Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: yonkers on July 14, 2004, 08:17:55 PM
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This is a lab experiment
Hydrobenzoin from benzil
Both on benzil rings
C=O
------> NABH4 H---OH H---OH
+
C=O H---OH HO---H
most outcome most stable
meso
Why is meso-hydrobenzoin obtained from the above procedure rather than more stable d,lhydrobenzoin or mixtures of meso and d,l ?
My prediction is Carbonyl groups are easily reduced by metal hydrides such as sodium boohydride NABH4. Hydride ion adds to carbonyl carbon and the alkoxide that is formed is protonated ,and the group is added by addind h- to h+
Aldhydes ketones and acyl halides can be reduced by sodium borohydride. also they react faster
thanks foe your help
yonkers
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what do you mean by "d,lhydrobenzoin"?
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the addition of two atoms of hydrogen to benzoin or four atoms of hydrogen to benzil gives a mixture of sterisomeric diols, of which the predominant isomer is the nonresolvable hydrobenzoin, the meso one, accompined by the enantiomer compound. it procededs rapidly.
by this i mean thre right one is more stable , the enantiomer, but the left is the predominant and almost always the outcome in the reaction... how could the more stable one not be the outcome and less stable is..this is what i meant by your question. its just the name of the product on the left
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The mostly likely reason for the meso product is due to the intramolecular hydrogen bonding while in the "cis" formation.
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I think the explanation is a little more subtle than the hydrogen bond. I would imagine that hydrogen bond is worth a lot less than the difference between the meso and DL forms, but I don't have the numbers to back that up. Also you could have H-bonding in the DL form and avoid the steric interactions between the two phenyl groups when the H-bond is in effect.
I would propose that after the first ketone is reduced the alkoxide that forms coordinated to the boron. This in turn can chelate to the adjacent carbonyl oxygen, locking the conformation. Then the addition of the second H- is blocked from attacking the opposite face of the first attack because of the phenyl ring. The second attack thus comes from the same side as the first.
If you build a model this explanation should make a lot more sense.
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I think the explanation is a little more subtle than the hydrogen bond. I would imagine that hydrogen bond is worth a lot less than the difference between the meso and DL forms, but I don't have the numbers to back that up. Also you could have H-bonding in the DL form and avoid the steric interactions between the two phenyl groups when the H-bond is in effect.
I would propose that after the first ketone is reduced the alkoxide that forms coordinated to the boron. This in turn can chelate to the adjacent carbonyl oxygen, locking the conformation. Then the addition of the second H- is blocked from attacking the opposite face of the first attack because of the phenyl ring. The second attack thus comes from the same side as the first.
If you build a model this explanation should make a lot more sense.
Here's the structure
http://www.kanto.com.tw/japanese/catalysis/2img/pic17.gif
The steric effects aren't so great, and there are countless stable compounds with steric interaction of greater degrees. Note that when referring to such reactions we are talking about the relative amounts of the products formed, despite the stability of one over the other; the species is not absolute, more transient, there is an interchange of the species formed.
Why are you searching for an absolute explanation...most explanations of chemical phenomena pertain to various factors and the hydrogen bonding is certainly a major factor, certainly one that can add significant favorability to the mesocompound.
I'm not an expert in boron chemistry, nevertheless, I suggest you point out some research to back it up, there are probably some "subtle" factors you have not considered in the formation of your suggested boron adduct, despite boron being a lewis acid.
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The hydrogen bonding in meso-hydrobenzoin requires a disfavorable steric interaction between the two phenyl groups. The hydrogen bonding in racemic hydrobenzoin (which is the picture posted above) does not have these interactions when the hydrogen bond is invoked. Those steric interactions have to be worth something, right? The hydrogen bond itself is essentially a wash, but the steric interaction makes the meso form higher in energy. I think that is the best explanation for why the racemic product is lower in energy than the meso product.
The product distribution, however, is not determined by the most stable product in this case, but by the lowest energy transition state. You would get the thermodynamically more stable (racemic) product if the reaction were reversible. However, hydride delivery from a borohydride reducing agent is an irreversible process so the lower energy transition state is what dominates.
You might look at the following articals for examples of chelation controlled additions:
Cram, J. Am. Chem. Soc. 1959, 81, 2748.
Nakata, Tetrahedron Lett. 1983, 24, 2653 and 2661.
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owever, hydride delivery from a borohydride reducing agent is an irreversible process so the lower energy transition state is what dominates.
Note that borohydride is somewhat of a catalyst, are you aware of what a catalyst does (borohydride is strong reagent)? This would indicate a REVERSIBLE reaction.
Once again, the hydrogen bonding is more than a significant factor... to say that "because of steric interactions, hydrogen bonding becomes difficult" is ridiculous. Once again, the steric interactions are NOT so influential in this case, I don't know why you are so obsessed about the steric interaction as to completely rule out the hydrogen bonding.
