See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is (4-methylcyclohexyl)methanol — a cyclohexane ring with a methyl group at C4 and a CH2OH (primary alcohol) group attached at C1 via a CH2 linker (i.e., the CH2OH is exocyclic). Step 2 - Reaction with PCC (excess) gives (A): PCC oxidizes primary alcohols to aldehydes and secondary alcohols to ketones. With excess PCC, the primary CH2OH is oxidized to an aldehyde (CHO). The ring has no secondary alcohol, so the only oxidation is CH2OH → CHO. However, looking more carefully at the starting material structure: it is drawn as a cyclohexane with a methyl at top (C1 of ring) and a CH2OH at C2 of the ring. With excess PCC: the primary alcohol (CH2OH) is oxidized to an aldehyde (CHO). So (A) = 4-methyl-cyclohexane-carbaldehyde (the CHO is exocyclic) with a methyl group on the ring. Actually re-examining: the structure shows a cyclohexane ring with methyl at C1 (top) and CH2OH at C2. PCC (excess) oxidizes the primary alcohol to aldehyde. Product (A) has both a ring ketone? No — there is no secondary OH on the ring. So (A) = aldehyde: (1-methyl-2-cyclohexyl)acetaldehyde equivalent — the CH2OH becomes CHO. Wait, re-reading: excess PCC could also mean it converts an aldehyde further only if there's a secondary alcohol nearby; since there's only a primary alcohol present, (A) = the corresponding aldehyde. Step 3 - Reaction with ethylene glycol (1 equiv), H+ gives (B): This is acetal protection. With 1 equivalent of ethylene glycol, one carbonyl is selectively protected as a cyclic acetal (1,3-dioxolane). Since (A) contains only one carbonyl (the aldehyde from oxidation of CH2OH), that aldehyde is protected as a 1,3-dioxolane acetal. So (B) = the aldehyde is protected as cyclic acetal, and the methyl group on the ring remains. Actually, let me reconsider the starting material more carefully. The structure shows a cyclohexane ring with: a methyl group at C1 (top of ring, indicated by the line going up) and a CH2OH group at C2. With excess PCC: primary alcohol → aldehyde. So A = 2-(1-methylcyclohex-2-yl)acetaldehyde... more simply, the CH2OH attached to the cyclohexyl carbon becomes CHO. (A) is an aldehyde. Step 4 - (B) with CH3MgBr, then H3O+ gives (C): The Grignard reagent CH3MgBr adds to the remaining carbonyl. But in (B) the aldehyde is protected as acetal. So what carbonyl does CH3MgBr react with? This means (A) must have had TWO carbonyls (the aldehyde from CH2OH oxidation AND a ketone from some ring oxidation), and (B) is the mono-protected product (aldehyde protected, ketone free). Then CH3MgBr adds to the ketone to give a tertiary alcohol. Re-examining starting material: cyclohexane ring with CH2OH at one carbon AND a methyl substituent. If the carbon bearing CH2OH also has an H (secondary position on ring), excess PCC would oxidize: (1) CH2OH → CHO (aldehyde), (2) the ring CH bearing CH2OH → C=O (ketone). So (A) has both an aldehyde (exocyclic CHO) and a ring ketone. With 1 equiv ethylene glycol/H+, the more reactive aldehyde is selectively protected as acetal. (B) = acetal-protected aldehyde + free ring ketone. Step 5 - CH3MgBr adds to the ring ketone in (B) to give tertiary alcohol (C): The ketone (C=O on ring) reacts with CH3MgBr to give a tertiary alcohol with methyl added. After H3O+ workup, (C) has: ring with C1(OH)(CH3) [tertiary alcohol, quaternary carbon] and C2 still bearing the protected aldehyde (acetal). Step 6 - NaBH4 reduces (D): NaBH4 is a mild reducing agent. It would not reduce an acetal. However, the acetal must be hydrolyzed first... but the question says NaBH4 directly. Under the reaction conditions, or perhaps during workup, the acetal hydrolyzes back to aldehyde, and NaBH4 reduces the aldehyde to primary alcohol. So (D) has: ring C1 with OH and CH3 (tertiary alcohol, from Grignard addition) and an exocyclic CH2OH (from reduction of aldehyde). The methyl group originally at C4 of the ring was actually the methyl at C1 which became part of the tertiary alcohol after Grignard addition. Final product (D): A cyclohexane ring where one carbon bears OH, CH3 (methyl), and CH2OH substituents — i.e., 1-methyl-1-(hydroxymethyl)cyclohexan-1-ol... but this is a quaternary carbon with OH, making it 1-(hydroxymethyl)-1-methylcyclohexanol. This matches option (b): cyclohexane ring with C1 bearing OH and CH2OH and methyl group (1-methyl-1-(hydroxymethyl)cyclohexanol). Why other options fail: (a) shows an open-chain CH(OH)CH3 side chain which would require different chemistry. (c) and (d) show 1-(hydroxymethyl)cyclohexanol without the methyl group on the quaternary carbon. Option (b) correctly shows the methyl group on the same carbon as OH and CH2OH, consistent with: original ring methyl (C4 position on original ring actually was on C1 of ring bearing CH2OH, making it a tertiary ring carbon) being converted to quaternary carbon after Grignard addition of CH3, then acetal deprotection and NaBH4 reduction of CHO to CH2OH. Therefore, the correct answer is B.