CH_3CH_2CH=CH_2 xrightarrow{B_2H_6 / H_2O_2, | vedclass

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(B) The reaction of $CH_3CH_2CH=CH_2$ with $B_2H_6$ followed by oxidation with $H_2O_2$ in the presence of $OH^-$ is known as hydroboration-oxidation.

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$CH_3CH_2CH_2CH_2OH$

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(B) The reaction of $CH_3CH_2CH=CH_2$ with $B_2H_6$ followed by oxidation with $H_2O_2$ in the presence of $OH^-$ is known as hydroboration-oxidation. This reaction follows anti-Markovnikov's rule,where the $OH$ group is added to the less substituted carbon atom of the double bond. Thus,the product $Z$ formed is $CH_3CH_2CH_2CH_2OH$ (butan$-1-$ol).

$CH_3CH_2CH=CH_2 \xrightarrow{B_2H_6 / H_2O_2, OH^{-}} Z$. What is $Z?$

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