I was searching for reaction of $\ce{PCl5}$ with carboxylic acids and found this mechanism: carboxylic acids to acyl chloride by PCl5

Here a corresponding acyl chloride is formed from a carboxylic acid.

Now I doubt if this is the final product as this acyl chloride's carbonyl group should also react further with $\ce{PCl5}$ just like it happens in Aldehydes and ketones.

So I searched for reaction of acyl halides with $\ce{PCl5}$ and I can't find any results showing the mechanism or even discussing what can be the product. I believe that further reaction should happen on the carbonyl group of the acyl halide forming gem-dichloride replacing the double bonded Oxygen. This direct replacement can't occur on the carbonyl group of the carboxylic acid as it's -OH group also have involvement (working as acid and releases H+ which changes the dynamics of the reaction) in the reaction while halide group of acid halides don't interfere that much and so acyl halides should react like aldehydes or ketones with $\ce{PCl5}$ as discussed in this answer by following mechanism: Ketone to gem-dichloride by PCl5

So With all things looking perfectly fine then why don't this reaction occurs? What are those factors preventing this reaction?

  • $\begingroup$ Actually, certain $\ce{MX_n}$ molecules can be strong halide abstractors, such that they can pull off the halide in $\ce{RCOX}$ to form an acylium cation. I don't think $\ce{PCl_5}$ is strong enough, but things like $\ce{SbF_5}$ and apparently $\ce{SbCl_5}$ are. $\endgroup$ Commented Sep 17, 2023 at 11:53
  • 2
    $\begingroup$ The carbonyl group of an acyl chloride will be less nucleophilic than a ketone because of the electron-withdrawing effect of the Cl making your first step of the second scheme less likely to occur $\endgroup$
    – Waylander
    Commented Sep 17, 2023 at 11:59
  • $\begingroup$ @Waylander if the carbonyl group can be enough nucleophilic for reaction with PCl5 in Carboxylic acids (EN Oxygen = 3.44) then it should also be in acid chloride (EN Chlorine = 3.16) $\endgroup$
    – D13G
    Commented Sep 18, 2023 at 7:46
  • $\begingroup$ @NicolauSakerNeto this looks like a fact rather than a comment for this question. In my understanding* $\endgroup$
    – D13G
    Commented Sep 18, 2023 at 7:51

2 Answers 2


The first step of the reaction with carboxylic acids shows the answer. Nucleophilic attack on the phosphorus is enabled by electrons from the second oxygen donating in to the central carbon.

Those electrons come from the free methyl group in your second mechanism via a keto/enol tautomerization, but they aren't present in most acyl chlorides. When they are present the reaction proceeds; eg. the reaction of niacin with phosphorus pentachloride.


One possibility is that the reaction mechanism springs a leak when you have a halogen attached to the carbonyl group.

When the chloride ion from the $\ce{O–\overset{-}{P}Cl5}$ function attacks the carbonyl carbon, either $\ce{POCl4-}$ or $\ce{POCl3}$ must act as a leaving group (depending on when the second chloride ion initiates its attack). If there is a better leaving group, the nucleophilic substitution might go the wrong way preventing the intended breakage of the carbon-oxygen bond.

An aldehyde or ketone presents no such problem because a hydride or alkide(?) ion is not going to get displaced in a nucleophilic reaction. But a halide ion could be displaced from an acyl halide instead of breaking the carbon-oxygen bond.

  • $\begingroup$ What I conclude from your answer is that $\ce{-Cl}$ is better leaving group than $\ce{-O-\overset{+}{P}Cl3}$, but then shouldn't $\ce{Cl-}$ not be able to push out $\ce{POCl3}$ in the second scheme? If it was relatively a good leaving group then it should be removed as soon as it attacks. $\endgroup$
    – D13G
    Commented Sep 18, 2023 at 7:36

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