Propan-1-ol can be prepared from propene by alcohol

H_{2} O / H_{2} S O_{4}

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H g (O A c)_{2} / H_{2} O

followed by

N a B H_{4}

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B_{2} H_{6}

followed by

H_{2} O_{2}

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C H_{3} C O_{2} H / H_{2} S O_{4}

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Solution

The correct option is C $B_{2} H_{6}$ followed by $H_{2} O_{2}$
When propene reacts with mercuric acetate followed by $N a B H_{4}$ , it gives secondary or tertiary alcohols, in accordance with Markownikoff’s rule. But, with diborane followed by $H_{2} O_{2},$ it gives primary alcohols which is in accordance with Anti-Markownikoff’s rule. $6 C H_{3} - C H = C H_{2} + B_{2} H_{6} H_{2} O_{2} - -- \to C H_{3} - C H_{2} - C H_{2} O H$

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Similar questions

1 – propanol can be prepared from propene by

P r o p a n - 1 - o l

Propan-1-ol can be prepared from propene by [AIIMS 2003]

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