Addition-Elimination Reactions with Acyl Chlorides and Anhydrides Revision Notes | A Level AQA Chemistry

1. Why Acyl Chlorides and Anhydrides React So Readily

Picture the carbonyl group in an acyl chloride: the carbon is bonded to an electronegative oxygen and an electronegative chlorine. Both atoms pull electron density away from the carbon, leaving it with a significant partial positive charge — $\delta^+$. That carbon is an irresistible target for any electron-rich species.

Ammonia ($\text{NH}_3$) and primary amines ($\text{RNH}_2$) both carry a nitrogen atom with a lone pair of electrons. That lone pair is the nucleophile — it attacks the electrophilic carbonyl carbon and sets off the addition-elimination sequence. The result is an amide, one of the most important functional groups in biochemistry and pharmaceutical chemistry.

Before working through the mechanism, it helps to see the overall transformation. When ethanoyl chloride reacts with ammonia, the product is ethanamide and hydrogen chloride is released as a by-product. When ethanoic anhydride reacts with ammonia, the product is again ethanamide, but the by-product is ethanoic acid rather than HCl. That difference in by-product is the key to understanding why the two reagents behave differently in practice.

2. Mechanism: Acyl Chlorides Reacting with Ammonia

The mechanism for acyl chlorides is the clearest starting point because the leaving group — chloride — is an excellent one. Work through the two stages carefully: addition first, then elimination.

Notice that two molecules of ammonia are consumed per molecule of acyl chloride: one acts as the nucleophile, and one acts as a base to mop up the HCl produced. This is why excess ammonia is used in practice. The overall equation is:

Worked example: Drawing the Mechanism: Acyl Chloride + Ammonia

Outline the mechanism for the reaction of ethanoyl chloride with ammonia. State the type of mechanism and identify the nucleophile, electrophile, and leaving group.

1. Nucleophile: N lone pair in NH3. Electrophile: carbonyl C in CH3COCl ($\delta^+$). Leaving group: Cl$^-$. — Always identify these three before drawing any arrows — it tells you where every arrow starts and ends.
2. Full curly arrow from N lone pair to carbonyl C. Second full curly arrow from C=O $\pi$ bond to O. This gives a tetrahedral intermediate: $\text{CH}_3\text{C}(\text{O}^-)(\text{NH}_3^+)(\text{Cl})$. — The C=O pi bond breaks so carbon can accommodate the new N-C bond without exceeding four bonds. Oxygen takes the electrons as a lone pair, gaining a negative charge.
3. Full curly arrow from C-Cl bond to Cl. Full curly arrow from O$^-$ lone pair to reform C=O. Cl$^-$ departs. Product at this stage: $\text{CH}_3\text{CONH}_3^+$ (protonated amide). — Chloride is a good leaving group because Cl$^-$ is stable. The C=O bond reforms, driving the elimination forward.
4. A second NH$_3$ molecule accepts H$^+$ from the N of the intermediate. Products: $\text{CH}_3\text{CONH}_2$ (ethanamide) + $\text{NH}_4^+\text{Cl}^-$ (ammonium chloride). — The second NH3 acts as a Brønsted-Lowry base. This step is why the reaction requires excess ammonia.
5. Nucleophilic addition-elimination. — Addition (N attacks C) then elimination (Cl$^-$ leaves and C=O reforms). Both stages are required for full marks in an exam question asking you to 'outline the mechanism'.

Answer: Mechanism type: nucleophilic addition-elimination. Product: ethanamide (CH3CONH2). By-product: ammonium chloride (NH4Cl). Two NH3 molecules are consumed per CH3COCl.

3. Mechanism: Acyl Chlorides Reacting with Primary Amines

The mechanism with a primary amine ($\text{RNH}_2$) is identical in type to the ammonia reaction — nucleophilic addition-elimination — but the nucleophile is now the nitrogen of the amine rather than ammonia itself. The product is an N-substituted amide (often called a secondary amide).

Take the reaction of ethanoyl chloride with methylamine ($\text{CH}_3\text{NH}_2$). The lone pair on nitrogen attacks the carbonyl carbon, the tetrahedral intermediate forms, chloride departs, and the C=O reforms. A second molecule of methylamine then acts as the base, accepting the proton from the nitrogen of the intermediate. The products are N-methylethanamide and methylammonium chloride.

The pattern is consistent: the amine nitrogen always attacks first, chloride always leaves, and a second amine molecule always mops up the proton. Recognising this pattern means you can write the mechanism for any primary amine reacting with any acyl chloride — the R groups change, but the electron flow does not.

4. Mechanism: Acid Anhydrides Reacting with Ammonia and Primary Amines

Acid anhydrides undergo the same nucleophilic addition-elimination mechanism as acyl chlorides, but the leaving group is now a carboxylate ion ($\text{RCOO}^-$) rather than chloride. This single difference has two important consequences: the reaction is less vigorous, and the by-product is a carboxylic acid (or its ammonium salt) rather than HCl.

The overall equations for the anhydride reactions are worth committing to memory, as examiners frequently ask you to state the by-product — and confusing it with HCl (the acyl chloride by-product) is one of the most common errors at A-level.

For a primary amine reacting with an acid anhydride, the mechanism is identical to the ammonia case: the amine nitrogen is the nucleophile, the carboxylate fragment is the leaving group, and a second amine molecule acts as the base. The product is an N-substituted amide and the by-product is the carboxylic acid (or its ammonium salt with the amine).

5. Comparing Acyl Chlorides and Acid Anhydrides: Reactivity and Practicality

Both reagents follow the same mechanistic pathway, but acyl chlorides react far more vigorously than acid anhydrides. The reason lies in the quality of the leaving group. Chloride ($\text{Cl}^-$) is a better leaving group than ethanoate ($\text{CH}_3\text{COO}^-$) because it is the conjugate base of the strong acid HCl and is therefore a weaker base, departing more readily. Ethanoate is the conjugate base of the weak acid ethanoic acid and is a stronger base, making it a poorer leaving group despite being stabilised by delocalisation of its negative charge.

In practical terms, acyl chlorides react violently with water and release corrosive HCl fumes, making them hazardous to handle. Acid anhydrides react more slowly, do not fume in the same way, and produce a carboxylic acid as the by-product — far less corrosive. This is why acid anhydrides are preferred in industrial and large-scale synthesis, including the manufacture of aspirin from ethanoic anhydride and 2-hydroxybenzoic acid.

6. Exam Technique and Common Mistakes

Mechanism questions on this topic follow predictable patterns in AQA papers. The command word is almost always 'outline the mechanism', which means you must show curly arrows, identify intermediates, and state the mechanism type. Partial credit is available for each correct arrow, so a systematic approach protects marks even if you make one error.

The most mark-winning habit is to draw curly arrows that start from a lone pair or a bond — never from an atom label or a partial charge symbol. An arrow starting from the N atom label rather than the lone pair on N will lose the arrow mark. Similarly, the arrow showing the C=O pi bond breaking must start from the middle of the double bond (or from the bond itself), not from the oxygen atom.

6a. Exam-Style Extended Response Practice

AQA A-level papers regularly include 4-6 mark questions asking you to 'outline the mechanism' or 'explain the difference in reactivity'. The long-answer question below mirrors that format. Attempt it before reading the marking points — the act of retrieving and organising your answer is where the learning happens.

Every mark in that question maps directly to a stage of the mechanism or a comparative point about reactivity. If you lost marks, trace back to the mechanism steps in Sections 2 and 4 and the comparison table in Section 5 — the answers are all there.