Question

Which linkage joins the monosaccharide units in sucrose?

• A. \(C_1 - C_4\) glycosidic linkage
• B. \(C_1 - C_2\) glycosidic linkage
• C. \(C_1 - C_6\) glycosidic linkage
• D. \(C_2 - C_4\) glycosidic linkage

Hint:

Use the fact that sucrose is non-reducing, so both anomeric carbons must be involved. Identify the anomeric carbon number in glucose and in fructose.

Answer 1

The Correct Option is B

Concept: Sucrose is a disaccharide composed of two monosaccharides:
• \(\alpha\)-D-glucose
• \(\beta\)-D-fructose These two units are connected through a glycosidic bond. The bond is formed between specific carbon atoms of the two sugars.

Step 1: Identify the carbon atoms involved in bonding. The glycosidic bond in sucrose is formed between: \[ C_1 \text{ of glucose} \] and \[ C_2 \text{ of fructose} \]

Step 2: Write the linkage type. Thus the linkage is: \[ C_1 - C_2 \] glycosidic linkage.

Step 3: State the final answer. Therefore, sucrose contains a \[ C_1 - C_2 \] glycosidic bond.

Answer 2

Concept:

• Sucrose is non-reducing because the anomeric carbon of each monosaccharide is used in the glycosidic bond.
• Finding those two anomeric carbons identifies the linkage positions.

Step 1: Identify the two monosaccharide units.
Sucrose contains $\alpha$-D-glucose and $\beta$-D-fructose.

Step 2: Locate each anomeric carbon.
The anomeric carbon of glucose is $C_1$.
Because fructose is a ketose, its anomeric carbon is $C_2$.

Step 3: Join the two anomeric centres.
The glycosidic oxygen connects glucose $C_1$ to fructose $C_2$. Since both anomeric groups are tied up, sucrose has no free reducing end.

Final Answer: $C_1-C_2$ glycosidic linkage, option B.