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I came across methylene blue's structure when it reacts with oxygen.

enter image description here

The left is the reduced state, and the right is the oxidized state.

However, when I looked at them carefully, the oxidation number of O2 changes normally, but I couldn't find the change of oxidation number in methylene blue.
( the middle N is -3, S is -2, the right N is -3, in both structures. The H is always +1.)

What am I missing here?

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    $\begingroup$ Count the number of atoms. It has changed. $\endgroup$ Commented Nov 23, 2021 at 13:42
  • $\begingroup$ 2H missing from methylene blue, and added to O2 to make H2O2. But their oxidation numbers are +1 on both molecules. Does that explain the oxidation of methylene blue? ( I know losing H is one kind of definition of oxidation, but I want to know how to justify it when it contradicts the oxidation number definition.) $\endgroup$
    – Wang
    Commented Nov 23, 2021 at 14:17
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    $\begingroup$ The concept of oxidation numbers is rarely used in organic chemistry. That being said, you surely can assign them, and the result won't contradict the overall picture that we already know. If you insist on doing that, then look again at those carbons, especially the one which is doubly bonded to N in the right structure, but not in the left one. $\endgroup$ Commented Nov 23, 2021 at 14:35
  • $\begingroup$ Thank you! I missed the change of carbon atoms! $\endgroup$
    – Wang
    Commented Nov 23, 2021 at 14:46

1 Answer 1

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You missed the carbon atoms, and specifically the carbon atoms in the central ring which are bonded to nitrogen and sulfur. When the methylene blue is in its reduced leuco state, each of these has one bond to the more electronegative atoms and none to hydrogen (the only other element in the molecule), thus you count out the oxidation state as +1 for all four of these carbons.

Now you shake the bottle and it turns blue. Note that in this form two of the central carbons now have two bonds to the nitrogen or sulfur, so they have been oxidized from +1 to +2. Which two carbons get oxidized depends on which of multiple contributing structures you draw, but the net result is always you oxidize two carbon atoms by one unit.

Half the available contributing structures have the two carbon atoms on the left side of the central ring oxidized, and half have the carbon atoms on the right half oxidized. Since the two halves are equivalent, if you are into fractional oxidation numbers you can say that "on average" the four central carbon atoms each go from +1 to +1.5.

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