Question

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? 

• A.
• B.
• C.
• D.

Hint:

- Markovnikov’s Rule helps determine which carbon gets the halide in addition reactions. - Friedel-Crafts Alkylation prefers tertiary alkyl halides for carbocation stability. - tert-Butylbenzene is resistant to oxidation because it lacks a benzylic hydrogen.

Answer 1

The Correct Option is A

Let's analyze the reactions step by step:

1. 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.
2. 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.
3. 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.