[pgk]

Thank you for your criticism that I fully accept.
But meanwhile can you explain the mechanism of ester hydrolysis with K2CO3 in methanol at room temperature during 1 hour?
(Isomerization of the Baylis-Hillman adducts using amberlyst-15 as a heterogeneous reusable catalyst: a simple and efficient stereoselective synthesis of (E)-cinnamyl alcohol derivatives, Indian Journal of Chemistry, 45B, 1729-1733, (2006)
http://nopr.niscair.res.in/bitstream/123456789/6587/1/IJCB%2045B%287%29%201729-1733.pdf)
And also, can you give a satisfactory explanation of the experimental data that are obtained by the OP in this post, apart the competition between amine substitution vs hydroxyl substitution that is favored by either the reactants lipophilicity or hydrophilicity, respectively?
PS1: Ionic reactions are very fast.
PS2: Chemistry is an experimental science, which means that you firstly work it in the lab before declaring in public “… it will not react in this way…”.

[wildfyr]

I think the way to settle this would be to find out from OP what the failure modes are. He never specifies whether he just gets back pure starting materials, or if he gets his amine back, but the 1-methoxy-4-methyl(1-ethoxy) benzene compound is present.

Rotovapping it down and NMR of the crude should give it away, the benzyl peak will shift, while the piperidine will stay the same if there is ethoxide attack. You can do a HCl extraction to remove the any version of the piperidine from the crude if its too messy to interpret.

PS If there was a strong base like KOH in this, I'd expect aldol condensation of that ketone to occur.

PPS pgk I think you are thinking of using K2CO3/MeOH for TMS-alkyne hydrolysis. Its a way to avoid fluorides or acids when deprotecting those compounds http://cssp.chemspider.com/article.aspx?id=100

In respect to the paper you cited, ester chemistry is much different from the irreversible reaction we are talking about at the benzylic position. Catalytic base and a huge excess of a MeOH can cause transesterification. You are transesterifying that acetate to give the alcohol and methyl acetate in that paper.

[pgk]

Aldol condensation may not be favored due to the steric hindrance of the ketone.

[wildfyr]

Edited my post after your response, but didnt see it sorry.

[clarkstill]

Thank you for your criticism that I fully accept.
But meanwhile can you explain the mechanism of ester hydrolysis with K2CO3 in methanol at room temperature during 1 hour?
(Isomerization of the Baylis-Hillman adducts using amberlyst-15 as a heterogeneous reusable catalyst: a simple and efficient stereoselective synthesis of (E)-cinnamyl alcohol derivatives, Indian Journal of Chemistry, 45B, 1729-1733, (2006)
http://nopr.niscair.res.in/bitstream/123456789/6587/1/IJCB%2045B%287%29%201729-1733.pdf)

This isn't hydrolysis, it's methanolysis (although the authors incorrectly name it the former). The by-product will be methyl acetate (not acetic acid), liberating the product as an alcohol. Alkoxide addition to esters is rapid and reversible, and in this case the product formation is driven thermodynamically by the excess of methanol present. You don't need to invoke some weird argument about adventitious water to explain the reactivity in this case.

[clarkstill]

In respect to the paper you cited, ester chemistry is much different from the irreversible reaction we are talking about at the benzylic position. Catalytic base and a huge excess of a MeOH can cause transesterification. You are transesterifying that acetate to give the alcohol and methyl acetate in that paper.

Just saw this - apologies to Wildfyr for re-stating a point he already made eloquently.

[rolnor]

Today I refluxed suspension of K2CO3 (283mg)in 20ml ethanol containing 2eq. water for 40minutes.
The solution was decanted and the solid was dried i vacuum and the remaining wite solid was 258mg.

[wildfyr]

Its OK . I think pgk has gotten the point. I hope OP comes back and find it worthwhile to run an NMR (or provide one he already took) so we can learn what the mode of failure was: low reactivity, or side reaction.

[pgk]

Sorry but I do not agree about methanolysis because as far as known, methanolysis of esters with K2CO3 demands several hours of reflux. Indicatively see:
Transesterification of vegetable oil with potassium cabonate/active carbon catalyst, Feed and Industrial Raw material: Industrial Materials and Biofuel, 5, 343-345, (2007)
http://gcirc.org/fileadmin/documents/Proceedings/IRCWuhan2007%20vol5/Pages_de_pages_from_vol-5-58.pdf
Microwave-assisted methanolysis of green coffee oil, Food Chemistry, 134, 999-1004, (2012)
https://core.ac.uk/download/pdf/82024355.pdf
In contrast to ester hydrolysis with KOH or NaOH in methanol that can occur within a few minutes at room temperature:
Facile Hydrolysis of Esters with KOH-Methanol at Ambient Temperature, Monatshefte für Chemie / Chemical Monthly, 135(1), 83-87, (2004)
https://link.springer.com/article/10.1007/s00706-003-0114-1
A simple method for the alkaline hydrolysis of esters, Tetrahedron Letters, 48(46), 8230-8233, (2007)
https://www.sciencedirect.com/science/article/pii/S0040403907018539

[pgk]

My personal experience with K2CO3 is that it works better when using it for phenols alkylation in solvents from open bottles than in freshly distilled solvents.

[rolnor]

Pgk, do you have any comment on my experiment? Its clear that very little KOH is formed under these conditions

[wildfyr]

My personal experience with K2CO3 is that it works better when using it for phenols alkylation in solvents from open bottles than in freshly distilled solvents.

I would greatly appreciate you running that experiment purposefully. Maybe just a simple phenol+bromoalkane and get conversion by GC or GC-MS to avoid workup artifacts.

[pgk]

K2CO3  +  H2O  →  KOH + KHCO3
KHCO3  → KOH + CO2
Assuming that starting from anhydrous 283mg/138 = 2.05 mmol K2CO3, a 65% hydrolysis leads to:
0.65x2.05x56 = 74.6 mg KOH
0.65x2.05x100 = 133.3 mg KHCO3
(1-0.65)x2.05x138 = 99.0 mg K2CO3
Or 306.9 mg solid, which after thermal degradation of 1.1 mmoles of KHCO3 (83% conversion) leads to:
2.43x56 = 136.2 mg KOH
0.23x100 = 23.0 mg KHCO3
0.71x138 = 99.0 mg K2CO3
Or, 258.2 mg of solid containing about 53% KOH, per mass.
Note that the above example is indicative because other combinations of KOH, KHCO3 and K2CO3 can also give this mass value (258 mg).

PS: I hope not being wrong in calculations that I have never done since the high school.

[pgk]

Thanks, “wildfyr” for your advice.

[rolnor]

I think that any formed KOH should go in solution under these conditions, thats why I made the experiment. So there is not much KOH formed at all if any. The small mass-loss from the K2CO3 is more likely coming from that some dissolves in the moist ETOH.
The solid from the reaction was in caracter identical to just K2CO3, it was not KOH.
Also if any formed KOH could react with benzylhalide it must be dissolved in the ETOH, dont you agree pgk?