See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The substrate is a quaternary carbon center at C1 of a cyclohexene ring bearing two ester groups: an acetate ester (CH3-CO-O-, i.e., the oxygen of the ring carbon is esterified with acetic acid) and a methoxycarbonyl ester (-C(=O)-OCH3). There is also a methyl group at C6 and a double bond in the ring (C2-C3 position). Step 2 - Reaction with LiAlH4: LiAlH4 is a strong reducing agent that reduces both esters to alcohols. - The acetate ester (CH3-C(=O)-O-C) upon reduction: LiAlH4 cleaves the acyl-oxygen bond of an ester, reducing -O-C(=O)-CH3 to give -OH (the alkoxide/alcohol on C1) and CH3CH2OH (ethanol, which is washed away). So the -OC(=O)CH3 group at C1 becomes -OH at C1. - The methoxycarbonyl ester (-C(=O)-OCH3) upon reduction: LiAlH4 reduces -C(=O)-OCH3 to -CH2OH. So the -C(=O)OCH3 group at C1 becomes -CH2OH. Step 3 - Product: After reduction of both ester groups, C1 bears: an -OH group (from acetate reduction), a -CH2OH group (from methoxycarbonyl reduction), the methyl-bearing cyclohexene ring carbon connectivity, and the ring itself. The double bond in the ring is not affected by LiAlH4 under normal conditions. The methyl group at C6 remains. This gives 1-(hydroxymethyl)-6-methylcyclohex-2-en-1-ol: a cyclohexene ring with a double bond, methyl at C6, OH at C1 (tertiary alcohol), and a CH2OH group at C1. Step 4 - Match to options: Option (c) shows a cyclohexene ring with double bond retained, methyl group on the ring, OH at the quaternary carbon C1, and a CH2OH substituent at C1. This matches exactly. Step 5 - Why other options fail: - (a) shows two OH groups directly on C1 (gem-diol), which would result from reduction of two carbonyl carbons directly attached to C1, not from an acetate ester reduction which gives OH on C1 by cleavage, but does not give a second direct OH from the methoxycarbonyl — actually option (a) lacks the CH2OH, so it is wrong. - (b) appears similar to (c) but shows a fully saturated ring (no double bond), so the alkene has been reduced, which LiAlH4 does not do. - (d) shows a ketone product, which is the opposite of reduction. Therefore, the correct answer is C.