[Sis290025]

I don't know if I am wording my inquiry correctly, but suppose one has the following molecules:

alkane
ether
aldehyde
ketone
carboxylic acid
ester
amide

How would you list them in increasing order of oxidized state? In other words, which which be the most oxidized?

If I surmise that the most oxidized forms have the most
C-O bonds and least C-H bonds then, in increasing order of oxidized molecules, will it be:

alkane, alcohol, ether, aldehyde, ketone, amide, ester, carboxylic acid?

If not, please let me know where I have went wrong.

Thank you.

[Wald_ron]

You aren't wording things correctly, and normally we think of things differently in Organic Chemistry.

An organic molecule is said to be oxidized if something more electronegative is added to the Carbon.

It is said to be reduced if something less electronegative than carbon is added to the carbon.

          oxidation
CH3-X    CH3-Y
          reduction

Where X is less electronegative than Carbon and Y is more electronegative than Carbon.

[macman104]

I guess I'm not sure about the ether/alcohol, but if we agree that a carboxylic acid is more oxidized than an ester (which I agree with), then analogously we can consider an alcohol more oxidized than an ether.  I would say this as they are essentially carbonyl-"less" derivatives of the acid/ester match.

[Cancer Boy]

In organic chemistry, oxidation levels are usually only assigned to one atom at a time. For example, the carbonyl carbon of an ester group is at oxidation level 3, whereas the other carbon of the group is at oxidation level 1. It doesn't really make sense to assign the entire group a single oxidation level. There are a couple of exceptions in which two adjoining carbon atoms can be assigned a single oxidation level (e.g. vicinal diols), but generally oxidation level only applies to one atom. To assign an oxidation level to a group containing more than one carbon like an ester or an amide, I would just go by the oxidation level of the most oxidised carbon (i.e. the carbonyl carbon). So...

Alkane < alcohol=ether < aldehyde=ketone < carboxylic acid=ester=amide.

[AC Prabakar]

I think we are confusing with "oxidation states" and  "oxidation levels".
In organic chemistry,Carbon atoms carrying functional groups can be classified by oxidation level.
According to the presence of functional groups in the given organic compound,five oxidation levels are possible.

Those are as follows:
1.Zero bond to hetero atom:Alkane oxidation state;Ex:-Alkanes
2.One bond to hetero atom:Alcohol oxidation state;Ex:-Alcohols,Amines,Ethers,Alkyl halides,Alkenes
3.Two bonds to hetero atom:Aldehyde oxidation state;Ex:-Aldehydes,ketones,Acetols,Alkynes
4.Three bonds to hetero atom:Carboxylic acid oxidation state;Ex:-Carboxylic acids,esters,amides,nitriles,acyl chlorides
5.Four bonds to hetero atom:Carbon dioxide oxidation state;Ex:-Carbon dioxide,CFC-12

In all these cases the oxidation state of carbon is "+4"

Now coming to your question,according to the above oxidation level classification,the order of the compound is as follows:

alkane, alcohol, ether,    aldehyde, ketone,     amide, ester, carboxylic acid

Highlighted in blue are belongs to "One bond to hetero atom:Alcohol oxidation state"(2 type in above definition)

Highlighted in orange are belongs to "Two bonds to hetero atom:Aldehyde oxidation state"(3 type in above definition)

The last in black (with out any colour) belongs to"Three bonds to hetero atom:Carboxylic acid oxidation state""( type in above definition)