See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is 4-ethynylcyclohexan-1-ol, which has a terminal alkyne (-C≡CH) at C1 of the cyclohexane and a hydroxyl (-OH) at C4. Step 2 - Reaction with NaNH2 (2 equivalents): NaNH2 is a strong base that deprotonates the terminal alkyne (pKa ~25) to form the acetylide anion. The second equivalent of NaNH2 deprotonates the alcohol (-OH, pKa ~16) to form the alkoxide anion. So both acidic protons are removed. Step 3 - Reaction with CH3CH2-I (ethyl iodide): The acetylide anion (more reactive/harder nucleophile) alkylates with ethyl iodide first (acetylide is a better nucleophile in SN2), converting the terminal alkyne -C≡CH into an internal alkyne -C≡C-CH2CH3. The alkoxide remains as an anion at this stage (it is less nucleophilic but the reaction sequence is stepwise). Wait - with 2 eq. NaNH2, both OH and terminal alkyne C-H are deprotonated. When CH3CH2-I is added, the acetylide (harder nucleophile, more reactive toward alkyl halides) reacts selectively, giving -C≡C-CH2CH3, while the alkoxide is still present. Step 4 - Reaction with CH3-I (methyl iodide): The alkoxide anion reacts with methyl iodide to give the methyl ether directly on C4: C4-OCH3. Step 5 - Reaction with H2/Pd-BaSO4 (Lindlar-type catalyst): This is a poisoned palladium catalyst that performs partial (syn) hydrogenation of the internal alkyne -C≡C-CH2CH3 to give the cis (Z) alkene: C1-CH=CH-CH2CH3 with Z geometry (cis addition of H2). Step 6 - Assemble the product: The product has a cyclohexane ring with a Z (cis) -CH=CH-CH2CH3 group at C1 and an -OCH3 group directly on C4. This matches option (b): cyclohexane ring with cis vinyl group CH=CHCH2CH3 at C1 and OCH3 at C4. Why other options fail: - (a) has CH2-O-CH3 (methoxymethyl) at C4, meaning -CH2-O-CH3, implying the oxygen was inserted via a different pathway; also the alkene is =CHCH2CH3 (exo-type), not an internal alkene. Incorrect. - (c) has =CHCH3 (only one carbon added) and CH2-OCH2CH3 at C4, implying ethyl group went to oxygen and methyl to alkyne - wrong order of alkylation. - (d) has =CHCH3 and O-CH2CH3 at C4, again wrong assignment of alkyl groups. The correct sequence places ethyl on the alkyne and methyl on the oxygen, giving a cis internal alkene and a methoxy group directly on the ring. Therefore, the correct answer is B.