Question

Propose a mechanism for the reaction of(a) 1-methylcyclohexanol with HBr to form 1-bromo-1-methylcyclohexane.

Accepted Answer

Step 1: Protonation of the alcohol group - The hydroxyl group (-OH) in 1-methylcyclohexanol is a poor leaving group. To make it a better leaving group, it is protonated by HBr, forming water (H2O) as a better leaving group. This step involves the transfer of a proton (H⁺) from HBr to the oxygen atom of the hydroxyl group. Step 2: Formation of a carbocation intermediate - After protonation, the water molecule leaves, resulting in the formation of a carbocation at the carbon atom that was originally bonded to the hydroxyl group. Since the hydroxyl group is attached to a tertiary carbon (1-methylcyclohexanol is a tertiary alcohol), the resulting carbocation is a tertiary carbocation, which is relatively stable due to hyperconjugation and inductive effects. Step 3: Nucleophilic attack by bromide ion - The bromide ion (Br⁻), which is generated from the dissociation of HBr, acts as a nucleophile and attacks the positively charged carbocation. This step involves the formation of a new C-Br bond, resulting in the product 1-bromo-1-methylcyclohexane. Step 4: Verify the stereochemistry - Since the reaction proceeds through a carbocation intermediate, the product may exhibit racemization if the carbon center is chiral. However, in this case, the product is not chiral, so no stereochemical considerations are necessary. Step 5: Summarize the mechanism - The overall reaction involves three key steps: protonation of the alcohol group, formation of a carbocation intermediate, and nucleophilic attack by the bromide ion. The final product is 1-bromo-1-methylcyclohexane.

Propose a mechanism for the reaction of
(a) 1-methylcyclohexanol with HBr to form 1-bromo-1-methylcyclohexane.

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Concetti chiave

Alcohol Reactivity

Nucleophilic Substitution Mechanism

Carbocation Stability