Nucleophilic addition–elimination — acyl chlorides

Step 1 of 3

1. 1.A lone pair on the oxygen of water attacks the δ+ carbonyl carbon.
2. 2.The C=O π bond breaks; both electrons move onto oxygen, giving a tetrahedral intermediate with O⁻.

The nucleophile adds to the carbonyl carbon and the π electrons move to O — a negatively charged tetrahedral intermediate.

Step 2 of 3

1. 1.A lone pair on O⁻ moves back down to reform the C=O double bond.
2. 2.As C=O reforms, the C–Cl bond breaks heterolytically and chloride leaves.

C=O reforms and Cl⁻ is eliminated, leaving a positively charged (protonated) intermediate.

Step 3 of 3

1. 1.A lone pair on the chloride ion removes the proton, forming HCl.
2. 2.The O–H bonding electrons return to the oxygen, giving the neutral carboxylic acid.

The proton is removed: ethanoic acid.