Question
An optically active compound having molecular formula C6H12O6 is found in two isomeric forms (A) and (B) in nature. When (A) and (B) are dissolved in water, they show the following equilibrium.
$(\text{A})\ \ \ \ \rightleftharpoons\ \ \ \ \text{Equilibrium mixture}\ \ \ \ \rightleftharpoons\ \ \ \ \text{(B)}\\ [\alpha]_{\text{D}}=111^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ 52.2^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 19.2^\circ$
  1. What are such isomers called?
  2. Can they be called enantiomers? Justify your answer.
  3. Draw the cyclic structure of isomer (A).

Answer

  1. Anomers.
  2. Since these anomers are not mirror images of each other so they cannot be called as enantiomers.
  3.  

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