This page guides you through the mechanism for the nucleophilic addition / elimination reaction between acyl chlorides (acid chlorides) and amines.

Ethanoyl chloride is taken as a typical acyl chloride. Similarly, ethylamine is taken as a typical amine. Changing either the acyl choride or the amine won't affect the mechanism in any way.

Important!  If you haven't already done so, it would help if you first read the page What is nucleophilic addition / elimination? before you go on.

This mechanism also looks extremely complicated! In fact, it isn't any more difficult than the mechanism for the reaction between acyl chlorides and ammonia. Make sure you understand that mechanism before you go any further with this one.

The reaction between ethanoyl chloride and ethylamine

The products of the reaction

Remember that the reaction produces N-ethylethanamide and ethylammonium chloride.

N-ethylethanamide is an N-substituted amide. Ethanamide (a simple amide) has the formula CH3CONH2. In an N-substituted amide, one of the hydrogens on the nitrogen has been substituted by a hydrocarbon group - which may be either an alkyl group (as here), or a benzene ring.

In the case we are discussing, an ethyl group has replaced one of the hydrogens on the nitrogen - hence N-ethylethanamide.

The other product is ethylammonium chloride. Ammonium chloride is NH4Cl. In ethylammonium chloride, one of the hydrogens on the nitrogen has been substituted by an ethyl group.

Ethylamine as a nucleophile

A nucleophile is a species (an ion or a molecule) which is strongly attracted to a region of positive charge in something else.

Nucleophiles are either fully negative ions, or else have a strongly - charge somewhere on a molecule.

Nitrogen is more electronegative than both hydrogen and carbon and so drags the bonding electrons towards itself. That produces a significant amount of negative charge on the nitrogen atom. The nitrogen also has an active lone pair of electrons. It is this which attacks the ethanoyl chloride.

Why are acyl chlorides attacked by nucleophiles?

The carbon atom in the -COCl group has both an oxygen atom and a chlorine atom attached to it. Both of these are very electronegative. They both pull electrons towards themselves, leaving the carbon atom quite positively charged. It is that carbon atom which is attacked by the lone pair on the nitrogen atom in an ethylamine molecule.

The ethanoyl chloride molecule is also planar (flat) around that carbon atom, and that leaves plenty of room for a nucleophile to attack either from above or below the plane of the molecule.

The mechanism

The reaction happens in two main stages - an addition stage, followed by an elimination stage.

In the addition stage, an ethylamine molecule becomes attached to the carbon in the ethanoyl chloride.

As the lone pair on the nitrogen approaches the fairly positive carbon in the ethanoyl chloride, it moves to form a bond with it. In the process, the two electrons in one of the carbon-oxygen bonds are repelled entirely onto the oxygen, leaving it negatively charged.

Notice that the nitrogen atom has gained a positive charge. The underlying reason for this is that when the lone pair forms a bond with the carbon, electrons are moving away from the nitrogen. What matters, though, is that you remember to show the positive charge in an exam. Think of it like this:

If you leave out the positive charge, two things are wrong with the equation:

  • The charges don't balance. You start with two overall neutral molecules and, if you forgot the positive charge, you would end up with a negative ion. There has got to be a positive charge somewhere to balance the negative one.

  • The nitrogen looks wrong! Nitrogen normally forms three bonds, but here it is forming four. Nitrogen can only form four bonds if it carries a positive charge. (A positively charged nitrogen atom has the same electronic structure as carbon - which normally forms four bonds.)

You can put both things right with a positive charge on the nitrogen.

The elimination stage stage of the reaction happens in two steps. In the first, the carbon-oxygen double bond reforms. As the electron pair moves back to form a bond with the carbon, the pair of electrons in the carbon-chlorine bond are forced entirely onto the chlorine to give a chloride ion.

Note:  There's nothing sophisticated going on in the switch of the ethyl group from left to right in the diagram. It just makes it look better in the final stages!

Finally, one of the hydrogens attached to the nitrogen is removed as a hydrogen ion. The electrons in the hydrogen-nitrogen bond move back onto the nitrogen, cancelling the positive charge.

The hydrogen ion might be removed in one of two ways. The first way is entirely consistent with what happens in the reactions between water or ethanol and acyl chlorides. The hydrogen is removed by the chloride ion.

The hydrogen chloride produced would at once react with any excess ethylamine present to form ethylammonium chloride.

The other (more likely) possibility is that the hydrogen ion gets removed directly by an ethylamine molecule. This forms an ethylammonium ion. The ethylammonium ion together with the chloride ion formed in the previous stage makes up the ethylammonium chloride produced in the reaction.

Note:  You probably don't really need to know both of these routes for the removal of the hydrogen ion. Choose for yourself. It doesn't matter very much, because both will happen.

The route involving the formation of hydrogen chloride is exactly in line with the water or alcohol mechanisms, but you do need to remember to react the HCl with excess ethylamine. You may well feel that this is less daunting than the other route.

The removal by another ethylamine molecule is probably the major route as long as the ethylamine is in excess - which it almost certainly will be. The reaction is normally done in the lab by dropping the ethanoyl chloride into concentrated ethylamine solution.

Don't worry too much about this. Most sources opt out of the problem altogether by quoting the final step simply as loss of hydrogen ion: " - H+ ", without making any attempt to explain what removes it.

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© Jim Clark 2000