But-1-ene reacts with diborane to give 'X'. O | vedclass

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(C) The reaction of $but-1-ene$ $(CH_3CH_2CH=CH_2)$ with diborane $(B_2H_6)$ is a hydroboration reaction,which follows anti-Markovnikov addition.This

Vedclass · Accepted Answer

$X = (CH_3CH_2CH_2CH_2)_3B, Y = CH_3CH_2CH_2CH_2OH$

Vedclass · Answer

(C) The reaction of $but-1-ene$ $(CH_3CH_2CH=CH_2)$ with diborane $(B_2H_6)$ is a hydroboration reaction,which follows anti-Markovnikov addition.This results in the formation of a trialkylborane '$X$'.$6CH_3CH_2CH=CH_2 + B_2H_6 \rightarrow 2(CH_3CH_2CH_2CH_2)_3B$ (where $X = (CH_3CH_2CH_2CH_2)_3B$).Subsequent oxidation of '$X$' with $H_2O_2$ in the presence of aqueous $NaOH$ replaces the boron atom with a hydroxyl group $(-OH)$,yielding a primary alcohol '$Y$'.$(CH_3CH_2CH_2CH_2)_3B + 3H_2O_2 \xrightarrow{NaOH} 3CH_3CH_2CH_2CH_2OH + B(OH)_3$.Thus,'$X$' is $(CH_3CH_2CH_2CH_2)_3B$ and '$Y$' is $butan-1-ol$ $(CH_3CH_2CH_2CH_2OH)$.

$But-1-ene$ reacts with diborane to give '$X$'. On oxidation of '$X$' with $H_2O_2$ in the presence of aqueous $NaOH$,it forms alcohol '$Y$'. What are '$X$' and '$Y$'?

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