Riboflavin (vitamin B2) behavior: The yellow pigment in egg yolk and milk becomes what on reduction with zinc in acid, and what upon re-oxidation?

Difficulty: Easy

Correct Answer: colorless on reduction with Zn in acid and regained its color on re-oxidation

Explanation:


Introduction / Context:
Riboflavin (vitamin B2) forms flavin coenzymes (FMN, FAD) that are yellow due to their conjugated isoalloxazine ring. The color changes with oxidation state provide a simple chemical demonstration of redox behavior.



Given Data / Assumptions:

  • Reduction performed with zinc in acidic solution.
  • Observation of visible color changes tied to electronic structure.
  • Re-oxidation occurs on exposure to air or oxidants.



Concept / Approach:
Oxidized flavins are yellow; the fully reduced form (leucoflavin) is colorless. Thus, reduction bleaches the pigment; re-oxidation restores the yellow color by re-establishing the chromophore's conjugation.



Step-by-Step Solution:
Start: oxidized riboflavin → yellow.Reduce with Zn/H+: riboflavin → leucoflavin (colorless).Expose to oxygen: leucoflavin → oxidized riboflavin (yellow returns).



Verification / Alternative check:
Laboratory demonstrations and spectroscopic data show diminished absorbance in the visible range upon reduction, recovering on re-oxidation.



Why Other Options Are Wrong:

  • Claiming colorless on oxidation reverses the real chemistry.
  • Suggestions of deeper color on reduction are contrary to the bleaching effect of forming leucoflavin.



Common Pitfalls:
Confusing semiquinone (blue/green radical) with fully reduced leucoflavin; complete reduction under the stated conditions yields colorless leuco species.



Final Answer:
colorless on reduction with Zn in acid and regained its color on re-oxidation

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