Q.Complete the following reactions:

Concept: Reactions of aniline and benzenediazonium chloride — carbylamine reaction, reductive deamination, salt formation/sulphonation, reduction by ethanol, bromination, acetylation, and replacement of the diazonium group by $-NO_2$.

Reasoning:

1. (i) A primary aromatic amine with chloroform and alcoholic KOH undergoes the carbylamine reaction, giving the foul-smelling phenyl isocyanide.
2. (ii) Hypophosphorous acid ($H_3PO_2$) reduces the diazonium salt, replacing $-N_2^+$ by $-H$: benzene.
3. (iii) Being basic, aniline is first protonated by conc. $H_2SO_4$ to give anilinium hydrogensulphate; on strong heating (453–473 K) this salt rearranges to sulphanilic acid (p-aminobenzenesulphonic acid, largely as its zwitterion).
4. (iv) Ethanol acts as a reducing agent toward the diazonium salt: the ring gets $-H$ (benzene) while ethanol is oxidised to ethanal ($CH_3CHO$).
5. (v) The free $-NH_2$ group activates the ring so strongly that bromine water gives 2,4,6-tribromoaniline (white precipitate) instantly.
6. (vi) Acetic anhydride acetylates the nitrogen: acetanilide. …

Concept & Intuition

Aniline ($C_6H_5NH_2$) is a primary aromatic amine. The lone pair on nitrogen makes it a nucleophile, a base, and a strong activator of the ring toward electrophilic substitution. The diazonium salt ($C_6H_5N_2^+Cl^-$) is the opposite kind of species: $-N_2^+$ is an excellent leaving group (departing as $N_2$ gas), so the diazonium group can be replaced by many other groups — the reagent decides which.

Let's go reaction by reaction.

(i) $C_6H_5NH_2 + CHCl_3 + \text{alc. KOH} \rightarrow$

This is the carbylamine reaction, given only by primary amines. The amine attacks dichlorocarbene ($:CCl_2$), generated in situ from chloroform and base; elimination of HCl then gives the isocyanide.

Product: $C_6H_5NC$ (phenyl isocyanide)

(ii) $C_6H_5N_2^+Cl^- + H_3PO_2 + H_2O \rightarrow$

Hypophosphorous acid is a mild reducing agent that replaces the diazonium group by hydrogen — reductive deamination.

Product: $C_6H_6$ (benzene), with $N_2$, $H_3PO_3$ and $HCl$

(iii) $C_6H_5NH_2 + H_2SO_4\ (\text{conc.}) \rightarrow$

Two stages, and both matter:

• Immediately (acid–base reaction): the basic amino group is protonated, giving the salt anilinium hydrogensulphate, $C_6H_5NH_3^+HSO_4^-$.
• On heating at 453–473 K: the salt rearranges (sulphonation "baking" process) to give sulphanilic acid — p-aminobenzenesulphonic acid — which exists largely as its zwitterion $p\text{-}{}^+H_3N\text{-}C_6H_4\text{-}SO_3^-$.

As the question prints no heating condition, the direct product of mixing is the anilinium salt; the full textbook sequence continues to sulphanilic acid on heating, and both stages should be shown.

Product: $C_6H_5NH_3^+HSO_4^-$; on heating (453–473 K) → sulphanilic acid (zwitterion)

(iv) $C_6H_5N_2^+Cl^- + C_2H_5OH \rightarrow$

Ethanol acts as a reducing agent here, not a nucleophile: the diazonium group is replaced by hydrogen while ethanol itself is oxidised to acetaldehyde. The net outcome parallels reaction (ii).

Product: $C_6H_6$ (benzene) $+\ CH_3CHO$ (ethanal)

(v) $C_6H_5NH_2 + Br_2(aq) \rightarrow$

The free $-NH_2$ group activates the ring so powerfully that bromine water substitutes all the ortho and para positions at once — no catalyst needed.

Product: $2,4,6$-tribromoaniline (white precipitate)

(vi) $C_6H_5NH_2 + (CH_3CO)_2O \rightarrow$

Acetic anhydride acylates the nitrogen (N-acylation), producing acetanilide.

Product: $C_6H_5NHCOCH_3$ (acetanilide / N-phenylethanamide) $+\ CH_3COOH$

--- …

Here is one clear solution method for each reaction, using the Reagent–Action–Product (RAP) approach: identify the reagent's chemical role (electrophile, nucleophile, acid, reducing agent, …), predict the site of attack (the amino group or the diazonium group), then write the stable product.

(i) $C_6H_5NH_2 + CHCl_3 + \text{alc. KOH} \rightarrow$

• Reagent Action: Carbylamine reaction (isocyanide test). $CHCl_3$ + strong base generates dichlorocarbene ($:CCl_2$).
• Step: the amine's lone pair attacks the carbene; dehydrohalogenation gives the isocyanide.
• Product: phenyl isocyanide

Final: $\boxed{C_6H_5NC}$

(ii) $C_6H_5N_2^+Cl^- + H_3PO_2 + H_2O \rightarrow$

• Reagent Action: reductive removal of the diazonium group. $H_3PO_2$ is a mild reducing agent.
• Step: $-N_2^+$ leaves as $N_2$; hydrogen takes its place on the ring.
• Product: benzene

Final: $\boxed{C_6H_6}$

(iii) $C_6H_5NH_2 + H_2SO_4\ (\text{conc.}) \rightarrow$

• Reagent Action: acid–base reaction first, then sulphonation on heating.
• Step 1 (immediate): the basic $-NH_2$ is protonated → anilinium hydrogensulphate, $C_6H_5NH_3^+HSO_4^-$.
• Step 2 (on heating at 453–473 K): the salt rearranges to sulphanilic acid ($p$-aminobenzenesulphonic acid, largely zwitterionic).

