Reaction of alcohols with PCl5 and PCl3

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.

You can see in the step, there is an electronegative element by the side of a slight positive hydrogen. This hydrogen inhibits intramolecular hydrogen bonding. So, it can be said that this hydrogen is not available for reaction and hence, the negatively charged chlorine attacks the carbon.

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.

You can see in the step, there is an electronegative element by the side of a slight positive hydrogen. This hydrogen inhibits intramolecular hydrogen bonding. So, it can be said that this hydrogen is not available for reaction and hence, the negatively charged chlorine attacks the carbon.