The product B formed in the following reactio | vedclass

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(D) Step $1$: The reaction of $1-(4-methylphenyl)prop-1-ene$ with $HCl$ follows Markovnikov's rule. The electrophilic addition of $H^+$ to the double

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$1-$($4$-methylphenyl)propyl isocyanide

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(D) Step $1$: The reaction of $1-(4-methylphenyl)prop-1-ene$ with $HCl$ follows Markovnikov's rule. The electrophilic addition of $H^+$ to the double bond forms a stable benzylic carbocation at the carbon adjacent to the benzene ring. Then,$Cl^-$ attacks this carbocation to form $1-(4-methylphenyl)-1-chloropropane$ as the major product $(A)$.Step $2$: The reaction of the alkyl halide $(A)$ with $AgCN$ is a nucleophilic substitution reaction. Since $AgCN$ is a covalent compound,the nitrogen atom acts as the nucleophile,leading to the formation of an isocyanide (isonitrile) as the major product $(B)$.Thus,the final product is $1-(4-methylphenyl)propyl$ isocyanide.

	$S_{N}1$	$S_{N}2$
$(a)$	Retention of configuration	Inversion of configuration
$(b)$	Racemisation	Inversion of configuration
$(c)$	Inversion of configuration	Retention of configuration
$(d)$	Racemisation	Retention of configuration

The product $B$ formed in the following reaction sequence is:

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