2-Butanol xrightarrow[Delta]{H^{+}} Products. | vedclass

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(C) The dehydration of $2$-butanol $(CH_3CH_2CH(OH)CH_3)$ in the presence of acid $(H^{+})$ and heat $(\Delta)$ follows the $E1$ mechanism.
This reac

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The dehydration of $2$-butanol $(CH_3CH_2CH(OH)CH_3)$ in the presence of acid $(H^{+})$ and heat $(\Delta)$ follows the $E1$ mechanism.
This reaction produces a mixture of alkenes: $1$-butene $(CH_2=CHCH_2CH_3)$,$cis$-$2$-butene,and $trans$-$2$-butene.
When these alkenes react with $Br_2$ in $CCl_4$,they undergo electrophilic addition to form vicinal dibromides.
$1$-butene reacts with $Br_2$ to form $1,2$-dibromobutane,which is a chiral molecule existing as a racemic mixture of $2$ enantiomers.
$cis$-$2$-butene reacts with $Br_2$ to form the meso isomer of $2,3$-dibromobutane ($1$ product).
$trans$-$2$-butene reacts with $Br_2$ to form a racemic mixture of $2$ enantiomers of $2,3$-dibromobutane.
Total distinct products formed are $2$ (from $1$-butene) + $1$ (meso from $cis$-$2$-butene) + $2$ (racemic from $trans$-$2$-butene) = $5$ products.

$2$-Butanol $\xrightarrow[\Delta]{H^{+}}$ Products. If all products formed in the above reaction undergo reaction with $Br_2(CCl_4)$ individually,then the number of products formed will be:

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