Consider two alternative mechanism for the chlorination of methane. Which step is crucial in deciding which mechanism is not likely to be observed?
Mechanism I Mechanism II
A ${C l}_{2} ⟶ {2 C l}^{∙}$
A ${C l}_{2} ⟶ {2 C l}^{∙}$
B ${C l}^{∙} + {C H}_{4} ⟶ H C l +^{∙} {C H}_{3}$
D ${C l}^{∙} + {C H}_{4} ⟶ {C H}_{3} C l + H^{∙}$
C $∙ C H_{3} + {C l}_{2} ⟶ {C H}_{3} C l + {C l}^{∙}$
E $H^{∙} + {C l}_{2} ⟶ H C l + {C l}^{∙}$

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Solution

The correct option is B B
According to the reaction, step B is crucial and it will tell whether mechanism is possible or not.

1. Initial reaction
$C l_{2} h ν - \to 2 C l ∙$

2. Chain propagation
$C l ∙ + C H_{4} \to H C l + C H_{3} ∙$
$C H_{3} ∙ + C l_{2} \to C H_{3} C l + C l ∙$

3. Chain termination reactions
$C l ∙ + C l ∙ \to C l_{2}$
$C H_{3} ∙ + C H_{3} ∙ \to C H_{3} - C H_{3}$
$C l ∙ + C H_{3} \to C H_{3} C l$

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Similar questions

A . C l_{2} \to 2 C l^{*}

B . C l^{*} + C H_{4} \to H C l +^{*} C H_{3}

C . ˙ C H_{3} + C l_{2} \to C H_{3} C l + C l^{*}

D . C l_{2} \to 2 C l^{*}

E . C l^{*} + C H_{4} \to C H_{3} C l + H^{*}

F . H^{*} + C l_{2} \to H C l + C l^{*}

C l_{2} h v - \to C l^{∙} + C l^{∙}

C l^{∙} + C H_{4} \to^{∙} C H_{3} + H C l

C l^{∙} + C l^{∙} \to C l_{2}

^{∙} C H_{3} +^{∙} C l \to C H_{3} C l

C l^{*} + C H_{4} \to H C l + H_{3} C^{*}

H_{3} C^{*} + C l_{2} \to H_{3} C C l + C l^{*}

C l^{*} + C H_{4} \to H_{3} C C l + H^{*}

H^{*} + C l_{2} \to H C l + C l^{*}

I : C H_{3} C H_{3} + C l^{*} \to C H_{3} C H_{2} C l + H^{*}

I I : C H_{3} C H_{3} + C l^{*} \to C H_{3} C H_{2}^{*} + H C l

I I I : C H_{3} C H_{2}^{*} + C l_{2} \to C H_{3} C H_{2} C l + C l^{*}

I V : 2 C l^{*} \to C l_{2}

(1) C l_{2} \to ∙ 2 C l (2) ∙ C H_{3} + ∙ C H_{3} \to C H_{3} - C H_{3} (3) ∙ C l + C H_{4} \to C H_{3} C l + ∙ C l (4) ∙ C H_{3} + ∙ C l \to C H_{3} - C l (5) ∙ C H_{3} + C l_{2} \to C H_{3} - C l + ∙ C l

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