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Offline xstrae

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doubt on oxidation number
« on: September 13, 2006, 12:29:27 PM »
Hi, I have been asked to find the oxidation state of:

 C in CH3COOH
    2x - 2(2) + 1(4) = 0   ;  x = 0

is that correct?

Offline Borek

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Re: doubt on oxidation number
« Reply #1 on: September 13, 2006, 01:13:07 PM »
OK

http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-failure

Scroll down to the end of the page. Oxidation numbers in organic compunds are crazy.
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Offline AWK

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Re: doubt on oxidation number
« Reply #2 on: September 14, 2006, 05:21:03 AM »
konichiwa2 ,you did it correctly.


Contrary to Borek, I think oxidation numbers problem in organic chemistry is a quite simple one.
You can calculate a mean oxidation number for for a specific kind of atoms in a  molecule, or oxidation numbers for each atom.

For sepatate atoms:
CH3  ON for C is -3 [x+3=0]
COOH ON for C is +3 [x-2(2)+1=0]
And mean is (-3+3)/2=0
C-C bonds do not contribute to ON in this method. Why?
Other example
CH3-CH2-COOH ON(for C) [ 3x+6-4=0] => x=-2/3
For separate atoms: -3 -2 and +3 respectively for C1, C2 and C3 (from the left)
mean from above numbers is (-3-2+3)/3=-2/3

Try to count oxidation numbers for all atoms  in CH3OOH (peroxide)
Use separate atom method


Just for checking)
 (C-2, H+1, O-1)
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Offline Borek

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Re: doubt on oxidation number
« Reply #3 on: September 14, 2006, 06:54:40 AM »
Contrary to Borek, I think oxidation numbers problem in organic chemistry is a quite simple one.
You can calculate a mean oxidation number for for a specific kind of atoms in a  molecule, or oxidation numbers for each atom.

I have never stated they are difficult, I just can't see any rationale behind.

Do they reflect any real matter property? Any real property of atoms in a molecule? Do they have any use other than accounting when doing redox reactions, redox reactions that can be done without introducing non-existing concepts?

If not, why do the students have to learn them? And learn how to assign them to every atom in organic molecules? For me that's a waste of time, art for the art's sake.
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Offline xstrae

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Re: doubt on oxidation number
« Reply #4 on: September 14, 2006, 10:19:45 AM »
ok thanks
Quote
Try to count oxidation numbers for all atoms  in CH3OOH (peroxide) Use separate atom method

I got that when I read that it is a peroxide. But how do i determine whether O in a given compound is a peroxide? If you hadnt told me it is a peroxide, i think I would have ended up with ON of C = 0

Offline enahs

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Re: doubt on oxidation number
« Reply #5 on: September 15, 2006, 12:02:57 AM »
Quote
I got that when I read that it is a peroxide. But how do i determine whether O in a given compound is a peroxide? If you hadnt told me it is a peroxide, i think I would have ended up with ON of C = 0

Then how would you have connected all the other atoms without breaking valencey and octet/duet rules?


Quote

If not, why do the students have to learn them? And learn how to assign them to every atom in organic molecules? For me that's a waste of time, art for the art's sake.

Since chemical reactions are defined as transfer of electrons it is very important.
Important in real synthesis (where you just do not remember a bunch of list of chemicals that react with each other). Determining mechanisms of reactions by following the electrons, etc etc. It is not art for arts sake, it does however have very limited to use at lower levels.





Offline Borek

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Re: doubt on oxidation number
« Reply #6 on: September 15, 2006, 02:31:54 AM »
Quote
If not, why do the students have to learn them? And learn how to assign them to every atom in organic molecules? For me that's a waste of time, art for the art's sake.

Since chemical reactions are defined as transfer of electrons it is very important.
Important in real synthesis (where you just do not remember a bunch of list of chemicals that react with each other). Determining mechanisms of reactions by following the electrons, etc etc. It is not art for arts sake, it does however have very limited to use at lower levels.

Can you show examples of organic reactions where ON help determine reaction mechanism? Can you elaborate on this "etc, etc" part?
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Offline xstrae

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Re: doubt on oxidation number
« Reply #7 on: September 15, 2006, 04:45:06 AM »
Quote
I got that when I read that it is a peroxide. But how do i determine whether O in a given compound is a peroxide? If you hadnt told me it is a peroxide, i think I would have ended up with ON of C = 0


Then how would you have connected all the other atoms without breaking valencey and octet/duet rules?

