Question

Which step is not important step in free radical bromination of propane?

• A. $C{H}_3\dot{C}HC{H}_3+B{r}^{{}^{\ast }}\to C{H}_{3}CHBrC{H}_3$
• B. $C{H}_3{C}^{\ast }HC{H}_3+B{r}_2\to C{H}_{3}CHBrC{H}_3+B\dot{r}$
• C. $C{H}_3^{\ast }CHC{H}_3+HBr\to C{H}_{3}CHBrC{H}_3+H^{\ast }$
• D. $C{H}_3^{\ast }C{H}_{2}C{H}_3+B\dot{r}\to C{H}_{3}CHC{H}_3+HB\dot{r}$

Answer 1

The correct option is C $C{H}_3^{\ast }CHC{H}_3+HBr\to C{H}_{3}CHBrC{H}_3+H^{\ast }$

Free radical bromination involves:

Initiation: Splitting or homolysis of a bromine molecule to form two bromine atoms, initiated by ultraviolet radiation or sunlight. A bromine atom has an unpaired electron and acts as a free radical.

$Br-Br\stackrel{uv}{\to }Br\ast \ast Br$

Chain propagation (two steps):

A hydrogen atom is pulled off from propane leaving an alkyl radical. The alkyl radical then pulls a $Br\ast$ from $B{r}_2$.

$C{H}_3{C}^{\ast }HC{H}_3+B{r}_2\to C{H}_{3}CHBrC{H}_3+B\dot{r}$

$C{H}_3^{\ast }C{H}_{2}C{H}_3+B\dot{r}\to C{H}_{3}CHC{H}_3+HBr$

Chain termination: recombination of two free radicals.

$C{H}_3\dot{C}HC{H}_3+B{r}^{{}^{\ast }}\to C{H}_{3}CHBrC{H}_3$

The following step is not involved in the mechanism of bromination of propane:

$C{H}_3^{\ast }CHC{H}_3+HBr\to C{H}_{3}CHBrC{H}_3+H^{\ast }$