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Chemistry Question 69 – JEE-MAIN 2026

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Identify the correct statements. A. Glucose exists in two anomeric forms. B. Anomers of glucose differ in configuration at C-1 in cyclic hemiacetal structure. C. Melting point of α - anomer of glucose is greater than β - anomer. D. Specific rotation of α - anomer is +19∘ while for β - anomer is +112∘ E. α and β - anomers of glucose are prepared by crystallization of saturated glucose solution at 303 K and 371 K respectively. Choose the correct answer from the options given below:

Recall that anomers are stereoisomers of cyclic saccharides that differ in configuration only at the anomeric carbon.

Step 1: Evaluate Statements A and B (Anomer Definition)✦ Active

Statement A: Glucose exists in two anomeric forms (α-D-glucose and β-D-glucose) in its cyclic hemiacetal structure. This is a fundamental property of monosaccharides. Thus, statement A is correct.

Statement B: Anomers are stereoisomers that differ in configuration at the anomeric carbon (C-1 for aldoses like glucose). Thus, statement B is correct.

Step 2: Evaluate Statements C and D (Physical Properties)○ Expand

Statement C: The melting point of α-D-glucose is 146∘C, while that of β-D-glucose is 150∘C. Therefore, the melting point of the β-anomer is greater than the α-anomer. Thus, statement C is incorrect.

Statement D: The specific rotation of pure α-D-glucose is +112.2∘, and that of pure β-D-glucose is +18.7∘. The values given in the statement are swapped. Thus, statement D is incorrect.

💡 Teacher's Secret Hint

Pay close attention to the specific values and their assignment to the correct anomer.

Step 3: Evaluate Statement E (Preparation Method) and Conclude○ Expand

Statement E: α-D-glucose crystallizes from a saturated aqueous solution at room temperature (around 303 K), while β-D-glucose crystallizes from a hot aqueous solution (above 371 K). Thus, statement E is correct.

Based on the evaluation, statements A, B, and E are correct. Therefore, the correct option is 'A, B and E Only'.

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