See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: When a molecule contains both a hydroxyl group and a nearby tertiary carbocation-stabilizing group, treatment with HBr can protonate one OH to form a carbocation, which can then be trapped intramolecularly by the other OH to form a cyclic ether (tetrahydropyran), rather than being trapped by bromide. Step 1 - Identify the substrate: The substrate is a 1,2-diol (on a cyclohexane framework) where one OH is a primary -CH2OH (unlabeled) and the other is a tertiary 18OH on a carbon also bearing a cyclohexenyl group. The tertiary position is highly stabilized. Step 2 - Protonation by HBr: HBr protonates the tertiary 18OH (the better leaving group as it forms a more stable tertiary carbocation) to give a tertiary carbocation with loss of H2-18O. Step 3 - Intramolecular cyclization: The primary hydroxyl oxygen (unlabeled -CH2OH) acts as an intramolecular nucleophile and attacks the tertiary carbocation. This forms a five- or six-membered cyclic oxonium ion, which loses a proton to give a cyclic ether (tetrahydropyran ring). The oxygen in the ring comes from the unlabeled OH, and the 18O is lost as water. Step 4 - Product: The product is a bicyclic compound where the oxygen bridge is formed from the unlabeled OH, consistent with option (a), which shows the cyclic ether with no 18O label on the ring oxygen. Step 5 - Why not (b): Option (b) shows the 18O in the ring, which would require the tertiary 18OH to be the nucleophile and the primary OH to leave - but primary carbocations are not stable; this pathway is disfavored. Step 6 - Why not (c) or (d): These show open-chain bromide products. In CCl4 (non-polar solvent) with an intramolecular oxygen nucleophile available, cyclization to the ether is faster than intermolecular bromide attack, especially when a favorable ring size results. The cyclic ether (tetrahydropyran, 6-membered) is strongly preferred. Step 7 - Conclusion: The major product is the cyclic ether formed by loss of the tertiary 18OH as water and cyclization of the unlabeled primary OH onto the tertiary carbocation, giving the bicyclic tetrahydropyran shown in option (a). Therefore, the correct answer is A.