The major product [C] of the following reacti | vedclass

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(C) Step $1$: $CH_2=CH-CHO$ reacts with $NaBH_4$ to give $CH_2=CH-CH_2OH$,which then reacts with $SOCl_2$ to form $CH_2=CH-CH_2Cl$ $[A]$.Step $2$: $CH

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(C) Step $1$: $CH_2=CH-CHO$ reacts with $NaBH_4$ to give $CH_2=CH-CH_2OH$,which then reacts with $SOCl_2$ to form $CH_2=CH-CH_2Cl$ $[A]$.Step $2$: $CH_2=CH-CH_2Cl$ reacts with benzene in the presence of anhydrous $AlCl_3$ (Friedel-Crafts alkylation) to give allylbenzene,$C_6H_5-CH_2-CH=CH_2$ $[B]$.Step $3$: The addition of $DBr$ to allylbenzene follows Markovnikov's rule. The electrophile $D^+$ adds to the terminal carbon $(CH_2)$ to form the more stable benzylic carbocation $(C_6H_5-CH^+-CH_2D)$,which is then attacked by $Br^-$ to form the final product $[C]$,$C_6H_5-CH(Br)-CH_2-CH_2D$.

The major product $[C]$ of the following reaction sequence will be

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