Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: clemi2310 on August 21, 2018, 10:09:48 AM
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Hello everyone,
I need some help regarding a reaction of N-alkylation of the piperidine 115 as mentionned in the attached scheme.
My classical conditions were the one above, piperidine + benzyl chloride + K2CO3 in EtOH activated by microwave at 80°C in 40 min.
In that way, I manage to substitute a large panel of benzyl chlorides including: benzyl chloride, 4-trifluorobenzylchloride, 4-bromobenzylchloride etc...
However, when I attempt to substitute the 4-methoxybenzyl chloride, I do not get my product. I guess the electron donor group enrich the molecule and make it less electrophilic.
Then I found a paper were they use the conditions B to substitute the 4-methoxybenzyl chloride to piperidine with DIEA, in DCM at TA overnight. But they do not give any justification on the choice of their procedure. Hopefully, it works on my scaffold.
Since I didn't do any optimisation or link between conditions A and B it is a bit difficult for me to explain why the reaction occurs in these conditions because I changed to many parameters.
There is a huge gap between the two conditions:
- polar protic solvent (EtOH) -> apolar aprotic (DCM)
- inorganic base (K2CO3) -> organic base (DIPEA)
- temperature
Do you have any suggestions please?
Thank you for your help !!
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P-methoxybenzyl chloride is a very rective Sn1 electrophile, maybe its to reactive for the conditions you originally use. It could even react with EtOH.
The other method with DCM is better.
If you have problems you could try reductive amination with p-methoxybenzaldehyde.
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Rolnor makes an excellent point, perhaps check for the presence of the ethoxy benzyl derivative in the first reaction? I certainly think giving the other one a shot is a good idea. If you don't have DIEA on hand, TEA would probably work in a pinch.
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The reason they use DIPEA is probably to avoid forming Q-salt.
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Piperidine is a much better nucleophile than TEA. Cyclic amines have some unique properties vs the straight chain ones.
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thank you all for you answers.
So you mean p-methoxybenzyl chloride is even more reactive than a benzyl chloride with a withdrawing group like, p-brombenzyl chloride ?
Because first conditions works with other kind of benzyl chloride (see file attached, even if yields are lower) and p-methoxybenzyl chloride is the only +M group
Indeed the second conditions work perfectly, I just wanted to find an explanation :)
Thank you !
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You could get better yield with the DCM-method for the other examples also.
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thank you all for you answers.
So you mean p-methoxybenzyl chloride is even more reactive than a benzyl chloride with a withdrawing group like, p-brombenzyl chloride ?
Because first conditions works with other kind of benzyl chloride (see file attached, even if yields are lower) and p-methoxybenzyl chloride is the only +M group
Indeed the second conditions work perfectly, I just wanted to find an explanation :)
Thank you !
I am having trouble following the discussion. Is it clear which mechanism is operating under each condition?
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No, it is not clear, could it not be a mixture of Sn1 and Sn2 i both cases?
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1). Whether Sn1 or SN2, it depends on the benzyl chloride substitution and not by the nature of the base (k2CO3 or DIPEA) E.g. benzyl chloride favors SN2, in contrast to p-MeO-benzyl chloride and p-Br-benzyl chloride that favor SN1.
2). a). K2CO3 (if not being completely fresh) is hygrospopic and contains both crystalline water, as well as humidity in bulk and also a small amount of KOH due to partial hydrolysis.
b). Absolute ethanol (if not being freshly prepared, super-dry) also contains a small amount of water.
As a consequence, hydroxide anions are formed during the reaction conditions that can competitively form the corresponding benzyl alcohol. However, the competitive alcohol formation is slow due to the low concentration of hydroxide anions, except in the case of p-MeO-benzyl chloride that is favored by the hydrophilicity of the methoxy- group.
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Thank you. I might imagine that the difference in solvent (ethanol versus DCM) might influence the mechanism also.
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Indeed, there is solvent influence on favoring SN1/E1 (polar solvents) vs SN2/E2 (non-polar solvents) but this reflects on the reaction rate (and indirectly, on the obtained yield), rather than the nature of the reaction mechanism.
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I dont understand this, that the hydrophilicity of the p-MeO-group should favor the reaction with hydroxide? It sounds very strange.
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There are two competitive reactions that occur simultaneously, hereby: Amine substitution of benzyl chloride and hydroxyl substitution of benzyl chloride. Generally, hydroxyl substitution of benzyl chloride is slow due to the low concentration of hydroxide anions, therein. But on the other hand, any increased hydrophilicity of the reaction substrate (benzyl chloride derivative) accelerates the hydroxyl substitution vs amine substitution, by helping the contact of the substrate with the (ethanol-) solvated hydroxyl anions.
