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JEE Advanced 2022Paper 2
Exam
JEE Advanced
Year
2022
Paper
Paper 2
Subject
Chemistry
Chapter
Biomolecules

JEE Advanced 2022 · Paper 2 · Chemistry · Biomolecules

Treatment of D -glucose with aqueous NaOH results in a mixture of...

Treatment of DD-glucose with aqueous NaOHNaOH results in a mixture of monosaccharides, which areTreatment of D -glucose with aqueous NaOH results in a mixture of monosaccharides, which areTreatment of D -glucose with aqueous NaOH results in a mixture of monosaccharides, which areTreatment of D -glucose with aqueous NaOH results in a mixture of monosaccharides, which areTreatment of D -glucose with aqueous NaOH results in a mixture of monosaccharides, which areTreatment of D -glucose with aqueous NaOH results in a mixture of monosaccharides, which are

Options

  1. ATotal enthalpy change for freezing of 1 mol of water at 10^ C to ice at -10^ C is ____ (Given : Δ_ fus H=x ~kJ / mol )
  2. BTotal enthalpy change for freezing of 1 mol of water at 10^ C to ice at -10^ C is ____ (Given : Δ_ fus H=x ~kJ / mol )
  3. CTotal enthalpy change for freezing of 1 mol of water at 10^ C to ice at -10^ C is ____ (Given : Δ_ fus H=x ~kJ / mol )
  4. DTotal enthalpy change for freezing of 1 mol of water at 10^ C to ice at -10^ C is ____ (Given : Δ_ fus H=x ~kJ / mol )

Correct answer

C. Total enthalpy change for freezing of 1 mol of water at 10^ C to ice at -10^ C is ____ (Given : Δ_ fus H=x ~kJ / mol )

Step-by-step solution

When glucose is treated with dilute NaOH, it gives a mixture of D-glucose, D-mannose and D-fructose. This reaction is known as Lobry de Bruyn-van Ekenstein rearrangement. Basic catalysed tautomerism through enediol intermediate takes place.Treatment of D -glucose with aqueous NaOH results in a mixture of monosaccharides, which are

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