09-21-01 05:14
No 215790
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 + (CH3CH2)2SO4 --> 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
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.
(Hive Bee)
09-21-01 05:21
No 215793
Would you happen to have the Title, Author, or ISBN of a reccomended title of said genre? Id be highly interested in purchasing a copy if I had that information.
Of COURSE we don't know what we're doing! That's why it's called research! Member: C_F Fan Club
(Distinctive Doe)
09-21-01 09:17
No 215844
Or maybe you could just leave the bromine there???
I'll bet you would have a totally psychedelic phenylethlamine/amphetamine!!!
Do Your Part To Win The War
(Newbee)
09-21-01 14:13
No 215909
I think this is already mentioned on Rhodium's site: 5-Hydroxy from 5-Bromo vanillin
../rhodium /mmda.me
(Hive Bee)
09-21-01 19:51
No 215987
Read it a little bit better, Karl. You'll see the difference. The addition of the phenol is the same method, but in this case you alkylate the 4 position before refluxing in NaOH. This way you don't have to alkylate both positions with the same group, and you can make some really useful aldehydes. And yes, it is a very common method used to replace halogens on the ring.
Foxy, never even considered that. Swic bets your right though. Hey, you could even do that with what you proposed with the 2-bromo-4,5-dimethoxybenzaldehyde. That would be cool!
(Newbee)
09-21-01 19:57
No 215988
Sorry...I missed your point
(Moderator)
09-21-01 20:25
No 215995
NaOH-solution, copper, 200°C: J.Amer.Chem.Soc. 51, 2195 (1929) and Chem. Ber. 105, 222 (1972). All other references use BuLi and H2O2.
(Hive Bee)
09-21-01 21:35
No 216011
Doesn't BuLi require protection of the aldehyde group?
(Of course, that might also be useful for the other synths. I beleive somebee over here noted that vanillin *needs* N2 atmosphere, otherwise you'll end up with large amounts of whatever-acid).
(Hive Bee)
09-22-01 03:20
No 216096
The only time swic has seen BuLi is in pihkal. Of course that is because he hasn't gone over enough literature yet. However he read that the reaction with NaOH usually is performed under high temperature and high pressure.
Example: chlorobenzene + 10%NaOH(aq) +370C & high pressure --> phenol
Lilienthal, so the addition of copper to this reaction omits the need for the high temps and high pressures right? Also, is it more common practice to use BuLi and H2O2, what swic means is that is that method more generally used and accepted for performing such reactions? Thanks.
(Stranger)
10-27-01 00:22
No 229145
Would there not be steric hinderance when alkylating hydroxyl group in middle of -Br and -OCH3??
Catastrophe, Butyl Lithium needs protection of groupe aldehyde.
(Stranger)
10-27-01 20:28
No 229321
AH! There would be no steric hinderance, very sorry.
The reasoning of this post looks to be mistaken. After reading this, SWIM doesn't understand why someone wants to take a hard way. Alkoxides are very good reagents for this reaction, but if you don't like to use them then after synthesis of 5-hydroxyvanillin, why not selectively alkylate the 4 and 5 positions?? A aldehyde as 4-EtO-3,5-MeO-benzaldehyde, can make from 5-hydroxyvanillin by using molar quantities of alkylating reagent...
1 mole 5-hydroxyvanillin, 1 mole methyl iodide, 1 mole Na2CO3 in methanol. Selectivity will be at the meta position, but there will be some dimethylated and some para-methylated aldehyde(you can use less then 1-molar for less of this). Now, you have syringaldehyde, finish by alkylating with ethyl halide.
This makes more logic to SWIM then what was said... unless someone likes to challenge it.
(Distinctive Doe)
10-27-01 22:52
No 229356
I think the point is that you need to buy the alkoxide or metal from a chem supply and they wanted to work around that.
EDIT:
Ooops sorry, I guess I read your question wrong.
Do Your Part To Win The War
(Stranger)
10-27-01 23:17
No 229365
Is that not what SWIM has proposed??? There are no alkoxides in what he/she said.