- Can be oxidized to a dicarboxylic acid with Br2 water
- despite the presence of – CHO does not give Schiff's test
- has 4-asymmetric carbon atom
- will coexist in equilibrium with 2 other cyclic structure
The Correct Option is A
Solution and Explanation
The compound shown is a monosaccharide with an aldehyde group (-CHO) and multiple hydroxyl (-OH) groups. When treated with Br$_2$ water, it is not oxidized to a dicarboxylic acid but rather forms a monocarboxylic acid, making option (1) incorrect.
Statement (2) is correct: Despite having a –CHO group, certain sugars like glucose do not respond to Schiff’s test.
Statement (3) is correct: There are 4 chiral (asymmetric) carbon atoms present in the structure.
Statement (4) is correct: The compound can exist in equilibrium with two other cyclic forms, $\alpha$-D-glucose and $\beta$-D-glucose.
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