Question

Which carbon atoms (numbered from $C_1$ to $C_6$) are involved in the formation of ring structure of glucose?

• A. $C_2$ and $C_5$
• B. $C_1$ and $C_3$
• C. $C_1$ and $C_4$
• D. $C_1$ and $C_5$

Hint:

Logic Tip: Remember the term "pyranose". Pyran is a 6-membered heterocyclic ring containing one oxygen. Since the oxygen forms one vertex of the hexagon, the remaining 5 vertices must be carbons. Connecting $C_1$ to $C_5$ exactly provides these 5 carbon atoms for the ring!

Answer 1

The Correct Option is D

Concept:
The cyclic structure of glucose is formed through an intramolecular nucleophilic addition reaction. The hydroxyl group on one of the carbon atoms attacks the electrophilic carbonyl carbon to form a stable cyclic hemiacetal.
Step 1: Identify the functional groups in open-chain glucose.
D-glucose is an aldohexose. Its open-chain structure consists of:

• An aldehyde group (-CHO) at carbon $C_1$.
• Hydroxyl groups (-OH) on carbons $C_2, C_3, C_4, C_5$, and $C_6$.

Step 2: Determine the thermodynamics of ring formation.
Six-membered rings (pyranose rings) are generally the most thermodynamically stable cyclic structures. For glucose to form a six-membered ring containing an oxygen atom, the reaction must occur between the $C_1$ carbonyl group and the hydroxyl group on $C_5$.
Step 3: Describe the hemiacetal formation.
The lone pair of electrons on the oxygen atom of the $C_5$ hydroxyl group attacks the partially positive carbonyl carbon at $C_1$. This forms a cyclic hemiacetal linkage, officially known as a glucopyranose ring, connecting $C_1$ and $C_5$ via an oxygen atom bridge.