# Can the nitrogen of an amide displace a primary chloride in a S_N_2 reaction in the presence of a strong base?

I was looking at the following reaction and am confused.

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?

• 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 – Waylander Apr 12 at 17:59
• 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. – Waylander Apr 12 at 18:44
• @Waylander The cyclisation where OH attacks? Why is that? – Mirte Apr 13 at 13:59
• OH attack would form a 9-membered ring, strongly disfavoured vs 6-membered. – Waylander Apr 13 at 17:47
• The other issue is whether the nucleophile is the N or the O of the amide anion. – user55119 Apr 13 at 23:08