4
$\begingroup$

As part of my A-level chemistry course, I have been taught that an acyl chloride will react violently with water to produce a carboxylic acid and hydrochloric acid. I have also been taught that an acyl chloride will react with an aqueous solution of ammonia to produce an amide and ammonium chloride. My question is, would the acyl chloride also react with the water in the aqueous ammonia solution? If so, this could decrease the yield of the amide, because some of the acyl chloride will react to produce the carboxylic acid.

I would appreciate any additional relevant details about the reaction, such as answers to the following questions (assuming that the acyl chloride does react with the water):

  • Is there an easy method to separate the amide from the carboxylic acid?
  • Would it be best to use a concentrated ammonia solution, so that there is plenty of ammonia for the acyl chloride to react with instead of reacting with the water?
  • Is the reaction with water negligible compared to the reaction with ammonia?

Thanks in advance!

$\endgroup$
4
$\begingroup$

Acyl halides react with water but usually not violently. The acyl halides are normally not soluble in water so the reaction is slow. Then, when the reaction proceeds, acid is formed and the pH decreases, reducing the reactivity of water.

Ammonia and amines are much more nucleophilic and react much faster with the acyl halide. This reaction is so much faster that it can be carried out in water. In this context it is called the Schotten–Baumann reaction, which normally proceeds with high yields.

An excess of ammonia is required at least to absorb the HCl formed, but normally a huge excess is used. The acyl halide is more expensive than ammonia so we want to have it completely transformed to the amide. The excess of ammonia will also help to dissolve the acid, if it is formed in the reaction.

The other points:

  • To separate acids and amides, extract with alkali. The acid will go to the aqueous phase and the amide to the organic phase. Or, if they are small, recrystallize it. The amides are less soluble than acids.

  • Huge excess of ammonia or simple amines is normally used.

  • The reaction of water is negligible under these conditions.

$\endgroup$
  • $\begingroup$ Thank you for your helpful answer, it all seems to make sense if the statement that "Acyl halides react with water but usually not violently. The acyl halides are normally not soluble in water so the reaction is slow." is correct, however I have seen several sources which appear to offer contradictory information e.g. "Ethanoyl chloride reacts instantly with cold water. There is a very exothermic reaction in which a steamy acidic gas is given off (hydrogen chloride) and ethanoic acid is formed." - chemguide.co.uk/mechanisms/addelim/water.html $\endgroup$ – Rational Function May 24 '18 at 11:34
  • 1
    $\begingroup$ Yes, but only because it is a very small acyl halide. When the size increases the solubility decreases and the reaction is slow. Then, once the reaction sets on, heat is generated, reaction rate increases, the solubility increases and you end up with a violent reaction. Therefore, if you have to react an acyl halide with water, you must use hot water and add the acyl halide slowly to it. $\endgroup$ – Raoul Kessels May 24 '18 at 11:39
  • 2
    $\begingroup$ Most acyl halides react slowly with cold water, acetyl chloride is one of the exceptions. Acyl halides derived from aromatic acids e.g. benzoyl chloride are really quite unreactive. Ammonia can be bought in solution in non-nucleophilic solvents (e.g. THF) to ensure no carboxylic acid is formed in the amide forming reaction. $\endgroup$ – Waylander May 24 '18 at 11:48
0
$\begingroup$

We could, of course, still get the carboxylic acid. Given the amide formed from ammonia, hydrolysis with hot sulfuric acid solution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.