`underset(" "CH_(2)OH)underset(|)underset(" "(CHOH)_(3))underset(|)overset(" "CH_(2)OH)overset(|)(" "C=O) overset(MaBH_(4))(rarr) A+B`
Fructose
The product A and B in the a above reaction are not
A. Diastereomers
B. `C-2` epimers
C. Anomers
D. Optically active hexahydroxy compounds

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1 Answers

Correct Answer - 3
`underset(" fructose")underset(" "CH_(2)OH)underset(|)overset(" "CH_(2)-OH)overset(|)overset(" "C=O)overset(|)(" "(CHOH)_(3)) overset(NaBH_(4))(rarr)underset((A))underset(" "CH_(2)OH)underset(|)overset(" "CH_(2)-OH)overset(|)overset(H-C-OH)overset(|)(" "(CHOH)_(3))+ underset((B))underset(" "CH_(2)OH)underset(|)overset(" "CH_(2)-OH)overset(|)overset(HO-C-H)overset(|)(" "(CH_(2)OH))`
A and B differ in configuration at `C-2` hence they are epimers, also they are non mirror images as `C-3,C-4,C-5` have some configurations, hence they are diastereomers. A and B both are optically active.

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