Question

What is one of the products of addition of $HBr$ to 2-Butene?
(A) 1-bromobutane
(B) 2-bromobutane
(C) 1,2-dibromobutane
(D) 2,3-dibromobutane

Answer 1

The above reaction involves addition of alkene for the formation of alkyl halide. Alkyl halides serve as valuable intermediates for synthesis of a variety of other classes of organic compounds. This reaction undergoes the electrophilic addition.

Complete step by step answer:
Electrophilic addition of symmetrical alkene is carried as:
Step 1: Initiation step:
$HBr\xrightarrow{{ionization}}{H^ + } + B{r^ - }$
First ionization of $HBr$ takes place and it gets dissociated into ${H^ + }$ and $B{r^ - }$ ions.
Step 2: Propagation step:
$C{H_3}$−$CH$=$CH$−$C{H_3}$ $+ {H^ + }\xrightarrow{{slow}}C{H_3}$−$\mathop C\limits^ + H$−$C{H_2}C{H_3}$
As the ${H^ + }$ ion attacks the $\pi$ bond, the formation of carbocation takes place, i.e.
formation of butyl carbocation takes place.
Step3: Termination step:
$B{r^ - } + C{H_3}$−$\mathop C\limits^ + H$−$C{H_2}C{H_3}$ $\xrightarrow[{Fast}]{{Nucleophilicattack}}C{H_3}$ $CH(Br)C{H_2}C{H_3}$
In this step, Nucleophilic addition takes place. This step is the fast step.in this step bromide
ion attack on carbocation and formation of 2-Bromobutane take place. Step 2 is the slow
step and is the rate determining step. Since 2-Butene is symmetrical alkene so there is only
one theoretical product is possible. The reactivity of addition of $HBr$ in case of alkene is
lesser than $HI$. But it is greater than $HCl$ because of the bond dissociation energy of
$HBr$ which lies in between the $HI$ and $HCl$.
This reaction involves a trans addition. The product formed in case of symmetrical alkene is
different from that of an unsymmetrical alkene.
Hence option B is correct.

Note:
In case of symmetrical alkene no such rule is applied, as it was in the case of an
unsymmetrical alkene. In the case of an unsymmetrical alkene, there is used the
markovnikov’s and anti markovnikov's rule. Antimarkovnikov’s rule is also known as peroxide
effect.