You seem to be ad hocking completely, I have come to the conclusion that you have no idea what you are talking about but just wish to carry out the argument for the sake of reputation.
I'll carry on with the discussion if you provide some solid, specific, and coherent evidence with your arguments.
Why don't we settle this, ask your professor or research the internet to invalidate my hydrogen bonding argument. It is frequently the case that intramolecular hydrogen bonding results in greatly increased stability. My suggestion relates to an frequently established explanation. You seem to be making up your own "subtle" and incoherent arguments, you need to rely on best and established explanations (such as you find frequently in a org chem book) and not posit your own vague theories. Note that we are NOT researchers arguing about a complex theory, the truth regarding this topic is out there and I'm sure that it is very, very, very simple.
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Before I go into some long explanation, look at the original question. It asks why the meso form predominates in the products despite the fact the meso form is higher in energy than the racemic form.
You are correct, there is a hydrogen bond. Yes it is significant. But both the meso and racemic forms can have a hydrogen bond. So what makes the meso form higher in energy than the racemic form? I haven't seen you give an explanation for that.
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Sodium borohydride is not a catalyst in this case. It is consumed in the reaction. Check you definition of catalyst.
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Hydrogen bonding is such a small effect, it will not decide stereochemistry. Chelation of the boron group is going to direct the formation of the meso compound.
Note, I haven't drawn the boron intermediate that would lead to this compound, but from previous experience I would expect it too. Perhaps someone could draw the boron intermediate to show how it directs formation of the meso compound
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You should read into what I write more carefully.
Why is meso-hydrobenzoin obtained from the above procedure rather than more stable d,lhydrobenzoin or mixtures of meso and d,l ?
This quote is a commentary from the original person who had asked the question...it implicates the questioner's suspicion, his/her original query to begin with. In this case, I'm guessing that she is wrong, she/he has underestimated the stability of the meso compound.
Sodium borohydride is not a catalyst in this case. It is consumed in the reaction. Check you definition of catalyst.
I apologize, I said "somewhat" of a catalyst (strong reagent)" and I wished to emphasize the latter; my doubts about your assertion of this being a irreversible reaction.
Hydrogen bonding is such a small effect, it will not decide stereochemistry. Chelation of the boron group is going to direct the formation of the meso compound.
What? By now Mitch, I'm sure you've heard of cases where the equilibrium is actually shifted due to the hydrogen bonding of the product and even small effect ionic interactions regarding acid base reactions where the conjugate base is stabilized due to these latter interactions. I've heard of countless examples where the equilibrium is shifted due to intramolecular hydrogen bonding, I'm surprised that both of you have not brought this in to case to begin with.
Perhaps you can search the internet to find some detailed explanation on your boron intermediate theory...if it is true as you say, than I'm more than positive that there will be lectures, articles, and even personal websites which one can easily find. I'm more open minded, just find it. None of you seem to be interested in finding the conclusion.
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I'm not an expert on this subject, thus I won't be surprised to find that I'm wrong. Nevertheless, none of you have provided to me an established evidence regarding this matter.
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But both the meso and racemic forms can have a hydrogen bond.
Really? Are you referring to INTRAmolecular hydrogen bonds.
Higher in energy? Shouldn't the more stable compound be lower in energy?
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movies, if you have done this experiment before (I have not) and you are certain about this matter, I would appreciate it if you would not suggest the proposals which you state to be a product of your own imagination...
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I have not personally run this experiment. However, reduction of this particular substrate with NaBH4 is reported in the paper that first reported the use of NaBH4 to reduce ketones. Here is the reference: J. Am. Chem. Soc. 71, 122 (1949). Therein, they report that the initial reduction gave an apparent mixture of the racemic and meso forms of hydrobenzoin, but after several recrystallizations they obtained pure meso-hydrobenzoin in 56% yield. Therefor the suggestion that there is a mixture of products is not simply an assertion, but an empirical fact. Furthermore, it seems to favor the meso form (there is no mention of recovered racemic hydrobenzoin).
I think I have deduced where your confusion lies GCT. The drawings in the initial question are deceptive in that they show a conformation for the racemic form that does not exhibit a hydrogen bond. However, there is free rotation about the C-C bond connecting the two PhCH(OH) carbons. If this is rotated to allow for a hydrogen bond, then I think you will see what I mean.
I will try to sketch some of these things out in ChemDraw and post them, but I've never tried that before.
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Alright, here goes.
