Question

The cholorination of methane may proceed according to one of the two alternative mechanisms (only propagation step are shown). Which statement about these alternative reaction is correct?
Mechanism A: $C{l}^{\ast }+C{H}_4\to HCl+H_3{C}^{\ast }$
$H_3{C}^{\ast }+C{l}_2\to H_{3}CCl+C{l}^{\ast }$
Mechanism B: $C{l}^{\ast }+C{H}_4\to H_{3}CCl+H^{\ast }$
$H^{\ast }+C{l}_2\to HCl+C{l}^{\ast }$

• A. Mechanism A is preferred because none of the steps is strongly endothermic
• B. Mechanism A is preferred because second steps is strongly exothermic
• C. Mechanism B is preferred because none of the steps is strongly endothermic
• D. Mechanism B is preferred because second steps is strongly exothermic

Answer 1

The correct option is A Mechanism A is preferred because none of the steps is strongly endothermic

Mechanism A: $C{l}^{\ast }+C{H}_4\to HCl+H_3{C}^{\ast }$
$H_3{C}^{\ast }+C{l}_2\to H_{3}CCl+C{l}^{\ast }$

Mechanism B: $C{l}^{\ast }+C{H}_4\to H_{3}CCl+H^{\ast }$
$H^{\ast }+C{l}_2\to HCl+C{l}^{\ast }$

In mechanism A, $C{H}_3^{\ast }$ forms which is more stable than $H^{\ast }$in mechanism B. So mechanism A is preferred over B. In A, none of steps is strongly endothermic which is favorable condition.