See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1: Identify V. Cyclopentanol treated with Na2Cr2O7 (an oxidizing agent) oxidizes the secondary alcohol to a ketone: cyclopentanone (V). Step 2: W is formed from cyclopentanol + PBr3. PBr3 converts the hydroxyl group to a bromide: bromocyclopentane (W). Step 3: X is formed from W + Mg/ether. This is a Grignard reagent formation: cyclopentylmagnesium bromide (X). Step 4: Y is formed from X + (1) V (cyclopentanone), (2) H3O+. The Grignard reagent (cyclopentylMgBr) attacks cyclopentanone to give, after aqueous workup, 1-cyclopentylcyclopentan-1-ol (a tertiary alcohol, Y). This is dicyclopentyl carbinol: (cyclopentyl)2CHOH... wait, cyclopentanone is a ketone, so attack of cyclopentyl Grignard on cyclopentanone gives a tertiary alcohol: 1-(cyclopentyl)cyclopentan-1-ol, i.e., both groups are cyclopentyl attached to a carbon bearing OH — that carbon is the former carbonyl carbon, giving (cyclopentyl)2C(OH) with no extra H... Actually cyclopentanone has the carbonyl in the ring, so the Grignard adds to give: the carbonyl carbon of cyclopentanone gets the cyclopentyl group added, breaking nothing in the ring. Product Y = 1-cyclopentylcyclopentan-1-ol (tertiary alcohol). Step 5: Z is formed from Y + CH3COCl (acetyl chloride). This is esterification of the tertiary alcohol with acetyl chloride to give the acetate ester: 1-cyclopentylcyclopentyl acetate, i.e., the tertiary alcohol is acetylated to give Z = 1-(cyclopentyl)cyclopentyl acetate (the tertiary acetate ester). The product Z is the acetate ester of 1-cyclopentylcyclopentan-1-ol, which is 1-cyclopentylcyclopentyl acetate. This corresponds to answer choice B. Therefore, the correct answer is B.