Pent-1-ene reacts with diborane to form X. X | vedclass

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Question

(B) The reaction of pent-$1$-ene $(CH_3CH_2CH_2CH=CH_2)$ with diborane $(B_2H_6)$ followed by oxidation with $H_2O_2$ in the presence of aqueous $NaOH

Vedclass · Accepted Answer

$CH_3CH_2CH_2CH_2CH_2OH$

Vedclass · Answer

(B) The reaction of pent-$1$-ene $(CH_3CH_2CH_2CH=CH_2)$ with diborane $(B_2H_6)$ followed by oxidation with $H_2O_2$ in the presence of aqueous $NaOH$ is known as hydroboration-oxidation.This reaction follows anti-Markovnikov addition of water ($H$ and $OH$) across the double bond.The terminal carbon atom gets the $-OH$ group,resulting in the formation of pentan-$1$-ol $(CH_3CH_2CH_2CH_2CH_2OH)$.

Pent-$1$-ene reacts with diborane to form $X$. $X$ on oxidation using $H_2O_2$ in the presence of aqueous $NaOH$ gives $Y$. Compound $Y$ is

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