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

$S{N^2}$- 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.