Which of the following reagents can be used to convert $1-\mathrm{pentyne}$ into an $\mathrm{aldehyde}$ via hydroboration-oxidation?

• A. 1. ${\mathrm{BH}}_3$ in THF, then $H_2$O, $2_{\mathrm{}}$ and $\mathrm{NaOH}$
• B. 1. $R_2$BH (such as disiamylborane), then $H_2$O, $2_{\mathrm{}}$ and $\mathrm{NaOH}$
• C. $O{3}_{\mathrm{}}$ followed by $\mathrm{Zn}$ and $H{2}_{\mathrm{}}$O
• D. $\mathrm{HgSO}{4}_{\mathrm{}}$ and $H{2}_{\mathrm{SO}}{4}_{\mathrm{}}$ in water

Answer: (B)

Verified step by step guidance

1. Identify the type of alkyne: 1-pentyne is a terminal alkyne, meaning the triple bond is at the end of the carbon chain.
2. Recall that hydroboration-oxidation of terminal alkynes with disubstituted boranes (like disiamylborane, R2BH) leads to the formation of aldehydes via anti-Markovnikov addition.
3. Understand why BH3 (borane) itself is not suitable: BH3 tends to add twice to terminal alkynes, giving a mixture of products and eventually ketones after oxidation, not aldehydes.
4. Recognize the reaction conditions needed: first, hydroboration with a bulky dialkylborane reagent (R2BH) to ensure regioselective addition to the terminal carbon, then oxidation with hydrogen peroxide (H2O2) in basic medium (NaOH) to convert the organoborane intermediate into the aldehyde.
5. Eliminate other reagents: O3 followed by Zn/H2O is ozonolysis, which cleaves the triple bond rather than converting it to an aldehyde; HgSO4 with H2SO4 leads to Markovnikov hydration giving ketones, not aldehydes.