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

To determine the identities of compounds X, Y, and Z, we will analyze the reactions and properties presented:

Step 1: Identify Alkene X

The alkene X is C₄H₈. Possible structures are 1-butene, 2-butene (cis and trans), and isobutene (2-methylpropene). Since X reacts with HBr to form a haloalkane Y, we apply Markovnikov's rule for the addition of HBr, where the bromine adds to the more substituted carbon. For isobutene (2-methylpropene), HBr predominantly forms tert-butyl bromide.

Step 2: Identify Y

When HBr reacts with isobutene, it forms tert-butyl bromide (2-bromo-2-methylpropane), C₄H₉Br. This reaction is supported by Markovnikov's rule.

Step 3: Identify Z

Y reacts with benzene in the presence of anhydrous AlCl₃ to undergo a Friedel-Crafts alkylation, forming a tert-butyl group attached to the benzene ring, creating tert-butylbenzene. This molecule is resistant to oxidation with KMnO₄ + KOH due to the stability of the tert-butyl group. The tert-butyl group does not have hydrogens on the alpha carbon, so oxidation by KMnO₄ is ineffective.

Therefore, X is isobutene (2-methylpropene), Y is tert-butyl bromide (2-bromo-2-methylpropane), and Z is tert-butylbenzene.

Answer 2

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.