In the given reaction sequence, the major product ' C ' is : C 8 H 10 ⟶ HNO 3 H 2 SO 4 A ⟶ Br 2 Δ B ⟶ alcoholic KOH C

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ChemistryAromatic HydrocarbonAROMATIC NUCLEPHIC SUBSTITUTION BY SNAE and SNEAHard2 minPYQ_2022

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In the given reaction sequence, the major product ' $C$ ' is :

$C_{8} H_{10} \underset{H_{2} {SO}_{4}}{\overset{{HNO}_{3}}{⟶}} A \underset{Δ}{\overset{{Br}_{2}}{⟶}} B \underset{KOH}{\overset{alcoholic}{⟶}} C$

Answer:

Solution:

In first step nitration of benzene ring will take place at para position, since ethyl group is ortho-para directing in nature.

Then halogenation reaction will take place at benzyl position followed by dehydrohalogenation reaction with alcoholic KOH.

Stream:JEESubject:ChemistryTopic:Aromatic HydrocarbonSubtopic:AROMATIC NUCLEPHIC SUBSTITUTION BY SNAE and SNEA

⏱ 2mℹ️ Source: PYQ_2022

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