:The reaction of $R C H = C H_{2}$ with $B_{2} H_{6}$ followed by oxidation with alkaline $H_{2} O_{2}$ gives

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

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R C O C H_{3}

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R C H_{2} C H_{2} O H

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R C H_{2} C H O

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Solution

The correct option is B $R C H_{2} C H_{2} O H$
The reaction of $R C H = C H_{2}$ with $B_{2} H_{6}$ followed by oxidation with alkaline $H_{2} O_{2}$ gives $R C H_{2} C H_{2} O H$ .

$R C H = C H_{2} + B_{2} H_{6} \to R C H_{2} C H_{2} (B_{2} H_{5}) H_{2} O_{2} - -- \to R C H_{2} C H_{2} O H$

Hydroboration oxidation reaction always follows anti markownikoff's rule. So answer is C.

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:The reaction of RCH=CH2 with B2H6 followed by oxidation with alkaline H2O2 gives

The correct option is B RCH2CH2OHThe reaction of RCH=CH2 with B2H6 followed by oxidation with alkaline H2O2 gives RCH2CH2OH.RCH=CH2+B2H6→RCH2CH2(B2H5)H2O2−−−→RCH2CH2OHHydroboration oxidation reaction always follows anti markownikoff's rule. So answer is C.

:The reaction of $R C H = C H_{2}$ with $B_{2} H_{6}$ followed by oxidation with alkaline $H_{2} O_{2}$ gives