Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: vivekrai on February 25, 2012, 10:51:50 AM
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I tried to look this thing on google but could not come up with anything more than that on Wikipedia which gives,
" Selective oxidation of alcohols
It is uncommon, but possible for NBS to oxidize alcohols. E. J. Corey et al. found that one can selectively oxidize secondary alcohols in the presence of primary alcohols using NBS in aqueous dimethoxyethane (DME).
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fthumb%2Ff%2Ff8%2FNBS_Oxidation_Corey.png%2F320px-NBS_Oxidation_Corey.png&hash=933b8a6e64619a596eda9ece78601e74e242316a)
However, Could the respected members here provide me with any more information on the kind of selectivity is shown or possible mechanism.
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There are a few reports about oxidations with bromine, and these seem to favour a mechanism involving hydride abstraction. Perhaps that could explain the selectivity in this case. (Corey gives no details in his paper)
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Perhaps that could explain the selectivity in this case. (Corey gives no details in his paper)
So, If any substituent is hindering the Hydrogen abstraction process, it would disallow the oxidation. I get this thanks.
But still we miss details about the selectivity of the -OH group to be reduced. Certainly because there must be substitutes for this type of reactions where we have to selectively oxidize an alcohol. Some common ones known to me are PCC,Jone's Reagent etc. But I'm unaware of their selectivity for the -OH groups?
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I don't have access to any literature, but if the process does proceed through hydride abstraction, you'll get a carbocation at a certain moment. Now if you had a choice between forming a secondary carbocation or a tertiary one (hydroxide is a substituent too), which one would be formed fastest?
The problem with most other oxidation procedures (I won't claim to know them all) is that they react faster with primary alcohols than with secondary ones. A more conventional way to obtain the product above would be oxidizing both alcohols to the carbonyl level, then use a selective reducing agent to convert the aldehyde back to the alcohol.
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I don't have access to any literature, but if the process does proceed through hydride abstraction, you'll get a carbocation at a certain moment. Now if you had a choice between forming a secondary carbocation or a tertiary one (hydroxide is a substituent too), which one would be formed fastest?
I think the tertiary one should form the fastest.
then use a selective reducing agent to convert the aldehyde back to the alcohol.
Okay. It is the reverse question then. Selective reducing agents are may be unknown to me. I know a way of protecting the the aldehydes or ketones via acetal formation and then carry out the further reaction. However, If there is anything useful you could tell me about the highlighted thing.
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While Willug provided a mechanism that could result in bromide as the electron sink, I don't think this happens. The electrons of oxygen are more basic and a more likely nucleophile. Whether they add directly to NBS or another form bromo compound, I don't know. Most oxidations follow the following scheme:
R2CHOH :rarrow: R2CHOX :rarrow: R2C=O + HX
X is the group accepting electrons. A base may pick up the proton rather than as indicated.
If that is true, then the selectivity may be proportional to the rate of formation of R2CHOBr. In this case, the electrons of a secondary alcohol should be more basic than those of a primary alcohol as predicted by alkyl groups being electron donors.
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I don't agree with that mechanism organopete - partly as I have the luxury of a university subscription, and partly as you seem to be suggesting hydride as a leaving group in the second part of your scheme (see attached).
I think the idea of carbocation stability (i.e. generated via a hydride abstraction) is a better way of explaining the selectivity here.
There are quite a few studies of this reaction (with bromine, not NBS, but I would suggest the two reagents are similar) in the literature, and most suggest hydride abstraction.
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I will chip in here.
Summary of post below: Both mechanisms have been argued in the lit, and I don't think anyone really knows. I lean towards hydride abstraction, but would like to see further discussion.
you seem to be suggesting hydride as a leaving group
1. Orgopete did not suggest hydride as a leaving group, he posted the "standard" general mechanism of alcohol oxidation involving elimination of HX across a C-O bond. It's like an E2 elimination, just that instead of...
