A) Ammonia
B) Methylamine
C) Dimethylamine
D) Trimethylamine
Correct Answer: A
Solution :
\[\underset{\begin{smallmatrix} {{\text{2}}^{\text{o}}}\text{amine is} \\ \text{most}\,\,\text{basic} \end{smallmatrix}}{\mathop{{{(C{{H}_{3}})}_{2}}NH}}\,>\underset{{{\text{1}}^{\text{o}}}\text{amine}}{\mathop{C{{H}_{3}}N{{H}_{2}}}}\,>\underset{{{\text{3}}^{\text{o}}}\text{amine}}{\mathop{{{(C{{H}_{3}})}_{3}}N}}\,>\underset{\begin{smallmatrix} \text{Ammonia is} \\ \text{least}\,\,\text{basic} \end{smallmatrix}}{\mathop{\overset{\cdot \,\,\cdot }{\mathop{N}}\,{{H}_{3}}}}\,\]
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