Grizli7
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Equivalent to moles conversion?
Hello all.
1. Did I convert the equivalent to moles correctly? My answer 1:3 to 5: 1.5min?
2. Can i change EtOH to DCM or EtOAc?
3. Can i use some boric acid or sulfur to prevent polymerization of phenylacetaldehyde (i dont have hydroquinone)?
Attachment: Oxydation of aromatic aldehydes.pdf (227kB) This file has been downloaded 86 times
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DraconicAcid
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1:3 is the same as 5:15, yes.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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bnull
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2. Apparently not. If it is what I'm thinking, you may use methanol (I was about to say acetone but, hey, we have a peroxide here...); I have no data
about DCM, and ethyl acetate may react with formic acid. I'd use EtOH azeotrope (96.5%) because it's cheap and a little more water is no problem.
3. The last phenylacetaldehyde I saw used citric acid as stabilizer. If citric acid will stand the conditions is a good question.
1. This one is a little messy. Well, you need 1 mole of peroxide to each 1 mole of aldehyde. The proportion in equivalents (according to the paper) is
Ar-CHO:HCOOH:H2O2 1:3--5 (or more):3. But formic acid is also working as solvent and the oxidizer is performic acid, generated
from the combination of peroxide and acid. The proportion in equivalents is more like Ar-CHO::HCOOH:H2O2 1:1:3, with the excess
HCOOH as solvent. The 3 equivalents of 30% peroxide amount to almost one equivalent of pure hydrogen peroxide (I guess that's why they wrote "a
minimum of 3 eq of cold 30%" peroxide solution). We convert the whole thing to moles and have 1:1:1, with the rest of formic acid as solvent.
Quod scripsi, scripsi.
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DraconicAcid
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Quote: Originally posted by bnull ![](images/xpblue/lastpost.gif) | The 3 equivalents of 30% peroxide amount to almost one equivalent of pure hydrogen peroxide (I guess that's why they wrote "a minimum of 3 eq of cold
30%" peroxide solution). |
No. One mole equivalent of hydrogen peroxide is one mole equivalent, regardless of the concentration. Moles of water are irrelevant.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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bnull
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Quote: Originally posted by DraconicAcid ![](images/xpblue/lastpost.gif) | No. One mole equivalent of hydrogen peroxide is one mole equivalent, regardless of the concentration. Moles of water are irrelevant.
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Oh, crap. You're right.
Quod scripsi, scripsi.
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Grizli7
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I understood correctly. Molar ratio 1:1:1 ? I came across information somewhere that the equivalent of hydrogen peroxide in oxidative reactions is
equal to half the molar mass, i.e. 17 g/mol.
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Grizli7
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Quote: Originally posted by bnull ![](images/xpblue/lastpost.gif) | 2. Apparently not. If it is what I'm thinking, you may use methanol (I was about to say acetone but, hey, we have a peroxide here...); I have no data
about DCM, and ethyl acetate may react with formic acid. I'd use EtOH azeotrope (96.5%) because it's cheap and a little more water is no problem.
3. The last phenylacetaldehyde I saw used citric acid as stabilizer. If citric acid will stand the conditions is a good question.
1. This one is a little messy. Well, you need 1 mole of peroxide to each 1 mole of aldehyde. The proportion in equivalents (according to the paper) is
Ar-CHO:HCOOH:H2O2 1:3--5 (or more):3. But formic acid is also working as solvent and the oxidizer is performic acid, generated
from the combination of peroxide and acid. The proportion in equivalents is more like Ar-CHO::HCOOH:H2O2 1:1:3, with the excess
HCOOH as solvent. The 3 equivalents of 30% peroxide amount to almost one equivalent of pure hydrogen peroxide (I guess that's why they wrote "a
minimum of 3 eq of cold 30%" peroxide solution). We convert the whole thing to moles and have 1:1:1, with the rest of formic acid as solvent.
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Citric acid is probably a great simple and cheap solution. I think the question of what will happen to it is not entirely important - after all, its
mission ends at the stage of storing the aldehyde before the reaction. Yes, there may be by-products, but that’s what recrystallization is for.
Won't methanol dissolve with water?
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bnull
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Quote: Originally posted by Grizli7 ![](images/xpblue/lastpost.gif) | I understood correctly. Molar ratio 1:1:1 ? I came across information somewhere that the equivalent of hydrogen peroxide in oxidative reactions is
equal to half the molar mass, i.e. 17 g/mol. |
Just consider (2) and (3). (1) is a proof that I'm more than capable to make an ass of myself.
