A) \[{{H}_{2}}O\]
B) \[N{{a}_{2}}C{{O}_{3}}+{{I}_{2}}\]
C) \[N{{H}_{3}}\]
D) \[HCl\]
Correct Answer: B
Solution :
\[C{{H}_{3}}OH\]and\[{{C}_{2}}{{H}_{5}}OH\]can be differentiated by using \[N{{a}_{2}}C{{O}_{3}}\]and \[{{I}_{2}},{{C}_{2}}{{H}_{5}}OH\]gives yellow precipitate of \[CH{{I}_{3}}\]whereas \[C{{H}_{3}}OH\]does not react with it. \[{{C}_{2}}{{H}_{5}}OH+4{{I}_{2}}+N{{a}_{2}}C{{O}_{3}}\xrightarrow{{}}\underset{\begin{smallmatrix} \text{Iodoform} \\ \text{(yellow}\,\text{ppt}\text{.)} \end{smallmatrix}}{\mathop{CH{{I}_{3}}}}\,\] \[+\,5NaI+HCOONa+3C{{O}_{2}}+{{H}_{2}}O\]
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