[Fireredburn1]

Hi guys

In the synthesis of Sildenafil (Viagra), one of the steps was to let [46] react with t-BuOH, where there will be a cyclization to give a ring product. I'm not exactly sure how this reaction work. How does the base make the two amide moiety react and cyclize? I do not have any knowledge that amides can actually react together to form an imine group.

Would greatly appreciate if someone could tell me how t-BuOH works in the reaction and the mechanism between the amide groups. Thank you!

[spirochete]

The tert butoxide is deprotonating one of the acidic hydrogens on the amide, which makes it more nucleophilic. It will then attack the carbonyl group. I think if you start the mechanism you will see that it is not that complicated. Just lots of proton transfers and whatnot.

[Fireredburn1]

The tert butoxide is deprotonating one of the acidic hydrogens on the amide, which makes it more nucleophilic. It will then attack the carbonyl group. I think if you start the mechanism you will see that it is not that complicated. Just lots of proton transfers and whatnot.

Why is the hydrogen atom on the amide acidic? Does it have something to do with resonance? I've always thought that amides were neutral?

[kriggy]

Yes, the carbonyl group is electron withdrawing so it makes N-H bond weaker. They are not acidic that much. pKa for formamide is 23,5 so they act more like bases than acids. However in presence of stronger base they can act as acids and deprotonate