Feasibility of Nucleophilic Substitutions
Short answer - No. See below for a workaround. 


Consider your acids and bases:


  
  Substituting an alcohol with ammonia directly
  


$$\ce{NH3 + CH3CH2OH &lt;=&gt; CH3CH2NH2+ + OH-}$$

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. 


  
  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.

$$\ce{NH4+ +CH3CH2OH &lt;=&gt; NH3 + CH2CH3OH2+}$$

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 $10^{-11}$.


  Workaround


If you are needing to do the transformation

$$\ce{ROH -&gt; RNH2}$$

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

$$\ce{CH3CH2OH -&gt;[\ce{TsCl,base}] CH3CH2OTs }$$
$$\ce{CH3CH2OTs -&gt;[\ce{NH3}] CH3CH2NH2}$$