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When vitamin C reacts with iodine, it will lose two of its electrons (oxidation) which iodine will accept (reduction). What I do not understand is what part of the vitamin C exactly is losing the electrons? Where is the change in oxidation number happening when it is oxidized to dehydroascorbic acid?

Half equations:

$$\ce{C6H8O6 → C6H6O6 + 2H+ + 2e-}$$

$$\ce{I2 + 2e- → 2I-}$$

I have looked for answers online, but I have not managed to find any, or at least none that I understand (high school student).

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A. M. Helmenstine states,"Triiodide oxidizes vitamin C to form dehydroascorbic acid." Wikipedia shows structural diagrams for the molecule in various states of hydrolysis.

Note that there is debate on representing the structure: "The actual structures of the various forms of dehydroascorbic acid have been known for at least a quarter of a century so that the continued use of the oversimplified tricarbonyl structure is hard to excuse."

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Typically, when organic molecules get oxidized or reduced, the oxidation numbers of carbon atoms changes, while those of hydrogen (+1) and oxygen (typically -2) don't change.

The oxidation number is a formal number (not measurable), and does not imply certain charges on certain atoms. The atoms in organic molecules are held together by covalent bonds, so the electrons are not "assigned" to a given atom.

One easy way to figure out the oxidation number of a carbon (with four bonds and no formal charge) in a given Lewis structure is to start at zero, add one for each bond with oxygen, and subtract one for each bond with hydrogen.

The Lewis structure of the reduced form has two carbons at an oxidation state of +1, and they change to +2 in the oxidized forms shown below:

![enter image description here

The image shows three forms of the oxidized vitamin C (with a nod to the other answer) but in all of them, the bottom two carbon atoms show two bonds to oxygen each (either two single bonds or one double bond) and no bonds to hydrogen.

The reactions that are shown, hydrolysis and hydration, don't change the oxidation state of the carbon atoms.

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An oxidation number suggests that an atom is enriched or depleted in electron density relative to the free (unbonded) atom. This from the definition of the free form of an atom as the reference oxidation state, with zero oxidation number. If you inspect examples in the Wikipedia page linked above you'll note that increasing the number of bonds (or equivalently bond order) between carbon and oxygen increases the oxidation state of carbon, since oxygen is more electronegative. Formally, due to its high electronegativity, bonded oxygen is assigned a -2 state. As oxidation numbers of individual atoms have to add up to the charge of the molecule, this implies that the carbon OS increases with increasing CO bonds or bond order.

enter image description here

In the oxidation of ascorbic acid (structure above from the Wikipedia) electrons involved in the CC double bond are transferred to oxygen to form CO bonds or increase the CO bond order, implying that the carbons are oxidized. Electrons in a CC double bond are however not necessarily shared equally between carbon, meaning one carbon could be assigned a formal charge of +1 relative to the other with OS -1. Therefore one interpretation is that one carbon (that nearest to the carbonyl) is oxidized while the state of the other formally does not change.

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