A) Glyceraldehyde
B) Gluconic acid
C) saccharic acid
D) Glucaric acid
Correct Answer: B
Solution :
\[C{{H}_{2}}OH-\underset{Glu\cos e}{\mathop{{{(CHOH)}_{4}}CHO}}\,\xrightarrow[B{{r}_{2}}/{{H}_{2}}O]{[O]}\] \[\underset{Gluconic\,acid}{\mathop{C{{H}_{2}}OH{{(CHOH)}_{4}}C{{H}_{2}}OH}}\,\]
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