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$C{{H}_{4}}+C{{l}_{2}}\xrightarrow{h\nu }C{{H}_{3}}Cl+HCl$ .
To obtain high yields of $C{{H}_{3}}Cl$, the ratio of $C{{H}_{4}}\text{ to C}{{\text{l}}_{2}}$ must be:
[A] High
[B] Low
[C] Equal
[D] Can’t be predicted

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Last updated date: 16th Jun 2024
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Answer
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HINT: Here, to answer this question you should remember that the given reaction proceeds via substitution mechanism. The reaction cannot be controlled by general physical means once started. The hydrogen atoms will be replaced by chlorine to give the product.

Complete step by step solution:
Here, the reagents given in the reaction are methane and chlorine. When we expose this mixture to ultraviolet light, a substitution reaction takes up giving us chloromethane as a product.
However, the reaction does not stop here as all the hydrogen in the methane can be replaced by chlorine and we can get any of chloromethane, dichloromethane, trichloromethane or tetrachloromethane.
Even maintaining the proportion of chlorine and methane will not give us the desired product but a mixture of products. And, if we add excess chlorine, we will get tetrachloromethane.
If we want chloromethane as the major product, we can carry on the reaction using excess methane so that the probability of chlorine radical hitting methane is higher. Although we will get a mixture of products, we can expect chloromethane yield to be higher.
We can write the reaction as- $C{{H}_{4}}+C{{l}_{2}}\xrightarrow{h\nu }C{{H}_{3}}Cl+HCl$
So, we can understand from the above discussion that $C{{H}_{4}}\text{ to C}{{\text{l}}_{2}}$ ratio should be higher to expect higher yield of $C{{H}_{3}}Cl$.

Therefore, the correct answer is option [A] High.

NOTE: Here, the overall reaction is $C{{H}_{4}}+C{{l}_{2}}\xrightarrow{h\nu }C{{H}_{3}}Cl+HCl$
However, as the reaction proceeds, due to unavailability of methane, the chlorine radical will hit on chloromethane and give us dichloromethane and the reaction will continue till all the hydrogen atoms are replaced.
$\begin{align}
& C{{H}_{4}}+C{{l}_{2}}\to C{{H}_{3}}Cl+HCl \\
& C{{H}_{3}}Cl+C{{l}_{2}}\to C{{H}_{2}}C{{l}_{2}}+HCl \\
& C{{H}_{2}}C{{l}_{2}}+C{{l}_{2}}\to CHC{{l}_{3}}+HCl \\
& CHC{{l}_{3}}+C{{l}_{2}}\to CC{{l}_{4}}+HCl \\
\end{align}$