An alkene X (\( C_4H_8 \)) on reaction with HBr gave Y (\( C_4H_9Br \)). Reaction of Y with benzene in the presence of anhydrous \( AlCl_3 \) gave Z which is resistant to oxidation with \( KMnO_4 + KOH \). What are X, Y, Z respectively?
An alkene X (\( C_4H_8 \)) on reaction with HBr gave Y (\( C_4H_9Br \)). Reaction of Y with benzene in the presence of anhydrous \( AlCl_3 \) gave Z which is resistant to oxidation with \( KMnO_4 + KOH \). What are X, Y, Z respectively?
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The Correct Option is A
Solution and Explanation
Let's analyze the reactions step by step:
- Step 1: The alkene X (\(C_4H_8\)) reacts with HBr. This is an addition reaction where HBr adds across the double bond of the alkene. The product Y (\(C_4H_9Br\)) must be an alkyl bromide.
- Step 2: Compound Y (\(C_4H_9Br\)) reacts with benzene in the presence of anhydrous \(AlCl_3\). This is a Friedel-Crafts alkylation reaction. The alkyl group from Y substitutes a hydrogen atom on the benzene ring, forming compound Z.
- Step 3: Compound Z is resistant to oxidation with \(KMnO_4\) - KOH. This means Z does not have any readily oxidizable groups like alkenes or primary/secondary alcohols. Since Z is formed from benzene, it must be a tert-butylbenzene, which is resistant to oxidation under these conditions.
Working backwards: Since Z is tert-butylbenzene, Y must be tert-butyl bromide, and X must be isobutylene (2-methylpropene), as it gives tert-butyl bromide on reaction with HBr.
Therefore, X is isobutylene, Y is tert-butyl bromide, and Z is tert-butylbenzene.
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