An alkene $\ce{CH_3CH=CH_2}$ is treated with $\ce{B_2H_6}$ in presence of $\ce{H_2O_2}.$ The final product formed is$:$
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Hydroboration$-$oxidation reaction follows anti-Markovnikov's addition of $\ce{H−OH}$ across $\ce{C=C}$ to give alcohol.
Thus an alkene $\ce{CH_3CH=CH2}$ when treated with $\ce{B_2H_{6}}$ in presence of $\ce{H_2O_2}$ will yield the final product as $\ce{CH_3CH_2CH_{2}OH}$
Thus an alkene $\ce{CH_3CH=CH2}$ when treated with $\ce{B_2H_{6}}$ in presence of $\ce{H_2O_2}$ will yield the final product as $\ce{CH_3CH_2CH_{2}OH}$
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