organic chemistry - Reaction of alcohols with PCl5 and PCl3

The reaction of alcohols $\ce{ ROH }$ with $\ce{ PCl5 }$ and $\ce{ PCl3 }$ yields an alkyl halide $\ce{RCl}.$ With $\ce{ PCl5 }$, the reaction is quite simple leading to the formation of $\ce{ RCl }$ and $\ce{ POCl3 }.$

But with $\ce{ PCl3 }$ a problem arises. Since $\ce{ PCl3 }$ has both a lone pair and vacant $\ce{3d}$ orbitals it can act both as a Lewis base and a Lewis acid.

In the first figure, $\ce{ PCl3 }$ accepts a lone pair showing its acidic character and expelling $\ce{Cl-}$ out. Now this is where I am getting confused. On one hand we have a Lewis base $\ce{ PCl3 }$ and on the other we have the "expelled" $\ce{Cl-}$? Now in this case why does $\ce{PCl3}$ take away the proton and not $\ce{Cl-}$?

And in the next step, why does $\ce{Cl-}$ attacks the carbon not the $\ce{H+}$?

Answer

Remember your general chemistry. In an acid-base reaction, the equilibrium favors the side of the reaction with the weaker acid/base pair. Strong bases have weak conjugate acids. Strong acids have weak conjugate bases. Weak bases have strong conjugate acids. Weak acids have strong conjugate bases. Consider the following:

$$\ce{HA + B- <=> A- + HB}$$

If $\ce{HA}$ is a stronger acid than $\ce{HB}$, then $\ce{A-}$ must be a weaker base than $\ce{B-}$.

Here is an example with real compounds:

$$\ce{HCl + NH3 <=> Cl- + NH4+}$$

Even though the right side of the reaction contains ions instead of neutral molecules, the right side is favored because $\ce{HCl}$ is a strong acid and $\ce{NH4+}$ is a weak acid.

When we study organic chemistry, we are given some rules of thumb to help us compare relative acidity:

• $\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y}$ is more electronegative than $\ce{Z}$ and they are in the same period.


• $\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y}$ is larger than $\ce{Z}$ and they are in the same group.


• $\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more resonance stabilization than $\ce{Z-}$ and if $\ce{Y-}$ and $\ce{Z-}$ are otherwise similar.


• $\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more inductive stabilization than $\ce{Z-}$ and if $\ce{Y-}$ and $\ce{Z-}$ are otherwise similar.


• $\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more $\ce{s}$-character than $\ce{Z-}$ and $\ce{Y-}$ and if $\ce{Z-}$ are otherwise similar.


• $\ce{YH2+}$ is always more acidic than $\ce{YH}$, though it is hard to compare $\ce{YH2+}$ and $\ce{YZ}$ or $\ce{YH}$ and $\ce{YZ2+}$.

With these rules in hand, it is sometimes challenging to remember that we also have an experimental measure of acid strength, and that there are only a limited number of "strong" acids (those acids which are stronger than $\ce{HSolvent+}$).

The $K_\mathrm{a}$ of $\ce{HCl}$ is not easily determinable, since it is more acidic than most protonated solvents. However, the Evans pKa table and other sources often estimate it at -7 or -8, with only $\ce{HBr}$, $\ce{HI}$, and the various "superacids" including such things as $\ce{HSbF6}$ being stronger.

The Evans table lists the $\mathrm{p}K_\mathrm{a}$ of various protonated phosphines. For example

• $\ce{CH3PH3+}$ is 2.7 in DMSO ($\ce{HCl}$ is 1.8 in DMSO).


• $\ce{Et3PH+}$ is 9.1 in DMSO

Would we expect $\ce{ROPCl2H+}$, which is the intermediate in your reaction, to be more or less acidic than the two reference phosphonium ions above? Both $\ce{RO}$ amd $\ce{Cl}$ are electron withdrawing by induction.

$\ce{Cl-}$ maybe could attack $\ce{H}$ and not $\ce{C}$, but the $\ce{HCl}$ that forms would be so acidic in comparison to everything else in the reaction that something else would take that proton away again, regenerating $\ce{Cl-}$. Once $\ce{RCl}$ forms, there is no other good nucleophile present that is capable of displacing the chloride group.