Hydroboration-oxidation of $C H_{3} C H = C H_{2}$ gives:

C H_{3} C H_{2} C H_{2} O H

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C H_{3} C H (O H) C H_{3}

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C H_{3} C H (O H) C H_{2} O H

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C H_{3} C O C H_{3}

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Solution

The correct option is A $C H_{3} C H_{2} C H_{2} O H$
A hydroboration-oxidation reaction is a two-step reaction that converts an alkene into neutral alcohol by the addition of water across the double bond, it follows Anti-Markovnikov's rule.
$C H_{3} - C H = C H_{2} 1. B H_{3} / T H F - -------- \to 2. O H^{-} / H_{2} O_{2} C H_{3} - C H_{2} - C H_{2} - O H$

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