Nucleophilic addition–elimination — acyl chlorides
Carbonyl & aromatic chemistry · ethanoyl chloride + water / alcohol / ammonia / amine · room temperature
Watch every step, with the electron flow and reasoning shown.
Reagents: ethanoyl chloride + water / alcohol / ammonia / amine · Conditions: room temperature
Step 1 of 3
- 1.A lone pair on the oxygen of water attacks the δ+ carbonyl carbon.
- 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.A lone pair on O⁻ moves back down to reform the C=O double bond.
- 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.A second molecule of water acts as a base and uses its lone pair to remove the proton.
- 2.The O–H bonding electrons return to the oxygen, giving the neutral carboxylic acid.
The proton is removed: ethanoic acid.
Product: ethanoic acid
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