the amino group give seven hydrogens, and one nitrogen fits perfectly. The reactions described are classic tests for a primary aromatic amine:

• Sparingly soluble in water – aniline is only slightly soluble.
• Forms a water‑soluble salt with mineral acid – the basic –NH₂ group gets protonated.
• Carbylamine reaction – with CHCl₃ and alcoholic KOH, a primary amine gives an isocyanide (foul smell).
• Hinsberg test – with benzenesulphonyl chloride (C₆H₅SO₂Cl), a primary amine yields a sulphonamide that is soluble in alkali (because the –SO₂–NH– group still has an acidic hydrogen).
• Diazotisation and coupling – with NaNO₂/HCl at low temperature, a primary aromatic amine forms a diazonium salt, which couples with phenol in alkaline medium to produce an orange azo dye.

Every clue points to aniline. Let’s walk through each step.

Step‑by‑Step Reasoning

1. Identify compound A from molecular formula and solubility.

C₆H₇N fits aniline (C₆H₅NH₂). Aniline is a colourless liquid (often turns brown on exposure to air) and is sparingly soluble in water because the hydrophobic benzene ring dominates over the polar –NH₂ group.

Thus A = aniline.

2. Reaction with mineral acid → compound B.

Aniline is basic. When treated with a mineral acid like HCl, it forms a salt:

This salt (anilinium chloride) is highly soluble in water.

So B = anilinium chloride (or any aniline salt, depending on the acid used).

3. Carbylamine reaction → compound C (obnoxious smell).

Primary amines react with chloroform (CHCl₃) and alcoholic KOH to form isocyanides (carbylamines), which have a penetrating, foul odour.

The product is phenyl isocyanide.

Thus C = phenyl isocyanide.

4. Hinsberg test → compound D (soluble in alkali).

Benzenesulphonyl chloride (C₆H₅SO₂Cl) reacts with a primary amine to form a sulphonamide that still has one acidic hydrogen on the nitrogen:

This N‑substituted benzenesulphonamide is soluble in aqueous alkali because the –SO₂–NH– group can lose a proton:

So D = N‑phenylbenzenesulphonamide (or benzenesulphonanilide).

5. Diazotisation → compound E. …

Method: Reaction-Based Deduction with Functional Group Analysis

This method uses characteristic chemical tests to identify an unknown organic compound by matching its reactions to known functional group behaviours.

Step 1 — Analyse molecular formula of ‘A’

• Formula: $C_6H_7N$
• Degree of unsaturation:

• Suggests an aromatic amine (aniline derivative).

Step 2 — Solubility behaviour

• ‘A’ is sparingly soluble in water → typical of aromatic compounds.
• On adding mineral acid, gives water-soluble ‘B’ → salt formation.

• So ‘A’ is a primary aromatic amine.

Step 3 — Carbylamine reaction (with $\text{CHCl}_3$ + alcoholic KOH)

• ‘A’ gives obnoxious smell → characteristic of isocyanide formation.
• Only primary amines give this test:

• ‘C’ is an isocyanide (carbylamine).

Step 4 — Hinsberg test (with benzenesulphonyl chloride)

• ‘D’ is soluble in alkali → confirms primary amine.
  • Primary amine: forms sulphonamide with acidic H, soluble in NaOH.
  • Secondary amine: forms sulphonamide without acidic H, insoluble.
  • Tertiary amine: no reaction.

Step 5 — Diazotisation and coupling

• ‘A’ with $\text{NaNO}_2$ + HCl at 0–5°C gives diazonium salt ‘E’.
• ‘E’ couples with phenol in alkaline medium to give orange azo dye ‘F’.
• Orange colour suggests aniline as the starting amine. …

Common Mistakes & How to Avoid Them

1. Mistake: Misidentifying the Obnoxious Smell Reaction (Carbylamine Test)

The Error: Students often confuse the carbylamine reaction (isocyanide formation) with the chloroform–NaOH reaction for aldehydes (haloform reaction). They may also forget that only primary amines give this test.

How to Avoid:

• Remember: Carbylamine reaction = $CHCl_3 + KOH$ (alcoholic) + primary amine → foul-smelling isocyanide.
• The formula $C_6H_7N$ has one nitrogen — check if it's primary, secondary, or tertiary.
• Key clue: The obnoxious smell is a dead giveaway for isocyanide formation → A must be a primary amine.

2. Mistake: Overlooking the Solubility of Compound D in Alkali

The Error: Students identify D as a sulphonamide but forget that only primary amine sulphonamides are soluble in alkali (due to acidic N–H). Secondary amine sulphonamides are insoluble.

How to Avoid:

• Hinsberg test rule:
  • Primary amine + benzenesulphonyl chloride → sulphonamide soluble in alkali (has acidic H on N).
  • Secondary amine → sulphonamide insoluble in alkali.
• Since D is alkali-soluble, A must be a primary amine.

3. Mistake: Confusing the Diazotisation and Coupling Reaction

The Error: Students may think E is a nitroso compound or miss that the orange dye is an azo dye formed by coupling with phenol.

How to Avoid:

• $NaNO_2 + HCl$ at low temperature (0–5°C) with a primary aromatic amine → diazonium salt (E).
• Diazonium salt + phenol in alkaline medium → azo dye (orange/red).
• Remember: The colour (orange) and the word "dye" point to an azo compound.

4. Mistake: Ignoring the Molecular Formula $C_6H_7N$

The Error: Students may propose aliphatic amines or aniline derivatives that don't match the formula.

How to Avoid:

• Calculate degree of unsaturation:
  • $C_6H_7N$: For $C_nH_mN$, unsaturation = $\frac{2n + 2 - (m - 1)}{2} = \frac{12 + 2 - 6}{2} = 4$
  • 4 degrees of unsaturation → likely an aromatic ring (benzene ring = 4 unsaturation).
• So A is an aromatic primary amine — the simplest is aniline ($C_6H_5NH_2$), but that's $C_6H_7N$? Wait: aniline = $C_6H_5NH_2$ = $C_6H_7N$ — yes!

5. Mistake: Forgetting That Aniline Is Sparingly Soluble in Water

The Error: Students may think aniline is water-soluble because it has an amino group.

How to Avoid:

• Aniline is sparingly soluble in water due to the large hydrophobic benzene ring.
• On adding mineral acid, it forms anilinium salt ($C_6H_5NH_3^+$), which is water-soluble (B).

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