Final: $\boxed{C_6H_5NH_3^+HSO_4^-}$, which on heating at 453–473 K gives $\boxed{p\text{-}{}^+H_3N\text{-}C_6H_4\text{-}SO_3^-}$ (sulphanilic acid)

(iv) $C_6H_5N_2^+Cl^- + C_2H_5OH \rightarrow$

• Reagent Action: reduction by ethanol. Ethanol is not acting as a nucleophile here — it donates hydrogen to the ring and is itself oxidised.
• Step: the diazonium group is replaced by $-H$; ethanol becomes ethanal.
• Product: benzene + acetaldehyde

Final: $\boxed{C_6H_6 + CH_3CHO}$

(v) $C_6H_5NH_2 + Br_2(aq) \rightarrow$

• Reagent Action: electrophilic aromatic substitution on a very strongly activated ring.
• Step: the free $-NH_2$ directs and activates so strongly that all three ortho/para positions are brominated at once.
• Product: 2,4,6-tribromoaniline (white precipitate)

Final: $\boxed{2,4,6\text{-}Br_3C_6H_2NH_2}$

(vi) $C_6H_5NH_2 + (CH_3CO)_2O \rightarrow$

• Reagent Action: N-acylation (nucleophilic acyl substitution on the anhydride).
• Step: the nitrogen lone pair attacks the carbonyl carbon; one acetyl group transfers to N.
• Product: acetanilide + acetic acid

Final: $\boxed{C_6H_5NHCOCH_3}$

(vii) $C_6H_5N_2^+Cl^- \xrightarrow[(ii)\ NaNO_2/Cu,\ \Delta]{(i)\ HBF_4}$ …

Here are the common mistakes students make with these specific reactions, along with the correct reasoning and how to avoid each.

General Mistake: Confusing Diazonium Salt Reactions

Students often mix up the conditions for replacing the diazonium group ($-N_2^+$) with different groups (Cl, Br, CN, OH, H, NO₂, F). The key is to read every reagent over the arrow — the nucleophile source and the catalyst/conditions together decide the product.

Reaction (i): $C_6H_5NH_2 + CHCl_3 + \text{alc. KOH} \rightarrow$

• Common Mistake: writing a simple substitution product (e.g. $C_6H_5NHCHCl_2$) or a nitrile ($-CN$).
• Why it's wrong: this is the carbylamine reaction (isocyanide test) — base deprotonates $CHCl_3$ to give dichlorocarbene ($:CCl_2$), which attacks the amine's lone pair. The product carries an $-NC$ group, not $-CN$.
• ✓ Correct Product: $C_6H_5NC$ (phenyl isocyanide)

Reaction (ii): $C_6H_5N_2^+Cl^- + H_3PO_2 + H_2O \rightarrow$

• Common Mistake: thinking $H_3PO_2$ is an oxidising agent, or writing a coupling product.
• Why it's wrong: $H_3PO_2$ (hypophosphorous acid) is a reducing agent — it replaces the diazonium group with a hydrogen atom.
• ✓ Correct Product: $C_6H_6$ (benzene)

Reaction (iii): $C_6H_5NH_2 + H_2SO_4\ (\text{conc.}) \rightarrow$

• Common Mistake: jumping straight to a sulphonation product and skipping the salt stage.
• Why it's wrong: the amine is basic, so the immediate reaction is protonation — anilinium hydrogensulphate forms first. Sulphanilic acid is the product of the subsequent heating stage (453–473 K), not of simple mixing.
• ✓ Correct Product: $C_6H_5NH_3^+HSO_4^-$ first; on heating at 453–473 K it rearranges to sulphanilic acid ($p\text{-}{}^+H_3N\text{-}C_6H_4\text{-}SO_3^-$). Show both stages.

Reaction (iv): $C_6H_5N_2^+Cl^- + C_2H_5OH \rightarrow$

• Common Mistake: writing the ether $C_6H_5OC_2H_5$ (phenetole) or a coupling product.
• Why it's wrong: ethanol acts as a reducing agent toward the diazonium salt (milder analogue of $H_3PO_2$), replacing $-N_2^+$ by $-H$ while being oxidised itself to acetaldehyde.
• ✓ Correct Product: $C_6H_6$ (benzene) $+ CH_3CHO$ (ethanal)

Reaction (v): $C_6H_5NH_2 + Br_2(aq) \rightarrow$

• Common Mistake: writing mono-bromination ($o$- or $p$-bromoaniline).
• Why it's wrong: the free $-NH_2$ group activates the ring so strongly that tribromination occurs instantly in bromine water — mono-bromination requires acetylating the amine first.
• ✓ Correct Product: $2,4,6$-tribromoaniline (white precipitate)

Reaction (vi): $C_6H_5NH_2 + (CH_3CO)_2O \rightarrow$

• Common Mistake: writing a ring-acylation (Friedel–Crafts) product.
• Why it's wrong: the nitrogen lone pair is a far better nucleophile than the ring — this is N-acylation, the standard protection reaction for aniline. …