Question

Select the chain propagation steps in the free-radical chlorination of methane.
(1)$C{l}_2\to 2C{l}^{\cdot }$
(2) $C{l}^{\cdot }+C{H}_4\stackrel{}{\to }C{H}_{3}C1+H^{\cdot }$
(3) $C{1}^{\cdot }+C{H}_4\stackrel{}{\to }C{H}_3^{\cdot }+HCI$
(4) $H^{\cdot }+C{l}_2\stackrel{}{\to }HCl+C{l}^{\cdot }$
(5) $C{H}_3^{\cdot }+C{l}_2\stackrel{}{\to }C{H}_{3}CI+C{l}^{\cdot }$

• A. 2,3,5
• B. 1,3,6
• C. 3,5
• D. 2,3,4

Answer 1

Answer: (C)

3,5

(1) $C{l}_2⟶2C{l}^{\cdot }$ [Initiation]

(2) $C{l}^{\cdot }+C{H}_4⟶C{H}_{3}Cl+H^{\cdot }$

(3) $C{l}^{\cdot }+C{H}_4⟶C{H}_3^{\cdot }+HCl$ [Propagation]

(4) $H^{\cdot }+C{l}_2⟶HCl+C{l}^{\cdot }$

(5) $C{H}_3^{\cdot }+C{l}_2⟶C{H}_{3}Cl+C{l}^{\cdot }$ [Propagation]

In the first propagation step (3), a $C{l}^{\cdot }$ combines with a hydrogen on $C{H}_4$ giving $HCl$ and ${}^{\cdot }C{H}_3$. In the $2^{nd}$ propagation step (5) $C{l}_2$ combines with ${}^{\cdot }C{H}_3$ to form $C{H}_{3}Cl$ and $C{l}^{\cdot }$.