JEE Main 2025 · 28 January · evening · Single correct · +4 / −1
Match List-I with List-II.
Choose the correct answer from the options given below :
| List - I (Saccharides) | List - II (Glycosidic-linkages found) |
|---|---|
| (A) Sucrose | (I) α 1 - 4 |
| (B) Maltose | (II) α 1 - 4 and α 1 - 6 |
| (C) Lactose | (III) α 1 - β 2 |
| (D) Amylopectin | (IV) β 1 - 4 |
- (a) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
- (b) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
- (c) (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
- (d) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
Answer and solution
Answer (a) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Solution
Chemistry ma’am · 3 steps
For each sugar, ask which carbons join and whether the anomeric carbon is α or β.
- Read the label. In a label like α 1–4, the Greek letter is the form of the sugar whose C1 forms the bond. The numbers are the two carbons joined through the oxygen bridge. In α 1–β 2, both anomeric carbons are tied up.
Saccharide Units joined Linkage Match (A) Sucrose C1 of α-D-glucose to C2 of β-D-fructose α 1–β 2 (III) (B) Maltose C1 of α-D-glucose to C4 of the next glucose α 1–4 (I) (C) Lactose C1 of β-D-galactose to C4 of glucose β 1–4 (IV) (D) Amylopectin α 1–4 along the chain, α 1–6 at each branch α 1–4 and α 1–6 (II) - Why these two decide. Sucrose is the only sugar here whose bond joins two anomeric carbons, so only it fits α 1–β 2. That is also why sucrose is non-reducing. Amylopectin is the only branched one, so only it needs the extra α 1–6 link.
- So. The matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II), which is option (a).
The idea to keep. Learn the partners and the anomer: maltose is α-glucose with glucose (α 1–4), lactose is β-galactose with glucose (β 1–4), sucrose is α-glucose with β-fructose (α 1–β 2). Branching always adds α 1–6.
Answer from the NTA final answer key. Solution by PrepWiser.