A) Fehling's solution
B) \[{{I}_{2}}/NaOH\]
C) Tollen's reagent
D) carbonic acid
Correct Answer: B
Solution :
Due to the presence of \[{{H}_{3}}C-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-\] group, acetaldehyde \[({{H}_{3}}C-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-H)\] and acetone \[({{H}_{3}}C-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}})\] react with iodine in sodium hydroxide to give yellow ppt of iodoform. \[C{{H}_{3}}CHO+3{{I}_{2}}+3NaOH\xrightarrow{{}}\] \[C{{l}_{3}}CHO+3NaI+3{{H}_{2}}O\] \[C{{l}_{3}}CHO+NaOH\xrightarrow{{}}\underset{Iodoform}{\mathop{CHI}}\,+HCOONa\] \[C{{H}_{3}}COC{{H}_{3}}+3{{I}_{2}}+3NaOH\xrightarrow{{}}\] \[C{{l}_{3}}COC{{H}_{3}}+3NaI+3{{H}_{2}}O\] \[C{{l}_{3}}CO{{H}_{3}}+NaOH\xrightarrow{{}}\underset{Iodoform}{\mathop{CHI}}\,+C{{H}_{3}}COONa\]
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