Acylation Reactions: Mechanisms and Applications Revision Notes | A Level AQA Chemistry

1. Why Acylation Matters

Acylation reactions sit at the heart of synthetic organic chemistry. Every time a chemist needs to attach an acyl group ($\text{RCO}-$) to a molecule — whether making a drug, a polymer, or a fragrance — they reach for an acyl chloride or an acid anhydride.

The classic example is aspirin. Millions of tonnes of aspirin are manufactured each year using an acylation reaction, and the choice of acylating agent — ethanoyl chloride versus ethanoic anhydride — has real industrial consequences. By the end of this lesson, you will be able to draw the mechanism, predict the products, and justify that industrial choice with precision.

2. The Acylating Agents: Structure and Reactivity

Both acyl chlorides and acid anhydrides contain a carbonyl carbon ($\text{C=O}$) that is strongly electrophilic. The carbonyl oxygen withdraws electron density through the $\pi$ bond, leaving the carbon electron-poor and vulnerable to attack by nucleophiles.

Ethanoyl chloride ($\text{CH}_3\text{COCl}$) is the simplest acyl chloride you need to know. Ethanoic anhydride ($\text{(CH}_3\text{CO)}_2\text{O}$) is the corresponding acid anhydride. Both react with the same set of nucleophiles — water, alcohols, ammonia, and primary amines — but at different rates and with different by-products.

The SMILES above shows ethanoyl chloride. Notice the $\text{C=O}$ bond and the $\text{C-Cl}$ bond on the same carbon — that carbon is the site of nucleophilic attack in every acylation mechanism.

3. The Nucleophilic Addition-Elimination Mechanism

Here is the core idea: a nucleophile attacks the electrophilic carbonyl carbon, forming a tetrahedral intermediate. Then the leaving group ($\text{Cl}^-$ or a carboxylate anion) departs, and the $\text{C=O}$ bond reforms. The net result is substitution at the carbonyl carbon — but the route goes through addition then elimination, which is why the mechanism is called nucleophilic addition-elimination.

This is distinct from nucleophilic substitution at a saturated carbon ($S_N1$ or $S_N2$), where no tetrahedral intermediate involving a $\text{C=O}$ bond is involved. The carbonyl group is what makes the addition-elimination pathway possible here.

The key curly arrow rule to remember: every arrow must start from an electron-rich site — a lone pair or a bond — and point towards an electron-poor site. Never draw a curly arrow starting from an atom label or a positive charge.

4. Reactions with Four Nucleophiles: Products and Equations

The same two-stage mechanism operates regardless of which nucleophile attacks. What changes is the identity of the nucleophile and therefore the product formed. Work through each case carefully — AQA expects you to write balanced equations and name the organic products.

Notice that ammonia and primary amines require two moles of the nucleophile per mole of acyl chloride. The first mole acts as the nucleophile and forms the amide; the second mole acts as a base, neutralising the $\text{HCl}$ produced to form the ammonium salt.

The same four nucleophiles react with acid anhydrides by an analogous mechanism, but the leaving group is now a carboxylate anion ($\text{RCOO}^-$) rather than $\text{Cl}^-$. The carboxylate then picks up a proton to give a carboxylic acid as the second product. For ethanoic anhydride reacting with an alcohol, for example: $\text{(CH}_3\text{CO)}_2\text{O} + \text{R'OH} \rightarrow \text{CH}_3\text{COOR'} + \text{CH}_3\text{COOH}$.

5. Drawing the Mechanism: Step-by-Step Walkthrough

AQA specifically requires you to outline the mechanism for acyl chloride reactions. The word 'outline' means you must show the key curly arrows, the tetrahedral intermediate, and the departure of the leaving group — but you do not need to show every proton transfer in full detail. Here is a model walkthrough for the reaction with an alcohol.

Worked example: Mechanism: Ethanoyl Chloride + Ethanol

Outline the nucleophilic addition-elimination mechanism for the reaction of ethanoyl chloride ($\text{CH}_3\text{COCl}$) with ethanol ($\text{CH}_3\text{CH}_2\text{OH}$).

