CH_2 = CH_2 + B_2H_6 xrightarrow[H_2O_2]{NaOH | vedclass

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(B) The reaction given is the hydroboration-oxidation of ethene. $1$. In the first step,$B_2H_6$ (diborane) adds across the double bond of ethene $(CH

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$CH_3CH_2OH$

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(B) The reaction given is the hydroboration-oxidation of ethene. $1$. In the first step,$B_2H_6$ (diborane) adds across the double bond of ethene $(CH_2=CH_2)$ to form a trialkylborane intermediate. $2$. In the second step,the trialkylborane is oxidized by hydrogen peroxide $(H_2O_2)$ in the presence of an aqueous base $(NaOH)$. $3$. This process results in the anti-Markovnikov addition of water across the double bond,yielding ethanol $(CH_3CH_2OH)$ as the final product.

$CH_2 = CH_2 + B_2H_6 \xrightarrow[H_2O_2]{NaOH} \text{Product}$.
The product in the above reaction is:

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$CH_2 = CH_2 + B_2H_6 \xrightarrow[H_2O_2]{NaOH} \text{Product}$.The product in the above reaction is: