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Cowley , Leigh H. Synthesis, structures and reactions of some metallocene alcohols. Journal of Organometallic Chemistry , 1 , 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. Quote from: movies on July 21, , 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.
Nakata, Tetrahedron Lett. 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. Sodium borohydride is not a catalyst in this case.
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