4
$\begingroup$

I was looking at the following reaction and am confused.

enter image description here

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?

$\endgroup$
  • 5
    $\begingroup$ 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 Apr 12 at 17:59
  • 1
    $\begingroup$ 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 Apr 12 at 18:44
  • $\begingroup$ @Waylander The cyclisation where OH attacks? Why is that? $\endgroup$ – Mirte Apr 13 at 13:59
  • 1
    $\begingroup$ OH attack would form a 9-membered ring, strongly disfavoured vs 6-membered. $\endgroup$ – Waylander Apr 13 at 17:47
  • 1
    $\begingroup$ @Waylander Your comments would make a nice answer, care to write it up? $\endgroup$ – Karsten Theis May 13 at 17:19
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. Apr 13 at 15:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.