Question

The reaction of one molecule of $HBr$ with one molecule of $1,3-$butadiene at ${40}^{\circ }C$ gives predominantly:

• A. $3-$bromobutene under kinetically controlled conditions
• B. $1-$bromo-$2$-butene under thermodynamically controlled conditions
• C. $3-$bromobutene under thermodynamically controlled conditions
• D. $1-$bromo$-2-$butene under kinetically controlled conditions

Answer 1

The correct option is B $1-$bromo-$2$-butene under thermodynamically controlled conditions

Solution:- (B) $1$-bromo -$2$-butene under thermodynamically controlled conditions

$C{H}_2=CH-CH=C{H}_2+HBr⟶\underset{1,2-\text{addition product}}{C{H}_3-\underset{\stackrel{|}{Br}}{CH}-CH=C{H}_2}+\underset{1,4-\text{addition product}}{C{H}_3-CH=CH-\underset{\stackrel{|}{Br}}{C{H}_2}}$

Under mild conditions (temprature $=80℃$), kinetic product is the $1,2$-addition product and under vigorous conditions, (temp. $=40℃$) thermodynamic product is the $1,4$-addition product.

Hence, $1$-bromo -$2$-butene is the major product under a given condition.