Match List-I with List-II.
\begin{tabular{|c|l||c|l| \hline List-I & Reaction of Glucose with & List-II & Product formed
\hline A. & Hydroxylamine & I. & Gluconic acid
B. & Br water & II. & Glucose pentaacetate
C. & Excess acetic anhydride & III. & Saccharic acid
D. & Concentrated HNO & IV. & Glucoxime
\hline \end{tabular
Choose the correct answer from the options given below:
A-IV, B-I, C-II, D-III
A-IV, B-III, C-II, D-I
A-I, B-III, C-IV, D-II
A-III, B-I, C-IV, D-II
Correct answer:A
Standard Method
Given: Match the reactions of glucose with their products.
Find: The correct correspondence between List-I and List-II.
Reaction with hydroxylamine: Glucose reacts with hydroxylamine to form an oxime called glucoxime.
Reaction with bromine water: Bromine water is a mild oxidising agent and oxidises the aldehyde group of glucose to form gluconic acid.
Reaction with excess acetic anhydride: All five hydroxyl groups of glucose are acetylated to form glucose pentaacetate.
Reaction with concentrated nitric acid: Both the aldehyde group and the primary alcohol group are oxidised to carboxylic acids to form saccharic acid.
Therefore, the final matching is
The correct option is A.
Assuming bromine water oxidises glucose completely to saccharic acid is incorrect because bromine water is only a mild oxidising agent. It oxidises only the aldehyde group to give gluconic acid. Use concentrated nitric acid for oxidation of both terminal groups.
Confusing the product with hydroxylamine as an oxidation product is wrong because hydroxylamine forms a derivative of the carbonyl group, namely an oxime. Therefore glucose gives glucoxime, not an acid, with hydroxylamine.
Forgetting that excess acetic anhydride acetylates all five hydroxyl groups leads to an incorrect match. The correct product is glucose pentaacetate, so count all available hydroxyl groups before matching.
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