Question

Which of the following will give silver mirror test?

• A. Glucose
• B. Fructose
• C. Starch
• D. Sucrose

Answer 1

Answer: (A)

Aldehyde group $(-CHO)$ reduces Tollen's reagent and gives silver mirror. Tollen's reagent is ammoniacal silver nitrate solution. Glucose contains $-CHO$ group hence it gives silver mirror test. $\underset{\begin{array}{c}|\\ C{H}_{2}OH\end{array}}{\stackrel{\begin{array}{c}CHO\\ |\end{array}}{{(CHOH)}_4}}+\underset{\begin{array}{c}Tolle{n}^{\prime }s\\ reagent\end{array}}{A{g}_{2}O}\to \underset{\begin{array}{c}|\\ C{H}_{2}OH\end{array}}{\stackrel{\begin{array}{c}COOH\\ |\end{array}}{{(CHOH)}_4}}+\underset{\begin{array}{c}silver\\ mirror\end{array}}{2Ag\downarrow }$