In the above reaction,product B is: | vedclass

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(A) The reaction proceeds in two steps:$1$. The reaction of $4$-hydroxybenzyl alcohol with $HCl$ under heating $(\Delta)$ leads to the substitution of

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$4-$(iodomethyl)phenol

Vedclass · Answer

(A) The reaction proceeds in two steps:$1$. The reaction of $4$-hydroxybenzyl alcohol with $HCl$ under heating $(\Delta)$ leads to the substitution of the alcoholic $-OH$ group with $-Cl$,forming $4$-(chloromethyl)phenol as product $A$.$2$. The reaction of $A$ with $NaI$ in acetone (Finkelstein reaction) proceeds via an $S_N2$ mechanism,where the $-Cl$ atom is replaced by an $-I$ atom to form $4$-(iodomethyl)phenol as product $B$.

In the above reaction,product $B$ is:

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