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Biomolecules question

2019 · Shift 1 · Q10
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Biomolecules question

2019 · Shift 1 · Q10

JEE AdvancedChemistryBiomoleculesMultiple correct+4 / −1
Which of the following statement(s) is(are) true?
  1. A
    The two six-membered cyclic hemiacetal forms of D-(+)- glucose are called anomers.
  2. B
    Oxidation of glucose with bromine water gives glutamic acid
  3. C
    Monosaccharides cannot be hydrolysed to given polyhydroxy aldehydes and ketones
  4. D
    Hydrolysis of sucrose gives dextrorotatory glucose and laevorotatory fructose
View written solutionFree

Correct answer: A, C, D

  1. Check statement A

    Glucose exists in cyclic hemiacetal forms due to intramolecular reaction of the aldehyde group with an –OH group.

    For D−(+)−glucose\mathrm{D-(+)-glucose}D−(+)−glucose, the two six-membered ring forms are:

    • α\alphaα-D-glucopyranose
    • β\betaβ-D-glucopyranose

    These differ in configuration at the anomeric carbon (C-1), so they are called anomers.

    Hence, A is true.

  2. Check statement B

    Bromine water is a mild oxidizing agent. It oxidizes the aldehyde group of glucose to a carboxylic acid.

    So, Glucose→Br2/H2OGluconic acid\text{Glucose} \xrightarrow{Br_2/H_2O} \text{Gluconic acid}GlucoseBr2​/H2​O​Gluconic acid

    It does not give glutamic acid.

    Hence, B is false.

  3. Check statement C

    Monosaccharides are the simplest carbohydrates. By definition, they cannot be hydrolysed further into simpler polyhydroxy aldehydes or ketones.

    Hence, C is true.

  4. Check statement D

    On hydrolysis, sucrose gives one molecule each of glucose and fructose: Sucrose+H2O→Glucose+Fructose\text{Sucrose} + H_2O \rightarrow \text{Glucose} + \text{Fructose}Sucrose+H2​O→Glucose+Fructose

    The glucose formed is dextrorotatory, and fructose is laevorotatory.

    Hence, D is true.

  5. Final selection

    True statements are:

    • A
    • C
    • D

  6. Comparison with stored correct answer

    Stored correct answer: A, C, D

    My derived answer matches exactly.

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