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Topic: Borohydride reduction of keytone  (Read 30500 times)

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Re:Borohydride reduction of keytone
« Reply #15 on: August 02, 2004, 12:30:52 AM »
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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Re:Borohydride reduction of keytone
« Reply #16 on: August 02, 2004, 01:35:15 AM »
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

-----

Corrected my spelling of "Newman"
« Last Edit: August 02, 2004, 02:07:33 AM by movies »

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Re:Borohydride reduction of keytone
« Reply #17 on: August 03, 2004, 10:32:38 PM »
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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Re:Borohydride reduction of keytone
« Reply #18 on: August 04, 2004, 05:04:41 AM »
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:



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:



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.

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