Which statement best describes hydroboration-oxidation of propene?

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Multiple Choice

Which statement best describes hydroboration-oxidation of propene?

Hydroboration-oxidation adds water across the double bond in a way that places the oxygen on the less substituted carbon, giving anti-Markovnikov orientation. In propene, the less substituted carbon is the terminal CH2 group, so boron attaches to that carbon while hydrogen adds to the more substituted internal carbon. After oxidation, the boron-carbon bond is replaced by a C–OH bond, producing a primary alcohol. So propene becomes 1-propanol. This contrasts with acid-catalyzed hydration, which follows Markovnikov selectivity and would yield 2-propanol. The term anti-hydration isn’t the process here, and halohydrin formation involves halogenation followed by hydration, not hydroboration-oxidation.

Hydroboration-oxidation adds water across the double bond in a way that places the oxygen on the less substituted carbon, giving anti-Markovnikov orientation. In propene, the less substituted carbon is the terminal CH2 group, so boron attaches to that carbon while hydrogen adds to the more substituted internal carbon. After oxidation, the boron-carbon bond is replaced by a C–OH bond, producing a primary alcohol. So propene becomes 1-propanol.

This contrasts with acid-catalyzed hydration, which follows Markovnikov selectivity and would yield 2-propanol. The term anti-hydration isn’t the process here, and halohydrin formation involves halogenation followed by hydration, not hydroboration-oxidation.

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