See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: 2-heptanone (CH3-CO-CH2CH2CH2CH2CH3) has 7 carbons with a ketone at C2. The strategy is to build a terminal alkyne with the right carbon count, then perform Markovnikov hydration (Hg2+/H2SO4/H2O) to give a methyl ketone. Step 1: Identify the target. 2-heptanone = CH3COCH2CH2CH2CH2CH3. It has 7 carbons, ketone at position 2. Step 2: Retrosynthetic analysis. Hydration of a terminal alkyne (Markovnikov) gives a methyl ketone: RC≡CH + H2O/Hg2+/H2SO4 → RC(=O)CH3. So we need 1-heptyne (CH3CH2CH2CH2CH2C≡CH) as Y to get 2-heptanone. Step 3: Build 1-heptyne. 1-heptyne has 7 carbons. We can make it by alkylating a terminal alkyne anion. If we start with ethyne (HC≡CH, 2 carbons) and treat with NaNH2, we get the acetylide HC≡C⁻Na⁺ (X). Then alkylate with n-C5H12Br (n-pentyl bromide, C5H11Br, 5 carbons): HC≡C⁻ + n-C5H11Br → HC≡C-C5H11 = 1-heptyne (Y, 7 carbons total). Hydration of 1-heptyne with H2O/Hg2+/H2SO4 gives 2-heptanone. Step 4: Verify option B. Ethyne (2C) + NaNH2 → HC≡CNa (X) + n-C5H12Br (n-pentyl bromide, which is C5H11Br, 5C) → 1-heptyne (Y, 7C) → hydration → 2-heptanone. This is correct. Step 5: Check other options. - Option (a): propyne (3C) + n-C4H9Br (4C) → 1-heptyne... wait, propyne is CH3C≡CH; NaNH2 deprotonates terminal H to give CH3C≡C⁻. Alkylation with n-C4H9Br gives CH3C≡C-C4H9 = 2-heptyne (internal alkyne). Hydration of internal alkyne gives a mixture of ketones, not exclusively 2-heptanone. Incorrect. - Option (c): 1-hexyne (6C) + CH2Br (1C) → 7C alkyne, but CH2Br (methylene bromide/CH2Br2 or bromomethane CH3Br?) is ambiguous; using CH3Br gives 1-heptyne, but n-CH2Br is not a standard reagent (it would be CH3Br for 1C addition). Also the reagent written is CH2Br which is unusual. This route is not the best/standard approach. - Option (d): 1-pentyne (5C) + C2H5Br (2C) → 1-pentyne has terminal alkyne CH3CH2CH2C≡CH; NaNH2 gives CH3CH2CH2C≡C⁻; + C2H5Br gives CH3CH2CH2C≡CC2H5 = 3-heptyne (internal alkyne). Hydration gives a mixture, not 2-heptanone. Incorrect. Only option B correctly produces a terminal alkyne (1-heptyne) that upon Markovnikov hydration gives exclusively 2-heptanone. Therefore, the correct answer is B.