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