I would love to conclude the photo essay with more details on the NaBH4 reduction of the beta-keto-dmt, but a break was needed, and of course 10 other projects started at once..
My problem is the solubility of both reagents (keto-DMT and NaBH4) in the reactions solvent, n-propyl alcohol.
I am told, we can not interchange this solvent as it would reduce the DMT to the alcohol.
1g of beta-keto-dmt added to 257ml of n-propyl alcohol.
Now if we are to dissolve NaBH4 (double the weight of beta-keto-dmt is needed) in the same solvent, we will find ourselves using 250-300ml of solvent to disolve 2g of NaBH4.
At this point we are using about 500ml of solvent to return an expected 49% yield of .49g of DMT.
So while doing a 45g reduction of 2,5-dimethoxy-beta-nitrostyrene to 1-(2,5-dimethoxyphenyl)-2-nitroethane, using NaBH4, I ask myself.. why can't one carry out this process with the beta-keto-DMT the same way??
The NaBH4 isn't even close to being fully disolved in the ethyl acetate, and the 2,5 DMNS is added as is.. recrystalized needles, directly to the NaBH4 suspension.
Is it possible the reduction of beta-keto-dmt can be carried out in a suspension similiar to that described when making (2,5-dimethoxyphenyl)-2-nitroethane?
Say you had 15g of the beta to work with, and you wanted to reduce it all at once. Given the info above, how would you go about it?
Is it necessary for a large amount of solvent to be present for this specific reduction to work??
Any suggestions, comments or ideas would be very welcome.