A) any aliphatic amine
B) any aromatic amine
C) any amine
D) any primary amine
Correct Answer: D
Solution :
Carbylamine is given by primary amine. \[\underset{\begin{smallmatrix} \text{Primary} \\ \text{amine} \end{smallmatrix}}{\mathop{R-N{{H}_{2}}}}\,+CH\,OH\to 3KCl+\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\text{Carbylamine} \\ (unpleasent\,smeell) \end{smallmatrix}}{\mathop{2{{H}_{2}}O}}\,\]
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