JEE Advanced 2023 Paper 2 Q04 Chemistry Organic Chemistry Isomerism & Stereochemistry Medium

JEE Advanced 2023 Paper 2 · Q04 · Isomerism & Stereochemistry

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

  1. A. Sucrose (glucose-$\alpha$-1,2-$\beta$-fructose)
  2. B. Maltose (glucose-$\alpha$-1,4-glucose)
  3. C. Cellobiose (glucose-$\beta$-1,4-glucose)
  4. D. Lactose (galactose-$\beta$-1,4-glucose)
Reveal answer + step-by-step solution

Correct answer:A

Solution

$X$ is non-reducing (cannot be oxidised by bromine water), so both anomeric carbons must be in the glycosidic bond — sucrose. Acid hydrolysis of sucrose gives $D$-glucose (dextrorotatory $+52.7°$) + $D$-fructose (laevorotatory $-92.4°$). Net rotation is laevorotatory — invert sugar.

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