brew - 12-7-2013 at 00:46
I stumbled upon the reaction of
2,5-dimethoxytetrahydrofuran (cis or trans) in acidic conditions, forming a 1,4dialdehyde. The acidic conditions was the use of acetic acid, and being
a catalyst, is reformed. Its no doubt simple for those experienced by I found the mechanism somewhat challenging. Hint, the intermediates are of
course the acetate anion, two methanol molecules and a water molecule - I just thought presenting cool simple yet tricky reactions could be
advantageous to a forum that is about learning chemistry oneself, and thought why not.
Now if this reaction started with the 2,5-dimethoxytetrahydrofuran, and included the acetic acid catalyst, along with aniline, the end product would
be a N-Phenylpyrrole, along with two molecules of methanol, and a molecule of water. Both are cool reactions, and worth having a go at. Directed at
the newbie. Cheers.
ps. at first glance I thought easy, but in fact I found it tricky and took abit of time working it out.
kavu - 12-7-2013 at 03:45
The reaction is quite straight forward, just double acetal hydrolysis. Another neat and rather simple mechanism proposal might be the reaction
sequence a,b. The role of acetic anhydride in b should also be explained:
[Edited on 12-7-2013 by kavu]
brew - 12-7-2013 at 22:27
I don't know how to put reaction schematics in a post,
I would really appreciate knowing how to post a mechanism. I use Symyx Draw 3.3, but by the look at the reaction shematic posted by yourself it seems
like a swift copy and paste. Any hints appreciated. As far as the reaction mech that I worked out, one of the methoxy group gets protonated, hence +
charged, and the ether electrons resonating, flicking of the methanol. The carbonyl(not really a carbonyl, just a double bond between the ether oxygen
and the carbon below) stays positively charged though, so anyway, the other methoxy gets protonated also+ charge again, then the methanol and acetic
causes a condensation reaction forming a water molecule which attacks the carbon alpha to the other carbonyl oxygen(not really as explained), and with
resonance again this time of the water molecule, breaks the bond between carbonyl oxygen(as well as and the carbon alpha to that position(not the
carbon attached to the double bond of the ether oxygen. Elimination of the other methoxy group occurs as the water molecule resonates and forms the
aldehyde. THe protons get plucked of, in order to regenerate the acidic catalyst - and voila, a dialdehyde. That's how I read it, and possibly wrong.
I'm learning, as well as abit rusty, forgotten names of many reaction types, but all in all, I felt it was a sweet reaction. Especially the Pyrole
synthesis. I didn't expect the addition of the nucleophile to attack the carbon between the ether and the oxygen of the methoxy group, causing, I
think it would be a 1,2,3 elimination, and a carbonyl group the amine can attack - forming an eventual water molecule that gets eliminated. Blew me
out when I saw the simplicity of it, amidst a real struggle in sorting it out. We are all at different levels of our understanding. So its all good.
Ill re-read what you have stated, and see if I can relate. Thanks a lot for responding, and presenting another reaction. Cheers brew
[Edited on 13-7-2013 by brew]
[Edited on 13-7-2013 by brew]
brew - 12-7-2013 at 23:17
As far as what u have posted Kavu, gut feeling, its more at intermediate level, but Ill have a look at trying to nut it out. Thx. Better to be
striving towards harder things, than maintaining the simple basics. I've been out of this for some time, and just felt the urge to start with some
basics. Cheers Brew.
ps if you could assist me in copy pasting, or presenting my own drawings, Id be forever grateful.
I definitely got to go over the basic terminology, as yes it is indeed a double acetal formation, then the water addition, resulting in the eventual
dialdehyde. Still thought that aspect was somewhat sweet.
[Edited on 13-7-2013 by brew]
[Edited on 13-7-2013 by brew]
brew - 13-7-2013 at 23:02
One quick question Kuta, the reaction conditions, appears to be implying that catalyst and the allylbromide, along with the tertbutyl salt, are one
reaction, 16 hours, then the acetone and anhydride at -10 to r.t. for another 16 hours, etc, Are they separate reactions, because the schematic just
has both a,b over a theoretical yield. Its interesting. As said, Im rusty, bigtime rusty, that's why I thought Id present some reactions, to get back
into the swing of things.
They must be together, If they were separate they wouldn't have a,b over the arrow, and theyd indicate a separate step. Just thinking out aloud
somewhat.
[Edited on 14-7-2013 by brew]