Consider the following reaction sequence:Viny | vedclass

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(C) $1$. Vinylbenzene $(C_6H_5-CH=CH_2)$ on oxidation with alkaline $KMnO_4$ followed by heating undergoes oxidative cleavage of the side chain to for

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Aromatisation of $n-$hexane

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(C) $1$. Vinylbenzene $(C_6H_5-CH=CH_2)$ on oxidation with alkaline $KMnO_4$ followed by heating undergoes oxidative cleavage of the side chain to form Potassium benzoate $(C_6H_5COOK)$,which on acidification gives Benzoic acid $(C_6H_5COOH)$. Thus,$X$ is Benzoic acid.$2$. Benzoic acid $(C_6H_5COOH)$ reacts with Sodalime $(NaOH+CaO)$ upon heating (decarboxylation) to form Benzene $(C_6H_6)$. Thus,$Y$ is Benzene.$3$. Benzene can be prepared by the aromatisation (dehydrocyclisation) of $n-$hexane $(C_6H_{14})$ in the presence of catalysts like $Cr_2O_3/Al_2O_3$ at high temperature and pressure.

Consider the following reaction sequence:
Vinylbenzene $\xrightarrow[\Delta]{KMnO_4+KOH} X$ $\xrightarrow[\Delta]{NaOH+CaO} Y$
'$Y$' can also be formed from:

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Consider the following reaction sequence:Vinylbenzene $\xrightarrow[\Delta]{KMnO_4+KOH} X$ $\xrightarrow[\Delta]{NaOH+CaO} Y$'$Y$' can also be formed from: