Given D-Glucose. The compound(s),which on rea | vedclass

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(B) The reaction of an aldose with $HNO_3$ (nitric acid) oxidizes both the terminal aldehyde group and the terminal primary alcohol group to carboxyli

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$C, D$

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(B) The reaction of an aldose with $HNO_3$ (nitric acid) oxidizes both the terminal aldehyde group and the terminal primary alcohol group to carboxylic acid groups,forming a dicarboxylic acid (saccharic acid).$D$-Glucose gives a product with $[\alpha]_{D} = +52.7^{\circ}$.The enantiomer of this product will have an equal and opposite optical rotation,i.e.,$[\alpha]_{D} = -52.7^{\circ}$.To obtain the enantiomer of the product formed from $D$-Glucose,we must start with the enantiomer of $D$-Glucose,which is $L$-Glucose,or any sugar that oxidizes to the enantiomer of the product formed from $D$-Glucose.Looking at the structures provided:Compound $(C)$ is $L$-Gulose.Compound $(D)$ is $L$-Glucose.Both $L$-Gulose and $L$-Glucose,upon oxidation with $HNO_3$,yield dicarboxylic acids that are enantiomers of the product formed from $D$-Glucose,thus exhibiting a rotation of $-52.7^{\circ}$.Therefore,the correct compounds are $(C)$ and $(D)$.

Given $D$-Glucose. The compound$(s)$,which on reaction with $HNO_3$ will give the product having degree of rotation,$[\alpha]_{D}=-52.7^{\circ}$ is (are)

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