R2HCCH2Br -> R2C=CH2 + HBr
...you have a bromate:
R2HCOBr -> R2C=O + HBr
If anyone is suggesting hydride as a leaving group willug, it's you, but I'm not sure I disagree with you.
2. I used to do lots of related hemiacetal -> lactone oxidations with bromine water on sugar derivatives. There is a body of evidence that suggests these oxidations proceed by hydride abstration. Hemicatetals are much more reactive than alcohols (primary or secondary) under these conditions, which is attributed to the presence of an extra lone pair compared to an alcohol. You can look at this as a more stable oxonium ion internediate in the hydride abstraction mechanism - i.e R(RO)C=O+H is better resonance stabilised than R2C=O+H.
In the context of 6-membered hemiacetals (lot's of lit here with sugar oxidation), beta-anomers are oxidised much faster than alpha-anomers. This seems to fit the hydride abstraction mechanism, since the beta-anomers bear an axial hydrogen and the C-H sigma* effectively overlaps with two O lone pairs. For an alpha-anomer (equatorial H) there is only effective overlap of one O lone pair with the C-H sigma*. In other words, the stereoelectronics of the beta-anomers should favour ejection of hydride compared to the alpha-anomers, and this parallels their reactivity with "Br+" reagents.
Granted, hemiacetals are not the same as alcohols, but the mechanistic pathways we're looking at are the same. Arguments for the selectivity of sugar oxidation invoking the bromate pathway have been made. Off the top of my head, Box et al around 2000, I can dig it out if anyone cares, who essentially argue that the beta-anomers have more nucleophilic O lone pairs than the alpha, and thus form bromate intermediates more readily.
3. I am not convinced that the basicity of a secondary alcohol vs a primary alcohol argument is a good one. If we were talking about protonation, then yes, but can we apply the reactivity with H+ to that with Br+? Would we not expect steric factors in the case of a larger electrophile, and therefore a less nucleophilic and less reactive secondary alcohol vs primary?
4. As far as I know, there is no proof of the mechanism. I have seen both mechanisms dismissed by different academics equally casually. Personally, I take the stance that hydride abstraction explains and predicts reactivity more satisfactorily than the bromate mechanism - but there is the problem that hydride abstraction is not a mainstream mode of reactivity for Br-X...
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Dan, thanks for clarifying my argument. I would like to follow up on the "hydride abstraction mechanism". Is there such a thing? I'm having a difficult time imagining a Br(+) reaction that can lead to H(-) abstraction.
I am inclined to agree that it isn't a kinetic formation of a hypobromite that leads to the selectivity. Selectivity has be to due to either a faster hypobromite formation or a faster elimination from the hypobromite. I hadn't thought correctly on the second option. If the rate limiting step is decomposition of the hypobromite to an carbonyl group, then the most electron rich CH may be the one most able to participate. That should be the secondary CH. That would account for ketone over aldehyde.
If this were applied to acetals, then it needn't be hydride abstraction, rather simple electron donation with proton abstraction (which I would favor). That would keep all of the Br(+) mechanisms in the same family. It is difficult for me to imagine that Br2 can form a bromonium ion, hypobromite, or perbromide with an alkene, water or bromide, but that it should just form a Br(+) in the presence of all of these sp3 oxygen atoms. It seems as though one of them should form a ROBr or R2OBr(+) somewhere.
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I am inclined to agree that it isn't a kinetic formation of a hypobromite that leads to the selectivity. Selectivity has be to due to either a faster hypobromite formation or a faster elimination from the hypobromite. I hadn't thought correctly on the second option.
Ok, good point to think about.
If the rate limiting step is decomposition of the hypobromite to an carbonyl group, then the most electron rich CH may be the one most able to participate.
I don't see why. Surely a more acidic, less electron rich proton would be abstracted more readily?
If this were applied to acetals, then it needn't be hydride abstraction, rather simple electron donation with proton abstraction (which I would favor).