1. Electrophile: carbonyl carbon of $\text{CH}_3\text{COCl}$ (electron-poor due to $\text{C=O}$ and $\text{C-Cl}$ polarisation). Nucleophile: oxygen of $\text{CH}_3\text{CH}_2\text{OH}$ (lone pair available). — The carbonyl carbon carries a partial positive charge ($\delta+$) because both O and Cl are electronegative. This makes it susceptible to nucleophilic attack.
2. Arrow 1: from lone pair on O of ethanol $\rightarrow$ carbonyl C. Arrow 2: from $\text{C=O}$ bond $\rightarrow$ O (giving $\text{O}^-$ on the carbonyl oxygen). Result: tetrahedral intermediate with $\text{-O}^-$, $\text{-OCH}_2\text{CH}_3$, $\text{-CH}_3$, and $\text{-Cl}$ all on the central carbon. — The pi bond of C=O breaks heterolytically. The electron pair moves onto the more electronegative oxygen, giving it a negative formal charge. The central carbon is now sp3 (tetrahedral).
3. The $\text{-O}^-$ on the former carbonyl oxygen accepts a proton (from the $\text{-OH}_2^+$ or solvent), giving a neutral $\text{-OH}$ group on the intermediate. — This step restores charge neutrality before the leaving group departs. In an outline mechanism, you may show this as a single proton transfer arrow.
4. Arrow 3: from lone pair on $\text{-OH}$ oxygen $\rightarrow$ C (reforms $\text{C=O}$). Arrow 4: from $\text{C-Cl}$ bond $\rightarrow$ Cl (Cl$^-$ departs as leaving group). — The C=O double bond reforms as the leaving group departs. Cl- is a good leaving group because the C-Cl bond is weak and Cl- is a stable anion.
5. $\text{CH}_3\text{COCl}_{(l)} + \text{CH}_3\text{CH}_2\text{OH}_{(l)} \rightarrow \text{CH}_3\text{COOCH}_2\text{CH}_3_{(l)} + \text{HCl}_{(g)}$ — The organic product is ethyl ethanoate (an ester). HCl is released as a gas — you would observe steamy fumes. The overall transformation is nucleophilic substitution at the carbonyl carbon, achieved via the addition-elimination route.

Answer: Organic product: ethyl ethanoate ($\text{CH}_3\text{COOCH}_2\text{CH}_3$). By-product: $\text{HCl}_{(g)}$. Mechanism type: nucleophilic addition-elimination.

The same four-arrow logic applies when the nucleophile is water, ammonia, or a primary amine. The only differences are: which atom carries the lone pair that attacks, and which by-product forms when the leaving group departs.

6. Acid Anhydrides: Same Products, Different Conditions

Acid anhydrides react with the same four nucleophiles by an analogous nucleophilic addition-elimination mechanism. The key difference is the leaving group: instead of $\text{Cl}^-$, a carboxylate anion ($\text{CH}_3\text{COO}^-$) departs, which then picks up a proton to give a carboxylic acid as the second organic product.

Acid anhydrides react more slowly than acyl chlorides with the same nucleophile. This is because the carboxylate leaving group ($\text{CH}_3\text{COO}^-$) is a weaker leaving group than $\text{Cl}^-$ due to its higher basicity. The conjugate acid of the carboxylate (ethanoic acid) has a much higher pKa (~4.75) than that of chloride (HCl, pKa ~−7), meaning the carboxylate has a far greater tendency to remain bonded rather than depart. Although delocalisation of the negative charge across two oxygen atoms stabilises the carboxylate anion, it is still a poorer leaving group than $\text{Cl}^-$ because of this greater basicity, which reduces the driving force for the elimination step.

7. Industrial Application: Aspirin Manufacture

Aspirin (acetylsalicylic acid) is made by acylating the phenolic -OH group of 2-hydroxybenzoic acid (salicylic acid). The acyl group CH3CO- is introduced by reacting salicylic acid with either ethanoyl chloride or ethanoic anhydride. The carboxylic acid group remains present in the product; it is the phenolic oxygen that is acylated.

• Ethanoic anhydride is safer to handle: it does not produce corrosive HCl fumes, only ethanoic acid as a by-product. Ethanoyl chloride reacts violently with moisture in the air, releasing HCl gas, which is hazardous to workers and corrodes equipment.
• Ethanoic anhydride is less expensive to produce on an industrial scale than ethanoyl chloride.
• Ethanoic anhydride reacts less vigorously, making the reaction easier to control in large-scale reactors — reducing the risk of runaway exotherms.
• The ethanoic acid by-product from ethanoic anhydride is easier to remove and less problematic to dispose of than HCl.
• Ethanoic anhydride is less corrosive to the reaction vessels and pipework used in industrial plant.

The overall equation for aspirin synthesis using ethanoic anhydride is: $\text{HOC}_6\text{H}_4\text{COOH} + \text{(CH}_3\text{CO)}_2\text{O} \rightarrow \text{CH}_3\text{COOC}_6\text{H}_4\text{COOH} + \text{CH}_3\text{COOH}$. The phenolic $\text{-OH}$ acts as the nucleophile, and the acyl group replaces the hydrogen on that oxygen.

8. Exam Technique and Common Mistakes

Mechanism questions on acylation are among the most reliably tested topics in AQA A Level organic chemistry. The marks are there to be won, but only if your curly arrows are drawn with precision. Here is what separates a full-mark response from a near-miss.

Acylation reactions — the nucleophilic addition-elimination of water, alcohols, ammonia, and primary amines with acyl chlorides and acid anhydrides — form the mechanistic and industrial core of this topic, and every spec requirement maps directly onto the exam questions you will face.