[movies]

The selectivity for axial versus equatorial alcohol is highly dependent on the system in question.  I forget exactly what your system looks like, however, so it might favor the equatorial alcohol.  In general, very bulky hydride sources are controlled by the steric hindrance associated with the approach of the hydride onto the carbonyl, that is to say, the hydride will approach from the sterically less hindered face of the carbonyl.  I don't know if that will work for you or not.

[taurean]

The selectivity for axial versus equatorial alcohol is highly dependent on the system in question.  I forget exactly what your system looks like, however, so it might favor the equatorial alcohol.  In general, very bulky hydride sources are controlled by the steric hindrance associated with the approach of the hydride onto the carbonyl, that is to say, the hydride will approach from the sterically less hindered face of the carbonyl.  I don't know if that will work for you or not.

Here is my substrate.  Please let me  know your opinion.

I guess it also depends on the reactive conformation of the substrate.

[movies]

I would predict that most reducing agents will come from the convex face of your molecule.  In this system it might be really difficult to get the other diastereomer because the other face is so significantly hindered.  If sodium borohydride gave some reasonable level of selectivity, my guess would be that one of the Selectride reagents would give the same major product, but in a higher ratio.  I suspect that two of your stereoisomers are resulting from epimerization of the stereocenter next to the ester in the reduced product.  It might be very difficult to control that.

In simple systems, Selectride definitely does give axial alcohols when it is favorable.  For example, consider the reduction of 4-tert-butylcyclohexanone.  Small reducing agents like LiAlH₄ generally give mostly equatorial alcohol (92:8 e:a), whereas large reducing agents like Selectride give mostly axial alcohol (7:93 e:a).  See: Brown, JACS 1972, 94, 7159.

One other system to consider is the Noyori Ru-BINAP system for hydrogenation of beta-ketoesters.  As I recall, it is selective for the reduction of beta-ketoesters over other alkenes in the molecule.  You might be able to squeeze some selectivity out of the system with a chiral ligand as well.

[taurean]

I would predict that most reducing agents will come from the convex face of your molecule.  In this system it might be really difficult to get the other diastereomer because the other face is so significantly hindered.  If sodium borohydride gave some reasonable level of selectivity, my guess would be that one of the Selectride reagents would give the same major product, but in a higher ratio.  I suspect that two of your stereoisomers are resulting from epimerization of the stereocenter next to the ester in the reduced product.  It might be very difficult to control that.

In simple systems, Selectride definitely does give axial alcohols when it is favorable.  For example, consider the reduction of 4-tert-butylcyclohexanone.  Small reducing agents like LiAlH₄ generally give mostly equatorial alcohol (92:8 e:a), whereas large reducing agents like Selectride give mostly axial alcohol (7:93 e:a).  See: Brown, JACS 1972, 94, 7159.

One other system to consider is the Noyori Ru-BINAP system for hydrogenation of beta-ketoesters.  As I recall, it is selective for the reduction of beta-ketoesters over other alkenes in the molecule.  You might be able to squeeze some selectivity out of the system with a chiral ligand as well.

Thanks for your suggestions.  Sodium borohydride gives some amount of selectivity.  I obtained four diastereomers (1,3-diols) with sod.borohydride,  of which two are formed from the epimerization of the stereocenter next to the ester, as you expected.  I would try hindered hydrides in the next few days and will let you know the result.

I am doing diastereoselective synthesis.  By doing enantioselective hydrogenation, using Ru-BINAP, I would loose half of the material.