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
$$C{H_3}Cl > \left( {C{H_3}} \right)C{H_2} - Cl > {\left( {C{H_3}} \right)_2}CH - Cl > {\left( {C{H_3}} \right)_3}CCl$$

Releted MCQ Question on
Organic Chemistry >> Alkyl and Aryl Halide

Releted Question 1

Chlorobenzene can be prepared by reacting aniline with :

A. hydrochloric acid
B. cuprous chloride
C. chlorine in presence of anhydrous aluminium chloride
D. nitrous acid followed by heating with cuprous chloride
Releted Question 2

The reaction conditions leading to the best yields of $${C_2}{H_5}Cl$$  are :

A. \[{{C}_{2}}{{H}_{6}}\text{(excess)}+C{{l}_{2}}\xrightarrow{\text{uv}\,\text{light}}\]
B. \[{{C}_{2}}{{H}_{6}}+C{{l}_{2}}\xrightarrow[\text{room temperature}]{\text{dark}}\]
C. \[{{C}_{2}}{{H}_{6}}+C{{l}_{2}}\text{(excess)}\xrightarrow{\text{uv}\,\text{light}}\]
D. \[{{C}_{2}}{{H}_{6}}+C{{l}_{2}}\xrightarrow{\text{uv}\,\text{light}}\]
Releted Question 3

$$n - $$  Propyl bromide on treatment with ethanolic potassium hydroxide produces

A. Propane
B. Propene
C. Propyne
D. Propanol
Releted Question 4

The number of structural and configurational isomers of a bromo compound, $${C_5}{H_9}Br,$$   formed by the addition of $$HBr$$  to 2 - pentyne respectively are

A. 1 and 2
B. 2 and 4
C. 4 and 2
D. 2 and 1

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Alkyl and Aryl Halide


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