Q.How will you convert 4-nitrotoluene to 2-bromobenzoic acid?

Concept: Aromatic Synthesis Route — the trick is to brominate first, while both original substituents are still on the ring. In 4-nitrotoluene the position ortho to $-CH_3$ (an ortho/para director) is the very same position that is meta to $-NO_2$ (a meta director), so both groups direct the incoming bromine to the same carbon.

The textbook's Solution, step by step:

1. Brominate 4-nitrotoluene with $Br_2$. Bromine enters ortho to the methyl group (which is simultaneously meta to the nitro group), giving 2-bromo-4-nitrotoluene.
2. Reduce $-NO_2$ to $-NH_2$ with $Sn/HCl$: 2-bromo-4-methylaniline.
3. Diazotise with $NaNO_2/HCl$ at 273–278 K, then treat the diazonium salt with $H_2O/H_3PO_2$ to replace $-N_2^+Cl^-$ by $-H$: 2-bromotoluene. …

The target, 2-bromobenzoic acid, needs bromine ortho to a carboxylic acid — but $-COOH$ is a meta director, so it can never direct bromine to its own ortho position. The insight of the textbook's Solution is that in the starting material the two substituents already present agree on exactly the right position, so the bromination should be done before anything else is changed.

1. Brominate while both original groups are on the ring.

Number the ring with $-CH_3$ at C-1 and $-NO_2$ at C-4. The methyl group (ortho/para director) favours C-2/C-6; the nitro group (meta director) favours the positions meta to itself, which are the same C-2/C-6. Both effects reinforce, so bromination gives 2-bromo-4-nitrotoluene cleanly:

2. Reduce the nitro group to an amine.

3. Remove the amino group via the diazonium salt.

Diazotise with $NaNO_2/HCl$ at 273–278 K, then treat with $H_2O/H_3PO_2$ (hypophosphorous acid), which replaces $-N_2^+$ by $-H$:

4. Oxidise the methyl group last.

$\text{2-bromotoluene} \xrightarrow{KMnO_4,\ OH^-\ (\text{then } H_3O^+)} \text{2-bromobenzoic acid}$ …

Method: Directing-Effect Audit Before Any Step

For a multi-step aromatic synthesis, list every substituent's directing effect in the starting material first, and ask whether the groups already present can place the new substituent where the target needs it. Only if they cannot do you start interconverting groups.

Step 1: Map the target onto the starting material

• Target: 2-bromobenzoic acid — $-COOH$ at C-1, $-Br$ at C-2 (ortho).
• Start: 4-nitrotoluene — $-CH_3$ at C-1 (future $-COOH$, by oxidation), $-NO_2$ at C-4.
• Needed: bromine at C-2, i.e. ortho to the methyl group.

Step 2: Audit the directors in 4-nitrotoluene

• $-CH_3$: ortho/para director → favours C-2/C-6.
• $-NO_2$: meta director → its meta positions are also C-2/C-6.
• Both groups direct to the same carbon. So bromination should be done immediately, before either group is touched.

Step 3: Forward synthesis (the textbook's Solution)

1. Brominate: $Br_2$ on 4-nitrotoluene → 2-bromo-4-nitrotoluene (Br enters ortho to $-CH_3$, meta to $-NO_2$).
2. Reduce: $Sn/HCl$ converts $-NO_2$ to $-NH_2$ → 2-bromo-4-methylaniline.
3. Diazotise: $NaNO_2/HCl$, 273–278 K → the diazonium salt.
4. Deaminate: $H_2O/H_3PO_2$ replaces $-N_2^+$ by $-H$ → 2-bromotoluene. …

These mistakes all come from re-ordering the steps of the synthesis without checking the ring geometry at each stage.

✓ The correct route (the textbook's Solution)

Bromination comes first because $-CH_3$ (ortho/para director) and $-NO_2$ (meta director) both point to the same carbon — the position ortho to methyl is the position meta to nitro.

✗ Mistake 1: Reduce the nitro group first, protect, and brominate ortho to the acetamido group

• The wrong route: $-NO_2 \to -NH_2$ ($Sn/HCl$), acetylate to $-NHCOCH_3$, brominate "ortho to the director", deprotect, deaminate.
• Why it's wrong — check the geometry: the acetamido group sits at C-4 and the methyl at C-1. Bromine ortho to $-NHCOCH_3$ means C-3, and C-3 is meta to the methyl group. After deamination this gives 3-bromotoluene, and oxidation gives 3-bromobenzoic acid — the wrong isomer.
• How to avoid: before swapping any group, ask what the existing groups already direct to. Here they already agree on C-2, so no protection strategy is needed at all.

✗ Mistake 2: Oxidising the methyl group before brominating

• Why it's wrong: once $-CH_3$ becomes $-COOH$, the ring carries two meta directors that disagree — $-COOH$ at C-1 favours C-3/C-5 while $-NO_2$ at C-4 favours C-2/C-6 — on a doubly deactivated ring. There is no longer any position both groups favour, so you cannot rely on getting the 2-bromo product cleanly. The textbook avoids the conflict entirely by brominating while $-CH_3$ and $-NO_2$ still reinforce each other.

✗ Mistake 3: Oxidising with $KMnO_4$ while the free amine is on the ring

• Why it's wrong: $KMnO_4$ attacks aromatic amines as well as side chains — a free $-NH_2$ would be destroyed. The amino group must be removed (diazotisation, then $H_3PO_2$) before the oxidation step.

✗ Mistake 4: Letting the diazonium salt warm up …