Why does methylene blue (which is initially blue) reduce to form a colorless solution?

I was told by my teacher that it has to do with the presence of an additional double bond on methylene blue, but I can't figure out or find any source on the internet that claims so.


The color change results from the reversible oxidation-reduction reaction of the methylene blue indicator. In alkaline solutions, glucose is oxidized to D-gluconic acid or alpha-D-gluconolactone.

$\ce{HOCH2(CHOH)4CHO + 3OH- -> HOCH2(CHOH)4CO2 + 2H2O + 2e-}$

In the course of this reaction, methylene blue is reduced from the blue (oxidized) form to the colorless (reduced) form.

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Shaking the flask dissolves $\ce{O2}$ in the solution, which oxidizes the indicator back to the blue (oxidized) form.

  • $\begingroup$ Okay, but does the presence of extra double bond has anything to do with colour change? $\endgroup$ – mathnoob123 Feb 27 '18 at 10:53
  • $\begingroup$ Can you expand on how the presence of double bond leads to colour change? $\endgroup$ – mathnoob123 Feb 27 '18 at 10:55
  • 4
    $\begingroup$ Sure ! The double bond changes its position in the product , which is due to tautomerisation. Now , since tautomers have different physical properties , they have different color. One reason for this is , the lone pairs of Nitrogen which were earlier undergoing conjugation at one side , now undergo conjugation from both sides , resulting in decreasing the electron density on N . Since , it becomes harder for N to donate electrons to the solvent , it now shows different spectral behaviour and hence different color . $\endgroup$ – Tanuj Feb 27 '18 at 11:00
  • $\begingroup$ Is the same reason applicable for the colour change on methyl orange due to breaking of double bond between the nitrogen ? $\endgroup$ – mathnoob123 Feb 27 '18 at 11:16
  • $\begingroup$ Yes ! You're correct. $\endgroup$ – Tanuj Feb 27 '18 at 11:18

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