As a comparison: Methoxy- decreases logP by 0.5, in contrast to bromine that increases logP by 0.6; which means that the difference in hydrophilicity between p-MeO-benzylchloride and p-Br-benzylchloride is ΔlogP (= 1.1) in power of 10.
Partition coefficients and their uses, Chemical Reviews, 71(6), 525-616, (1971)
https://pubs.acs.org/doi/10.1021/cr60274a001
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I do not understand at all, how much hydroxide can be formed, 1%? Will this greatly influence the yield?
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Please, do not confuse the solvent polarity with the reagents hydrophilicity/lipophilicity. To detail:
1). Polar solvents favor polar reactions, as both these competitive reactions are.
2). The reagents molecules must collide, in order to react. But before their collision, they must contact each other. Thus, similarity in their hydrophilicity/ lipophilicity helps their contact and favors their collision/ reaction (if having moderate kinetic energy), regardless the polarity of the solvent.
3). All the above are not just words because phase transfer catalysis (where, a cationic surfactant or a crown ether are used as catalysts), is based on this principle.
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Did you read my last post, I edited it?
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Ethanol is highly hygroscopic and quickly absorbs humidity from the air, up to 4.4 % w/w (azeotrope mixture). Higher absorption of humidity causes evaporation of ethanol, in order to keep constant the azeotrope mass ratio at EtOH/H2O = 95.6/4.4 w/w, which corresponds to a molar ratio EtOH/H2O = 8.5/1.0.
On the other hand, addition of molecular sieves simply delays water absorption and does not really dehydrates ethanol because molecular sieves can absorb water, up to 25% per their mass. Thus and roughly, addition of molecular sieves in ethanol at 4% w/w, can absorb 1% w/w of humidity, only.
In other words, almost all K2CO3 has been hydrolyzed to KOH during the reaction procedure, if using absolute ethanol from a (old and often opened) half-empty bottle (with or without molecular sieves), as a solvent in reactions involving K2CO3 and at dilutions that are in accordance with the ordinary laboratory practice.
PS: Sorry but I was off-line during the weekend.
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What? Sieves can completely dry ethanol down to 10ppm or so. See that sieve paper I always post.
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Yes but it depends from which side you look at the point:
Agreed! Molecular sieves can absorb water, up to 25% w/w. As an example, ethanol containing water 1% w/w can effectively be dehydrated by addition of 4% w/w molecular sieves and ethanol containing water 2% w/w can effectively be dehydrated by addition of 8% w/w molecular sieves and so on.
But if the above ethanol samples are kept in half-empty bottles that are often opened, (humid) air is renewed and the so penetrating, additional amount of moisture cannot be trapped by the contained and already saturated molecular sieves.
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I think all of us in this discussion including OP understand that if you're going to bother to dry something to the degree that sieves can, you should store it in a shlenk flask and withdraw solvent under a positive pressure dry inert gas flow to replace the volume you are withdrawing. That's the point of a shlenk flask. Isn't this standard procedure?
Typically 10-20% v/v 3A sieves are a good place to start for drying (and keeping dry) hygroscopic solvents.
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1). I fully agree.
2). Thinking twice, my phrase “addition of molecular sieves simply delays water absorption and does not really dehydrates ethanol” is wrong and can lead to confusions. Sorry.
The right is: “addition of molecular sieves 3 Å can effectively dehydrate ethanol but precautions are needed during often handling flasks of the so dehydrated ethanol”.
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So, if I need some KOH I can heat K2CO3 to 80°C i absolute ethanol for 40 min.?
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As explained above, a bottle labeled “Ethanol Absolute” does not obligatory mean that it really contains absolute ethanol. Consequently, the yield of K2CO3 hydrolysis depends on the water content of the said “absolute ethanol”. At the azeotropic mass ratio (EtOH/H2O = 95.6/4.4 and which is considered to be the maximum water content) up to 33.7 gr K2CO3 per 100 gr of the alcohol, can be hydrolyzed upon heating, giving a mixture of KOH and KHCO3, which is getting more rich in KOH within temperature increase (due to CO2 release), up to the azeotrope point = 78.1oC.
Hint: The use of a condenser is obligatory, otherwise the contained water will azeotropically be evaporated, prior to the complete hydrolysis of K2CO3.
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By the way, the water content can satisfactory explain the facile hydrolysis of esters with K2CO3 in methanol(1),(2), which has been posted in this forum in the past (3), with rather unsatisfactory explanations.
