Question

Match the two columns: based on the reaction Glucose + 'X' → 'P' List-I (Reagent-X) A. Br\(_2\)/water
B. Acetic anhydride (excess)
C. Conc. HNO\(_3\)
D. NH\(_2\)OH
List-II (Product-P) (i) Glucose oxime
(ii) Saccharic acid
(iii) Glucose pentaacetate
(iv) Gluconic acid

• A. A – iv, B – ii, C – iii, D – i
• B. A – ii, B – iv, C – iii, D – i
• C. A – ii, B – iii, C – iv, D – i
• D. A – iv, B – iii, C – ii, D – i

Hint:

In organic reactions, acetic anhydride is used to form esters, while Br\(_2\)/water and NH\(_2\)OH are used for functional group interconversions.

Answer 1

The Correct Option is D

To solve this matching question, we need to understand the reactions of glucose with different reagents to determine the products formed.

1. Reagent A is Br\(_2\)/water. When glucose reacts with Br\(_2\)/water, the aldehyde group is oxidized to a carboxylic acid, forming gluconic acid. Therefore, the product is Gluconic acid (iv).
2. Reagent B is Acetic anhydride (excess). When glucose reacts with excess acetic anhydride, all hydroxyl groups are acetylated, resulting in glucose pentaacetate. Therefore, the product is Glucose pentaacetate (iii).
3. Reagent C is Conc. HNO\(_3\). When glucose is treated with concentrated nitric acid, it undergoes oxidation at both ends to form saccharic acid. Therefore, the product is Saccharic acid (ii).
4. Reagent D is NH\(_2\)OH, which is hydroxylamine. Glucose reacts with hydroxylamine to form an oxime, specifically glucose oxime. Therefore, the product is Glucose oxime (i).

Based on the above reasoning, the correct matching is:

• A – iv (Gluconic acid)
• B – iii (Glucose pentaacetate)
• C – ii (Saccharic acid)
• D – i (Glucose oxime)

Thus, the correct answer is: A – iv, B – iii, C – ii, D – i.