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NCERT Exemplar · Q15

Q.Three cyclic monosaccharide structures (labelled I, II and III) are drawn as Fischer projections closed by a ring oxygen, each ending in a CH2OH group. Reading the ring carbons from the top (anomeric) carbon downward: in structure I the OH groups are on the right (anomeric), right, left, right; in structure II they are on the left (anomeric), right, left, right; in structure III they are on the left, left, left, right, left. Which of these structures are anomers of one another?
(A) I and II
(B) II and III
(C) I and III
(D) III is an anomer of I and II
(A) I and II
(B) II and III
(C) I and III
(D) III is an anomer of I and II

Uttarakhand UbseMCQ· 1mImportance★★★★★
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Two cyclic sugars are anomers only if they differ at the anomeric carbon and nowhere else. Structures I and II are identical at C-2, C-3 and C-4 and differ solely at the anomeric carbon, so they are anomers; III differs at multiple centres and cannot be an anomer of either.

Concept

The anomeric carbon is the ring carbon that carries the newly formed hemiacetal OH (the carbon derived from the open-chain carbonyl). Alpha- and beta-anomers of a given sugar are identical at every other stereocentre and differ only in whether that anomeric OH points to the right (in one) or the left (in the other) of the Fischer projection.

Comparison

  • Structure I: OH orientations right (anomeric), right, left, right.
  • Structure II: OH orientations left (anomeric), right, left, right. …

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