Consider the reaction I & II
I. $C_{2} H_{6} O \to_{443 K}^{C o n c . H_{2} S O_{4}} A \to_{}^{H B r} B$
II. $B + C_{6} H_{6} \to_{}^{A n h . A l C l_{3}} Z$
The correct statements about Z are
(A) It is anisole
(B) Hyper conjugation is possible in it
(C) The substituent in it is a meta directing group
(D) On oxidation with $K M n O_{4} / O H^{-}$ followed by hydrolysis gives a product in which the substituent is meta directing group

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A & B only

B & C only

B & D only

A & D only

answer is C.

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Detailed Solution

Consider the reaction I & II
I. ${CH}_{3} {CH}_{2} OH \to_{443 K}^{Conc . H_{2} {SO}_{4}} {CH}_{2} = {CH}_{2} \to_{}^{HBr} {CH}_{3} - {CH}_{2} Br$
II. ${CH}_{3} - {CH}_{2} Br + C_{6} H_{6} \to_{}^{Anh . {AlCl}_{3}} {CH}_{3} - {CH}_{2} - C_{6} H_{6}$

III. ${CH}_{3} - {CH}_{2} - C_{6} H_{5} \overset{{KMnO}_{4} / {OH}^{-}}{\to} C_{6} H_{5} - COOH$
The correct statements about Z are
(A) It isEhtyl alcohol
(B) Hyper conjugation is notpossible in it
(C) The substituent in it is a ortho and para directing group
(D) On oxidation with $K M n O_{4} / O H^{-}$ followed by hydrolysis gives a product in which the substituent is meta directing group