This is what I would have done : x + 1(3) + -2(2) + 1(1) = 0 ; x = 0
How do I figure out that the O is in the peroxide state?

Offline AWK

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Re: doubt on oxidation number
« Reply #8 on: September 15, 2006, 05:25:02 AM »
[
This is what I would have done : x + 1(3) + -2(2) + 1(1) = 0 ; x = 0
How do I figure out that the O is in the peroxide state?

You hit the right nail on the head.
Having known nothing about structure of a compound you can calculate oxidation numbers nearby at your pleasure taking into account only two rules (1.ON for elements=0, and sum of all ONS is eqaul to 0 for compounds or is equal to charge of ion), and counting taken and lost electrons you can solve any redox equation. This is an aproach which has nothing common with quantum theory but always efficient. Moreover you use math on the level of primary school.

For compound CH4O2 formal ON for C is 0 (using +1 as ON for H and -2 as ON fot O). But when you know its structure CH3OOH ON for C is -2.
Decomposition of CH4O2 leads us to the reaction
CH4O2 = CH2O + H2O
which is non-redox reaction for ON of C=0 or a redox reaction (disproportionation) for ON of C=-2. In the later case you can learn something on a mechamism (chemistry) of this reaction (which Borek put in question). Note, this approach is for slightly more advanced chemists.
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Offline Borek

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Re: doubt on oxidation number
« Reply #9 on: September 15, 2006, 06:38:26 AM »
For compound CH4O2 formal ON for C is 0 (using +1 as ON for H and -2 as ON fot O). But when you know its structure CH3OOH ON for C is -2.
Decomposition of CH4O2 leads us to the reaction
CH4O2 = CH2O + H2O
which is non-redox reaction for ON of C=0 or a redox reaction (disproportionation) for ON of C=-2. In the later case you can learn something on a mechamism (chemistry) of this reaction (which Borek put in question). Note, this approach is for slightly more advanced chemists.

When you know structure of the compound you know it is peroxide, that's enough to suspect what is going on during decomposition. I rest my case.
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Offline enahs

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Re: doubt on oxidation number
« Reply #10 on: September 15, 2006, 08:59:14 AM »

Can you show examples of organic reactions where ON help determine reaction mechanism? Can you elaborate on this "etc, etc" part?

It is not limited to organic chemistry and synthesis.
However, the easiest example to point to (because I am lazy) is electrochemistry (making batteries).

Offline Borek

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Re: doubt on oxidation number
« Reply #11 on: September 15, 2006, 09:57:13 AM »

Can you show examples of organic reactions where ON help determine reaction mechanism? Can you elaborate on this "etc, etc" part?

It is not limited to organic chemistry and synthesis.
However, the easiest example to point to (because I am lazy) is electrochemistry (making batteries).

You are mistaking balancing redox reaction with determining reaction mechanism.

But let's start from the beginning. Define oxidation number.
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Offline enahs

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Re: doubt on oxidation number
« Reply #12 on: September 15, 2006, 10:18:15 AM »
Quote
You are mistaking balancing redox reaction with determining reaction mechanism.

No I am not. I am just trying to illustrate the uses of oxidation numbers without writing a paper or listing 20 reactions.


I do not understand what your problem with them is. As a chemist you want to understand what is happening in a chemical reaction. Oxidations numbers are useful for that. Knowing what gained and lost electrons is very important. You might see no use for it to get a good grade on the test, but just by the definition of the word it is clearly important in understanding what is taking place in a chemical reaction.


Offline Borek

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Re: doubt on oxidation number
« Reply #13 on: September 15, 2006, 11:11:31 AM »
I do not understand what your problem with them is. As a chemist you want to understand what is happening in a chemical reaction. Oxidations numbers are useful for that. Knowing what gained and lost electrons is very important. You might see no use for it to get a good grade on the test, but just by the definition of the word it is clearly important in understanding what is taking place in a chemical reaction.

For the oxidation numbers to be usefull they have to reflect some matter property. They don't. Please provide definition of oxidation number.
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Offline enahs

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Re: doubt on oxidation number
« Reply #14 on: September 15, 2006, 11:58:53 AM »
Quote

For the oxidation numbers to be usefull they have to reflect some matter property. They don't. Please provide definition of oxidation number.

You are wanting me to say “It is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic.” (taken directly from Wikipedia). And you want to pounce on me about the hypothetical part about it not being a real “matter property”. However, by doing it for both sides it does show you the actual loss and gain of electrons, and in that case it is your precious “matter property”.

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