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When sugar ‘X’ is boiled with dilute $H_{2} S O_{4}$ in alcoholic solution, two isomers ‘A’ and ‘B’ are formed. ‘A’ on oxidation with $H N O_{3}$ yields saccharic acid where as ‘B’ is laevorotatory. The compound ‘X’ is :

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Starch

Sucrose

Maltose

Lactose

answer is B.

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Detailed Solution

$C_{12} H_{22} 0_{11} + d i l . H_{2} S O_{4} \overset{A l C .}{\to} \underset{\begin{array}{l} G l u \cos e \\ (d) \end{array}}{C_{6} H_{12} O_{6}} + \underset{\begin{array}{l} F r u c t o s e \\ (l) \end{array}}{C_{6} H_{12} O_{6}}$

$C_{6} H_{12} O_{6} + H N O_{3} \to \underset{S a c c h a r i c a c i d}{C_{6} H_{12} O_{8}}$

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