Question

(i) Write a short note on the cyclic structure of glucose.
(ii) What is the difference between a nucleoside and a nucleotide? (2+2=4)

Hint:

Glucose: C-5 \( -OH \) closes onto the C-1 \( -CHO \) to give a six-membered pyranose hemiacetal, creating the anomeric carbon and the \( \alpha \)/\( \beta \) anomers. Nucleoside = base + sugar; nucleotide = base + sugar + phosphate.

Answer 1

Part (i): Cyclic structure of glucose.

Step 1: Why an open chain is not enough.
The open-chain (Fischer) form of glucose is an aldohexose with a \( -\text{CHO} \) group at C-1. However, several observations (glucose does not give the Schiff test, does not form the hydrogensulphite addition product with \( \text{NaHSO}_3 \), and exists in two crystalline forms) cannot be explained by the open chain alone.

Step 2: Formation of the ring.
The \( -\text{OH} \) group on C-5 attacks the aldehyde carbon (C-1) intramolecularly. This forms a stable six-membered ring (a cyclic hemiacetal) containing one oxygen atom, called the pyranose ring.

Step 3: The anomeric carbon and anomers.
During ring closure, C-1 becomes a new chiral (asymmetric) centre called the anomeric carbon. Depending on the orientation of the newly formed \( -\text{OH} \) at C-1, two forms arise: \( \alpha\text{-D-glucose} \) (\( -\text{OH} \) at C-1 below the plane, on the right in Fischer) and \( \beta\text{-D-glucose} \) (\( -\text{OH} \) at C-1 above the plane, on the left). These are called anomers.

Step 4: Conclusion.
The two anomers interconvert in solution through the small amount of open-chain form, a process called mutarotation. The cyclic pyranose structure therefore explains the missing aldehyde reactions and the existence of two forms of glucose.

Part (ii): Nucleoside versus nucleotide.

Step 5: Nucleoside.
A nucleoside is formed when a nitrogenous base (purine or pyrimidine) is joined to a pentose sugar (ribose or deoxyribose) at C-1' of the sugar through an N-glycosidic bond. Composition: base + sugar (no phosphate).

Step 6: Nucleotide.
A nucleotide is formed when a phosphoric acid group is attached (esterified) to the C-5' \( -\text{OH} \) of the sugar in a nucleoside. Composition: base + sugar + phosphate.
\[ \boxed{\text{Nucleoside} = \text{base} + \text{sugar}; \quad \text{Nucleotide} = \text{base} + \text{sugar} + \text{phosphate}} \]
Thus a nucleotide is simply a phosphorylated nucleoside, and nucleotides are the building blocks that link together to form nucleic acids (DNA and RNA).