Question

The reaction conditions leading to the best yield of $C_2{H}_{5}Cl$ are:

• A. $C_2{H}_6\left(Excess\right)+{Cl}_2\underset{}{\overset{U.Vlight}{\to }}$
• B. $C_2{H}_6{Cl}_2\left(Excess\right)\underset{roomtemp.}{\overset{Dark}{\to }}$
• C. $C_2{H}_6+{Cl}_2\left(Excess\right)\underset{}{\overset{U.Vlight}{\to }}$
• D. $C_2{H}_6+{Cl}_2\underset{}{\overset{U.Vlight}{\to }}$

Answer 1

The correct option is A $C_2{H}_6\left(Excess\right)+{Cl}_2\underset{}{\overset{U.Vlight}{\to }}$

The reaction proceeds through free radical mechanism. Since ethane is present in excess, mono-substituted product will be the major product. Thus, in this reaction, H atom of ethane will be replaced with Cl atom to form ethyl chloride.