Question

Which will undergo fastest $S_{N}2$ substitution reaction when treated with $NaOH?$

• A. $H-\underset{Br}{\underset{|}{\stackrel{H}{\stackrel{|}{C}}}}-C{H}_2-C{H}_2-C{H}_3$
• B. $H-\underset{C_2{H}_5}{\underset{|}{\stackrel{C{H}_3}{\stackrel{|}{C}}}}-Br$
• C. $C{H}_3-\underset{C{H}_3}{\underset{|}{\stackrel{C{H}_3}{\stackrel{|}{C}}}}-Br$
• D. $H_5{C}_2-\underset{H}{\underset{|}{\stackrel{C{H}_3}{\stackrel{|}{C}}}}-Br$

Answer 1

The correct option is A $H-\underset{Br}{\underset{|}{\stackrel{H}{\stackrel{|}{C}}}}-C{H}_2-C{H}_2-C{H}_3$

$H-\underset{Br}{\underset{|}{\stackrel{H}{\stackrel{|}{C}}}}-C{H}_2-C{H}_2-C{H}_3$ is a primary alkyl halide. It will undergo fastest $S_{N}2$ substitution reaction when treated with NaOH.

Primary alkyl halides prefer substitution through $S{N}^2$ mechanism.

Secondary alkyl halides prefer substitution through either $S{N}^1$ or $S{N}^2$ mechanism. It depends on the strength of nucleophile and solvent polarity.

Tertiary alkyl halides prefer substitution through $S{N}^1$ mechanism.

$C{H}_3-\underset{C{H}_3}{\underset{|}{\stackrel{C{H}_3}{\stackrel{|}{C}}}}-Br$ are tertiary alkyl halides.
$H_5{C}_2-\underset{H}{\underset{|}{\stackrel{C{H}_3}{\stackrel{|}{C}}}}-Br$ is a secondary alkyl halide.