Which route converts benzene into 4-nitrobenzoic acid (as the major product at the aromatic stage, ignoring separation of minor isomers)?

A(i) $HNO X_{3}$ , $H X_{2} SO X_{4}$ ; (ii) $CH X_{3} Cl$ , $AlCl X_{3}$ ; (iii) hot alkaline $KMnO X_{4}$ , then $H X_{3} O X^{+}$
B(i) $CH X_{3} COCl$ , $AlCl X_{3}$ ; (ii) $HNO X_{3}$ , $H X_{2} SO X_{4}$ ; (iii) $NaOI$ then $H X_{3} O X^{+}$
C(i) $CH X_{3} Cl$ , $AlCl X_{3}$ ; (ii) $HNO X_{3}$ , $H X_{2} SO X_{4}$ ; (iii) hot alkaline $KMnO X_{4}$ , then $H X_{3} O X^{+}$
D(i) $CH X_{3} Cl$ , $AlCl X_{3}$ ; (ii) hot alkaline $KMnO X_{4}$ , then $H X_{3} O X^{+}$ ; (iii) $HNO X_{3}$ , $H X_{2} SO X_{4}$

Answer: C

The para relationship needs the o/p director (

- CH X_{3}

) present during nitration, and oxidation to

- COOH

must be last. Nitrating first blocks Friedel-Crafts alkylation because of the deactivated ring. Oxidising before nitration, or acylating first, places the nitro group meta.

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Which route converts benzene into 4-nitrobenzoic acid (as the major product at the aromatic stage, ignoring separation of minor isomers)?

Answer: C

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