Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: xshadow on November 26, 2019, 04:25:31 AM
-
Hi,I have some doubt about this mechanism:
(https://i.imgur.com/CFj2SZE.jpg)
I wrote my doubts below the mechanism
Thankss!!
-
It is a multi-stage synthesis optimized for one-pot operation. CH3O- and OH- are too reactive and will mainly cause unwanted side reactions in the methyl 2,3-dibromopropionate (take into account: substitution, elimination).
As for B - the compound at this point of the synthesis contains one OH phenolic group.
-
It is a multi-stage synthesis optimized for one-pot operation. CH3O- and OH- are too reactive and will mainly cause unwanted side reactions in the methyl 2,3-dibromopropionate (take into account: substitution, elimination).
As for B - the compound at this point of the synthesis contains one OH phenolic group.
I don't understand why in the first step K2CO3 deprotonstes only ONE of the phenolic goup...they have both the same acidity...why is only one deprotonated?
.
Thanks
-
Compare pKas of catechol and carbonic acid.
-
Removing a second proton would place two negative charges in close proximity. I don't have a number in front of me, but the second pKa of catechol is probably a good deal higher.
-
Compare pKas of catechol and carbonic acid.
Wikipedia says:
pka1: 9.45
pka2:12.8
pka for H2CO3 are:
pka1: 6.4
pka2: 10.2
In solution I put CO32- and Cathecol acid...the acid base reaction should be (I think)
1) R(OH)2 + CO32- <---> R(OH)O- + HCO3-
2) R(OH)O- + HCO3- <---> ROO2- + H2CO3
now I have to check where equilibrium is shifted: towords products ,reagents or in "the middle"...in order to do this I have to look at the pka(s ) (right?) of the acidic species in product and then in reagent and compare them
For 2) I have:
Reactant acid: R(OH)O- pka= 10.2
Product acid: H2CO3 pka= 6.4
Equilibrium 2 all shifted toward the reagents
Is it correct this reasoning??
THANKS!!
-
Removing a second proton would place two negative charges in close proximity. I don't have a number in front of me, but the second pKa of catechol is probably a good deal higher.
Thanks
My reply to AWK is also for your post
Is it correct my reasoning?
Thank you too
-
Phenols only react with carbonates to form bicarbonates. The bicarbonates still do not react with phenols.
Two moles of carbonate are needed for this reaction.
The synthetic procedure uses an excess of potassium carbonate (up to 3 moles per mole of catechol).
-
Because the pka(s)I suppose...
Thanks!
--------------------------------------------------------------
For the other doubt?
After the deprotonation of Phenolic group I have a species similar to RO- :
RO- could also attack the C-carbonylic of ester?
(Instead of attacking the C-Br and gives Sn)
Because I've studied this type of reactions too:
A RO- thath attacks a carbonyl:
(https://i.imgur.com/6I2JHBg.jpg)
Is Sn1/Sn2 faster than mucleophile attack to carbonyl??
Thnaks all!
-
https://www.academia.edu/34301107/A_Synthesis_Involving_the_Reaction_of_a-Halo_Michael_Acceptors_with_Catechol
-
https://www.academia.edu/34301107/A_Synthesis_Involving_the_Reaction_of_a-Halo_Michael_Acceptors_with_Catechol
I saw a bit of this study..
And it says:
""""It seemed likely to us
that, under the conditions employed for these reactions
(KzC03, 55", 16 hr), elimination of hydrogen bromide
from 8 and 9 to give the a-bromo olefinic esters 3 and 5
would occur more rapidly than would nucleophilic displacement
of bromide ion by catechol monoanion. ""
So It's saying that the first step ISN'T a Sn2 attacck of deprotonate catechol on halide (the alpha one to carbonyl)
But in my textbook there is that mechanism where I have instead a Sn2 attack!!
I'm a bit confused
-
If the reaction mechanism is not studied precisely, it can very often be explained in several ways.
The work comes from almost 50 years ago. Maybe something new has happened in the meantime.
Or maybe the author of your textbook simply prefers a different mechanism.
-
If the reaction mechanism is not studied precisely, it can very often be explained in several ways.
The work comes from almost 50 years ago. Maybe something new has happened in the meantime.
Or maybe the author of your textbook simply prefers a different mechanism.
So we can say the mechanism of this reaction us not complete clear and there should be different interpretation of the mechanism??!
.thanks
-
It is commonly stated: we cannot PROVE that any mechanism actually was followed.* Instead, we propose all of the possible mechanisms that we can think of, then we use experiments to try to DISPROVE them. In the end, we expect only one possible mechanism will remain, and this becomes the established mechanism.
But new data can upset things. You probably learned about SN1 and SN2. But did you know there is another second order mechanism (SNET) that is sometimes followed? It gives a few discrepancies from traditional SN2 that were overlooked at first.
So while there are well-established mechanisms that have stood the test of time, they might not actually be completely true.
*(This was more true back in the early days of mechanistic chemistry - now we have advanced techniques like scanning tunneling microscopy and time resolved x-ray crystallography where you might be able to see the atoms actually reacting. These can't always be applied, and often can't exactly replicate the reaction conditions. Also, although we observe one set of atoms following one mechanism, there are 10^23 individual reactions for a mole - is this truly the ONLY way they react? Or did we see a 1 in a million outlier? )
-
If a paper has a proposed mechanism, I tend to treat it as just that: a reasonable hypothesis, which may or may not be true.
-
These are the times when I realize that chemistry is not like math . True or false. Here can also be a ""possible"
Thanks!!
-
The likely reaction mechanism can be calculated.
But first you need to know chemistry well.