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Hydroboration-oxidation of propene CH$_3$-CH=CH$_2$ on the lab gives:
AAcetone, the ketone by Markovnikov-style addition only
BPropan-2-ol, the Markovnikov product of hydration here
CPropanal, the aldehyde by full oxidation of the alkene
DPropan-1-ol, the anti-Markovnikov terminal alcohol product
Answer & Solution
Correct answer: D. Propan-1-ol, the anti-Markovnikov terminal alcohol product
Hydroboration-oxidation gives anti-Markovnikov: H adds to internal C, BH$_2$/OH ends up on the terminal C, giving propan-1-ol.
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