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My textbook states that when reacting an acid chloride (for example, acetyl chloride) with an amine, we need to take two moles of amines because when the $ \ce{-NH-R}$ group displaces the $\ce{-Cl}$ group, the newly formed second degree amide is susceptible to attack by the $\ce{Cl-}$ ion produced and hence the second mole of amine reacts with the $\ce{Cl}$ to pacify it. How is this possible, when $\ce{NH2-}$ is a stronger base than $\ce{Cl-}$?

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  • $\begingroup$ Actually, it's not a textbook. It's my professor's lecture notes from school. $\endgroup$
    – CannedOrgi
    Commented May 30, 2021 at 12:34

2 Answers 2

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I am not sure what the book is talking about, but I am sure that one amine molecule attacking the acid chloride will form a certain byproduct:

$\ce{R^{(1)}-C(O)-Cl + NHR^{(2)}R^{(3)} -> R^{(1)}-C(O)-NR^{(2)}R^{(3)} + \color{blue}{HCl}}$

The HCl acts as a protic acid and can react with another amine molecule. To get a stable mixture of products you need one mole of amine to combine with the carbonyl group and another to neutralize the HCl:

$\ce{R^{(1)}-C(O)-Cl + 2 NHR^{(2)}R^{(3)} -> R^{(1)}-C(O)-NR^{(2)}R^{(3)} + [NH2R^{(2)}R^{(3)}]^+ + Cl^-}$

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  • $\begingroup$ Yeah, that's what the book said. So, if we didn't use another amine molecule, what would happen? Also, when we react esters with amines, why do we not need to use 2 molecules of amine? $\endgroup$
    – CannedOrgi
    Commented May 30, 2021 at 12:32
  • $\begingroup$ If you add $1$ amine molecule for each molecule of $\ce{RCOCl}$, the yield of amide will be only $50$%. $\endgroup$
    – Maurice
    Commented May 30, 2021 at 14:38
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Another way to put this is that the primary product is not the amide itself, but its protonated form. In this, the leaving group is not a charged amide-ion, but a neutral amine, which is "stable" enough to be displaced by a chloride ion (the reverse of the addition reaction). The second molecule of amine is needed to deprotonate the positively charged species into a neutral molecule, in which the leaving group would in fact be the negatively charged, "unstable" amide-ion.

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  • $\begingroup$ Oh yeah, it's not the charged ion. That's where I was going wrong. Thanks! $\endgroup$
    – CannedOrgi
    Commented May 30, 2021 at 17:18

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