Identify a ketose sugar. - Solved MCQ | Agrigyan

• Asymmetric (Chiral) Carbons in Monosaccharides:
• Glyceraldehyde: The simplest aldose ($C_3H_6O_3$, an aldotriose) possessing exactly 1 asymmetric carbon atom (at $\text{C-2}$).
• Glucose: An aldohexose possessing 4 asymmetric carbon atoms (at $\text{C-2, C-3, C-4, C-5}$).
• Fructose: A ketohexose possessing 3 asymmetric carbon atoms (at $\text{C-3, C-4, C-5}$).
• Dihydroxyacetone: A ketotriose that contains no (0) asymmetric carbon atoms (it is an achiral, optically inactive sugar).
• Exam Context: Among the sugars that contain asymmetric centers, glyceraldehyde contains the minimum number ($1$).

Correct option: (1) Glyceraldehyde

• Equilibrium Mutarotation of D-Glucose: In an aqueous solution, open-chain D-glucose spontaneously cyclizes via intramolecular hemiacetal formation into two interconverting anomers:
• $\beta$-D-glucopyranose: Comprises approximately $64\%$ (about $2/3$) of the mixture because all bulky hydroxyl groups and the $-\text{CH}_2\text{OH}$ group occupy energetically favored equatorial positions.
• $\alpha$-D-glucopyranose: Comprises approximately $36\%$ (about $1/3$) of the mixture, with its anomeric $\text{C-1}$ hydroxyl group in the axial orientation.
• Open-chain (Linear) form: Less than $0.1\%$ at equilibrium.

Correct option: (1) a-D-glucopyranose

• Anomeric Carbon of Glucose: The anomeric carbon is the former carbonyl carbon atom that becomes a new chiral center upon cyclic hemiacetal/hemiketal ring closure:
• In D-glucose (an aldohexose), the carbonyl group is located at Carbon-1 ($\text{C-1}$).
• Attack by the hydroxyl oxygen at $\text{C-5}$ on the aldehyde carbon ($ ext{C-1}$) creates the pyranose ring and makes Carbon 1 the anomeric carbon (giving rise to $\alpha$ and $\beta$ anomers).
• Note: The official examination option refers to Carbon 1 (OCR corrected from Carbon 4).

Correct option: (1) Carbon 1