Q.Convert

Concept: Aromatic Synthesis via Diazonium Salts — the diazonium group can either be swapped for another substituent at the very same ring position, or deleted (replaced by H), depending on the reagent chosen.

1. 3-Methylaniline → 3-Nitrotoluene In 3-methylaniline the $-NH_2$ (C-1) and $-CH_3$ (C-3) are already meta to each other — exactly the relationship the product needs between $-NO_2$ and $-CH_3$. So the amino group only needs to be replaced in place, not removed: diazotise with $NaNO_2/HCl$ at 273–278 K, convert the diazonium chloride into the stable diazonium fluoroborate with $HBF_4$, then heat the fluoroborate with aqueous $NaNO_2$ in the presence of copper ($\Delta$). The $-N_2^+$ group is replaced directly by $-NO_2$ at the same carbon, giving 3-nitrotoluene.
2. Aniline → 1,3,5-Tribromobenzene …

The Core Concept: The Diazonium Route

The amino group ($-NH_2$) is a powerful ortho/para director and a strong activator. This makes it excellent for placing substituents at specific positions — but it also means aniline cannot be nitrated or mono-brominated cleanly. Diazonium chemistry resolves this: the amino group does its directing work (or simply marks a position), and is then either converted into the substituent needed at that very carbon, or removed entirely.

(i) 3-Methylaniline → 3-Nitrotoluene

In 3-methylaniline, the methyl group sits at C-3 relative to the amino group at C-1 — they are already meta to each other, which is exactly the $-NO_2$/$-CH_3$ relationship the target needs. So the cleanest route is not to nitrate the ring at all: it is to convert the existing amino group directly into a nitro group at the position it already occupies.

1. Diazotise the amine.

Treat 3-methylaniline with $NaNO_2$ and dilute HCl at 273–278 K:

2. Convert the diazonium chloride into the fluoroborate.

Treat the diazonium salt with fluoroboric acid; the stable, sparingly soluble diazonium fluoroborate separates:

3. Replace $-N_2^+$ with $-NO_2$.

Heat the diazonium fluoroborate with aqueous $NaNO_2$ in the presence of copper — nitrite displaces the diazonium group:

Because the substitution happens at the same carbon the amino group occupied, the methyl group never moves and the new nitro group inherits the meta relationship. The product is 3-nitrotoluene.

(ii) Aniline → 1,3,5-Tribromobenzene

The target is a benzene ring with three bromines in a 1,3,5 pattern and no nitrogen. The free amino group is the perfect tool: it activates the ring so strongly that bromine water substitutes all three ortho/para positions almost instantly.

1. Brominate free aniline directly.

Add excess bromine water at room temperature — the reaction is immediate and gives a white precipitate:

$C_6H_5NH_2 + 3Br_2 \xrightarrow{H_2O} 2,4,6\text{-}Br_3C_6H_2NH_2 + 3HBr$ …

Aromatic Synthesis Route — Two Clear Methods

(i) 3-Methylaniline → 3-Nitrotoluene

Method: Diazotisation → diazonium fluoroborate ($HBF_4$) → replacement of $-N_2^+$ by $-NO_2$ ($NaNO_2/Cu$, $\Delta$)

Why this works:

In 3-methylaniline the amino group (C-1) and the methyl group (C-3) are already meta to each other — the exact geometry the product needs between $-NO_2$ and $-CH_3$. So no new substitution on the ring is required at all: the amino group is converted in place into a nitro group. No nitration step, no protection step, no isomer problem.

Steps:

1. Diazotise Treat 3-methylaniline with $NaNO_2$ + dilute HCl at 273–278 K to form the diazonium salt:

2. Form the diazonium fluoroborate Treat the diazonium chloride with $HBF_4$; the stable fluoroborate separates:

3. Replace $-N_2^+$ by $-NO_2$ Heat the fluoroborate with aqueous $NaNO_2$ in the presence of copper. Nitrite displaces the diazonium group at the same carbon:

The methyl group never moves, so the product keeps the meta relationship.

Final product: $\boxed{\text{3-nitrotoluene}}$

(ii) Aniline → 1,3,5-Tribromobenzene

Method: Direct tribromination → Diazotisation → Reduction (replacement of $-N_2^+$ by $-H$)

Why this works:

The free $-NH_2$ group is strongly activating and ortho/para-directing, so bromine water substitutes all three ortho/para positions of aniline at once — giving 2,4,6-tribromoaniline directly. Deleting the amino group afterwards leaves the three bromines in the 1,3,5 pattern.

Steps:

1. Brominate aniline directly Add excess bromine water at room temperature; the reaction is instantaneous:

Product: 2,4,6-tribromoaniline (white precipitate). No acetylation/protection step is used — protection would moderate the ring and stop bromination at the mono stage.

2. Diazotise $NaNO_2$ + dilute HCl at 273–278 K → the diazonium salt. …

These two conversions test whether you can tell when to swap the amino group for another substituent and when to delete it — and when a protecting group helps versus when it ruins the synthesis.

(i) 3-Methylaniline → 3-Nitrotoluene

✗ Common Mistake 1: Direct nitration of 3-methylaniline

Why it's wrong:

The free $-NH_2$ group is a strongly activating ortho/para director, so nitration would go ortho/para to the amino group — not where the target needs it — and the amine itself is prone to oxidation in the nitrating mixture. More fundamentally, no nitration is needed at all: the product's $-NO_2$ belongs at the very carbon the $-NH_2$ already occupies.

✗ Common Mistake 2: The protect–nitrate–deaminate detour

Why it's wrong — check the geometry:

Acetylating to 3-methylacetanilide and then nitrating does not save this route, because $-NHCOCH_3$ is an ortho/para director (the textbook's own bromination of acetanilide gives the para product as the major one — the amide group does not direct meta). Nitration would therefore land ortho/para to the acetamido group at C-1 — e.g. at C-4, which is ortho to the methyl at C-3. After hydrolysis and deamination the nitro group would sit ortho to the methyl group (a 2-nitrotoluene-type product), not meta. The detour is longer and delivers the wrong isomer.

✗ Common Mistake 3: Using $H_3PO_2$ on the diazonium salt

Why it's wrong:

$H_3PO_2$ replaces $-N_2^+$ with plain $-H$ — that deletes the nitrogen entirely and just gives toluene. In this part the nitrogen position must become a nitro group, not vanish.

✓ Correct Route

1. Diazotise 3-methylaniline ($NaNO_2$/HCl, 273–278 K).
2. Form the diazonium fluoroborate with $HBF_4$.
3. Heat with $NaNO_2/Cu$, $\Delta$ — the diazonium group is replaced by $-NO_2$ at the same carbon → 3-nitrotoluene. The original meta relationship between $-CH_3$ and the nitrogen position is preserved automatically.

(ii) Aniline → 1,3,5-Tribromobenzene

✗ Common Mistake 1: Acetylating (protecting) the amine before bromination

Why it's wrong:

Protection is the tool for mono-bromination: the acetamido group moderates the ring precisely so that bromination stops at one position (mainly para) — that is why acetanilide, not aniline, is brominated when p-bromoaniline is the target. Here the target needs three bromines, so protection defeats the whole plan. The free amine's full activation is exactly what delivers 2,4,6-tribromoaniline in one step.

✗ Common Mistake 2: Stopping at 2,4,6-tribromoaniline

Why it's wrong:

The target contains no nitrogen. The amino group must still be removed after it has done its directing work.

✗ Common Mistake 3: Using $NaNO_2/Cu$ on the diazonium salt

Why it's wrong: …