The reaction of one molecule of $$HBr$$ with one molecule of $$1,3-$$butadiene at $${40}^{o}C$$ 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

Solution

Solution:- (B) $$1$$-bromo -$$2$$-butene under thermodynamically controlled conditions
$$C{H}_{2}=CH-CH=C{H}_{2} + HBr \longrightarrow \underset{1,2-\text{addition product}}{C{H}_{3}-\underset{\overset{|}{Br}}{CH}-CH=C{H}_{2}} + \underset{1,4-\text{addition product}}{C{H}_{3}-CH=CH-\underset{\overset{|}{Br}}{C{H}_{2}}}$$
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.