A) Tollen reagent
B) Fehling solution
C) \[NaOH/{{I}_{2}}/{{H}^{+}}\]
D) \[NaOH/NaI/{{H}^{+}}\]
Correct Answer: C
Solution :
Methyl ketones when treated with \[NaOH/{{I}_{2}}/{{H}^{+}}\] give iodoform and an acid (with one carbon less than the parent ketone). \[\underset{\begin{smallmatrix} bu\tan e-22-one \\ (methyl\,ketone) \end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}}}}\,+NaOH+{{I}_{2}}\xrightarrow[\begin{smallmatrix} -NaI \\ -{{H}_{2}}O \end{smallmatrix}]{{}}\] \[CH{{I}_{3}}+C{{H}_{3}}C{{H}_{2}}COONa\xrightarrow[{}]{{{H}^{+}}}\underset{propanoic\,acid}{\mathop{C{{H}_{3}}C{{H}_{2}}COOH}}\,\]
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