Our next goal is to create rings III and IV which are directly attached to our current molecule. Highlighted in red is are the atoms from our molecule thus far. In blue is also the ethyl ester carbon from our current molecule, and the purple bond will be one created from the veratryl group (the di-methoxy benzene substituent) and the ethyl ester. Rings VI and VII will come after this.
Below are listed the next few (very general) steps of this synthesis to create ring III.
Overview:
The focus today will be on the borohydride reduction. While similar to the lithium aluminum hydride covered in a previous post, this chemical is an extremely common reducing group and is weaker than the aluminum hydride. The choice if reagent is simple, we want to reduce the double bond of the imine, but not other functional groups (such as the ester!). Use of LiAlH4 would certainly reduce the ester (as that is what it is used for) or the amide (hidden in the tosylate protecting group). Interestingly enough, sodium borohydride can actually be stored in water (crazy huh?). Even though it does form some hydrogen gas initially, a basic solution quickly forms, which is stable. You can purchase it as a basic solution if you really want to. Solvents for reduction of sodium borohydride are typically done in methanol or ethanol which have weakly acidic protons (pka ~ 20-25, so they are to no extent deprotonated in aqueous solutions near pH=7). If you are really into the specifics, there is a lot of work done on these reactions which can be nicely summed up in this article.
The hard part about all of these reducing agents is knowing the strength, which honestly seems greatly ambiguous in undergraduate organic. Actually, as an interesting aside, sodium borohydride can reduce esters under particular conditions. Using a lewis acid (such as CaCl2) or by heating to 65-80 C. In reality, esters can be reduced at room temperature if you let the reaction stand long enough, but typically these reactions are performed quickly (< 1 hour) and so reduction of the ester is not a problem.
Last thing to note about reductions using these hydride reagents, the cation really matters!! If you are hurrying to get a reaction done and you are all out of sodium borohydride, but you happen to have lithium borohyride, you should really just wait to get the correct reagent. This is because the cation that coordinates to the carbonyl group plays a key role in the activation of the carbonyl toward nucleophilic attack by the hydride. This is why adding such lewis acids like CaCl2 or LiBr (of which the Ca2+ or the Li+ ion coordinates with the carbonyl) leads to increased activity.
How does this mechanism look like though? After all, we form an alcohol, but we really don't have any spare protons lying around. The actual reaction is shown below, and the protons to form the amine (or alcohol if you are reducing a ketone) come after the aqueous work up, which always follows these reactions. There is actually a lot going on in solution. Since we are usually using methanol or ethanol, these can react slowly with the borohydride forming alkyl borates (alkoxy groups on a boron).
Workup:
I'm always so surprised at how easy Woodward makes chemistry look. Keep in mind that I'm sure a lot of these procedures took lots of time to figure out, but once you figure it out, it makes the synthesis seem so elegant sometimes. Anyway, let's see what Woodward did. First of all, he dissolved the compound in ethanol. But wait, didn't he crystallize the previous compound from methanol?! Yup he did, so chances are when he put this compound into ethanol, it didn't dissolve either. Since ethanol is more non-polar (but still very polar), what is the average chemist to do? What most chemists tend to do...use brute force (i.e. heat the darn thing). After all, it must be soluble in hot solvent since it was recrystallized from hot solvent. Will we worry about reducing the ester? It's a little more of a worry, but still, the sodium borohydride reacts much faster with the imine than the ester that unless we let this reflux this reaction for a long period of time with a large excess of borohydride, we really don't have to worry about it. So after slowly adding the borohydride (exothermic) and heating for 1 hour, a clear solution had formed. Doing what Woodward does best, precipitating product while quenching reaction with the water to donate the final proton to the now amine, gave a solid (yellow and impure) product. Notice how all the borate salts formed will have been washed away in the water. However, any remaining unreacted starting material (or the very very small amount of reduced ester) has not been separated from the product (hence the yellow color). Recrystallization from chloroform (good solvent) and methanol (poor solvent) was able to purify the compound to give colorless crystals.
See you next time where we delve into the ever-so-common acetylation reactions.
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