Can $\ce{NH3}$ substitute an alcohol? The OH group will depart as a weaker base - some O is more electronegative - so will be a better leaving group than amine. And what about using ammonium ions - such as in an ammonium salt? They will donate an $\ce{H+}$ to protonate the alcohol and make it a good leaving group, then show substitution followed by deprotonation to form an amine? How is the the feasibility decided in case of nucleophilic substitution reactions? Is there any rule of thumb to know what happens, and what doesn't?


Short answer - No. See below for a workaround.

Consider your acids and bases:

  1. Substituting an alcohol with ammonia directly

$$\ce{NH3 + CH3CH2OH <=> CH3CH2NH2+ + OH-}$$

If this were an acid-base reaction (and it is, at least in the Lewis sense), then which side of this reaction would be favored by equilibrium? Which side has the stronger acid-base pair? Hydroxide is a stronger base than ammonia, so this equilibrium would favor the reactants.

  1. Using an ammonnium salt

The first step of this mechanism would be a proton transfer to form an activated leaving group and the $\ce{NH3}$ nucleophile.

$$\ce{NH4+ +CH3CH2OH <=> NH3 + CH2CH3OH2+}$$

Again, which side of this equilbrium is favored? Ammonia is definitely the stronger base compared with the alcohol. Estimates of the pKa of $\ce{ROH2+}$ are in the vicinity of $-2$, while the pKa of $\ce{NH4+}$ is $9.25$. The equilbrium constant of this reaction is on the order of $10^{-11}$.


If you are needing to do the transformation

$$\ce{ROH -> RNH2}$$

there is a workaround in converting the alcohol into a molecule like the tosylate ester, which can then react with ammonia:

$$\ce{CH3CH2OH ->[\ce{TsCl,base}] CH3CH2OTs }$$ $$\ce{CH3CH2OTs ->[\ce{NH3}] CH3CH2NH2}$$


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