I verified the aldehyde again. It has 0.01% of citric acid.
And so does ethanol. The only function of the alcohol seems to be the
removal of formic acid from the aromatic acid produced. The authors wrote Quote: | In cases where the carboxylic acid did not precipitate even after the addition of H2O, the mixture was repeatedly diluted with EtOH and
concentrated under reduced pressure. |
Could it be because an azeotrope containing ethanol-formic acid--about which I've found no information--is formed and boils out of the solution, thus
reducing the solvabilityNote of formic acid toward the aromatic acid? It is a wild guess, of course.
Note: The ability of the solvent to dissolve a substance. I couldn't find the right word for it. Maybe
solvancy, but it looks too close to solvency to be useful. If there is a word for that ability, please tell me.
Quod scripsi, scripsi.
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Grizli7
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Well, let’s assume that we’ve sorted out the stabilization and isolation. Once again I want to clarify about the molar ratio. Check whether my
reasoning is correct or not. It seems to me that the situation is this - from the point of view of an ideal reaction, we have a molar ratio of 1: 1:
1. The problem is that in practice the aldehyde will not dissolve in such a small amount of HCOOH, so the author took an excess of 2 moles from above,
moreover, in his paper he indicates that there may be a situation when this is not enough, so he indicates the figure 5 or more. Considering that
HCOOH 1eq=1mol. We have a molar ratio of aldehyde to acid of 1:3 (let's take the minimum value suggested by the author). All that remains is to deal
with peroxide:
According to the author of the paper, he took 3 equivalents as a minimum (he indicates that this is the minimum amount). If we accept that 1
equivalent of peroxide = 1 mole, then he took 3 moles.
As I assume, due to the fact that the aldehyde needs an excess of acid to dissolve, and then this acid needs to be converted into a peracid and have
an excess of oxidizing agent in relation to the aldehyde, the author of the paper offers the following figures:
Molar ratio 1:3:3 or 1:5:5.
If the aldehyde does not dissolve in this amount of acid, then the molar ratio may look like 1:6:6.
Now the question is: 1 equivalent of peroxide is exactly equal to 34 g/mol or half of it 17 g/mol?
By the way, a few more questions on the topic:
a) What will the finished product be like if semicarbazone gets into the reaction?
b) What happens if you perform Jones oxidation?
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bnull
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I would increase only the HCOOH because it is the solvent. They wrote "a minimum of 3 eq of 30% aq" peroxide but didn't give instructions on when to
increase this quantity. Make it 4 eq of peroxide, I don't believe it will prefer to stick a --OH into the benzene ring rather than attack the more
attractive --CHO of the side chain.
a) How did the semicarbazone get into the solution? by the way, what semicarbazone is it?
b) Jones oxidation of what, the aldehyde? You're not thinking about doing an one-pot synthesis with a Jones and a peroxide oxidation in the same
solution, right?
Quod scripsi, scripsi.
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Grizli7
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Well 1eq of H2O2 =1 mole H2O2 right?
a) I made a mistake with semicarbazone - for some reason I decided that it would be a high-boiling fraction when preparing phenylacetaldehyde, which
could partially enter into the reaction if the separation was poor.
b) Yes the aldehyde and of course it will be another procedure. Although it would probably be fun to do everything in one pot. I just thought that
since the Jones oxidation of primary alcohols to carboxylic acids passes through aldehydes, perhaps this method will work in this case. It’s just
going to happen from the middle of the process. But it’s problematic to recalculate the amount of oxidizing agent - probably also equimolar.
And one more question about the paper seems to me to be contradictory. On the one hand, oxidation is carried out under relatively mild conditions of
0-4C. On the other hand, the author gives 100% perhydrol as an alternative. What does peroxide concentration result? Reaction will be faster and if we
take not such an extreme substance as 100% perhydrol, but for example 60%, then can we assume that the reaction will take place at the same
temperature, say, in 6 hours instead of 12.
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Texium
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It is if you're using 1 mole of starting material. Equivalents are
relative. Moles are absolute. If you're using 0.6 moles of your aldehyde, then one equivalent = 0.6 moles.
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bnull
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Quote: Originally posted by Grizli7 ![](images/xpblue/lastpost.gif) | And one more question about the paper seems to me to be contradictory. On the one hand, oxidation is carried out under relatively mild conditions of
0-4C. On the other hand, the author gives 100% perhydrol as an alternative. What does peroxide concentration result? Reaction will be faster and if we
take not such an extreme substance as 100% perhydrol, but for example 60%, then can we assume that the reaction will take place at the same
temperature, say, in 6 hours instead of 12. |
I'll tackle once more the problem of the equivalents and why the thing looks weird without making an ass of myself again.
