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Topic: Borohydride reduction of keytone  (Read 30613 times)

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yonkers

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Borohydride reduction of keytone
« on: July 14, 2004, 08:17:55 PM »
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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Re:Borohydride reduction of keytone
« Reply #1 on: July 14, 2004, 09:01:57 PM »
what do you mean by "d,lhydrobenzoin"?
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yonkers

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Re:Borohydride reduction of keytone
« Reply #2 on: July 14, 2004, 10:04:47 PM »
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

GCT

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Re:Borohydride reduction of keytone
« Reply #3 on: July 18, 2004, 06:02:08 PM »
The mostly likely reason for the meso product is due to the intramolecular hydrogen bonding while in the "cis" formation.
« Last Edit: July 19, 2004, 08:41:10 PM by GCT »

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Re:Borohydride reduction of keytone
« Reply #4 on: July 21, 2004, 07:58:20 PM »
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.

GCT

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Re:Borohydride reduction of keytone
« Reply #5 on: July 31, 2004, 08:10:31 PM »
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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Re:Borohydride reduction of keytone
« Reply #6 on: August 01, 2004, 03:37:23 PM »
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.

GCT

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Re:Borohydride reduction of keytone
« Reply #7 on: August 01, 2004, 05:15:57 PM »
Quote
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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Re:Borohydride reduction of keytone
« Reply #8 on: August 01, 2004, 06:45:38 PM »
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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Re:Borohydride reduction of keytone
« Reply #9 on: August 01, 2004, 07:07:26 PM »
Sodium borohydride is not a catalyst in this case.  It is consumed in the reaction.  Check you definition of catalyst.

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Re:Borohydride reduction of keytone
« Reply #10 on: August 01, 2004, 10:21:02 PM »
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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GCT

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Re:Borohydride reduction of keytone
« Reply #11 on: August 01, 2004, 10:46:59 PM »
You should read into what I write more carefully.

Quote
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.



Quote
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.

Quote
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.

GCT

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Re:Borohydride reduction of keytone
« Reply #12 on: August 01, 2004, 10:49:01 PM »
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.

GCT

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Re:Borohydride reduction of keytone
« Reply #13 on: August 01, 2004, 10:57:07 PM »



Quote
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?
« Last Edit: August 01, 2004, 11:28:11 PM by GCT »

GCT

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Re:Borohydride reduction of keytone
« Reply #14 on: August 01, 2004, 11:33:32 PM »
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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