I have never seen specifics but always ran across sources saying that reductions that use hydrides require less hydrogen than non-hydride reductions.
I know for instance the normal reduction of Nitro groups to Amine groups precedes through 6 steps, consuming 6 hydrogen, 4 of the hydrogen being
required for converting the released Oxygen to water.
If the hydrogen requirement is less for hydride-based reductions, then how do these mechanisms differ? Nicodem - 14-3-2014 at 08:58
You can easily calculate the stoichiometry of the redox reactions, if you count the exchanged electrons by comparing the oxidation state changes.
There are plenty of online resources that explain how to do this, step by step for the beginners. Once you learn this - it takes no more than a few
minutes of reading and a few hours of practicing - you can calculate the stoichiometry of any given reaction.Metacelsus - 14-3-2014 at 13:46
A hydride ion delivers up to 2 electrons. This is the same amount as a molecule of diatomic hydrogen.
Electra - 14-3-2014 at 15:56
I cannot calculate the stichiometry if I do not know the balancing equation. Do the oxygens leave the nitro to form water in a hydride-based
reduction, as they would in a non-hydride reduction?
Much literature I have come acrossed implies that only 4 hydride's are needed for a hydride based reduction of nitro compounds, while non-hydride
reductions require 6. How does this work? The only compound I could think of forming would be OH.
If sodium formate were being used as a hydride donor in a transfer hydrogenation, does it just become sodium carbonate? I recall reading this in
literature somewhere but cannot remember specifically. If this were the case, something like this would happen:
2HCOONa + R-NO2 -> Na2CO3 + R-NH2 + CO2
Though this equation does not balance. Still a missing Oxygen...
[Edited on 14-3-2014 by Electra]Nicodem - 15-3-2014 at 01:19
I cannot calculate the stichiometry if I do not know the balancing equation.
You don't need to know in advance what the balanced equation is. You balance it by counting the transfered electrons (for which you only need to know
what the main product is). That is part of the redox equation stoichiometry calculation. It is a method, so all it takes is learning it. Nobody can
force you that.
Quote:
Do the oxygens leave the nitro to form water in a hydride-based reduction, as they would in a non-hydride reduction?
Well, what becomes of the oxygens depends on the reaction, but this is irrelevant to your question. What you need to find out is which atoms change
their oxidation states and by how much (for this you need to define at least one main product). Oxygens do not change their oxidation state in the
nitro group reductions. The nitrogen atom does and one of the atoms in the reducing reagents (which is hydrogen in the case of complex hydride
reagents based reductions).
Quote:
Much literature I have come acrossed implies that only 4 hydride's are needed for a hydride based reduction of nitro compounds, while non-hydride
reductions require 6. How does this work? The only compound I could think of forming would be OH.
It works because it is the electron transfer that does the reduction. The number of atoms is not directly relevant as such, at least not in advance -
it becomes relevant trough multiplication to achieve a predefined number of transfered electrons and/or bonding change, a phenomenon called
stoichiometry (each atom transfers a defined number of electrons). For example, if the reduction is performed by direct electron transfer from/to an
electrode (as in electrochemistry), you would not be counting atoms, you would be counting electrons, which is actually charge when multiplied by the
stoichiometric amounts (or current per given time). In the first step of the method, it is not much different with "purely chemical" redox reactions.
Quote:
If sodium formate were being used as a hydride donor in a transfer hydrogenation, does it just become sodium carbonate?
Yes, taking two electrons from the aqueous formate anion makes it a carbonate or CO2, depending on the reaction conditions. Basic
stoichiometry tells you that. All you need to do, is counting electrons.
Quote:
2HCOONa + R-NO2 -> Na2CO3 + R-NH2 + CO2
Though this equation does not balance. Still a missing Oxygen...
Because you did not apply stoichiometric rules to the balancing. You can't pick up a random number without accounting for the actual electron
transfers.
Go trough some on-line training site or a school textbook and learn the basics first. You will only get more and more confused, if you skip the very
basics of chemistry and trying to first understand advanced concepts. It is like trying to run before you learn to walk.Electra - 15-3-2014 at 16:24
Nicodem,
With all due respect, if I had been able to derive understanding of this particular hydride transfer from various literature by following electron
transfer, then I would not be posting here asking for help in understanding it. Telling me to do again what I have already attempted 5x isn't going to
arrive me to a new solution. If I just wanted to perform the reaction, I don't need to understand the mechanism, just that I have to add 4 eq. hydride
source. I am interested in the theory above all.
No matter how much I look at the electron transfers that are taking place here, there is something I am obviously overlooking and I am asking for help
in understanding it.... I do not understand what products are formed from these 2 excess hydrogen. Nicodem - 16-3-2014 at 01:54
I don't think anybody can help you unless you specify exactly which part of the oxidation state determination method and which parts of the
redox reaction balancing method you do not understand. If you are unable to count the transfered electrons, then you obviously do not know how to
determine the oxidation states. If you can't do that, then you can not even start solving redox equations. So focus on oxidation states first.
And a personal rant about your attitude, which you can either consider or ignore: You keep asking extraordinarily vague questions, talk about
imaginary "hydride sources", give no reference to the reaction you want to balance, say nothing about what skills you lack to solve the problem by
yourself, and at the end you mock on us by saying that you only attempted solving it 5 times before you asked (5 times, seriously?!). Please do not
come back before you mastered the redox balancing method from at least three different textbooks and solved >100 equations. If you think basic
chemistry is as simple as attempting 5-times before calling for help, then you would better reconsider your interests. Some of us invested >20
years of learning, not just 5 minutes for 5 attempts!Electra - 16-3-2014 at 11:30
Can you suggest me to a literature source that can detail for me the following mechanism, except based on hydride transfer? Explaining my quandary any
further seems futile.
Here for instance they detail each mechanistic step with each electron transfer for a non-hydride reduction.
Can you suggest me to a literature source that can detail for me the following mechanism, except based on hydride transfer?
The page does not show the mechanism, but shows the general stepwise reduction pathway from the nitro to the amino group (the mechanism cannot be
general and depends on the reagent used).
For the complex hydride based reagents, this same general pathway is identical. Also the number of electron transferred is exactly the same
(this can never be different). That is what I and Cheddite Cheese already told you. The electron transfer counting for an oxidation state change in
the substrate is independent from the reagent used as long as the product is the same: the reduction from the nitro group to the amino group will
always require 6 electrons, no mater what reagent or conditions you use.
I keep telling you to just read some textbook for students about how the oxidation states are determined. It is very easy and you can learn it mater
or weeks, even days if you put some effort into it.
