# Ochem Oxidation number question

Discussion in 'DAT Discussions' started by LaughingGas, 07.30.12.

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1. ### LaughingGas

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From Datqvault explanation:
The carboxylic acid has the most oxidized state for carbon. The oxidation number for a carboxylic acid is +3, for an aldehyde, ether, or alcohol is +1, and for a ketone is +2. Esters have the same oxidation states as carboxylic acids.

How do you calculate or assume that?

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3. ### chax1

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I'm not 100% sure, but I believe you can look at it as the number of bonds a carbonyl to an oxygen minus the number of bonds to a Hydrogen. So like carboxylic acid will be 3 because its a double bond to one oxygen and a single bond to another oxygen The carbonyl of a ketone has a double bond to O and no H bonds. Aldehyde is +1 because it has a hydrogen and double bond to oxygen (2-1) where a ketone has no hydrogen (2-0). This is based off of my understanding from chad's videos where he said in organic an oxidation increases the # of bonds to oxygen and a reduction increases number of bonds to H.
4. ### Sandstorm

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I agree with this, but as far as DAT goes, you simply need to know which is highest priority. Carboxylic acids are highest, followed by ketones and aldehydes (I would be shocked if they had a question that made you choose between the two), followed by alcohols, then ethers.

In summary:

1) Carboxylic acid
2) aldehyde/ketone
3) alcohol
4) ether

Of note, the above is for IUPAC naming. If you are assigning r/s configuration, go by molecular weight.

So if we are talking about oxygen containing compounds alone, and you had both an alcohol and and ether bound to the chiral carbon, the ether would have higher priority because you look first at the first atom bound to the chiral carbon. In the case of alcohols and ethers, both have an oxygen. So then you look at the next atom bound to that group. Alcohol would have a hydrogen bound, ether would be another carbon, so the ether would have higher priority in this case.