Question

In a \[S{N^2}\]substitution reaction of the type $R - Br + C{l^ - }\xrightarrow{{DMF}}R - Cl + B{r^ - }$
Which one of the following has the highest relative rate?
A) \[\begin{array}{*{20}{c}}
  {C{H_3}}& - &{\begin{array}{*{20}{c}}
  {C{H_3}} \\
  | \\
  C \\
  | \\
  {C{H_3}}
\end{array}}& - &{C{H_2}Br}
\end{array}\]
B) $\begin{array}{*{20}{c}}
  {C{H_3}}& - &{C{H_2}}& - &{Br}
\end{array}$
C) $\begin{array}{*{20}{c}}
  {C{H_3}}& - &{C{H_2}}& - &{C{H_2}}& - &{Br}
\end{array}$
D) \[\begin{array}{*{20}{c}}
  {C{H_3} - CH - C{H_2} - Br} \\
  | \\
  {C{H_3}}
\end{array}\]

Answer 1

Hint:\[S{N^2}\]- Bimolecular nucleophilic substitution reaction.
In this reaction, two molecules participate, and an intermediate is formed called transition state. The formation of two molecules as products. \[S{N^2}\] is a single step reaction. Intermediate being unstable, decomposes to give substituted product.

Complete step by step answer:
The relative reactivity of alkyl halide towards \[S{N^2}\] reactions is as follows:
\[Primary > Secondary > Tertiary\]
\[ - C{H_3} > \]\[\begin{array}{*{20}{c}}
  {C{H_3}} \\
  | \\
  { - CH > } \\
  | \\
  {C{H_3}}
\end{array}\]\[\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  {C{H_3}} \\
  | \\
  { - C} \\
  | \\
  {C{H_3}}
\end{array}}&{ - C{H_3}}
\end{array}\]
However if the primary alkyl halide or the nucleophile base is sterically hindered the nucleophile will have difficulty getting the back side of the alpha carbon as a result of this elimination product will be predominant. Here $\begin{array}{*{20}{c}}
  {C{H_3}}& - &{C{H_2}}& - &{Br}
\end{array}$ is the least hindered. Hence, it has the highest relative rate towards \[S{N^2}\] reaction.
\[S{N^2}\] mechanism is followed in case of primary and secondary alkyl halides i.e. \[S{N^2}\] reaction is favoured by small groups on the carbon atoms attached to halogens. SO,
$C{H_3} - X > R - C{H_2} - X > {R_2}CH - X > {R_3}C - X$
Primary is more reactive than secondary and tertiary alkyl halides.
\[S{N^2}\]order: $methyl > ethyl > isopropyl > tert > butyl > alkyl > benzyl$
The concentration of two molecules changes simultaneously in its rate determining step. Hence, this mechanism is called bimolecular nucleophilic substitution of \[S{N^2}\]. At first it undergoes an intermediate transition state
$R - Br + C{l^ - }\xrightarrow{{DMF}}R - Cl + B{r^ - }$
DMF used as a catalyst (dimethyl formamide).
$R - Br + C{l^ - }\xrightarrow{{slow}}\left[ {\mathop {Cl}\limits^{\delta - } \cdots \mathop R\limits^{\delta + } \cdots \mathop {Br}\limits^{\delta - } } \right] \to R - Cl + B{r^ - }$
$\left[ {\mathop {Cl}\limits^{\delta - } \cdots \mathop R\limits^{\delta + } \cdots \mathop {Br}\limits^{\delta - } } \right]$ is the transition state.


Note:
The method by which the substitution reaction takes place depends upon the nature of the alkyl group and the ionizing power of the solvent.
If \[R\] is a group of high \[ + I\] effect then \[R - X\] bond breaks easily.
Similarly, the reaction takes place by \[S{N^2}\] mechanism in methyl and ethyl halide.