A) acetic acid
B) saccharic acid
C) glyceraldehyde
D) gluconic acid
Correct Answer: D
Solution :
\[\underset{Glu\cos e}{\mathop{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}\,+[O]\xrightarrow[{}]{B{{r}_{2}}/{{H}_{2}}O}\] \[\underset{Gluconic\,acid}{\mathop{C{{H}_{2}}OH.{{(CHOH)}_{4}}COOH}}\,\]
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