See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Hydration of an alkene (acid-catalyzed or via oxymercuration) adds water across a C=C double bond. For an alcohol to be prepared by hydration of an alkene, there must exist an alkene such that addition of water (following Markovnikov's rule or anti-Markovnikov) gives that alcohol. This requires that the carbon bearing the OH group must have been part of a C=C double bond, meaning it must have had at least one hydrogen on it OR the adjacent carbon must have had a hydrogen — essentially, the carbon bearing OH or the adjacent carbon must be able to form a double bond. Step 1: Analyze option (a) — pentan-1-ol (CH3CH2CH2CH2CH2OH). The OH is on C1. This could be prepared by hydroboration-oxidation (anti-Markovnikov hydration) of pent-1-ene. So this CAN be prepared. Step 2: Analyze option (b) — 2-ethylbutan-1-ol. The OH is on a primary carbon (C1), with the adjacent carbon (C2) bearing an ethyl group and a hydrogen. Since C2 has a hydrogen, a double bond between C1 and C2 is possible (2-ethylbut-1-ene exists), and anti-Markovnikov hydration (hydroboration-oxidation) would give this alcohol. So this CAN be prepared. Step 3: Analyze option (c) — 4-methylpentan-1-ol. OH is on C1 of a chain with a methyl branch at C4. The adjacent carbon (C2) has hydrogens, so a C1=C2 double bond (4-methylpent-1-ene) exists. Anti-Markovnikov hydration gives this primary alcohol. So this CAN be prepared. Step 4: Analyze option (d) — 2,2-dimethylpropan-1-ol (neopentyl alcohol, (CH3)3C-CH2OH). The OH is on C1 (primary carbon). The adjacent carbon C2 is a quaternary carbon bearing three methyl groups and no hydrogen. For hydration of an alkene to give this alcohol, we would need a double bond at C1=C2. However, C2 has no hydrogen, so it cannot form a double bond (no H to remove to generate the alkene via elimination, and neopentyl system has no viable alkene precursor that would give a primary OH at C1 without rearrangement). The alkene that would need to exist is (CH3)3C=CH2 (3,3-dimethylbut-1-ene... wait, that is actually 2-methylbut-1-ene analog). Actually the required alkene would be (CH3)2C=CH2 with an extra methyl — i.e., 2,2-dimethylprop-1-ene... but that compound does not exist as the carbon would be pentavalent. In fact, neopentyl alcohol requires C(CH3)3-CH2OH: the C2 is C(CH3)3, a quaternary carbon with no H, so no alkene C1=C2 can be formed. Any other alkene hydration would not give OH at the neopentyl position due to rearrangements. Therefore, 2,2-dimethylpropan-1-ol CANNOT be prepared by simple hydration of an alkene. Therefore, the correct answer is D.