See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Formation of intermediate I: CH3CH2C≡CH is treated with NaNH2 in liquid NH3. NaNH2 is a strong base that deprotonates the terminal alkyne. However, CH3CH2C≡CH is an internal-looking alkyne but actually has the triple bond at C1 (terminal). Wait — CH3CH2C≡CH is 1-butyne (but-1-yne), which has a terminal alkyne proton. NaNH2 deprotonates it to give the acetylide anion: CH3CH2C≡C⁻ Na⁺. This is intermediate I. Step 2 - Formation of intermediate J: Intermediate I (CH3CH2C≡C⁻) reacts with cyclobutanone (the four-membered ring ketone shown) in Et2O. The acetylide anion acts as a nucleophile and attacks the carbonyl carbon of cyclobutanone in a nucleophilic addition reaction. This opens the carbonyl to give an alkoxide: cyclobutane ring with the carbonyl carbon now bearing an -OH (after workup) and the -C≡C-CH2CH3 (but-1-yn-1-yl) group attached. The product J is the alkoxide (before protonation): 1-(but-1-yn-1-yl)cyclobutan-1-olate. Step 3 - Protonation with H⁺: Treatment with H⁺ protonates the alkoxide to give the alcohol. Product K is 1-(but-1-yn-1-yl)cyclobutan-1-ol, i.e., cyclobutanol with a quaternary carbon at C1 bearing OH and the -C≡C-Et group. Wait — the cyclic ketone shown is cyclobutanone (four-membered ring with =O). The acetylide adds to give 1-(1-butynyl)cyclobutan-1-ol. But option (b) shows a five-membered ring (cyclopentane). Let me re-examine the structure: the image shows a square with =O, which is cyclobutanone (4-membered). But option (b) shows a cyclopentane ring. Actually, looking more carefully at the image description — the ring drawn could be cyclopentanone (5-membered). A pentagon-shaped ring with =O is cyclopentanone. The nucleophilic addition of CH3CH2C≡C⁻ to cyclopentanone gives 1-(but-1-yn-1-yl)cyclopentan-1-ol after protonation. So K = 1-(but-1-yn-1-yl)cyclopentan-1-ol: a cyclopentane ring with a quaternary C1 bearing -OH and -C≡C-CH2CH3 (i.e., -C≡C-Et). This matches option (b): cyclopentane ring with OH and -C≡C-Et on the same (quaternary) carbon. Why other options fail: - Option (a): Shows a cyclopentan-1-one (ketone retained), meaning no nucleophilic addition occurred — incorrect; H⁺ workup does not regenerate the ketone from an alkoxide. - Option (c): Open-chain structure, no ring — does not account for the cyclopentanone ring. - Option (d): Shows -C≡C-CH3 instead of -C≡C-Et (propyne unit instead of butyne unit) — wrong acetylide. Therefore, the correct answer is B.