See image — Alcohols Phenols and Ethers Chemistry Question
Question
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💡 Solution & Explanation
Concept: The reaction of alcohols with HBr proceeds via protonation of the -OH group followed by departure of water to form a carbocation, which then reacts with Br-. The rate of this reaction depends on the stability of the carbocation intermediate formed after protonation and loss of water. Step 1 - Identify the type of alcohol in each option: (a) Secondary alcohol on a bicyclic (cyclopentane-cyclohexane fused) system - forms a secondary carbocation. (b) Tertiary alcohol on a bicyclic (cyclopentane-cyclohexane fused) system - the carbon bearing -OH also has a -CH3 group and is at the ring junction, making it a tertiary carbon - forms a tertiary carbocation. (c) Secondary alcohol on a methylcyclohexane ring (trans-4-methylcyclohexan-1-ol type) - forms a secondary carbocation. (d) Secondary alcohol on a methylcyclohexane ring - forms a secondary carbocation. Step 2 - Apply reactivity rules: Alcohols react with HBr faster when they can form more stable carbocations. The order of carbocation stability is: tertiary > secondary > primary. Therefore, tertiary alcohols react fastest with HBr. Step 3 - Identify option B: Option (b) has both -OH and -CH3 attached to the same carbon at the ring junction of the bicyclic system, making it a tertiary alcohol. Upon protonation and loss of water, it forms a tertiary carbocation, which is the most stable among the four options. Step 4 - Why other options fail: (a) Secondary alcohol - slower than tertiary. (c) Secondary alcohol on methylcyclohexane - slower than tertiary. (d) Secondary alcohol on methylcyclohexane - slower than tertiary. Option (b) forms the most stable (tertiary) carbocation intermediate, so it reacts with HBr at the fastest rate. Therefore, the correct answer is B.