Swic was thinking of a way to get rid of those pesky alkoxides in those vanillin synths, and it really is quite simple. After reading some literature here's what he got:
First take your 5-bromovanillin and react it with your alkyl halide (or DMS, DES). This will alkylate your hydroxy group at the 4 position...
5-Br-4-OH-3-MeO-benzaldehyde + RX --> 5-Br-4-
RO-3-MeO-benzaldehyde
The alkylation will not effect the halogen at the 5 position. Next, simply reflux your aldehyde in NaOH to replace the halogen on the ring...
5-Br-4-RO-3-MeO-benzaldehyde + NaOH(aq) --> 5-
OH-4-RO-3-MeO-benzaldehyde
The hydroxylation will not interfer with the alkyl groups at the 3 & 4 positions. If you substitute a larger/complex alkyl group at the 4 position (isopropyl,allyl) swic thinks the basic reflux might destroy it. But it might very well be workable. There should be no problem with methoxy, ethoxy, butoxy, or propoxy substituents.
With this, finish up your 5 position by reacting with another alkyl halide...
5-OH-4-RO-3-MeO-benzaldehyde + RX --> 5-
RO-4-RO-3-MeO-benzaldehyde
Really the only downside is that there is no way to get syringaldehyde. But who cares? You alkylate it anyways.
Example --> 3,5 Dimethoxy-5-ethoxybenzaldehyde
5-bromo-4-hydroxy-3-methoxybenzaldehyde + (CH
3CH
2)
2SO
4 --> 5-bromo-4-ethoxy-3-methoxybenzaldehyde
5-bromo-4-ethoxy-3-methoxybenzaldehyde + NaOH(aq) --> 5-hydroxy-4-ethoxy-3-methoxybenzaldehyde
5-hydroxy-4-ethoxy-3-methoxybenzaldehyde + (CH
3)
2SO
4 --> 5-methoxy-4-ethoxy-3-methoxybenzaldehyde
Yields should be good, (swic thinks > 80%) on each step.
For bees beginning their own O-Chem schooling, swic recomments buying a reaction book. It is a book with a lot of drawings in it just going over reactions, and this way you'll see what you can and cannot do on that magnificant ring. It's really fun when you can start applying what you see.
Ugh, swic sounds like a nerd.