Question

What is the major product formed in the given reaction? 

• A. A
• B. B
• C. C
• D. D

Hint:

In oxidative cleavage reactions, bromine reacts with hydroxyl groups to form carboxyl groups, especially under mildly acidic or neutral conditions. This is a key reaction when dealing with sugars, and it is commonly used to form dicarboxylic acids.

Answer 1

The Correct Option is C

This reaction involves bromine ($Br_2$) in an aqueous medium with a pH of 6. The specific reaction here is an oxidative cleavage reaction typical for sugars. Let's break down the steps involved: 1. Initial Step:
The given sugar molecule is a cyclic structure with multiple hydroxyl ($OH$) groups attached to the carbon atoms. Under normal conditions, bromine in the presence of water and at pH 6 reacts with the hydroxyl groups of the sugar molecule.
2. Reaction Mechanism:
Bromine is an oxidizing agent, and when it reacts with hydroxyl groups in the presence of water, it typically oxidizes them. At pH 6, which is mildly acidic, the hydroxyl groups undergo oxidation, converting them into carboxyl groups ($COOH$).
3. Cleavage of the Sugar Ring:
The reaction proceeds by the cleavage of the sugar ring at positions where the oxidation occurs. In this case, bromine induces the cleavage of the bond between the carbon atoms, where oxidation of the hydroxyl groups occurs. This leads to the formation of carboxyl groups on two carbon atoms in the sugar structure.
4. Final Product:
The final product, which is option (C), shows a sugar molecule with two carboxyl groups ($COOH$) replacing the hydroxyl groups that were originally present at positions 1 and 2 on the sugar ring. The remaining hydroxyl groups are unaltered in this process.
Thus, the correct product is option (C), where two hydroxyl groups are oxidized to carboxyl groups, resulting in a dicarboxylic acid form of the sugar.