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I've been reviewing redox reactions and have been using the OIL RIG acronym to classify oxidations vs. reductions. Generally this makes sense, and many resources I've looked at repeat the "Oxidation Is Loss - Reduction Is Gain of electrons" statement.

But, I also know that aldehydes can be oxidized to carboxylic acids: $$ \ce{RCHO -> RCO2H} $$

The compound $\ce{RCHO}$ does not lose electrons when oxidized to $\ce{RCO2H}$ -- in fact, it gains a whole new oxygen atom. So, there appears to be a net gain of electrons here. Does OIL RIG not always hold, or am I misunderstanding something?

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    $\begingroup$ Take a look at the functional group carbon. Upon the reaction taking place, it loses a $\ce{C-H}$ bond and gets a $\ce{C-O}$ bond. Oxygen is more electronegative than hydrogen is relative to carbon, so in a way, oxygen pulls electrons away from carbon, so there's an oxidation. $\endgroup$ – Pritt Balagopal Jun 28 '17 at 5:13
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    $\begingroup$ My previous comment didn't mention oxidation numbers, so have a read about it here: en.m.wikipedia.org/wiki/Oxidation_state $\endgroup$ – Pritt Balagopal Jun 28 '17 at 5:18
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    $\begingroup$ Roughly speaking, compounds don't get oxidized. Elements do. Think of it this way for a while. Then turn back to the usual way. $\endgroup$ – Ivan Neretin Jun 28 '17 at 6:04
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OILRIG always holds, but you need to be careful at what you are actually looking at. When we say a compound gets oxidised, that is technically not true. In this framework, only atoms get oxidised or reduced. In the reaction $$\ce{R-CHO + [O] -> R-COOH}$$ we usually omit the reduction part, because it's always the same: the oxygen.

To identify the whole redox process, use oxidation numbers. Keep in mind that these are simply a bookkeeping tool and the partial charges of the atoms involved might be quite different. Setting $\ce{R {=} CH3}$ or the above reaction it boils down to this: $$\ce{ \overset{+1}{H}_3\overset{-3}{C}-\overset{\color{\red}{+1}}{C}\overset{+1}{H}\overset{-2}{O} + [\overset{\color{\green}{0}}{O}] -> \overset{+1}{H}_3\overset{-3}{C}-\overset{\color{\red}{+3}}{C}\overset{-2}{O}\overset{\color{\green}{-2}}{O}\overset{+1}{H} }$$

Now you can see that the carbonyl carbon increases its oxidation state from $\color{\red}{+1}$ to $\color{\red}{+3}$, so formally it lost $\pu{\color{\red}{2} e^-}$. The oxidising agaent, indicated with $\ce{[O]}$, at the same time decreases its oxidation state from $\color{\green}{0}$ to $\color{\green}{-2}$, therefore formally gaining $\pu{\color{\green}{2} e^-}$. The redox reaction is complete.

Let's look at a real world example, the Baeyer-Villiger oxidation:

Baeyer-Villiger oxidation with oxidation states

As you can see, the ketone gets oxidised, more precisely the carbonyl carbon gets oxidised as increases its oxidation state. On the other side, the peroxide oxygens get reduced as they decrease their oxidation state.

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I actually got confused with this before as well. But when you start to think of things based on the perspective you have to look at them from, it becomes easier to understand. It's all a matter of separating the meaning of oxidation and reduction when it comes to organic chemistry. Simply put, in organic chemistry what is meant by "oxidation" is simply an increase in an atom's number of bonds to oxygen, while the opposite implies reduction. For example, when you oxidize an alcohol, the C-O single bond becomes a C=O double bond and so in essence carbon's bonds to oxygen increased by one and thus the carbon is oxidized.

In relation to electrons, I think it's not a loss of electrons in the singular sense like how we could look at it in general chemistry, but more of a shift of electron densities when the bonds change. Here it says that when atoms lose bonds to hydrogen and gain bonds to other heteroatoms, it is generally an oxidative process, considering that H is the least electronegative. So when an alcohol gets converted to a ketone there is a net loss of electron "density" on the carbon atom - synonymous to the loss of electrons.

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  • $\begingroup$ While this is all true and the IUPAC even states that oxidation is the gain of oxygens or loss of hydrogens it doesn't explain the actual question: How is it a loss of electrons? $\endgroup$ – Martin - マーチン Jun 28 '17 at 6:47
  • $\begingroup$ I think it's not a loss of electrons in the singular sense like how we could look at it in general chemistry, more of a shift of electron densities when the bonds change. $\endgroup$ – tumblewush Jun 28 '17 at 7:20
  • $\begingroup$ Here it says that when atoms lose bonds to hydrogen and gain bonds to other heteroatoms, it is generally an oxidative process, considering that H is the least electronegative. So when an alcohol gets converted to a ketone there is a net loss of electron "density" on the carbon atom - synonymous to the loss of electrons. $\endgroup$ – tumblewush Jun 28 '17 at 7:21
  • $\begingroup$ You make an excellent point and I encourage you to include it into your answer, which would make it a lot stronger and would offer another insight then just my oxidation number approach. $\endgroup$ – Martin - マーチン Jun 28 '17 at 7:34

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