Skip to content
NCERT Exemplar · Q4

Q.Anomers are two cyclic forms of the same monosaccharide that differ in configuration only at the anomeric carbon (the carbon that was the carbonyl carbon before ring closure); all other carbons keep the same configuration. Four candidate pairs of sugar structures are described in the options. Which pair represents anomers?

(i) A pair of open-chain aldohexoses drawn as Fischer projections (CHO at the top, CH2OH at the bottom): the left one has its OH groups on the right, left, right, right at carbons 2-5 (D-glucose) and the right one has OH on the left, left, right, right (D-mannose).
(ii) A pair of open-chain aldohexose Fischer projections (CHO top, CH2OH bottom): the left has OH on right, left, right, right (D-glucose); the right has OH on left, right, left, left.
(iii) A pair of cyclic (ring-oxygen) sugar structures in which, reading down from the top ring carbon, the OH groups sit right, left, right at carbons 2-4 in BOTH structures; the two differ only in the orientation of the OH on the top (anomeric) carbon - on the right in the first and on the left in the second.
(iv) A pair of cyclic (ring-oxygen) structures whose OH groups differ at several carbons, not only at the top (anomeric) carbon.
Rajasthan RbseMCQ· 1mImportance★★★★★
34% · 37/110 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Anomers are the two cyclic forms (alpha and beta) of a sugar that differ in configuration at ONLY the anomeric carbon. That immediately eliminates the two open-chain pairs and the cyclic pair that differs at several carbons, leaving one correct answer.

Concept

When an aldose or ketose closes into a ring, the former carbonyl carbon (C-1 in an aldose) becomes a new stereocentre called the anomeric carbon. Its new OH can point to either side, giving the alpha- and beta-anomers. By definition, anomers are cyclic structures that are identical at every carbon except the anomeric one.

Why the open-chain pairs fail

  • (i) shows two open-chain aldohexoses (D-glucose and D-mannose). They have no anomeric carbon at all and differ at C-2, so they are C-2 epimers, not anomers.
  • (ii) likewise shows two open-chain aldohexoses; open-chain forms cannot be anomers.

Comparing the cyclic pairs …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.