A) \[HCOOH\]
B) \[C{{H}_{3}}COOH\]
C) \[C{{H}_{3}}COC{{H}_{3}}\]
D) \[C{{H}_{3}}OH\]
Correct Answer: A
Solution :
HCOOH reduces Tollens reagent to Ag mirror. \[HCOOH+\underset{Tollens\text{ }reagent}{\mathop{2{{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}}}\,+2O{{H}^{-}}\xrightarrow[{}]{{}}\] \[\underset{silver\text{ }mirror}{\mathop{2Ag\downarrow }}\,+C{{O}_{2}}\uparrow +4N{{H}_{3}}+2{{H}_{2}}O\]
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