The two forms of hydrobenzoin with Newman projections showing the hydrogen bonding states:
http://www.chemicalforums.com/~movies/hydrobenzoin.gif
and the boron chelate:
http://www.chemicalforums.com/~movies/boronchelate.gif
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Corrected my spelling of "Newman"
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I researched over the internet and found some sources implicating the importance of the steric hinderance and emphasizing the confusing regarding the hydrogen bonds. Suppposedly this experiment is significant in that it is one of the few exceptions in which the hydrogen bond explantion does not account for the conclusion. It's too bad I never took any of the org. labs. This peculiar experiment is signicfincatin in that the hydrogen bond explanation does not account for the conclusion...ALTHOUGH I'm not absolutely sure of this, I'll take your word (movies) for it, since I belive that you have experience regarding this matter. I just wish you could have stated it as os earlier instead of stating as your own hypotheiss and the product of your own imagination. This would have saved a lot of time.
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The reason I stated in that way was indeed because I have not seen the boron chelate explanation used in this particular case. However, this type of chelation is a common reactivity for boron. For example, much of the literature on asymmetric aldol reactions using chiral auxiliaries relies on the use of boron as a Lewis acid to order the transition state. For an example of this, see “Stereoselective Aldol Condensations via Boron Enolates” by D. A. Evans. J. Am. Chem. Soc. 103, 3099 (1981). Here is a link to the pdf, it should work if your school has access: http://pubs.acs.org/cgi-bin/archive.cgi/jacsat/1981/103/i11/pdf/ja00401a031.pdf
In the case of these aldol reactions boron chelates to both an alkoxide and a carbonyl carbon (for other examples of chelate controlled reactivity with things other than boron, see the references I cited in a post above). Diastereoselective carbonyl additions are very well known (see http://www2.haverford.edu/wintnerorganicchem/ specifically point 17.07 on that page). The extension of these to the present case of borohydride reduction was a connection I made myself. I didn’t think it was necessary to provide all of this background, which is significantly more complex, in order to suggest a possible explanation for the experiment at hand.
I think that there should also be some clarification about reversible and irreversible reactions and the effect of a catalyst in these cases. In theory all reactions are reversible, that is to say with enough energy you can always go from products to reactants and back again by the same pathway. However, if the reaction conditions do not provide enough energy to perform the reverse process, then the reaction can be considered irreversible. There is a brief description on this website: http://www.wordiq.com/definition/Chemical_reaction.
Put graphically, the energy profile for a reaction is roughly like this:
(https://www.chemicalforums.com/~movies/generalEdiagram.gif)
So if a particular reaction provides enough energy to get to the transition state and then on to products, but the product is much more stable than the reactants, then the amount of energy to perform the reverse reaction is so large that the reaction is essentially irreversible under the reaction conditions. This is a loose paraphrasing of “transition state theory” without all of the math. If you want to know more about the math, I recommend this website: http://www.engin.umich.edu/~cre/03chap/html/transition/#II or if you like more pictures, then this website: http://www.rpi.edu/dept/chem-eng/Biotech-Environ/Projects00/enzkin/transition.htm. For a reduction of a ketone or an aldehyde to an alcohol, you can make a first order approximation of the energy difference between products and reactants by looking at the bond energies of the bonds in question (a handy table can be found here: http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm ). A carbon-oxygen double bond in a ketone is worth about 178 kcal/mol. The bonds in H-C-O-H group (a reduced ketone) total to 295.5 kcal/mol. That’s a big difference. A hydrogen bond is worth only about 5 kcal/mol at most, according to this site: http://www.web-books.com/MoBio/Free/Ch2C3.htm. I don’t think I am stretching too much when I say that borohydride reduction is an irreversible process. If the process is irreversible, then there won’t be any equilibration to the more thermodynamically stable product, which agrees with the experimental evidence I cited above (the reported reduction of benzil). In most irreversible cases the lower energy transition state is the factor that dictates the products, the products themselves have little to do with it.
In a catalyzed reaction the only change is that the reaction follows a different pathway with a lower energy transition state (see again: http://www.wordiq.com/definition/Chemical_reaction ). The energy profile then looks something like this:
(https://www.chemicalforums.com/~movies/catalyzedEdiagram.gif)
Notice that the activation energy for both the forward and reverse reactions are lowered, but the energy of the products and reactants are unchanged (my drawing sucks). However, the same reversibility rules still apply. If the products are much more stable than the transition state, then the reverse reaction will be very slow, possibly immeasurably slow, in which case the reaction is irreversible. A good example of a catalytic reaction that is irreversible is hydrogenation of olefins catalyzed by palladium on carbon.
I didn’t intend to be misleading in my suggestion of a mechanism for the formation of the less stable product in the initial question in this thread. However, I also didn’t think it necessary to provide the same level of detail that I have in this post. There is a rather substantial amount of research that would seem to support the kind of reactivity I suggested initially, so I really didn’t expect anyone to suggest that it was all in my imagination. I will try to be more thorough in the future.