A) Glucose
B) Fructose
C) Starch
D) Sucrose
Correct Answer: A
Solution :
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. \[\underset{\begin{smallmatrix} | \\ C{{H}_{2}}OH \end{smallmatrix}}{\overset{\begin{smallmatrix} CHO \\ | \end{smallmatrix}}{\mathop{{{(CHOH)}_{4}}}}}\,+\underset{\begin{smallmatrix} Tollen's \\ reagent \end{smallmatrix}}{\mathop{A{{g}_{2}}O}}\,\xrightarrow[{}]{{}}\underset{\begin{smallmatrix} | \\ C{{H}_{2}}OH \end{smallmatrix}}{\overset{\begin{smallmatrix} COOH \\ | \end{smallmatrix}}{\mathop{{{(CHOH)}_{4}}}}}\,+\underset{\begin{smallmatrix} silver \\ mirror \end{smallmatrix}}{\mathop{2Ag\downarrow }}\,\]
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