But surely proton abstraction from an acetal is far less favourable than for an alcohol... this would involve build up of negative charge alpha to two oxygen atoms.
Maybe I have misunderstood your point.
It is difficult for me to imagine that Br2 can form a bromonium ion, hypobromite, or perbromide with an alkene, water or bromide, but that it should just form a Br(+) in the presence of all of these sp3 oxygen atoms. It seems as though one of them should form a ROBr or R2OBr(+) somewhere.
These species are undoubtedly present, I agree. The problem is that everything is in equilibrium, and we don't know which species are are the productive ones.
Here is an interesting paper: http://www.nrcresearchpress.com/doi/pdf/10.1139/v69-108
and here: http://www.springerlink.com/content/h5634p2724305gk8/
and: http://www.springerlink.com/content/q6745934xkw4v076/
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It is always difficult to prove a mechanism. Kinetics will tell us something about the rate limiting step. Then, those that study kinetics also try to find conditions in which they can alter the rate limiting steps in different ways. That can be useful information, but it doesn't necessarily tell you something about the reaction in which that rate limiting effect is no longer limiting. For example, I note the NRC paper refers to mercuric acetate suppresses the acid catalyzed formation of bromine from NBS. The other papers suggested mechanisms consistent with the kinetics of mercuric acetate oxidations. I am not sure what the mechanistic ramifications of this are. (I could only read the abstracts of the other papers as I do not have access.)
Re: proton abstraction
I agree. Only acidic protons are abstracted. An acetal CH is not particularly acidic hydrogen. The acidity must be increased. I have written a possible mechanism below. It does have some awkwardness to it because the oxygen is behaving differently than we are accustomed to seeing. I have treated it as though it were simply another source of electrons. A neighboring BrO(+) group should draw electrons and I argue preferentially from the most electron rich source. The transfer of electrons to the oxygen will increase its acidity. It will go from electron rich to electron poor. I argue this is similar to hydride migration in which the most electron rich group can migrate to form a new and more stable carbocation. In this case, I am showing the neighboring oxygen as a pushing atom.
I don't remember this chemistry all that well, so I just guessed as to how it results. I also don't know if this is what was meant with the beta acetal reactions. I did presume the stereochemistry would be such that the pushing electrons would be trans to the CH bond. I liken this to an elimination. I am also presuming the non-bonded electrons should enable the transfer of electrons in a similar manner to the alpha-effect of hydrazine or peroxide.
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That's not exactly what I meant, but related, scheme below (first part).
I view your hydride migration mechanism as an intramolecular hydride abstraction. Not something that had occurred to me, an interesting idea. I think the stereoelectronics of this mechanism applied to alcohols is dodgy though (but I think it looks legit for acetals because you have the extra lone pair.
In the second part of the scheme below, I have drawn three potential mechanisms. As I understand it, you favour mechanism 3. However, I cannot see a good reason why 3 would be slower if one of the R groups was H - I would expect it to be faster.
The papers I posted are essentially examining mechanisms 1 and 2. I am not a fan of 2, as it involves hydride abstraction from OH - to my eye this is dodgy territory. Mech 1 involves proton abstraction from OH - not unusual - and hydride abstraction from C-H, which is not unusual in general (though I grant you unusual for a "Br+" acceptor).
As I say, mech 3, analogous to Jones oxidation, is to me the most attractive on paper. My issue with it is that I can't convince myself that it explains the selectivity. Mech 1 looks a bit odd, but does appear to explain the data.
I hope you have some more thoughts on this, because the mechanism of these oxidations has always interested me.
Re: mercuric acetate - this is to scavenge bromide in order to study oxidation pathways by NBS without interference from Br2 oxidation. The authors claim that having studied both modes in isolation, they believe the mechanisms to be analogous, but that Br2 oxidation is 104 times faster - hence the need to suppress Br2 formation when using NBS in order to study NBS oxidation kinetics.
