Oxidation Number on Sulfur in DMSO?
Lloyd R. Parker
: We have a raging debate in our laboratory -- What is
: the oxid. num. on sulfur in dimethyl sulfoxide?
: The molecular formula is CH3SOCH3.
: We have two answers - zero and positive 4.
: If anyone could weigh in with their opinion
: and justifications, it would be greatly
: appreciated :-)
:
: Jeffrey C. Wallace
: Indiana University
: Department of Chemistry
: Jwal...@ezmail.ucs.indiana.edu
I say the S is +2. Since S and C have the same electronegativity (2.5),
using the Lewis structure method of finding oxidation numbers, the C and
the S split the electron pair in the bond; the O gets the electron pairs
in the double bond. This leaves the S with one electron in each C--S
bond and the unshared pair of electrons on the S itself. 4 electrons as
compared to 6 valance electrons in a S atom results in an oxidation
number of +2.
Jeff E. Janes
: We have a raging debate in our laboratory -- What is
: the oxid. num. on sulfur in dimethyl sulfoxide?
:
: The molecular formula is CH3SOCH3.
:
: We have two answers - zero and positive 4.
:
: If anyone could weigh in with their opinion
: and justifications, it would be greatly
: appreciated :-)
:
I would have to agree, the answer is zero and positive 4.
the electrons in the bond go the the most electronegative,
carbon and sulfur are alost equal in EN. In fact, on the Pauling
scale, S=2.58 > C=2.55, but on the Allred-Rochow scale,
S=2.44 < C=2.50. Who gets the electrons is a toss-up.
For the S-O bond, O clearly gets the electrons. I was about
to considere whether the bond is best called single or double,
but just realized that either way the O has -2 oxidation state,
formal charges not oxidation depends on the nature of the bond.
Now the 6 H's are clearly +1, the Carbons must be the same by
symmetry (I hope) and are either -4 or -2, depending on whether
carbon or sulfur win the tug-of-war. this leaves Sulfur at
-2+6+2*(-2 or -4)+S= charge on DMSO (0) so sulfur is 0 or +4.
My guess is sulfur wins, leaving C's at -2 and S at 0. I guess this
because
1) My intuition is that C likes being positive more than negative.
2) The structure with a single S-O bond leaves S with + formal charge,
which might cause it to pull harder on the C's electons.
Of course, in this case deciding on the oxidation state is more a
matter of book-keeping than chemistry, and I would say pick whichever
is more convenient for your purposes. (Sometimes I think that that is
*always* the case with oxidation states)
If anyone read this far, and you think I really screwed up, please
let me know before my Inorganic chemistry exam next Wednesday.
--
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
+ Jeff Janes at Michigan Tech,no logical University +
+ jej...@mtu.edu +
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Bob Samuel
>We have a raging debate in our laboratory -- What is
>the oxid. num. on sulfur in dimethyl sulfoxide?
>
>The molecular formula is CH3SOCH3.
>
>We have two answers - zero and positive 4.
>
>If anyone could weigh in with their opinion
>and justifications, it would be greatly
>appreciated :-)
It's 4 - 2 for the oxygen and 1 for each methyl. The sulphur atom
contributes 2 electrons to the S=O bond and 1 each to the S-C bonds. It's
not like Ni(PF3)4, for example, where Ni has oxidation state 0 and all the
bonding electrons come from the PF3.
--------------------------------------------------------------------------
Bob Samuel rsa...@cix.compulink.co.uk
Genzyme Pharmaceuticals and Fine Chemicals Tel: +44 1440 703522
Haverhill, Suffolk, CB9 8PU, UK Fax: +44 1440 707783
--------------------------------------------------------------------------
Bob Samuel
>Oops! slight miscalculation there - the series dimethylsulfide dimethylsulf-
>oxide, dimethylsulfone and dimethylsulfate are better thought of as -2, 0
>2 and 6 oxidation no's. Why? Take the sulfate as the most oxidized, then
>reduce it by 4 e to the sulfone, 2 e to the sulfoxide, and 2e to the sulfide.
>That would tend to favor the 0 people, I believe.
But these four compounds do not form a series. Dimethyl sulphide,
sulphoxide and sulphone clearly do, but the sulphate does not have the
methyl groups attached directly to the carbon, and should not be
considered as part of a group with the others. The oxidation states of S
in the first three compounds are 2, 4 and 6 respectively, and it's also 6
in the sulphate.
NE...@cunyvm.cuny.edu
agreement on this - carbon, phosphorus, and hydrogen have almost the same
electronegativities, but it is quite arbitrarily taken that C>P>H. Sulfur is
definitely more electronegative than all of these. So, in DMSO, the six
hydrogens are each +1, the oxygen is -2, the sulfur is zero, and each carbon
is -2. Any relation of these numbers to actual charges is unlikely, because
the oxidation number FORMALISM treats ANY electronegativity difference, no
matter how minute, as absolute. The FORMAL CHARGE on the sulfur is zero if
you have a S=O double bonded resonance structure, and +1 if you have a single
bond.
Really, isn't PRACTICAL chemistry more important?*
Jeffrey C. Wallace
the oxid. num. on sulfur in dimethyl sulfoxide?
The molecular formula is CH3SOCH3.
We have two answers - zero and positive 4.
If anyone could weigh in with their opinion
and justifications, it would be greatly
appreciated :-)
Gordon Miskelly
Oops! slight miscalculation there - the series dimethylsulfide dimethylsulf-
oxide, dimethylsulfone and dimethylsulfate are better thought of as -2, 0
2 and 6 oxidation no's. Why? Take the sulfate as the most oxidized, then
reduce it by 4 e to the sulfone, 2 e to the sulfoxide, and 2e to the sulfide.