Let's assume you have one mole of aldehyde, which means one equivalent. The authors state that you need 3--5 equivalents of formic acid (who acts both
as solvent and precursor to the oxidiser), which means 3--5 moles, and formic acid is not consumed (it looks like a carrier) in the course of the
reaction. If we're to follow the ideal reaction, we have aldehyde:formic acid:peroxide as 1:1:1, with the excess formic acid as solvent. The issue is
why 3 equivalents of peroxide if you need only 1 equivalent to oxidise the aldehyde to acid?
The reaction that produces performic acid (see below) is reversible and those Frenchpeople were shifting the equilibrium to the right by using an
excess peroxide. It seems I can't use LaTeX in the forum. Very well.
Quote: | H2O2 + HCOOH <=> HCOOOH + H2O |
The use of 3 equivalents of peroxide--that bugged my mind for a while--make more sense now. The authors assumed that everyone knew that the formation
of performic acid is reversible. The more peroxide, the more performic acid.
The suggested use of 100% perhydrol seems to be more appropriate if the water content difficults the dissolution of the aldehyde in the solution
formic acid/water/hydrogen peroxide. In the case of phenylacetaldehyde, whose solubility in water is about 2 g/L, I think that a more concentrated
peroxide (your 60%, for example) would be better. Less water, more aldehyde in solution.
There's no contradiction. The reaction time depends on the formation of performic acid, who in its turn depends on the quantity of peroxide. I think
you can cheat a little by adding catalytic amounts of sulfuric acid to the solution of 60% peroxide and formic acid (say, a couple of drops or even 1
mL/L). It will speed up the formation of performic acid and may reduce the reaction time. Try it small scale first.
Quote: Originally posted by Grizli7 ![](images/xpblue/lastpost.gif) |
a) I made a mistake with semicarbazone - for some reason I decided that it would be a high-boiling fraction when preparing phenylacetaldehyde, which
could partially enter into the reaction if the separation was poor. |
It looks like you're protecting the --CHO. So there is semicarbazone mixed with the aldehyde. Is it right?
Quote: Originally posted by Grizli7 ![](images/xpblue/lastpost.gif) |
b) Yes the aldehyde and of course it will be another procedure. Although it would probably be fun to do everything in one pot. I just thought that
since the Jones oxidation of primary alcohols to carboxylic acids passes through aldehydes, perhaps this method will work in this case. It’s just
going to happen from the middle of the process. But it’s problematic to recalculate the amount of oxidizing agent - probably also equimolar.
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You can't do peroxide and Jones in one-pot because Jones kills the peroxide, and if your Jones has acetone as solvent, well, there's peroxide and
sulfuric acid and acetone in the same place at the same time and Jones is quite exothermic... Don't do that, please.
It's not problematic, not at all. Calculate the quantities of chromium trioxide and sulfuric acid for a full Jones oxidation and divide by 2. If it is
4 moles of CrO3 to oxidise 3 moles of alcohol to carboxylic acid, it'll be 2 moles of CrO3 to oxidise 3 moles of aldehyde.
See Jones oxidation this way: chromium goes from +6 to +3, losing 3 electrons in the process. If you have 2 Cr+6, the number of electrons
changing places is 6. 2 CrO3 equals 1 Cr2O6 (for practical purposes). It loses 3 O and 6 electrons, or 3
O2-, and becomes Cr2O3. As each O2- converts one aldehyde molecule to carboxylic acid, you need 3 aldehyde
molecules for the 3 O2-. The 3 O2- came from 1 Cr2O6, which is equal to 2 CrO3. Finally (and
thankfully), it is 2 moles CrO3 for each 3 moles of aldehyde.
Quod scripsi, scripsi.
B. N. Ull
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Grizli7
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There were thoughts on the -CHO protection, but I never really came to a solution, I think the authors, using a low temperature of 0-4C, believe that
this will be enough.
Well, thanks to everyone who took part in clarifying some issues. Especially bnull as the one who showed the most involvement. Now all that remains
is to conduct a real experiment and get the result.
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bnull
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It is also much safer to work at lower temperatures. See D. Swern (of Swern oxidation fame), "Organic Peracids", https://doi.org/10.1021/cr60140a001.
You're welcome.
Quod scripsi, scripsi.
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