JEE Advanced 2025 — Paper 2 — CHEM-III solution

JEE Advanced2025Paper 2CHEM-III

A linear octasaccharide (molar mass = 1024 g mol $^{- 1}$ ) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is 58.26% (w/w) of the total amount of the monosaccharides produced in the hydrolyzed products. The number of ribose unit(s) present in one molecule of octasaccharide is _______.

Use : Molar mass (in g mol $^{- 1}$ ): ribose = 150, 2-deoxyribose = 134, glucose = 180; Atomic mass (in amu): H = 1, O = 16

Your answer

Solution

Complete hydrolysis of a linear octasaccharide requires $8 - 1 = 7$ water molecules (one per glycosidic bond):

Octasaccharide $+ 7 H_{2} O \to$ ribose units $+$ 2-deoxyribose units $+$ glucose units.

Mass balance: total mass of monosaccharides $= 1024 + 7 \times 18 = 1024 + 126 = 1150$ g per mole of octasaccharide.

The 2-deoxyribose constitutes 58.26% (w/w):

$mass of 2-deoxyribose = 0.5826 \times 1150 = 669.99 \approx 670$ g

Number of 2-deoxyribose units: $670/134 = 5$ .

Let the number of ribose units be $x$ and glucose units be $y$ , with $x + y = 8 - 5 = 3$ and the mass balance:

$5 \times 134 + 150 x + 180 y = 1150$

$670 + 150 x + 180 y = 1150 \Rightarrow 150 x + 180 y = 480$

With $x + y = 3$ , substitute $y = 3 - x$ : $150 x + 180 (3 - x) = 480 \Rightarrow 540 - 30 x = 480 \Rightarrow x = 2$ .

$∴$ Number of ribose units $= 2.00$ .

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