[Exploring]

Hi,
I have a symmetrical molecule with two imine bonds (both bonds are identical).  I would like to reduce only one bond, but not both. My question is, if I use two equivalents of my molecule and just one of NaBH4, the odds of getting a high yield of a monoreduced  imine bond are very high > 50% ??.  I thought that I could also have a mixture of the non reacted starting material, and the fully reduced imine bonds (besides the mono reduced one).  How can I maximize the chances of getting my target molecule?.    Thnsk a lot

[Dan]

You can minimise double reduction by using an excess of the bis(imine) starting material, then hopefully you can recover and recycle the excess in your purification.

I would avoid NaBH₄ because the stoichiometry is complicated by its reaction with the solvent (usually an alcohol), but you can fiddle with that experimentally to optimise it.

I would probably try Na(AcO)₃BH instead. If you mix Na(AcO)₃BH and the bis(imine) 1:1, the statistical distribution would be 1:2:1 bis(imine)/desired product/bis(amine) - i.e. 50% yield.

If you mix them 1:2 (excess starting material), the statistical formation of the undesired bis(amine) falls to only 6% (or 1/8) based on the limiting reagent: i.e. the ratio would be 8:7:1 bis(imine)/desired product/bis(amine).

Note these are statistical ratios based on the assumption that the hydride will react with the desired product at the same rate as the starting material, which is unlikely to be true in reality.

[Exploring]

thnsk a lot!,
unfortunalety I only have NaBH4... I´ll try my best wth this one.

You can minimise double reduction by using an excess of the bis(imine) starting material, then hopefully you can recover and recycle the excess in your purification.

I would avoid NaBH₄ because the stoichiometry is complicated by its reaction with the solvent (usually an alcohol), but you can fiddle with that experimentally to optimise it.

I would probably try Na(AcO)₃BH instead. If you mix Na(AcO)₃BH and the bis(imine) 1:1, the statistical distribution would be 1:2:1 bis(imine)/desired product/bis(amine) - i.e. 50% yield.

If you mix them 1:2 (excess starting material), the statistical formation of the undesired bis(amine) falls to only 6% (or 1/8) based on the limiting reagent: i.e. the ratio would be 8:7:1 bis(imine)/desired product/bis(amine).

Note these are statistical ratios based on the assumption that the hydride will react with the desired product at the same rate as the starting material, which is unlikely to be true in reality.

[orgopete]

I agree with Dan. Although the structures have not been provided, reaction of the imminent and amine are possible. Low pH conditions may tie up the amine. Low pH stable hydrides may be the best conditions.

Since you have NaBH4, then it can be reacted with acetic acid to generate NaHB(OAc)3 in situ.