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