A) \[C{{l}_{2}}\xrightarrow{hv}2C{{l}^{\bullet }}\]
B) \[C{{H}_{4}}+C{{l}^{\bullet }}\xrightarrow{{}}C{{H}_{3}}Cl+{{H}^{\bullet }}\]
C) \[C{{H}_{4}}+C{{l}^{\bullet }}\xrightarrow{{}}CH_{3}^{\bullet }+HCl\]
D) \[CH_{3}^{\bullet }+C{{l}^{\bullet }}\xrightarrow{{}}C{{H}_{3}}Cl\]
Correct Answer: C
Solution :
In the propagation reaction, the \[C{{l}^{\bullet }}\] attacks the \[C{{H}_{4}}\] molecule and converts it into another free radical\[\overset{\bullet }{\mathop{C}}\,{{H}_{3}}\]. The \[\overset{\bullet }{\mathop{C}}\,{{H}_{3}}\] thus produced reacts with a molecule of \[C{{l}_{2}}\] and forms another new free radical\[\overset{\bullet }{\mathop{C}}\,l\]. These sequence of reactions, (i) and (ii) is repeated over and over again and the chain gets propagated. Thats why these reactions are known as chain propagation reactions. \[C{{H}_{4}}+C{{l}^{\bullet }}\xrightarrow{{}}\underset{\begin{smallmatrix} methyl \\ free\,\,radical \end{smallmatrix}}{\mathop{\overset{\bullet }{\mathop{C}}\,{{H}_{3}}}}\,+HCl\] ... (i) \[\underset{\begin{smallmatrix} free \\ radical \end{smallmatrix}}{\mathop{\overset{\bullet }{\mathop{C}}\,{{H}_{3}}}}\,+C{{l}_{2}}\xrightarrow{{}}C{{H}_{3}}Cl+\underset{\begin{smallmatrix} chlorine \\ free\,\,radical \end{smallmatrix}}{\mathop{C{{l}^{\bullet }}}}\,\] ... (ii)
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