Author Topic: Need help  (Read 2850 times)

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Barium

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Need help
« on: April 15, 2003, 11:25:00 AM »
I'm in desperate need of help with balancing a reaction.

Na2S2O6 + -NO2 + 6 H+ --> Na ??? + -NH2 + 2H2O

I have no idea what the oxidation products are therefore I'm stuck.
Help  :-[


GC_MS

  • Guest
this?
« Reply #1 on: April 15, 2003, 12:00:00 PM »
Na2S2O6 + 2 RNO2 + 10H+ --> 2 NaHSO4 + 2 RNH2 + 2 H2O

As dithionites are oxidized to hydrogensulfites, I thought that dithionates might be oxidized to hydrogensulfates.


Megatherium

  • Guest
The reduction half reaction is: R-NO 2 + 6 H +
« Reply #2 on: April 15, 2003, 12:10:00 PM »
The reduction half reaction is:
R-NO2 + 6 H+ + 6 e-  __>  R-NH2 + 2 H2O

The oxidation half reaction is:
S2O62- + 2 H2__>  2 SO42- + 4H+ + 2 e-

Hence, the total reaction is:
R-NO2 + 3 S2O62- + 4 H2__>  R-NH2 + 6 SO42- + 6 H+

EDIT: GC_MS, you got to it first. You're right:  NaHSO4 will bee the product (screwed up the oxidation half reaction, lost track of the damned sodium ions LOL  :) )

Then the oxidation half reaction should be:


Na2S2O6 + 2 H2__>  2 NaHSO4 + 2 H+ + 2 e-

By multiplying the oxidation half reaction by three and adding it to the reduction half reaction, we obtain:
 R-NO2 + 3 Na2S2O6 + 4 H2__> R-NH2 + 6 NaHSO4

Megatherium

  • Guest
Megatherium is a bit confused over here ...
« Reply #3 on: April 15, 2003, 12:59:00 PM »
Megatherium is a bit confused over here ... the mass balances of both equations seem to bee correct ...
GC_MS: Na2S2O6 + 2 RNO2 + 10H+ --> 2 NaHSO4 + 2 RNH2 + 2 H2O
Mega : R-NO2 + 3 Na2S2O6 + 4 H2O  --> R-NH2 + 6 NaHSO4

Is Megatherium doing too many drugs & is he losing his mind ?

HELP ...

hypo

  • Guest
what about the electrons?
« Reply #4 on: April 15, 2003, 01:17:00 PM »
GC_MS has 10 positive charges on the left, but none on the right!?
that isn't ok, is it?

Megatherium

  • Guest
Indeed, you 're right: it isn't.
« Reply #5 on: April 15, 2003, 01:53:00 PM »
Indeed, you 're right: it isn't.

3 x (Na2S2O6 + 2 H2O -->  2 NaHSO4 + 2 H+ + 2 e-)  (ox)
   R-NO2 + 6 H+ + 6 e-  -->  R-NH2 + 2 H2O  (red)

What is nice here is that the reaction seems to be auto-catalyzed: the protons that are created in the oxidation are used in the reduction.  However, I think the addition of some acid (1 eq. during the reactiton) would be required since: R-NH2 + H+  -->  R-NH3+


Oeff, what a relief: I can continue with my old habits  ;)

Barium

  • Guest
Whoops
« Reply #6 on: April 15, 2003, 02:05:00 PM »
I mistook dithionate S2O62- for dithionite S2O42-. Feeling really silly  :-[
Dithionite is the reducing agent I'm curious about. But what is the oxidation product?

Bisulfate, HSO4- or
Sulfate, SO42-


GC_MS

  • Guest
shit
« Reply #7 on: April 15, 2003, 02:29:00 PM »

Oeff, what a relief: I can continue with my old habits




Yes, recalculated after hypo said my ion balance was way out. Have something like 2 S2O62- + 4 H2O + RNO2 --> 6 HSO4- + RNH2
Shit, and that while wasn't on my drugs...  ::)




starlight

  • Guest
oxidation product of dithionite
« Reply #8 on: April 15, 2003, 04:53:00 PM »
when dithionite is oxidised by Fe3+ or CrO42-the oxidation product is SO32- (sulfite)

not sure what the oxidation product is when reducing nitro groups tho'.

Megatherium

  • Guest
When dithionite is oxidised by Fe3+ or ...
« Reply #9 on: April 15, 2003, 05:52:00 PM »
Dithionite is the reducing agent I'm curious about. But what is the oxidation product?

Bisulfate, HSO4- or
Sulfate, SO42-




When dithionite is oxidised by Fe3+ or CrO42-the oxidation product is SO32- (sulfite)

not sure what the oxidation product is when reducing nitro groups tho'.




It seems indeed more reasonable to expect that the oxidation product here will be SO32- instead of SO42- or HSO4-.

I can't imagine that R-NO2 would be a better oxidant than CrO42-, and has the power to oxidate the sulfur from Na2S2O4 to an oxidation state +VI (like in SO42-).  It is indeed pretty sure that the oxidation state from the sulfur in dithionite will change from +III to +IV in the oxidation half reaction.

Hence, we get for the oxidation half reaction:
S2O42- + 2 H20 --> 2 SO32- + 4 H+ + 2 e-

The reduction half reaction remains the same (cfr supra), so after multiplying the oxidation half reaction by three and adding it to the reduction half reaction, we obtain:
R-NO2 + 3 S2O42- + 4 H20  -->  R-NH2 + 6 SO 32- + 6 H+, which can be rewritten as:
R-NO2 + 3 Na2S2O42- + 4 H20  -->  R-NH2 + 6 NaHSO 3

OK, enough redox chemistry for today already.  Now, its time for a good joint  :) .

Barium

  • Guest
Thank you all
« Reply #10 on: April 15, 2003, 07:03:00 PM »
This will hopefully be very beneficial to all of us. Just give me some time to get dirty in the lab  :P


Megatherium

  • Guest
No prob. Are you going to reduce a ...
« Reply #11 on: April 15, 2003, 07:53:00 PM »
No prob.  Are you going to reduce a aryl-2-nitropropane with the Na2S2O4?  That would bee rather cool  :) .

GC_MS

  • Guest
Naah
« Reply #12 on: April 16, 2003, 10:30:00 AM »
Naaaaaah, no way he is going to try that cheap OTC product on nitroalkanes  ;)  Might be nice thing if it worked. It works for phenyl-1-propanone -> phenyl-1-propanol.