By no means does Ag2O *have* to be used. It was used by Hoffmann, perhaps because the elimination was performed simply by adding base after alkylation and heating until the elimination product distilled off. Perhaps Ag+ helps to further destabilize the ammonium ion by making insoluble AgI, which speeds elimination, but I don’t believe there is anything magical about Ag2O.

Now, my question arises “What if we add a -M, -I functional group to the adjacent carbon. Will the carbonion be stabilised and form the Zaitsev product even when the leaving group is NR3 ?”

Hi Johann – that is a perfectly good explanation, and possibly a more rigorous one. The positively charged nitrogen is a strong electron withdrawing group, which increases acidity of the beta hydrogens. However the basicity of the tertiary hydrogen will be less, since it is made more electron-rich by its adjacent alkyl groups. The secondary will be more acidic. Also, there is a statistical contribution since there are two secondary hydrogens and only one tertiary hydrogen.

My professor told us that in the last example, the Hofmann product would be major because the trimethylamine is positively charged, and because the tertiary CH is less “acidic” than the secondary CH (which means that the sec. gives its H+ more easily), we would have the Hofmann product (because in thisn case the positive charge would make this difference stronger).

Would you say that it is also a valid explanation? (He didn’t mention the steric reason of the leaving group though.)

Thanks in advance!

POCl3 gives a phosphate leaving group. Like the trialkylammonium leaving group, It is not unknown that the phosphate leaving group can have unfavorable interactions with the rest of the carbon chain leading to “anti Zaitsev” products instead of the normal “Zaitsev” product. One other example where it has been observed is in eliminations adjacent to the steroid nucleus, as shown by Djerassi and co-workers. See Giner, J.-L. et. al. J Org Chem. 1989 54 369. DOI . https://pubs.acs.org/doi/pdf/10.1021/jo00263a020 .
“Dehydration of 22-
(S)-alcohol 7 gave the 20(22) E olefin 8. However, dehydration of 22(f?)-alcohol 9 did not produce the 20(22) Z
olefin, but rather stigmasterol or brassicasterol i-methyl
ether (6) together with a product thought, based on the
mass spectrum, to be the 22-chloride. This result can be
rationalized on the basis of unfavorable steric interactions
in the transition state leading to the 20*2® product (Figure
3)”.
Figure 3 shows a Newman projection rationalizing the product.

https://drive.google.com/open?id=1v-yTJP15O7qASQfIIYIEEUSKZo8hGirB

It’s from a reliable source (chem.libretexts) but I can’t bring myself to accept the outcome. Any hints towards the probable mechanism please?
Thanks in advance.

Gary – it is an honor to have you as a reader. Thank you for stopping by.