JEE Main 2022 · 24 Jun · Shift 1 · Chemistry Q73

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ChemistrySingle correct+4 / -1Question 73 of 90

In the given reaction sequence, the major product 'C' is :
$\mathrm{C_8H_{10}\xrightarrow[H_2SO_4]{HNO_3}A\xrightarrow[\Delta]{Br_2}B\xrightarrow[KOH]{alcoholic}C}$

Benzene ring bearing a

\mathrm{C(NO_2)=CH_2}

group (structure shown)

\mathrm{O_2N}

-C

_6

_4

\mathrm{CH=CH_2}

(para; structure shown)

4-nitrostyrene (para-nitro vinylbenzene)

\mathrm{O_2N}

-C

_6

_4

\mathrm{C(NO_2)=CH_2}

(para; structure shown)

Benzene ring with

\mathrm{NO_2}

and

\mathrm{CH=CH_2}

on adjacent carbons (ortho; structure shown)

2-nitrostyrene (ortho-nitro vinylbenzene)

Answer and worked solution

Official answer: B

A. $\mathrm{C_8H_{10}}$ is ethylbenzene; the ethyl group is ortho/para directing and nitration gives mainly 4-nitroethylbenzene.

B. $\mathrm{Br_2}$ with heat brominates the benzylic carbon: $\mathrm{O_2N\text{-}C_6H_4\text{-}CHBr\text{-}CH_3}$ .

C. Alcoholic KOH removes HBr: $\mathrm{O_2N\text{-}C_6H_4\text{-}CH=CH_2}$ , 4-nitrostyrene.

(B)

Topics:Aromatic hydrocarbons and electrophilic substitution Elimination reactions

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JEE Main 2022 · 24 Jun · Shift 1, Chemistry question 73