Question:

Which of the following is correct structure of $\alpha$-anomer of maltose? 

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When identifying carbohydrate structures: first, identify the monosaccharide units. Second, determine the linkage points (e.g., 1$\rightarrow$4, 1$\rightarrow$6). Third, determine the stereochemistry of the linkage ($\alpha$ = down/axial, $\beta$ = up/equatorial). Finally, check the stereochemistry of the free anomeric carbon if it's a reducing sugar.
Updated On: Jan 5, 2026
  • A

  • B

  • C

  • D

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The Correct Option is A

Solution and Explanation

Step 1: Basic structure of maltose Maltose is a disaccharide composed of two D-glucose units. Both glucose units exist in the pyranose (six-membered ring) form. Step 2: Type of glycosidic linkage In maltose:
The anomeric carbon (C1) of the first glucose unit
is linked to the C4 hydroxyl group of the second glucose unit
Thus, maltose has a (1$\rightarrow$4) glycosidic linkage. Step 3: Meaning of $\alpha$-(1$\rightarrow$4) linkage The term $\alpha$-(1$\rightarrow$4) means:
The glycosidic bond formed at C1 of the first glucose
has the $\alpha$-configuration
For D-glucose, $\alpha$-configuration means the substituent at C1 is oriented downwards (axial) relative to the ring. Step 4: Meaning of $\alpha$-anomer of maltose Maltose is a reducing sugar, because the second glucose unit has a free anomeric carbon (C1). The term $\alpha$-anomer of maltose refers to the configuration of this free anomeric carbon:
For the $\alpha$-anomer, the -OH group on C1 of the second glucose
must point downwards
Step 5: Structural features to check A correct structure of $\alpha$-maltose must show:
Two D-glucopyranose rings
$\alpha$-(1$\rightarrow$4) glycosidic bond
Free anomeric carbon on the second glucose
Downward (-OH) orientation at this free anomeric carbon
Step 6: Identification of the correct option In Option (A):
Both rings are D-glucose pyranose units
The linkage is correctly from C1 to C4
The glycosidic bond is $\alpha$ (downward)
The free anomeric -OH on the second glucose is also $\alpha$
Hence, option (A) correctly represents \[ \alpha\text{-D-glucopyranosyl-(1$\rightarrow$4)-}\alpha\text{-D-glucopyranose} \] Final Answer: \[ \boxed{\text{Option (A)}} \]
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