N - alkylation of piperidine with substituted benzyl chloride

No, it is not clear, could it not be a mixture of Sn1 and Sn2 i both cases?

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.

I do not understand at all, how much hydroxide can be formed, 1%? Will this greatly influence the yield?

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

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 !

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.

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.

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.

Quote from:

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?

Thank you.  I might imagine that the difference in solvent (ethanol versus DCM) might influence the mechanism also.

You could get better yield with the DCM-method for the other examples also.

Piperidine is a much better nucleophile than TEA. Cyclic amines have some unique properties vs the straight chain ones.

The reason they use DIPEA is probably to avoid forming Q-salt.

I dont understand this, that the hydrophilicity of the p-MeO-group should favor the reaction with hydroxide? It sounds very strange.