Question
In $${S_N}2$$ reactions, the correct order of reactivity for the following compounds :
$$C{H_3}Cl,C{H_3}C{H_2}Cl,{\left( {C{H_3}} \right)_2}CHCl$$ and $${\left( {C{H_3}} \right)_3}CCl$$ is :
A.
$$C{H_3}Cl > {\left( {C{H_3}} \right)_2}CHCl > C{H_3}C{H_2}Cl > {\left( {C{H_3}} \right)_3}CCl$$
B.
$$C{H_3}Cl > C{H_3}C{H_2}Cl > {\left( {C{H_3}} \right)_2}CHCl > {\left( {C{H_3}} \right)_3}CCl$$
C.
$$C{H_3}C{H_2}Cl > C{H_3}Cl > {\left( {C{H_3}} \right)_2}CHCl > {\left( {C{H_3}} \right)_3}CCl$$
D.
$${\left( {C{H_3}} \right)_2}CHCl > C{H_3}C{H_2}Cl > C{H_3}Cl > {\left( {C{H_3}} \right)_3}CCl$$
Answer :
$$C{H_3}Cl > C{H_3}C{H_2}Cl > {\left( {C{H_3}} \right)_2}CHCl > {\left( {C{H_3}} \right)_3}CCl$$
Solution :
Steric congestion around the carbon atom undergoing the inversion process will slow down the $${S_N}2$$ reaction, hence less congestion faster will the reaction. So, the order is
$$C{H_3}Cl > \left( {C{H_3}} \right)C{H_2} - Cl > {\left( {C{H_3}} \right)_2}CH - Cl > {\left( {C{H_3}} \right)_3}CCl$$