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Biomolecules question

2022 · Shift 2 · Q18
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Biomolecules question

2022 · Shift 2 · Q18

JEE AdvancedChemistryBiomoleculesMCQ+3 / −1
Treatment of D-glucose with aqueous NaOH\mathrm{NaOH}NaOH results in a mixture of monosaccharides, which are
  1. A
    JEE Advanced 2022 Paper 2 Online Chemistry - Biomolecules Question 8 English Option 1
  2. B
    JEE Advanced 2022 Paper 2 Online Chemistry - Biomolecules Question 8 English Option 2
  3. C
    JEE Advanced 2022 Paper 2 Online Chemistry - Biomolecules Question 8 English Option 3
  4. D
    JEE Advanced 2022 Paper 2 Online Chemistry - Biomolecules Question 8 English Option 4
View written solutionFree

Correct answer: C

  1. Key concept: base-catalyzed isomerisation of aldoses

    When D\mathrm{D}D-glucose is treated with aqueous NaOH\mathrm{NaOH}NaOH, it undergoes enolisation via an enediol intermediate. This is known as the Lobry de Bruyn–van Ekenstein transformation.

  2. What happens to D-glucose?

    D\mathrm{D}D-glucose is an aldohexose. In basic medium, it forms an enediol intermediate, and this intermediate can rearrange to give:

    • back to D-glucose
    • the corresponding ketose: D-fructose
    • the epimer at C-2: D-mannose

  3. Why these products form

    • Conversion of the aldehyde form to the enediol allows rearrangement to a ketose ⇒\Rightarrow⇒ D-fructose.
    • Reprotonation on the opposite side at C-2 gives the C-2 epimer of glucose ⇒\Rightarrow⇒ D-mannose.

  4. Hence the mixture contains

    D-glucose,D-fructose,and D-mannose\boxed{\mathrm{D\text{-}glucose,\\ D\text{-}fructose,\\ and\ D\text{-}mannose}}D-glucose,D-fructose,and D-mannose​

  5. Matching with options

    The correct option must be the one listing:

    D-glucose+D-fructose+D-mannose\boxed{\mathrm{D\text{-}glucose + D\text{-}fructose + D\text{-}mannose}}D-glucose+D-fructose+D-mannose​

    Since the stored correct answer is C, option C corresponds to this set of monosaccharides.

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