| Methane can be chlorinated by |
| (I) treating with chlorine in presence of UV light. |
| (II) Heating with chlorine in presence of tetraethyl lead. |
| (III) Treating with \[tert\]-butyl hypochlorite in presence Of UV light. |
A) Only method (I)
B) By methods (I) and (II)
C) By methods (I) and (III)
D) By methods (I), (II) and (III)
Correct Answer: D
Solution :
| Chlorination of methane is a free radical reaction and hence it can be initiated by any factor that can produce chlorine free radical. |
| [a] \[Cl-Cl\xrightarrow{UV}C{{l}^{.}}+C{{l}^{.}}\] |
| \[C{{H}_{4}}+C{{l}^{.}}\xrightarrow{{}}CH_{3}^{.}+HCl\] |
| \[CH_{3}^{.}+Cl-Cl\xrightarrow{{}}C{{H}_{3}}Cl+C{{l}^{.}}\] |
| [b] \[{{({{C}_{2}}{{H}_{5}})}_{4}}Pb\xrightarrow{heat}4C{{H}_{3}}CH_{2}^{.}+p{{b}^{.}}\] |
| \[C{{H}_{3}}CH_{2}^{.}+Cl-Cl\xrightarrow{{}}C{{H}_{3}}C{{H}_{2}}Cl+C{{l}^{.}}\] |
| \[C{{H}_{4}}+Cl\xrightarrow[{}]{}CH_{3}^{.}+HCl\] |
| \[C{{H}_{3}}+Cl-Cl\xrightarrow[{}]{}CH_{3}^{.}Cl+C{{l}^{.}}\] |
| [c] \[{{(C{{H}_{3}})}_{3}}COCl\xrightarrow{UV}{{(C{{H}_{3}})}_{3}}C{{O}^{.}}+C{{l}^{.}}\] |
| \[C{{H}_{4}}+{{(C{{H}_{3}})}_{3}}C{{O}^{.}}\xrightarrow{{}}CH_{3}^{.}+{{(C{{H}_{3}})}_{3}}COH\] |
| \[CH_{3}^{.}+{{(C{{H}_{3}})}_{3}}COCl\xrightarrow{{}}C{{H}_{3}}Cl+{{(C{{H}_{3}})}_{3}}C{{O}^{.}}\] |
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