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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 $\ce{Cl-}$?

enter image description here

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

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    $\begingroup$ What do you think is the stronger base, PCl3 or chlorid? $\endgroup$ – Martin - マーチン Jul 7 '15 at 10:41
  • $\begingroup$ It should be chloride ion as its having 4 lone pairs and importantly acting as a base, chloride ion becomes neutral while Phosphorous becomes electron deficient. $\endgroup$ – yasir Jul 7 '15 at 11:28
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    $\begingroup$ I think you are a little off on that thought. Let me rephrase the question: What do you think is the weaker acid HPCl3 or HCl? $\endgroup$ – Martin - マーチン Jul 7 '15 at 11:51
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    $\begingroup$ i don't think that HPCl3 even exists!!if you are asking hypothetically then, if we knock off a proton from both the acids, the anion $\ce{PCl3-}$ is less stable than $\ce{Cl-}$ because of the electro- negativity and size difference.so then $\ce{HPCl3}$ would be a weaker acid than $\ce{HCl}.$ $\endgroup$ – yasir Jul 7 '15 at 12:55
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    $\begingroup$ Oh sorry my mistake, I of course meant $\ce{H+PCl3}$. But to cut things a little shorter here. HCl is a strong acid, and chloride is a terrible base. This molecule will be (at least in most protic solvents) dissociated. On the other hand a hydrogen phosphorous bond has quite some strength, and it is much more covalent and it is a much stronger base. But the real driving force of the whole reaction is the oxophilie of phosphorous. $\endgroup$ – Martin - マーチン Jul 7 '15 at 13:00
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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.

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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.

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