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In lecture we discussed the nature of the nucleophile for the attack of the carbonyl.

LiAlH4 acts as a better nucleophile (hydride) than NaBH4. I know there is no hydroboration! So I would have expected that LiAlH4 gives more Felkin Anh product as the NaBH4 gives.

However LiAlH4 gives a Felkin Anh : Anti Felkin Anh ratio of 3:1 and NaBH4 gives a ratio of 5:1.

detailed reaction Source: Carreira, E.M.; Classics in Stereoselective Synthesis

Is there any good rational?

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Lithium aluminum hydride is such a reactive nucleophile that it reacts readily with any conformation of the ketone. The reduced nucleophilicity of sodium borohydride allows for some discrimination of the conformations.

Note that the difference in selectivity between aluminum hydride and borohydride small. A large boost in stereoselectivity can be achieved by increasing steric bulk of the reducing agent (L-selectride gives a 10-fold increase in selectivity).

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  • $\begingroup$ Because NaBH4 is more sterically demanding than LiAlH4 the selectivity is higher for NaBH4? $\endgroup$ – laminin Oct 27 '14 at 16:16
  • $\begingroup$ I expect that NaBH4 and LiAlH4 to be similar sterically. L-Selectride (lithium triisobutylborohydride) is much more sterically demanding. $\endgroup$ – jerepierre Oct 27 '14 at 16:39
  • $\begingroup$ I mean there is though a small difference between 75% and 83% selectivity. Do you believe this two ratios (3:1 and 5:1) are random? I guess the experimentators have used the same measuring conditions so there must be a rational. $\endgroup$ – laminin Oct 27 '14 at 16:44
  • $\begingroup$ No, those are not random. I think LiAlH4 is less selective because it is more reactive. $\endgroup$ – jerepierre Oct 27 '14 at 16:49
  • $\begingroup$ yeah...this seems to be a good explanation. $\endgroup$ – laminin Oct 27 '14 at 16:52

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