I was looking at the following reaction and am confused.

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It appears that $\ce{N}$ is attacking the $\ce{C-Cl}$ and $\ce{Cl}$ is leaving in a $\mathrm{S_N2}$ reaction. I know that $\ce{Cl}$ is a good leaving group, but $\ce{N}$ isn't a particularly soft nucleophile so I wouldn't expect it to be great at attacking a saturated carbon.

Additionaly there is $\ce{MeO-}$ in solution, even with intramolecular cyclisation being quick wouldn't there be a realitive amount of attack by $\ce{MeO-}$.

So am I missing something? Is this not as simple an $\mathrm{S_N2}$ as I have made out? Why is sodium methoxide ($\ce{NaOMe}$) a good choice of base?

  • 7
    $\begingroup$ This is not an amine, it is an amide anion. Methoxide is a strong enough base to deprotonate an amide, this will happen preferentially, and an intramolecular cyclisation is strongly favoured over intermolecular. Sodium hydride would be a better choice of base $\endgroup$
    – Waylander
    Commented Apr 12, 2019 at 17:59
  • 2
    $\begingroup$ I suspect the reason NaOMe was chosen is because of the free -OH elsewhere in the molecule. There will be some deprotonation of this, but the cyclisation is strongly disfavoured. $\endgroup$
    – Waylander
    Commented Apr 12, 2019 at 18:44
  • 1
    $\begingroup$ OH attack would form a 9-membered ring, strongly disfavoured vs 6-membered. $\endgroup$
    – Waylander
    Commented Apr 13, 2019 at 17:47
  • 1
    $\begingroup$ The other issue is whether the nucleophile is the N or the O of the amide anion. $\endgroup$
    – user55119
    Commented Apr 13, 2019 at 23:08
  • 4
    $\begingroup$ @Waylander Your comments would make a nice answer, care to write it up? $\endgroup$
    – Karsten
    Commented May 13, 2019 at 17:19

2 Answers 2


Well it is difficult to compare an intra-molecular reaction with an extra-molecular one because it really depends on the conditions. However, unless you have a very high concentration of one reactant, the intramolecular way will always be favored.

In your question, you consider that "N" is not a very good "attacking group". Well this is wrong. "N" is always happy to attack any nucleophilic group, even if it becomes "N+". However, a "CON-" is much more nucleophilic than just "CONH".

Sodium methoxide here just serves at deprotonating CONH to activate it, but MeONa (sodium methoxide) is not nucleophilic enough to interfere.

  • $\begingroup$ By "attack any nucleophilic group" do you mean "be a nucleophilic attacking group" or "attack an electrophilic group" or something? In this reaction the amide anion is the nucleophile. $\endgroup$
    – Curt F.
    Commented Apr 13, 2019 at 15:42

In accordance with what @SteffX said, it really depends on the conditions and every reaction has its own story. However a couple of factors favour the formation of the given product above others. There include the fact the MeOH might help to generate the "activated Nucleophile" anion at amide, being a polar protic solvent, it provides higher solvation to the other groups like OH, permitting the amide to take the reaction head on. Also, the greater favourability of a six membered intramolecular attack is a contributing factor. Amide anions are an unconventional nucleophile but in absence of stronger competitors, it can do the job. Hope that helps!!


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