A) Fehlings reagent
B) Tollens reagent
C) Shifts reagent
D) Iodoform test
E) None of these
Correct Answer: D
Solution :
Formaldehyde and acetaldehyde can be distinguish by iodoform test. i.e. \[C{{H}_{3}}-\overset{O}{\mathop{\overset{|\,\,|}{\mathop{C}}\,}}\,-H\xrightarrow{{{I}_{2}},\,\,NaOH}\]\[H-\overset{O}{\mathop{\overset{|\,\,|}{\mathop{C}}\,}}\,-{{O}^{-}}+\underset{\begin{smallmatrix} \,\,\,\,\,Iodoform \\ (yellow\,ppt) \end{smallmatrix}}{\mathop{CH{{I}_{3}}}}\,\]
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