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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: ptryon on March 04, 2013, 07:23:17 AM
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If ethanal is converted to ethanoic acid we would describe this as oxidation because more bonds to oxygen are formed / oxygen is added to the molecule. This process should also therefore be consistent with the electrochemical definition of oxidation (the loss of electrons). However, I cannot see how this is the case:
Ethanal has 18 valence electrons (including lone pairs)
Ethanoic acid has 24 valence electrons (including lone pairs)
To me it appears that ethanoic acid has actually been reduced as the number of electrons is really increasing.
If you try and argue that the carbon containing the oxygen atoms has been oxidized- in both cases (ethanal and ethanoic acid) this carbon has eight electrons. You might argue that they are further from this carbon atom when we add more bonds to the electronegative oxygen- however, this does not constitute a loss of electrons (the carbon becomes delta plus, not plus). In both cases (ethanal and ethanoic acid), the carbon to which the oxygen atoms are affixed has no formal charge, so I can't even see how it fits the definition if we analyze the molecule using formal charges.
..any thoughts???
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Because of electronegativity Hydrogen has positive charge and Oxygen has negative charge. We get the oxidation number from both molecules.
CH3CHO: left Carbon has -3 because of 3 Hydrogen atoms: Right carbon has +1 because of 2-negative charge from oxygen and one positive charge from hydrogen. C-C bonds are 0.
CH3COOH: left Carbon as above no change, right carbon 3 bonds to oxygen gives +3. The hydrogen is bonded to a oxygen.
So we have an increase of the oxidation number from +1 to +3, with other words loose of 2 electrons what means oxidation.
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Hey,
Thank you for your answer. That does help... but it leads me to two further questions:
(1) In your analysis you treat each carbon separately and consider the hydrogen and oxygen atoms that they are joined to... but each carbon atom is also joined to the other carbon atom. Why did you not account for this? (I am not sure how you would account for it, or whether its even important- but this is what made me hesitate from the oxidation state analysis)
(2) Where do these two electrons actually go from? (what bond in the molecule?) To me it seems like the number of electrons in the molecule actually increases.
Any more thoughts would be appreciated ;-)
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"Oxidation Numbers in the Study of Metabolism"
Ronald Bentley, James Franzen, and Thomas G. Chasteen
BIOCHEMISTRY AND MOLECULAR BIOLOGY EDUCATION
Vol. 30, No. 5, pp. 288–292, 2002
I find that it is sometimes helpful to sum the oxidation numbers of individual carbon atoms, but I am not sure that it is necessary to do so. Consider the reaction catalyzed by propanediol dehydratase, which is the conversion of propane-1,2-diol to propionaldehyde (propanal), as discussed in the reference above. One carbon increases in oxidation number and another decreases, but there is no net oxidation or reduction of the whole molecule.
EDT
I may have misunderstood your question. When a carbon atom is joined to another carbon atom, both atoms have the same electronegativity, and one of the atoms cannot be said to own both electrons, unlike a C-O bond, for example.
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I may have misunderstood your question. When a carbon atom is joined to another carbon atom, both atoms have the same electronegativity, and one of the atoms cannot be said to own both electrons, unlike a C-O bond, for example.
So are you suggesting that the carbon-carbon bond in e.g. 1,1,1-trichloroethane is not polar? Or, alternatively, if all carbon atoms are the same how do you account for regioselectivity in electrophilic aromatic substitution in for instance, aniline, or nitrobenzene?
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No. In order to assign oxidation numbers, one needs rules to decide who owns the electrons, and using the electronegativities of the atoms is pretty standard, unless I am mistaken. Perhaps the rules could be expanded to include cases such as the one you provide. There would be some sacrifice in simplicity, however.
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Hey,
Thank you for your answer. That does help... but it leads me to two further questions:
(1) In your analysis you treat each carbon separately and consider the hydrogen and oxygen atoms that they are joined to... but each carbon atom is also joined to the other carbon atom. Why did you not account for this? (I am not sure how you would account for it, or whether its even important- but this is what made me hesitate from the oxidation state analysis)
(2) Where do these two electrons actually go from? (what bond in the molecule?) To me it seems like the number of electrons in the molecule actually increases.
Any more thoughts would be appreciated ;-)
C-C bonds are 0. I wrote in my answer above. The other discussion doesnt help. In H2
, O2 ,N2 and so one, the oxidation number is also 0. By definition bonds between to same atoms are 0. That doesnt mean in 1,1,1-Trichlorethane is no polarity.