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Since you are showing the alpha and beta anomers of pyranoses oxidize at different rates, you indicate the rate limiting step is different for them. What you need to do is rationalize why. I could imagine using an argument as I had in my scheme for the conversion of an acetal to a benzoate (the scheme I thought was being discussed).
The other schemes do not make sense to me. The hydride elimination certainly is not operational as bromine is not involved. Surely the rate limiting step cannot be a unimolecular loss of hydride. Sugars just don't do that.
Scheme 0
I do not like bimolecular concerted reactions unless you can have really compelling reasons, a metal aligning groups in a directed aldol for example. I think there should be a great entropic price to pay to get a collision with this orientation. In this reaction, bimolecular bromine is acting as a base. Bromide is not a particularly strong base so I find it difficult to accept bromine to coordinate with two protons. Then, for me, it is like adding insult to injury, if the protons are coordinating with the electrons of bromine, then the electrons holding the proton attacks the bromine. Shouldn't the electron pair to which the proton is directed repel an electron pair from jumping from the carbon?
Scheme 1 and 2 (and 3)
These are really just variants of scheme 0.
Scheme 3
I don't know if this is supposed to be a variant of the intermolecular hypobromite reaction I suggested or not. I can't really tell from the curved arrows. If it is supposed to be intermolecular, then why? I can't think of why succinimide would be a base or why bromide could not be a leaving group. Isn't starting alcohol a stronger base?
I don't mean to seem pedantic, but bromine reacts with nucleophiles like enolates, hydroxide, alkenes (most electron rich react fastest), phenols, anilines, etc. You can chlorinate aniline with bleach and acid. I presume the aniline attacks the chlorine atom. Can't the electrons of an alcohol attack bromine?
If you wish to deliver hydride, I think you need to have at least some coordination of the electron pair and then the proton can simply follow the electron pair in a transfer. That is what I attempt to do in writing my mechanisms. I assumed the hypobromonium ion made the oxygen electron deficient. The neighboring electrons (with a proton attached) could be attracted to the oxygen. This is how I presume most rearrangement reactions take place. Even the decomposition of BH4(-) with acid. Don't you think protonation occurs on the electron pair? This should form BH5 which can decompose to BH3 + H2. BH5 is the analog of CH5(+), but easier to form as boron holds its electrons less tightly.
If you are thinking as I think, then it isn't simply because I can write a mechanism that gives it plausibility. I need to generate a force to justify the groups to become coordinated with one another.
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Since you are showing the alpha and beta anomers of pyranoses oxidize at different rates, you indicate the rate limiting step is different for them.
How so? The rate limiting step could be the same, just a different rate constant. Not all SN2 reactions proceed at identical rates, but that doesn't mean they're not all SN2 reactions.
The other schemes do not make sense to me. The hydride elimination certainly is not operational as bromine is not involved. Surely the rate limiting step cannot be a unimolecular loss of hydride.
Sorry, it was shorthand, I should have been clearer. The hydride attacks the Br of Br-X, and X- is ejected in a bimolecular reaction.
I don't mean to seem pedantic, but bromine reacts with nucleophiles like enolates, hydroxide, alkenes (most electron rich react fastest), phenols, anilines, etc. You can chlorinate aniline with bleach and acid. I presume the aniline attacks the chlorine atom. Can't the electrons of an alcohol attack bromine?
Of course they can, I have already said that.
If you wish to deliver hydride, I think you need to have at least some coordination of the electron pair and then the proton can simply follow the electron pair in a transfer. That is what I attempt to do in writing my mechanisms. I assumed the hypobromonium ion made the oxygen electron deficient. The neighboring electrons (with a proton attached) could be attracted to the oxygen.
Ok, I think I see what you mean now, see scheme below, mech 1 - is that it? 1,2 hydride shifts are essentially intramolecular hydride abstractions, right? I like it, my only reservation was that it doesn't look like the O lone pair can participate, but I suppose not really a problem.
