Question

The following alkyl halide can be prepared by the addition of HBr to two different alkenes. Draw the structures of both (reddish-brown 5 Br).

Short answer

The structure of alkenes is as follows;

${\text{H}}_{\text{3}}{\text{C-CH}}_{\text{2}}{\text{-CH=CH-CH}}_{\text{3}}$

${\text{H}}_{\text{3}}{\text{C-CH}}_{\text{2}}{\text{-CH}}_{\text{2}}{\text{-CH=CH}}_{\text{2}}$

Step-by-step solution

Addition reaction

The reaction in which two or more molecules combine to form a single molecule is called an addition reaction.

Addition of HBr to alkene to form alkyl halide

The structure of the given alkyl halide is ${\text{H}}_{\text{3}}{\text{C-CH}}_{\text{2}}{\text{-CH}}_{\text{2}}\text{-CH}\left(\text{Br}\right){\text{-CH}}_{\text{3}}$. Hence there are two possibilities by which the given alkyl halide is formed;

$$\begin{gathered} {\text{(a)H}}_{\text{3}}{\text{C-CH}}_{\text{2}}{\text{-CH=CH-CH}}_{\text{3}}\text{+HBr}\to {\text{H}}_{\text{3}}{\text{C-CH}}_{\text{2}}{\text{-CH}}_{\text{2}}\text{-CH}\left(\text{Br}\right){\text{-CH}}_{\text{3}} \\ {\text{(b)H}}_{\text{3}}{\text{C-CH}}_{\text{2}}{\text{-CH}}_{\text{2}}{\text{-CH=CH}}_{\text{2}}\text{+HBr}\to {\text{H}}_{\text{3}}{\text{C-CH}}_{\text{2}}{\text{-CH}}_{\text{2}}\text{-CH}\left(\text{Br}\right){\text{-CH}}_{\text{3}} \end{gathered}$$

Using the above two alkene structures, we can get the alkyl halide resultant.