Re: ORGLIST - orglist Digest - V01 #167

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From: Jacob Zabicky (jzabicky$##$etseq.urv.es)
Date: Thu Aug 06 1998 - 08:41:28 EDT



>From: "Dr. Michael Fassbender" <mifa$##$nac.ac.za>
>Subject: RE: ORGLIST: Oxidation States of Carbon
>
>
>|I hope I'm not missing the question but cellulose is essentially
>|(CH2O)n. Is there something I don't get?
>
>Right, and this results in an effective C oxidation state of zero !
>
>Examples:
>
>methane: CH4 (state -4)
>methanol: CH3OH (state -2)
>methanal: HCHO (state 0)
>methanic acid: HCOOH (state +2)
>
>Cheers,
>

Well... almost. (CH2O)n would be the case of glucose. In polyglucose one
molecule of water is lost in binding consecutive units. Thus it is rather
(C6(H2O)5)n. Furthermore, the convention in organic chemistry for cases
like this is that the C-H and C-C bonds are truely covalent, thus, the
oxidation state of the C and H atoms is 0. Also by convention O atoms in
these compounds have a fixed oxidation state of -2. So a progressively
oxidised series would appear as follows for C. The totally hydrated forms
are presented to see
more easily the progression steps of oxidation. These consist of inserting
an O atom between C and H, forming a C-O-H group. The double headed arrows
represent equilibrium arrows.

state 0 methane: CH4
state +1 methanol: CH3OH
state +2 methanal: CH2(OH)2 <--> H2CO + H2O
state +3 methanoic acid: CH(OH)3 <--> HCOOH + H2O
state +4 carbonic acid: C(OH)4 <--> O=C(OH)2 + H2O <--> CO2 + 2H2O

If you write down the formula of glucose or cellulose (see below) you can
see five C atoms bonded to one oxygen atom (state +1) and one C atom double
bonded to one O atom in glucose or singly bonded to two O atoms in
cellulose (state +2). Thus, the average oxidation state of C according to
the above conventions is (5x1 + 2)/6 = 7/6 = 1.167.

The conventions mentioned above may be a little bit confusing in some
cases, as illustrated by the following pair of questions.

1. Given the formula of cellulose (stereochemistry not included), what
would be the average oxidation state of the H atoms?

      O-H
      |
    H-C-H
      | H H H H
      | | | | |
   (--C---C---C---C---C---O--)n
      | | | | |
      | O-H O-H O-H |
      | |
      O---------------'

2. As we saw above, in the list from methane through carbonic acid, loss of
water molecules does not alter the oxidation state of the C atoms. In view
of this, what would be the average oxidation state of C in ethene? The
following reaction, depicting the dehydration of ethanol, may be helpful
(or add to the confusion):

CH3CH2OH ---> CH2=CH2 + H2O

All the best,

Jacob Zabicky


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Department of Mechanical Engineering Private No.
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