JEE Advanced 2023 · Paper 2 · Chemistry Q4

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ChemistrySingle correct+3 / -1Question 38 of 51

A disaccharide X cannot be oxidised by bromine water. The acid hydrolysis of X leads to a laevorotatory solution. The disaccharide X is

Haworth structure of sucrose: α-D-glucopyranose C1 joined through an O bridge to C2 of β-D-fructofuranose (a 1→2 link between the two anomeric carbons; no free hemiacetal OH).

Haworth structure: a fructofuranose unit joined through its anomeric C2 to C4 of a glucopyranose unit whose anomeric C1 carries a free OH (a reducing disaccharide).

Haworth structure of lactose: β-D-galactopyranose (C4-OH drawn up) joined β(1→4) to D-glucopyranose whose anomeric C1 carries a free OH (reducing).

Haworth structure of a trehalose-type disaccharide: two D-glucopyranose units joined 1→1 through their anomeric carbons (non-reducing; both units are glucose).

Answer and worked solution

Official answer: A

Not oxidised by bromine water. Bromine water oxidises a free hemiacetal (aldose) group. So X must be non-reducing: both anomeric carbons are tied up in the glycosidic bond.

Laevorotatory hydrolysate. Hydrolysis must give a laevorotatory mixture. Glucose + fructose (invert sugar) qualifies, because fructose is strongly laevorotatory.

Checking the options.

(A) Sucrose: α-D-glucopyranosyl-(1→2)-β-D-fructofuranoside. It is non-reducing and its hydrolysis gives invert sugar. ✓
(B) Has a free anomeric OH on the glucose unit, so it is reducing.
(C) Lactose (β-galactosyl-(1→4)-glucose) is reducing.
(D) Is 1→1-linked glucose (trehalose type). It is non-reducing but hydrolyses only to glucose, which is dextrorotatory.

Topics:Carbohydrates

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JEE Advanced 2023 · Paper 2, Chemistry question 4