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Q.
In the reaction, $ C{{l}_{2}}+C{{H}_{4}}\xrightarrow{hv}C{{H}_{3}}Cl+HCl $ presence of a small amount of oxygen
BVP MedicalBVP Medical 2012
Solution:
During chlorination of methane, a small amount of oxygen acts as radical inhibitor. $ {{O}_{2}} $ combines with $ ^{*}C{{H}_{3}} $ free radical to form methyl peroxy free radicals which are much less reactive than $ C{{H}_{3}} $ free radicals and hence the reaction slows down. After the inhibitor has been consumed, the reaction proceeds normally. $ \underset{\begin{smallmatrix} methyl \\ radical \end{smallmatrix}}{\mathop{^{*}C{{H}_{3}}+O-O}}\,\xrightarrow{{}}\underset{\begin{smallmatrix} methyl \\ peroxy\,radical \end{smallmatrix}}{\mathop{C{{H}_{3}}-O-{{O}^{\bullet }}}}\, $