Which gives maximum yield of $C_{2} H_{5} C l$ ?

C_{2} H_{6} + C l_{2} e x c e s s h v . l i g h t - --- \to C_{2} H_{5} C l + H C l

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C_{2} H_{6} e x c e s s + C l_{2} h v - - \to l i g h t C_{2} H_{5} C l + H C l

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C_{2} H_{6} + C l_{2} h v - - \to l i g h t C_{2} H_{5} C l

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C_{2} H_{6} + C l_{2} D a r k - -- \to C_{2} H_{5} C l + H C l

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Solution

The correct option is B $C_{2} H_{6} e x c e s s + C l_{2} h v - - \to l i g h t C_{2} H_{5} C l + H C l$
The reaction of $C_{2} H_{6}$ and $C l_{2}$ in presence of $h ν$ light can give monochlorination as well as $d i$ and $t r i -$ chlorination.
If we want the only monochlorinated product, we need to make sure that $C l_{2}$ is used for monochlorination only. This is possible when $C_{2} H_{6}$ is taken in excess.

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Similar questions

Condition for maximum yield of $C_{2} H_{5} C l$ is

C_{2} H_{5} C l

Condition for maximum yield of $C_{2} H_{5} C l$ is

C_{2} H_{5} C l

C_{2} H_{6} + C l_{2} \to H C l + C_{2} H_{5} C l

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