A) \[-COOH\]group
B) a basic group
C) a ketonic group
D) an aldehyde group
Correct Answer: D
Solution :
Glucose gives silver mirror due to the presence of aldehydic group. \[\underset{glucose}{\mathop{{{C}_{5}}{{H}_{11}}{{O}_{5}}.CHO}}\,+\underset{Tollen's\text{ }reagent}{\mathop{2Ag{{(N{{H}_{3}})}_{2}}.OH}}\,\xrightarrow[{}]{{}}\] \[\underset{gluconic\text{ }acid}{\mathop{{{C}_{5}}{{H}_{11}}{{O}_{5}}.COOH}}\,+\underset{\begin{smallmatrix} silver \\ mirror \end{smallmatrix}}{\mathop{2Ag}}\,+4N{{H}_{3}}+{{H}_{2}}O\]
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