JEE Chemistry 2023 Question with Solution

Given: Sugar $[X]$ is shown in open-chain form.

Find: The correct six-membered pyranose representation.

From the solution, sugar $[X]$ is identified as D-glucose. D-glucose cyclizes to give a six-membered glucopyranose ring. The required identification depends on the orientation of the $-OH$ group at the anomeric carbon $C-1$.

• In $\alpha$-D-glucopyranose, the anomeric $-OH$ is oriented downward.
• In $\beta$-D-glucopyranose, the anomeric $-OH$ is oriented upward.

The solution explicitly states that the correct option is A. Although the answer key says (2), the solution is the primary source here.

Conclusion: The correct option is A.

Given: The sugar is D-glucose and the question asks for its six-membered pyranose form.

Find: Which diagram matches the required pyranose form.

To distinguish the cyclic forms, inspect the substituent at the anomeric carbon formed after ring closure.

For D-glucose:

• six-membered ring formation gives glucopyranose
• downward anomeric $-OH$ corresponds to $\alpha$ form
• upward anomeric $-OH$ corresponds to $\beta$ form

The provided explanation on the solution's concludes with The Correct Option is A. It also contains an internal inconsistency by naming $\beta$-D-glucopyranose but then referring to Diagram 1 as correct. Since the answer label is stated explicitly on the solution, we take A as the final answer.

Therefore, the correct representation is Option A.