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Question
Three cyclic structures of monosaccharides are given below which of these are anomers.
(I) | ![]() |
(II) | ![]() |
(III) | ![]() |
Options
I and II
II and III
I and III
III is anomer of I and II
Solution
I and II
Explanation:
This behaviour could not be explained by the open-chain structure (I) for glucose. It was proposed that one of the –OH groups may add to the –CHO group and form a cyclic hemiacetal structure. It was found that glucose forms a six-membered ring in which –OH at C – 5 is involved in ring formation. This explains the absence of –CHO group and also existence of glucose in two forms as shown below.
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