An alkene $C H_{3} C H = C H_{2}$ is treated with $B_{2} H_{6}$ in presence of $H_{2} O_{2}$ . The final product formed is:

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

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

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

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(C H_{3} C H_{2} C H_{2})_{3} B

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Solution

The correct option is C $C H_{3} C H_{2} C H_{2} O H$
Hydroboration-oxidation reaction follows anti-Markovnikov's addition of $H - O H$ across $C = C$ to give alcohol.
Thus an alkene $C H_{3} C H = C H_{2}$ when treated with $B_{2} H_{6}$ in presence of $H_{2} O_{2}$ will yield the final product as $C H_{3} C H_{2} C H_{2} O H$

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Similar questions

A

H_{2} O_{2}

N a O H

H_{3} C - C H = H_{3} C | C - C H_{2} - C H_{3}

B_{2} H_{6}

H_{2} O_{2}

C H_{3} C H = C H_{2} \to C H_{3} C H_{2} C H_{2} O H

{C H}_{3} C H = {C H}_{2}

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