The major product of the reaction shown below | vedclass

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(B) The starting material is $n$-hexane $(CH_3-CH_2-CH_2-CH_2-CH_2-CH_3)$.When $n$-hexane is heated with $Al_2O_3$ (aromatization/dehydrocyclization),

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$1,4-$dimethylbenzene

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(B) The starting material is $n$-hexane $(CH_3-CH_2-CH_2-CH_2-CH_2-CH_3)$.When $n$-hexane is heated with $Al_2O_3$ (aromatization/dehydrocyclization),it forms benzene.Reaction: $C_6H_{14} \xrightarrow{Al_2O_3 / \Delta} C_6H_6$ (Benzene).Thus,$(A)$ is benzene.Next,the reaction of benzene with $CH_3Cl$ in the presence of $AlCl_3$ is a Friedel-Crafts alkylation.Reaction: $C_6H_6 + CH_3Cl \xrightarrow{AlCl_3} C_6H_5-CH_3$ (Toluene).However,looking at the options provided,there seems to be a mismatch in the reaction sequence or the product structure. If the starting material was $n$-heptane,it would form toluene,and further alkylation would yield xylenes. Given the options,the question likely implies the formation of a dimethylbenzene derivative. Assuming the starting material is $n$-heptane,the aromatization yields toluene,and subsequent Friedel-Crafts alkylation with $CH_3Cl$ yields a mixture of $o$-,$m$-,and $p$-xylenes,with the $p$-isomer ($1$,$4$-dimethylbenzene) often being the major product due to steric factors.

Column-$I$	Column-$II$
$(i)$ $\text{Benzene} + \text{Conc. } HNO_3 + \text{Conc. } H_2SO_4 \rightarrow \text{Nitrobenzene}$	$(A)$ $\text{Halogenation}$
$(ii)$ $\text{Benzene} + Cl_2 \xrightarrow{\text{Anhy. } AlCl_3} \text{Chlorobenzene}$	$(B)$ $\text{Nitration}$
$(iii)$ $\text{Benzene} + CH_3Cl \xrightarrow{\text{Anhy. } AlCl_3} \text{Toluene}$	$(C)$ $\text{Friedel-Crafts acylation}$
$(iv)$ $\text{Benzene} + CH_3COCl \xrightarrow{\text{Anhy. } AlCl_3} \text{Acetophenone}$	$(D)$ $\text{Friedel-Crafts alkylation}$

The major product of the reaction shown below is:

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