Question

The bromides formed by the cleavage of ethers A and B with HBr respectively are:

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

In acidic cleavage of ethers, the stability of the formed carbocation determines the cleavage pattern. Aryl ethers form aryl bromides, while tert-alkyl ethers follow SN1 cleavage.

Answer 1

The Correct Option is D

Step 1: Understanding Acidic Cleavage of Ethers 
- Ethers react with hydrogen bromide (HBr) via acidic cleavage, forming alkyl and aryl bromides.
- The cleavage mechanism depends on the stability of carbocations formed during the reaction.
Step 2: Cleavage of Ether A (Phenyl Ethyl Ether) 
1. The ether oxygen is protonated by HBr, making it a better leaving group.
2. Since phenyl carbocation is unstable, the bond breaks to form:
- Phenyl bromide (\( C_6H_5Br \)) (as benzyl carbocation is not formed).
- Ethanol (\( C_2H_5OH \)), which further reacts with HBr to form ethyl bromide (\( C_2H_5Br \)).
Step 3: Cleavage of Ether B (tert-Butyl Propyl Ether) 
1. The tert-butyl carbocation (\( (CH_3)_3C^+ \)) is highly stable, so the ether undergoes SN1 cleavage.
2. This results in:
- tert-Butyl bromide (\( (CH_3)_3CBr \)) (from stable carbocation).
- Propyl bromide (\( C_3H_7Br \)) (via nucleophilic substitution).
Step 4: Evaluating the Given Options 
- Option (1): Incorrect, as the structures do not match the cleavage products.
- Option (2): Incorrect, as it does not correctly represent the products.
- Option (3): Incorrect, as tert-butyl bromide formation is missing.
- Option (4): Correct, as it correctly represents phenyl bromide and ethyl bromide for A, and tert-butyl bromide and propyl bromide for B.
Thus, the correct answer is Option (4).