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Q.
The alkene $R-CH=CH_2$ reacts readily with $B_2 H_6$ and formed the product $B$ which on oxidation with alkaline $H_2 O_2$ produces
AIPMTAIPMT 1995Hydrocarbons
Solution:
Here, $\frac{1}{2}$ mole of $B _{2} H _{6}$ react with alkene by
Markownikoff's addition and form trialkyl Borone called Hydroboration, $H _{2} O _{2} / OH ^{-}$gives oxidation. So, trialky borone oxidise in alcohols and reaction is also called Hydroboration-oxidation.
$3 R - CH = CH _{2} \xrightarrow{ B _{2} H _{6}}\underset{'B' }{\left( R - CH _{2}- CH _{2}\right)_{3}}- B$
$\xrightarrow{ H _{2} O _{2} / H ^{+}}3 R - CH _{2}- CH _{2}- OH$