Which compound on oxidation with acidic KMnO_ | vedclass

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(D) Oxidation of alkylbenzenes with acidic $KMnO_4$ yields Benzoic acid,provided that the alkyl group attached to the benzene ring contains at least o

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All of these

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(D) Oxidation of alkylbenzenes with acidic $KMnO_4$ yields Benzoic acid,provided that the alkyl group attached to the benzene ring contains at least one benzylic hydrogen atom.$1.$ $1-$phenylethanol $(C_6H_5CH(OH)CH_3)$ contains a benzylic hydrogen and can be oxidized to Benzoic acid.$2.$ Ethylbenzene $(C_6H_5CH_2CH_3)$ contains two benzylic hydrogens and is readily oxidized to Benzoic acid.$3.$ $1-$phenylethanone $(C_6H_5COCH_3)$ is a ketone and is generally resistant to oxidation by $KMnO_4$ under standard conditions,but the question asks which compound gives Benzoic acid. Among the choices,$1-$phenylethanol and Ethylbenzene are clear candidates. However,since the question implies a single correct option and $1-$phenylethanol and Ethylbenzene both undergo oxidation to Benzoic acid,the most appropriate answer in a multiple-choice context where 'All of these' is present is often evaluated based on the reactivity of the alkyl side chain. Given the options,Ethylbenzene is the classic textbook example for this reaction.

List-$I$ (Chemical Reaction)	List-$II$ (Reagent used)
$(a)$ $CH_{3}COOCH_{2}CH_{3} \rightarrow CH_{3}CH_{2}OH$	$(i)$ $CH_{3}MgBr / H_{3}O^{+}$ ($1$ equivalent)
$(b)$ $CH_{3}COOCH_{3} \rightarrow CH_{3}CHO$	$(ii)$ $H_{2}SO_{4} / H_{2}O$
$(c)$ $CH_{3}C \equiv N \rightarrow CH_{3}CHO$	$(iii)$ $DIBAL-H / H_{2}O$
$(d)$ $CH_{3}C \equiv N \rightarrow CH_{3}COCH_{3}$	$(iv)$ $SnCl_{2}, HCl / H_{2}O$

Which compound on oxidation with acidic $KMnO_4$ gives Benzoic acid?

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