A) 3
B) 4
C) 2
D) 1
Correct Answer: A
Solution :
\[\underset{\begin{smallmatrix} methyl \\ amine \end{smallmatrix}}{\mathop{C{{H}_{3}}N{{H}_{2}}}}\,\xrightarrow[-HI]{C{{H}_{3}}I}\underset{\begin{smallmatrix} dimethyl \\ amine \end{smallmatrix}}{\mathop{{{(C{{H}_{3}})}_{2}}NH}}\,\xrightarrow[-HI]{C{{H}_{3}}I}\] \[\underset{\begin{smallmatrix} trimethyl \\ \downarrow C{{H}_{3}}I \\ \underset{tetramethyl\text{ }ammonium\text{ }iodide}{\mathop{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{(C{{H}_{3}})}_{4}}{{N}^{+}}{{I}^{-}}}}\, \end{smallmatrix}}{\mathop{{{(C{{H}_{3}})}_{3}}}}\,\,\underset{amine}{\mathop{N}}\,\] Hence, three molecules of \[C{{H}_{3}}I\] is used.
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