Carbohydrates Reactions
Both D-glucose and D-fructose give the same osazone when treated with excess phenylhydrazine, despite differing in their carbonyl functional groups. Which observation does this identical osazone formation establish about the two sugars?
Select the correct option:
Solution
They share the same configuration at all carbons except C1 and C2
Osazone formation is a diagnostic reaction in sugar chemistry because phenylhydrazine reacts only at the C1 and C2 carbons, effectively erasing the stereochemical and functional distinctions there. When glucose, an aldose, and fructose, a ketose, each react with three molecules of phenylhydrazine, the reaction consumes the carbonyl at C1 and the adjacent C2 position, forming the same bis-phenylhydrazone (osazone). The fact that the two sugars yield an identical osazone proves that they must be identical in configuration at every carbon below C2, namely at C3, C4, C5, and C6, while differing only at C1 and C2. Therefore the first option is correct. They are not enantiomers, since enantiomers would differ at all stereocentres and would not be an aldose-ketose pair. Neither sugar contains a free carboxylic acid group in this context, so that option is false. The claim of completely different stereochemistry is contradicted by the very fact that the osazones match. This reflects the NCERT and classical reasoning about epimers and osazones. As a consistency check, glucose, fructose, and mannose all give the same osazone, agreeing that only the top two carbons distinguish them.
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About This Question
- Subject
- chemistry
- Chapter
- biomolecules
- Topic
- carbohydrates reactions
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
They share the same configuration at all carbons except C1 and C2
Osazone formation is a diagnostic reaction in sugar chemistry because phenylhydrazine reacts only at the C1 and C2 carbons, effectively erasing the stereochemical and functional distinctions there. When glucose, an aldose, and fructose, a ketose, each react with three molecules of phenylhydrazine, the reaction consumes the carbonyl at C1 and the adjacent C2 position, forming the same bis-phenylhydrazone (osazone). The fact that the two sugars yield an identical osazone proves that they must be identical in configuration at every carbon below C2, namely at C3, C4, C5, and C6, while differing only at C1 and C2. Therefore the first option is correct. They are not enantiomers, since enantiomers would differ at all stereocentres and would not be an aldose-ketose pair. Neither sugar contains a free carboxylic acid group in this context, so that option is false. The claim of completely different stereochemistry is contradicted by the very fact that the osazones match. This reflects the NCERT and classical reasoning about epimers and osazones. As a consistency check, glucose, fructose, and mannose all give the same osazone, agreeing that only the top two carbons distinguish them.
This hard difficulty chemistry question is from the chapter biomolecules, covering the topic of carbohydrates reactions. It appeared in the 2025 exam.
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