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Sorry to be awkward but I am still confused over a couple of things:
1 I know that atoms in an element have an oxidation state of zero (by definition) as they will have equal electronegativity. I am not sure that this also applies to atoms of the same element joined together in a compound. In this case, for instance, we would predict that the C-C bond in both ethanal and ethanoic acid would be polar, so I am not convinced that it is valid to assume they have an oxidation state of zero.
2 I am still unsure whether (and if so how) this transformation constitutes oxidation in terms of electron transfer. Notice the oxidized molecule, overall, has more electrons (implying a net gain of electrons rather than a loss which would be expected in oxidation). Specifically, where are the electrons lost from?
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The whole thing is a model made by human.
If you have C-C and judge them both zero, so is it also possible in CCl3-CH3 to give left C +3,45 and right C -2,55. But for calculation its easier to transfer 0.45 from one C to the other so we +3 and -3. The real world probably looks different.
There not more electrons. Its covalent andsometimes polarized bonds in which two orbitales have 2 electrons.
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Good point about oxidation states being a model developed by humans. Even if we assume that the carbons have a zero oxidation state with respect to one another I would still argue that (in total) ethanoic acid has more electrons than ethanal. Check out their lewis structures if you're not convinced and you'll find :
Ethanal has 18 valence electrons (including lone pairs)
Ethanoic acid has 24 valence electrons (including lone pairs)
Oxidation requires a loss of electrons. The difference in oxidation state is simply a result of the carbon atom having more bonds to oxygen; but as far as I can see, it does not result from a loss of electrons either from the molecule or even from this carbon atom...
Any thoughts?
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I think to count the electrons of the whole molecule makes no sense, only one C get a change from Aldehyde to Acid.
R-CHO has an Oxidation number from +1, in R-COOH it is +3. So its an oxidation from +1 to +3.
Redox equation: R-CHO + H2O => R-COOH + 2 H+ + 2 e-
These Electrons are used to reduce the oxidation chemical like Chromate, Permanganate, etc. to make it as whole redox reaction.
Or more simple example. We have Methane CH4. Oxidation number of C = -4.
This will be burnt and create CO2 oxidation number + 4 for C.
In your model we have 8 Electron in methane and 16 in carbon dioxide with lone pair.
The burning is an oxidation and we have a transfer of 8 Electrons.
CH4 => C4+ + 4 H+ + 8 e-
2 O2 + 8 e- => 4 O2-
CH4 + 2 O2 => CO2 + 2 H2O
It also model, we wouldn't have hydrogen ions like in solution.
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Hey, thank you for your time on this question. Your answer really helps. You gave the following half equation for the oxidation process:
R-CHO + H2O => R-COOH + 2H+ + 2e-
There is no doubt that this is oxidation with respect to electron transfer, because electrons are produced and transferred to the oxidizing reagent... but, would I be right thinking that technically the electrons come from water and not from the aldehyde? Hence, 2H+ ions are formed and one oxygen atom is also formed- the oxygen atom is added to the aldehyde to form the acid (I accept in real life it may not follow these steps, but for the sake of conceptual simplicity I have broken my thinking down like this). If I am correct (and this may seem pedantic) but wouldn't it therefore be more correct to say that water is being oxidized rather than ethanal?
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Why should be water oxidized? H2O: Hydrogen is +1 and Oxygen is -2. There is no change to Oxidation number if the Oxygen is connected to the Carbon later. Also the Hydrogen is the same, it doesn't matter to which element its connected later on. The only change is at Carbon.
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Because, if I am not mistaken, the electrons actually come from the water molecule. Hence the hydrogen atoms in the water molecule are actually uncharged at the start, but end up as H+ ions
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But how do you explain the oxidizing of alcohol
R-CH2OH => R-CHO + 2 H+ + 2 e-
There is no water. And in water you have everytime H+ and OH- Ions. The Redoxreaction takes place in acidic or alkaline environment.
Also in inorganic:
Example Permanganate : MnO4- + 8 H+ + 5 e- => Mn2+ + 4 H2O Reduction
Oxidation pick one of our organic molecules.
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In the case of ethanoic acid, the electrons that are brought in belong to oxygen. Oxygen is more electronegative; therefore, it owns the electrons in this formalism. In biochemistry and organic chemistry, it is natural to focus on carbon and look for reagents that oxidize or reduce certain carbon-containing compounds. Besides the reference I already provided, there are some other resources one could consult:
http://pubs.acs.org/doi/pdf/10.1021/ed079p965
http://web1.uct.usm.maine.edu/~newton/Chy251_253/Lectures/OxidationLevels/OxiationLevels.html
Halkides CJ, Journal of Chemical Education 2000;77(11):1428-1432. http://pubs.acs.org/doi/pdf/10.1021/ed077p1428