The reagent which converts alkylbenzenes to carboxylic acids is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(KMnO4 – KOH/Δ, HCl\)
Alkylbenzenes are oxidised to aromatic carboxylic acids by heating with alkaline potassium permanganate, which converts the side chain into –COOK; acidification with HCl then gives the acid:
\( \displaystyle \ce{C6H5CH3} \xrightarrow[\Delta]{\ce{KMnO4}-\ce{KOH}} \ce{C6H5COOK} \xrightarrow{\ce{HCl}} \ce{C6H5COOH} \)
Why the other options are incorrect
Option (1) \(CrO_2Cl_2\) — Incorrect. Chromyl chloride oxidises a methyl group only to –CHO (Etard reaction), giving benzaldehyde.
\( \ce{C6H5CH3} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{CrO2Cl2}, \ce{CS2}} \ce{C6H5CHO} \)
Option (2) \(CrO_3\) — Incorrect. Chromic oxide in acetic anhydride gives benzylidene diacetate, which is hydrolysed to benzaldehyde.
\( \ce{C6H5CH3} \xrightarrow{\ce{CrO3}, (\ce{CH3CO})_2\ce{O}} \ce{C6H5CH(OCOCH3)2} \xrightarrow{\ce{H3O+}} \ce{C6H5CHO} \)
Option (3) Anhydrous \(AlCl_3/CuCl\) — Incorrect. With CO and HCl this is the Gattermann–Koch reaction, which introduces –CHO into benzene.
\( \ce{C6H6 + CO + HCl} \xrightarrow{\text{anhyd. }\ce{AlCl3}/\ce{CuCl}} \ce{C6H5CHO + HCl} \)
NCERT (2026-27 edition):
From Alkylbenzenes
