# How do I calculate the yield when an amine reacts with an acid halide?

Apr 20, 2017

You could calculate this straightforwardly; of course there is a catch.....

#### Explanation:

Acid halides react with added bases such as amines........

$R C \left(= O\right) C l + 2 R N {H}_{2} \rightarrow R C \left(= O\right) N H R + \left\{R N {H}_{2} \cdot H C l\right\}$

Normally we add TWO equiv of base; one to react at the carbonyl centre, and t'other to remove the hydrogen halide as the ammonium salt. Are you with me?

If the amine were really expensive, say something you had to synthesize over a few steps, or it was an optically active amine, you could use ONE equiv of so-called non-nucleophilic base, e.g. $E {t}_{3} N$. This tertiary amine is a non-nucleophilic base. It is still basic, and will exchange with the hydrogen on the nitrogen centre, however, if we used one equiv of this stuff we would need only the ONE equiv of $R N {H}_{2}$:

$R C \left(= O\right) C l + R ' N {H}_{2} + E {t}_{3} N \rightarrow R C \left(= O\right) N H R ' + {\left\{E {t}_{3} N H\right\}}^{+} C {l}^{-}$

Given stoichiometric amine (i.e. 2 equivs), the yield is:

"%Yield"=(1/2*"Moles of amine")/"Moles of amide"xx100%.

I think it is fairly clear in your scenario that you ARE adding 2 equiv of aniline, so you adjust your yield calculation accordingly.