That would tend to favor the 0 people, I believe.
Somewhere there must be a S(O)2(Me)(OMe) to complete the series, oh yes -
methyl methanesulfonate oxid no. 4.....
>Re Jeff Jones reply - 0 and 4 are good answers, but so is 2. Why? Because
>if we were to consider the series dimethylsulfide dimethylsulfoxide,
>dimethylsulfone, and dimethylsulfate, and were to say that each was being
>oxidized relative to the preceding compound, then one possible accounting
>would be oxidation no's 0,2,4 and 6 (the max we like to consider for S). All
>of which reminds me why I far prefer half reactions for electron book-keeping... In Jeff Jones' language, 2 would mean that we regarded C and S as having
>equal electronegativities. Oh, I love formalisms.
Gordon Miskelly
Daniel F Schmidt
by Jeffrey C. Wallace@nicke
>the oxid. num. on sulfur in dimethyl sulfoxide?
>
>The molecular formula is CH3SOCH3.
>
>We have two answers - zero and positive 4.
>
>If anyone could weigh in with their opinion
>and justifications, it would be greatly
>appreciated :-)
I would say zero, as both methyl groups SHOULD be +1 and O is most
likely -2. For it to be 4, each methyl group would have to be -1,
assuming that the O was still -2, which would not happen, OR the O would
have to be -6, which is VERY unlikely, at least to my knowledge.
I)aniel Schmidt
AKA
[-(-<Vector>-)-]
la...@inland.com
> We have a raging debate in our laboratory -- What is
> the oxid. num. on sulfur in dimethyl sulfoxide?
>
> The molecular formula is CH3SOCH3.
>
> We have two answers - zero and positive 4.
>
> If anyone could weigh in with their opinion
> and justifications, it would be greatly
> appreciated :-)
How about an empirical approach: how does DMSO react with nucleophiles? IMO,
a nucleophile will seek the positive end of a polar bond. So if it combines
with sulfur that would suggest an oxidation state of +4 for sulfur, whereas if
it combines with carbon the apparent oxidation state of sulfur is 0.
--OL
la...@inland.com
Mat Dowell
(Lloyd R. Parker) wrote:
I would ask 'why do you need to know the oxidation number?'.
Using the language of Valence Bond Theory, you can view the O-S bond as
being dative. Me-S-Me can be oxidised (so raising of the O.S. of S in the
positive sense is implicit) to Me-SO-Me in several ways; the bond results
from the donation of one electron pair from the sulphur to the oxygen and
probably some form of p-pi to d-pi overlap from the oxygen back to the
sulphur.
Whether the S to O bond is single or double is a contentious point, though,
as I'm sure you know. This is where the problem lies as the bonding cannot
be effectively described by Valence Bond Theory (and Lewis structure). If
the bond is considered as single, then the molecule is an ylid, with "S+"
and "O-" and the sulphur is in O.S. 2. If it is considered as a formal
double bond, the oxygen has to be considered both as a two-electron
acceptor AND a two-electron donor, rendering the sulphur in O.S. 0 again...
Your question highlights the issue of conventions and makes the inadequacy
of Valence Bond Theory glaringly obvious. It is possible to construct a
conventions that give rise to an O.S. of 0, +2 or +4. It is for you to
decide whether you treat a methyl group as being +1, 0 or -1. I would opt
for 0, for the reason cited in the reply before mine. Similarly, you have
to decide whether you treat oxygen as 0 or -2. I would opt for -2, so I
would say, if forced, that the O.S. is 2. But...
I'm a Molecular Orbital Theory man, so the question only interests me
because it can be used to knock down Valence Bond Theory. The true picture
is probably that the bonds between C and S are neutral, whilst the bond
between S and O is fairly polarised. Thus the sulphur is, to a reasonable
degree, oxidised, the extent of which is best measured by NMR shift data.
Apologies for any incoherences - I'm hung over, big time.
--
-- Mat Out (mj...@hermes.cam.ac.uk)
NE...@cunyvm.cuny.edu
comparison of elements in different compounds. I was taught as an undergraduate
that the order for certain elements was conventionally set as S>C>P>H , even
though the electronegativities in one table or another - Pauling's are not the
only set - might be equal. Since oxidation number is not an experimentally
determined quantity, it matters little what assumptions you make, so long as
there is consistency within the General Chemistry courses ! However, the
oxidation number of S in DMSO is the same whether you take the resonance form
(H3C)2S=O or (H3C)2S(+)-O(-) You give the S a (+) for every bond to oxygen,
whether sigma or pi, so the contribution from oxygen along with the sulfur
formal charge is +2 in both cases. Students should, I think, be told that both
oxidation number and formal charge are artificial concepts that have value,
but suffer from invalid and opposite assumptions. Oxidation number assumes that
ALL electronegativity differences, no matter how small (sulfur is perhaps a
little more electronegative than carbon, with 2.55 vs 2.50) shall be taken as
producing 100% polarisation. Formal charge assumes that ALL electronegativity
differences are zero ! The actual charges will be somewhere in between.
For example, carbon monoxide by formal charge is (-):C:::O:(+) with a CO triple
bond, but the oxidation numbers are (-2) for oxygen and (+2) for carbon. The
observed dipole moment is about ZERO (0.04D), although the bond length is right
for a triple bond.