Question

Which of the following gives a silver mirror test?
A. Formic acid
B. $\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{C}\mathrm{H}\mathrm{O}\mathrm{H}\mathrm{C}{\mathrm{H}}_3$
C. Tartaric acid
D. Glucose

Answer 1

Hint: We know that the silver mirror test is generally known as Tollen’s’ reagent test. It is usually used in the laboratory to show the difference between the organic compounds that are aldehyde and ketones. The silver mirror test mainly used for testing the aldehyde compounds.

Complete step by step solution
The Tollen’s reagent is the alkaline solution of ammoniacal silver nitrate and it is generally used to give the test of aldehyde. The silver ions in the presence of hydroxide ions are involved in it.
As we all know, the Tollen’s’ reagent is a solution of colourless silver ion that is $\mathrm{A}{\mathrm{g}}^{+}$. A terminal alpha hydroxy ketone generally gives the positive Tollens' test for the reason that Tollens' reagent oxidizes the alpha hydroxy ketone to an aldehyde compound.
The Tollen’s test or silver mirror test is not given by the ketones. In the ketone, the acetone does not give the silver mirror test.
The general reaction of the silver mirror test as shown below.
$$\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{H}\mathrm{O}+2\mathrm{A}\mathrm{g}(\mathrm{N}{\mathrm{H}}_3{{)}_2}^{+}+3\mathrm{O}{\mathrm{H}}^{-}\to \mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}{\mathrm{O}}^{-}+2\mathrm{A}\mathrm{g}\left(\mathrm{p}\mathrm{p}\mathrm{t}\right)+4\mathrm{N}{\mathrm{H}}_3+2{\mathrm{H}}_2\mathrm{O}$$
Hence, the correct option for this question is all A, B,C and D that is formic acid, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{C}\mathrm{H}\mathrm{O}\mathrm{H}\mathrm{C}{\mathrm{H}}_3$, tartaric acid and glucose.
The silver mirror test for formic acid is shown below.
$$\mathrm{H}\mathrm{C}\mathrm{H}\mathrm{O}+2\mathrm{A}\mathrm{g}(\mathrm{N}{\mathrm{H}}_3{)}_2+3\mathrm{O}{\mathrm{H}}^{-}\to \mathrm{H}\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}+2\mathrm{A}\mathrm{g}\left(\mathrm{p}\mathrm{p}\mathrm{t}\right)+2{\mathrm{H}}_2\mathrm{O}+4\mathrm{N}{\mathrm{H}}_3$$
The silver mirror test of tartaric acid is shown below.
$${\mathrm{C}}_4{\mathrm{H}}_6{\mathrm{O}}_6+\mathrm{A}{\mathrm{g}}^{+}+2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{C}}_2{\mathrm{H}}_2{\mathrm{O}}_4+\mathrm{C}{\mathrm{O}}_2+\mathrm{H}\mathrm{C}{\mathrm{O}}_2\mathrm{H}+6\mathrm{A}\mathrm{g}\left(\mathrm{p}\mathrm{p}\mathrm{t}\right)+6{\mathrm{H}}^{+}$$
The silver mirror test of glucose is shown below.
$${\mathrm{C}}_5{\mathrm{H}}_{11}{\mathrm{O}}_5\mathrm{C}\mathrm{H}\mathrm{O}+2\mathrm{A}\mathrm{g}(\mathrm{N}{\mathrm{H}}_3{)}_2+\mathrm{O}{\mathrm{H}}^{-}\to {\mathrm{C}}_5{\mathrm{H}}_{11}{\mathrm{O}}_5\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}+2\mathrm{A}\mathrm{g}\left(\mathrm{p}\mathrm{p}\mathrm{t}\right)+2{\mathrm{H}}_2\mathrm{O}+4\mathrm{N}{\mathrm{H}}_3$$

Note:
The aldehydes are those compounds which have $-\mathrm{a}\mathrm{l}$ as a functional group while ketone are those compounds which have $-\mathrm{o}\mathrm{n}\mathrm{e}$ as a functional group. They aldehyde and ketone both are affected by the temperature.