1). Total syntheses of (±)-crinine and (±)-buphanisine, Tetrahedron Letters, 28(5), 503-506, (1987)
https://www.sciencedirect.com/science/article/pii/S0040403900957666
2). 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
3). Hydrolysis of ester with sat. K2CO3 in MeOH, Chemical Forums, (2011)
http://www.chemicalforums.com/index.php?topic=47635.0
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When I buy pure or absolute ethanol from a reputable company like Sigma or BDH they better damn mean it.
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No, they don’t.
A sealed bottle of “Absolute Ethanol” purchased from a reliable supplier (and accompanied with a certificate of analysis), indeed contains absolute ethanol. The real problem is the “often-opened and often-used, half-empty” bottles in the lab, especially when handled by different chemists.
PS: Attention because "pure ethanol" is the azeotropic ethanol, in some European countries.
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So pgk, if you heat a suspension of K2CO3 in dry ETOH in an open flask to 80°C for 40 minutes it will be almost completely converted to KOH? Yes or No?
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It depends on much dry your “dry ethanol” is.
So, yes if you add K2CO3 in amounts that respect the stoichiometric ratio of the water content; but you need more than 90 min to ensure the complete thermal decomposition of the simultaneously formed KHCO3 and not in an open flask because the contained water will azeotropically be removed, prior the complete hydrolysis of K2CO3 (you have to reflux in a flask equipped with a cooling condenser).
Note: The temperature will never rise above 78.4oC, which is the boiling point of ethanol and a borderline temperature for the (fast) thermal decomposition of KHCO3.
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Hereinafter I will take your posts with a grain of salt pgk, this simply wrong, K2CO3 will not react in this way at 80°C in ethanol in so short time, even if there is water in stochiometric amount present. Sorry, this is nonsense if you ask me.
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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…”.
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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.
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Aldol condensation may not be favored due to the steric hindrance of the ketone.
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Edited my post after your response, but didnt see it sorry.
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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.
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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.
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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.
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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.
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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
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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.
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Pgk, do you have any comment on my experiment? Its clear that very little KOH is formed under these conditions
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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.
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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.
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Thanks, “wildfyr” for your advice.
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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?
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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.
What about pKa values? The first reaction you've drawn is an acid-base reaction in reverse: the pKa of NaHCO3 is about 10 (according to the Evans table), while the pKb of KOH is 14. So the reaction as you've drawn it will lie far to the left, with an equilibrium constant of 10^-4.
Can you cite any examples of the reactions you propose?
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My experiment shows that this "hydrolysis" clearly do not take place under these conditions.
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1). I agree.
“rolnor” is right and I was wrong. Indeed KOH is soluble in warm ethanol. In this case, let’s assume that 258 mg of solid correspond to a mixture of KHCO3 and unhydrolyzed KHCO3. Thus:
100x + 138y = 258
x + y = 2.05
The above equations system gives that x = 0.66 mmoles KHCO3 and 1.4 mmoles K2CO3. Or, 0.66x56 = 37 mg KOH have been formed and removed by solvent decantation, which corresponds to a hydrolysis yield = 32%.
2). If well understood, anhydrous ethanol plus the stoichiometric amount of water were used. Thus:
2x2.05x18 = 74 mg H2O, which may help but it is unlike that can completely dissolve 283-258 = 25 mg K2CO3 in ethanol.
3). “clarkstill” is right too. This is an equilibrium reaction with an acid-base equilibrium constant = 4.3x10^-7 at room temperature that corresponds to a hydrolysis constant 10^-14/4.3x3x10^-7 = 2.3x10^-8, which means that the equilibrium is 4.3x10^-7/2.3x10^-8 = 5.3 folds favored to the left and which corresponds to a hydrolysis ratio (x + y = 1 and x = 5.3y) = 0.158 or a hydrolysis yield = 15.8% that can easily be doubled by heating, due to the temperature influence in the equilibrium constant.
4). But please, calm down. Everybody (and not chemists only) knows that NaOH is not massively produced during cleaning with aqueous soda ash (Na2CO3) or during cooking with baking soda (NaHCO3) and thus, all the above calculations (if not being incorrect) might be simple coincidences.
5). Regardless if the above calculations are wrong or not, the fact is that the hydrolysis is an equilibrium reaction (as “clarkstill” denotes) and therefore, the hydrolysis ratio of K2CO3 is completely different in presence of a benzyl chloride or an ester that reacts with KOH and continuously removes the equilibrium to the right.
6). In other words, a chemical reaction with a low yield (like this one), may not have a synthetic/preparative interest but as a side reaction, it can be more than enough to significantly decrease the reaction yield or even, it can completely minimize it.
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CORRECTION
6). In other words, a chemical reaction with low yield (like this one), may not have a synthetic/preparative interest but as a side reaction, it can be more than enough to significantly decrease the yield of the ‘main’ reaction or even, it can completely minimize it.