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