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NCERT Exemplar · Q58

Q.Predict the missing reagents (labelled 1, 3 and 5) and the missing products (labelled 2 and 4) in the following reaction sequence. p-nitrotoluene (a benzene ring with –CH3 and –NO2 para to each other) is treated with reagent 1 to give p-toluidine (–CH3 and –NH2 para). p-Toluidine is treated with (CH3CO)2O / pyridine to give p-methylacetanilide (–CH3 and –NHCOCH3 para). p-Methylacetanilide is treated with HNO3/H2SO4 to give product 2. Product 2 is treated with reagent 3 to give 4-methyl-2-nitroaniline (–CH3 para to –NH2, with –NO2 ortho to the –NH2). 4-methyl-2-nitroaniline is treated with NaNO2/HCl to give compound 4. Compound 4 is treated with reagent 5 to give m-nitrotoluene (a benzene ring with –CH3 and –NO2 meta to each other).

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The sequence reduces the nitro group to an amine, protects it as the acetanilide, nitrates ortho to nitrogen, hydrolyses the protecting group, diazotises the amine, and finally removes it by deamination — shifting the substitution pattern to give m-nitrotoluene.

Reagent 1 – reduction

p-Nitrotoluene → p-toluidine requires reduction of –NO2 to –NH2. Reagent 1 = Sn/HCl (equivalently Fe/HCl or H2/catalyst).

Product 2 – nitration of p-methylacetanilide

p-Methylacetanilide has –CH3 and –NHCOCH3 para. On nitration (HNO3/H2SO4), the strongly directing acetamido group sends –NO2 to the position ortho to itself. Product 2 = 4-methyl-2-nitroacetanilide (–CH3, –NHCOCH3 para, –NO2 ortho to –NHCOCH3).

Reagent 3 – hydrolysis

Product 2 → 4-methyl-2-nitroaniline means the acetyl protecting group is removed. Reagent 3 = H3O+ (dilute acid hydrolysis; aqueous base then acidification also works).

Compound 4 – diazotisation

4-Methyl-2-nitroaniline + NaNO2/HCl (273–278 K) → the diazonium salt. Compound 4 = 4-methyl-2-nitrobenzenediazonium chloride. …

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