I have run into an exercise problem while working on some homework for my O-Chem I class. I am completely stuck, and have assembled models trying to
think out all the possible solutions I can come up with given my limited knowledge on substitution rxns.
I've attached the full exercise. The actone in the rxn is supposed to act as the nucleophile while at the same time the H+ in sol. is supposed to help
acetone become the nuceophile in the rxn... Isn't acetone at its present state already a suitable nu-? Wouldn't an H+ form an enol? And is Carbon 2
what is acting as the electrophile? However if that is the case how on Earth do I end up forming the end-product?
This teacher loves to pass out so called extra credit problems to work out over a week period but always asks troll questions he knows damn well a
beginner o-chem student would never be able to answer.
Hope you guys can help! Thank you all.
Attachment: glucose-acetone-rxn.pdf (444kB) This file has been downloaded 765 timesNicodem - 29-6-2013 at 00:49
This teacher loves to pass out so called extra credit problems to work out over a week period but always asks troll questions he knows damn well a
beginner o-chem student would never be able to answer.
Well, why don't you demand the reference for that reaction? Or even better, find it yourself. You are the student after all and if you don't look for
the reference, then how would you know what is the exact reaction procedure and what else happens? Without this data you cannot propose a
reasonable mechanism - you can only propose a probable mechanism (which is just fiction anyway, as you just make up all the
assumptions).
Acid catalyzed aldol reactions are well known and you can find the mechanism even at the Wikipedia entry, but this is no proof the reaction works on
this specific system. So, of course, there is always the possibility that your teacher just "invented" this problem, but in such case it is still your
duty to check the literature and provide evidence that he is misleading you.Chemstudent - 29-6-2013 at 09:26
I have not even begun to learn any named reactions in class. Our professor has just barely gotten done talking about the rudimentary concepts behind
E1/E2/SN1/SN2 reactions. I can only apply what I know about the inductive effect and Electrophiles/Nucleophiles. So far know just that has been enough
but this problem I got stuck on, furthermore he does in fact pull problems from his own personal research at times to illustrate concepts but often it
is way too advanced for anyone at our point in the lectures to comprehend.
Anyhow, after looking through enough google search, it seems I should begin learning the Acid-Catalyzed Aldol, and figure out how to apply it to this
problem. Lastly, what on Earth kind of glucose is this compound? The BnO substituents are unusual, I normally see alcohols.
Thank you again.
and sorry there is no additional ref. info other than what I have provided. This was what I was given.Nicodem - 30-6-2013 at 12:53
Lastly, what on Earth kind of glucose is this compound? The BnO substituents are unusual, I normally see alcohols.
"BnO" is shorthand for benzyloxy: Bn = benzyl (PhCH2-).
Quote:
and sorry there is no additional ref. info other than what I have provided. This was what I was given.
That's why I suggested you to find some reference. You are a student after all, aren't you? You can't just be complacent with what you are given from
a professor, if you want to actually learn something.Chemstudent - 30-6-2013 at 13:14
Okay, so this is indeed an acid-catalyzed Aldol reaction. The acetone forms an enolate ion because of the H+ ions in solution. The enolate can then
use it's Pi bond (on the alpha carbon) to react with carbon 2 and kick off the OAc group or bond with the methyl group on OAc and form a single
(acetic anhydride looking) substituent (=O/OH).
Is my thinking on the right track? So far my inquiries have taken me this far. Lastly, why cannot the Acetone directly act as the Nucleophile using
the lone pairs on O? Why must it form the enolate first to be an effective nucleophile? sonogashira - 30-6-2013 at 14:17
The enolate can then use it's Pi bond (on the alpha carbon) to react with carbon 2 and kick off the OAc group or bond with the methyl group on OAc and
form a single (acetic anhydride looking) substituent (=O/OH).
You want to have the lone pair of the ring oxygen (1) making a double bond to carbon (2) and displacing the protonated (as C=OH plus sign)
acetate. That is your 'oxonium intermediate'. The enol will the react with the ring C (2) with the double bond going back to O (1).Chemstudent - 30-6-2013 at 15:49
Okay, so Ring Oxygen (1) deprotonates Carbon 2 by pulling the sigma C-H electrons. A pi bond forms between Ring Oxygen and C2, leaving us with a
negative charge on oxygen.
Now, since the entire solution is highly acidic, the H+ that left from C2 bonds with the Oxygen 1 from the OAc subsituent which essentially kicks off
the OAc group giving an enthanoic acid molecule floating about.
Now the Enolate formed from Acetone (enolate formed because of acidic condition?) come in and bonds with C2? I then assume the Pi bond on Central
Oxygen (1) breaks, and a pi bond reforms on the oxygen of the enolate?
Sorry guys, I'm trying to peice this out, and all the supplementary Aldol videos show a strong base being involved which is not the case in this rxn.
I am confused as to how Anhydrous HCl is involved.
