Which of the following alcohols can be synthesized selectively by hydroboration-oxidation of an alkene?

• A. ${\mathrm{CH}}_3\mathrm{CH}{2}_2\mathrm{CH}\left(\mathrm{OH}\right)\mathrm{CH}{3}_3$ (3-pentanol)
• B. ${\mathrm{CH}}_3\mathrm{CH}{2}_2\mathrm{CH}{2}_2\left(\mathrm{OH}\right)$ (1-pentanol)
• C. ${\mathrm{CH}}_3\left(\mathrm{OH}\right)\mathrm{CH}{2}_2\mathrm{CH}{2}_2\mathrm{CH}{3}_3$ (2-methyl-1-butanol)
• D. ${\mathrm{CH}}_3\mathrm{CH}\left(\mathrm{OH}\right)\mathrm{CH}{2}_2\mathrm{CH}{3}_3$ (2-pentanol)

Answer: (C)

Verified step by step guidance

1. Recall that hydroboration-oxidation of an alkene adds water (H and OH) across the double bond in an anti-Markovnikov fashion, meaning the OH group attaches to the less substituted carbon of the double bond.
2. Identify the structure of the alkene that would lead to each alcohol upon hydroboration-oxidation. The OH group will be on the less substituted carbon, so the product alcohol will have the OH on the terminal or less substituted carbon.
3. Analyze each alcohol option to determine if the OH is on the less substituted carbon (primary alcohol) or on a more substituted carbon (secondary or tertiary alcohol). Hydroboration-oxidation selectively forms primary alcohols from terminal alkenes.
4. For example, 1-pentanol has the OH on the first carbon (a primary alcohol), which is consistent with hydroboration-oxidation of 1-pentene. In contrast, 2-pentanol or 3-pentanol have the OH on more substituted carbons, which are typically formed by acid-catalyzed hydration, not hydroboration-oxidation.
5. Conclude that the alcohols with the OH on the less substituted carbon (primary alcohols) can be synthesized selectively by hydroboration-oxidation, while those with OH on more substituted carbons cannot.