Question

Which of the following alkyl halides is used as a methylating agent?
A.\[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Br}}\]
B. \[{{\rm{C}}_6}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\]
C.\[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{Cl}}\]
D. \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\]

Answer 1

Hint: Methyl group is a compound possessing one carbon atom which forms bonds with three hydrogen atoms. The chemical formula that represents the methyl group is\[{\rm{C}}{{\rm{H}}_{\rm{3}}}\]. Methyl group is obtained from alkyl halide.

Complete Step by Step Solution:
Let’s first understand what a methylating agent is. Any compound which has the capability of adding a methyl group to a compound is termed a methylating agent.
Let’s discuss the characteristics of a methylating agent.
(1) A methylating must have \[{\rm{C}} - {\rm{X}}\] (X is any halogen, Cl, Br, F, I) bond. So, that bond becomes polar, and the bond breaks.
(2) In the C-X bond, C should be of \[{\rm{C}}{{\rm{H}}_{\rm{3}}}\] group. So carbocation or methyl radical forms after the breaking of the bond.

Let’s find the alkyl halide from the options which are used as a methylating agent.
The structure of the \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Br}}\] is \[{\rm{C}}{{\rm{H}}_3} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{Br}}\] , where, Br is bonded to the \[{\rm{C}}{{\rm{H}}_{\rm{2}}}\] group. This violates the second characteristics of the methylating agent. So, it cannot act as a methylating agent.

In \[{{\rm{C}}_6}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\](Chlorobenzene), due to the resonance the C-Cl bond acquires stability. So, it cannot act as a methylating agent. Therefore, option B is incorrect.

In \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{Cl}}\], C-X bond is present and also Cl atom is bonded to \[{\rm{C}}{{\rm{H}}_{\rm{3}}}\] group. Therefore, it can act as a methylating agent.
The structure of the \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\] is \[{\rm{C}}{{\rm{H}}_3} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{Cl}}\] , where, Cl is bonded to the \[{\rm{C}}{{\rm{H}}_{\rm{2}}}\] group. This violates the second characteristics of the methylating agent. So, it cannot acts as methylating agent.
Hence, option C is right.

Note: It is to be noted that, best methylating agent is \[{\rm{C}}{{\rm{H}}_{\rm{3}}} - {\rm{I}}\] because Iodine is the largest among Fluorine, Chlorine and Bromine atoms. And this results in the largest bond length. And due to large bond length, the breaking of \[{\rm{C}}{{\rm{H}}_{\rm{3}}} - {\rm{I}}\] is easy.