1 and 3 are both hydride-based mechanisms, and explain why the reaction is faster for secondary alcohols (and hemiacetals).
Mech 2 is the proton abstraction, Jones-oxidation-style, mechanism that I feel does not explain the selectivity of the reaction.
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I meant rate.
For mechanism 1 and 3, we should be able to compare the strength of the bases from the acidities of the conjugate acids. If we compare the acidity of CH5(+) with CH3OH2(+), I believe CH3OH2(+) is more acidic. I argue that bromine is more likely to react with the lone pair of the oxygen than the bonded pair of the CH bond.
That doesn't prove the mechanism occurs in that manner. It just suggests that if bromine were to react as an electron acceptor, it should react with the oxygen. If that was unproductive, then another reaction may take place. The deprotonation of the alpha hydrogen in a Claisen condensation is the thermodynamically less favored reaction. Mostly addition occurs on the carbonyl group, but it does not lead to product. I would argue that in order for the CH bond to lead to oxidation, the formation of a hypobromite should be reversible and unproductive.
If it were me, and I wanted to know definitely whether a hypobromite could explain the different rates, I would look for examples of mechanistically similar reactions or via formation of a bona fide intermediate and measuring the rates. Certainly chromic acid oxidations show different rates. It has been speculated the rates are due to release of steric hindrance. Could be, I don't know. Are there any solvolysis experiments that are similar? In a way, this could be similar to a pinacol rearrangement. Do they show any stereoelectronic effects with a similar electron release?
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For mechanism 1 and 3, we should be able to compare the strength of the bases from the acidities of the conjugate acids. If we compare the acidity of CH5(+) with CH3OH2(+), I believe CH3OH2(+) is more acidic. I argue that bromine is more likely to react with the lone pair of the oxygen than the bonded pair of the CH bond.
I assume you meant CH5+ is more acidic, I agree with you. I do not doubt that hypobromites will form...
That doesn't prove the mechanism occurs in that manner.
I would argue that in order for the CH bond to lead to oxidation, the formation of a hypobromite should be reversible and unproductive.
Agreed
In a way, this could be similar to a pinacol rearrangement. Do they show any stereoelectronic effects with a similar electron release?
Agree with the pinacol analogy, I was also thinking along the lines of Hofmann/Beckmann rearrangement chemistry. The problem with the pinacol rearrangement is that (as I understand it) the RDS is loss of water to form a carbocation, whereas I think we will agree that an analogous intermediate would not form the Br-X oxidation (scheme).
I'm not sure about the validity of comparing Hofmann/Beckmann rearrangement kinetics either... what do you think?
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Dan,
Thanks for picking up my obvious error on the acidity note.
Re: stereoelectronic effects
Clearly we can find them in many reactions. The Hoffmann rearrangement is sensitive to the orientation of the oxime and can give different products from each. I believe the pinacol does also, but I am not familiar with any bona fide results to quote.
If anyone is paying attention, you can discern that I like to group things together. While I wouldn't normally link the stereoelectronic demands of a Hoffmann rearrangement with a hypobromite oxidation, I would if that was the best I could do. I think very carefully about how a pair of electrons react. How are the non-bonded electrons of hydronium ion different from hydroxide or water? They have the same charge (-2). I find it comes to distance which is compatible with the inverse square law.
Clearly, I was being methodical in suggesting a pinacol rearrangement as it placed a leaving group on a carbon beta to migrating group. You simply expanded this to consider other mechanistically similar reactions. I think when this is done in a general case such as this, then we should find most reactions have some kind of precedence or expectations as to what might occur. The suitability, it seems to me, would be the similarity of the driving forces in a reaction. If you have a migration due to a pushing from a lone pair of electrons, a quite different result may (not will) occur in the absence of the lone pair.