Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: chiddler on February 28, 2012, 04:24:19 PM
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Rather than LiAlH4 or NaBH4, why not use NaH?
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I think part of the answer is that sodium hydride is a much better base than it is a reducing agent, so molecules tend to condense.
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I believe its because the hydrogen has its electrons in an S orbital which, by definition, is non directional, which makes it a terrible nucleophile. LAH, on the other hand, has covalent bonds to the hydrogen, imparting directionality to the electrons of the Al-H bond, making them more nucleophilic. Anybody else care to weigh in?
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Clayden's Organic Chemistry says it's because hydride's 1s orbital is too small to efficiently overlap with the carbonyl π* orbital. NaBH4 and LiAlH4 react through their σ orbitals, which are comparable in size to the π*.
The implication is that hydride can much more effectively overlap with low-energy C-H σ* orbitals.
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well i'm satisfied with the responses.
thanks very much!
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In more practical terms, NaH will simply deprotonate it at the alpha hydrogen - it does not reduce. I have done this reaction, it is useful when you have a recalcitrant enolate.
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In more practical terms, NaH will simply deprotonate it at the alpha hydrogen - it does not reduce. I have done this reaction, it is useful when you have a recalcitrant enolate.
I apologise for what is probably a really simple question but what do you mean by "recalcitrant enolate"??