Carboxylic Acids and Derivatives: Question 8

Worked solution

Part (a): Preparing hexanoyl chloride with PCl5

Carboxylic acids are converted into the much more reactive acyl chlorides using phosphorus(V) chloride, phosphorus(III) chloride or sulfur dichloride oxide. With solid $\text{PCl}_5$, added directly to the carboxylic acid at room temperature:

$\text{CH}_3(\text{CH}_2)_4\text{COOH} + \text{PCl}_5 \rightarrow \text{CH}_3(\text{CH}_2)_4\text{COCl} + \text{POCl}_3 + \text{HCl}$

Observation: the mixture reacts vigorously even without heating, giving off steamy white fumes of hydrogen chloride gas. Other two products: phosphorus oxychloride, $\text{POCl}_3$ (a liquid, which must be separated from the acyl chloride, e.g. by distillation), and hydrogen chloride gas, $\text{HCl}$, which escapes from the mixture.

Part (b): Reaction with excess concentrated aqueous ammonia

Hexanoyl chloride reacts rapidly with ammonia by nucleophilic addition-elimination, in the same general way as with water, an alcohol or an amine, but here the nucleophile is unsubstituted ammonia, so the product is a simple (primary) amide rather than an N-substituted one:

$\text{CH}_3(\text{CH}_2)_4\text{COCl} + 2\text{NH}_3 \rightarrow \text{CH}_3(\text{CH}_2)_4\text{CONH}_2 + \text{NH}_4\text{Cl}$

Organic product: hexanamide. Why excess ammonia is needed: the first mole of ammonia attacks the acyl chloride to give hexanamide and HCl. If no further ammonia were present, this HCl would react with (protonate) the basic ammonia or even the amide product, lowering the yield of hexanamide. Using an excess means the second mole of ammonia instead reacts with the HCl by-product, forming the solid salt ammonium chloride, $\text{NH}_4\text{Cl}$, and leaving the hexanamide product unreacted.

Part (c): The addition-elimination mechanism

The reaction proceeds in two distinct mechanistic steps:

1. Nucleophilic addition. The lone pair of electrons on the nitrogen atom of an ammonia molecule attacks the carbonyl carbon of hexanoyl chloride. This carbon is strongly electrophilic because it is flanked by two highly electronegative atoms, oxygen (double-bonded) and chlorine (single-bonded), both of which withdraw electron density from it by induction. Attack by the nitrogen lone pair breaks the C=O pi bond, pushing its electrons onto the oxygen and forming a negatively charged tetrahedral intermediate, in which the former carbonyl carbon is now bonded to four groups: the alkyl chain, the oxygen (now $\text{O}^-$), the chlorine, and the incoming nitrogen.

2. Elimination. This tetrahedral intermediate is unstable. The C-Cl bond breaks heterolytically, with both electrons leaving on the chlorine to give a chloride ion, while the C=O double bond re-forms. A proton is then lost from the positively charged nitrogen (readily removed by a second ammonia molecule acting as a base), leaving the neutral amide, and the chloride ion combines with a proton to give hydrogen chloride.

Chlorine is an excellent leaving group (a weak conjugate base of the strong acid HCl), which is why this second step occurs so readily, and it is this combination of a strongly electrophilic carbonyl carbon and a good leaving group that makes acyl chlorides react by addition-elimination far more readily, and without any catalyst, than the parent carboxylic acid could.

Final answers

• (a) $\text{CH}_3(\text{CH}_2)_4\text{COOH} + \text{PCl}_5 \rightarrow \text{CH}_3(\text{CH}_2)_4\text{COCl} + \text{POCl}_3 + \text{HCl}$; steamy white fumes of HCl observed; the other two products are phosphorus oxychloride and hydrogen chloride.
• (b) $\text{CH}_3(\text{CH}_2)_4\text{COCl} + 2\text{NH}_3 \rightarrow \text{CH}_3(\text{CH}_2)_4\text{CONH}_2 + \text{NH}_4\text{Cl}$; hexanamide is formed; excess ammonia is needed so the second mole can react with the HCl by-product, forming ammonium chloride.
• (c) Nucleophilic addition of the ammonia lone pair to the electrophilic carbonyl carbon gives a tetrahedral intermediate, followed by elimination of the chloride ion to reform the C=O bond; the carbonyl carbon is especially electrophilic due to the electron-withdrawing oxygen and chlorine, and chloride is a good leaving group.