AirCowPeaCock - 19-1-2012 at 09:28
With an acid catalyst....Benzaldehyde + Dimethylether => 3-methoxybenzaldehyde?
Moiety - 20-1-2012 at 11:44
I doubt it.
Aromatic systems are very hard to react, and aldehydes are very easy. Your best bet would probably be to make 4 methoxy toluene from toluene, then
oxidize it.
Just a guess with no sources.
ThatchemistKid - 20-1-2012 at 11:56
First off you are suggesting that the protonated positively charged ether will be attacked by the benzene ring through elecrophilic aromatic
substitution. that would not lead to methoxide but to a methyl group added to the benzene ring and methanol as a leaving group, methyl cations are
high energy species and formation of them under most reaction conditions is...well basically not happening. Secondly you are trying to react a
benzaldehyde , which is a benzene ring with an electron withdrawing group on it via a EAS mechanism..which would require harsh conditions and say..
maybe a lewis acid as is seen in Friedels craft chemistry. More electron rich / larger aromatic systems such as anthracene and naphthalene can be
reacted with protonated alcohols and the like ( alcohol + sulfuric or other strong acid) to yield the alkylated product, but it is difficult to
control the regio-selectivity.
AirCowPeaCock - 20-1-2012 at 13:35
I thought It was a long shot.. Im guessing a base catalyst wound work either but, maybe.. my goal is to go from Cinnamaldehyde to benzaldehyde to
vanillin, cinnamaldehyde to benzaldehyde is easy, but I have no clue how to go from benzaldehyde to vanillin, It seems it would be easy, because of
the similar structure, vanillin is just benzaldehyde with a methoxide and OH group..
kavu - 21-1-2012 at 00:06
In synthetic organic chemistry a seemingly simple structural difference can be difficult to achieve. For example there are only a handful of versatile
C-C bond forming reactions. Conversion of benzaldehyde to vanillin is a multistep reaction. Methoxide group can be achieved by running a Williamson
ether synthesis with a corresponding phenol and alkyl halide. 3,4-dihydroxybenzaldehyde (also known as protocatechualdehyde) and iodomethane react to
form vanillin in basic solution. Synthesis of the aldehyde has been described in literature. Reimer-Tiemann reaction of catechol is one of the most
accessible to an amateur. A downside is that Reimer-Tiemann tends to have low yields and nasty dark oily side products. Methoxy can also be added by
using corresponding aryl bromide and alkoxide base in the presence of a transition metal catalyst. The following is a common lab experiment: http://valhalla.chem.udel.edu/vanillin.html. Synthesis of such precursors has been discussed:
http://www.sciencemadness.org/talk/viewthread.php?tid=10702#... Starting from benzaldehyde and working your way up is a taunting task. Aldehydes
tend to react with all sorts of nucleophiles and get oxidized easily.
AirCowPeaCock - 31-1-2012 at 12:44
I see that...I'm t5rying to see how I can get from benzaldehyde to 3,4-dihydroxybenzaldehyde..every thing I can find and think of would destroy the
aldehyde functional group..or so I think. BTW that vanillin link is dead. What do you think about benzaldehyde and 95% sulfuric acid to form the
sulfonic acid, then the addition of sodium hydroxide? Then Id go methylbromide and [insert transition metal here, maybe Ni or Fe] methoxide? Thanks
alot for the help!
UnintentionalChaos - 31-1-2012 at 13:17
It's not going to work. You're trying to sulfonate a deactivated aromatic ring with an oxidation sensitive group on it. Hot sulfuric acid is
moderately oxidizing. Even if that worked, you need to fuse the product with molten NaOH or NaOH/KOH eutectic to get the sulfonic acid to react. This
is known to destroy the aldehyde functionality. http://www.orgsyn.org/orgsyn/orgsyn/prepContent.asp?prep=cv3...
Reaction of a phenoxide with an alkyl halide does not need transition metal catalysis. Only the reaction of an alkoxide with an aryl halide. In the
first case, the reaction is an SN2. In the latter, SN2 is impossible and you need an organometallic intermediate. Copper is traditional.
3-methoxybenzaldehyde is not all that easy a ring to produce. To substitute at the 3-position, you need an m-directing substituent at the 1 position.
carboxylic acids and derivatives thereof are acceptable, as are aldehydes and nitro groups.
Direct nitration of benzaldehyde seems to work: http://www.orgsyn.org/orgsyn/orgsyn/prepContent.asp?prep=cv3...
this can be reduced to m-aminobenzaldehyde, diazotized, and heated to make the phenol, which can be methylated with MeI and base.
[Edited on 1-31-12 by UnintentionalChaos]
AirCowPeaCock - 31-1-2012 at 15:55
That's going to produce a slew of byproducts, and I don't know of I really want to distil the 3,4-dinitrobenzealdehyde due to the warning with a
single nitration! Regardless. I have not a significant clue what temperature it boils at, if its going to boil before degrading atall! Any ideas
for how to get a more pure product? Isn't it obnoxious when you can't find physical data on your chemicals?
GreenD - 1-2-2012 at 10:47
Man, Air cow - I'm in no position to say this but utfse. This kind of stuff is discussed everywhere.
AirCowPeaCock - 1-2-2012 at 12:22
Holy shit! I found data! I never find data on this kind of thing--usually I find someone who is selling it but not a single piece of useful data!
Doesn't say when it breaks down though, I suppose I could find that myself though, or at-least when the crude mixture decomposes.