3.3.9 Carboxylic Acids and Derivatives

Recognise –COOH as the carboxylic-acid group and –COO– as the ester link between an acyl side and an alkyl/aryl side. Be able to read both from displayed, structural and skeletal formulae.

When naming/drawing an ester, keep track of which carbon belongs to the carbonyl-containing acid-derived part and which group is attached through oxygen.

When you call a carboxylic acid weak, remember that ‘weak’ means only partial ionisation in water: RCOOH ⇌ RCOO⁻ + H⁺. It can still react readily with bases and carbonates, and a carbonate reaction releases CO₂.

Do not confuse “weak” with “unreactive”. Weak describes the extent of ionisation in water, not whether an acid can neutralise an alkali or carbonate.

A carboxylic acid and an alcohol form an ester and water in a reversible acid-catalysed reaction. The acid supplies the acyl part of the ester; the alcohol supplies the group attached through oxygen.

Write the equilibrium equation and identify the ester from both reactants. If you hydrolyse that ester, the same two components are recovered under the appropriate acid conditions.

Many esters are useful as solvents and some have characteristic smells used in perfumes and flavourings; ester-containing additives can also act as plasticisers. The application follows from physical properties such as volatility, solvency and intermolecular interactions.

You do not need one property to explain every ester use. Match the property to the particular application the question gives you.

If you draw a fat or vegetable oil, think of propane-1,2,3-triol (glycerol) with all three OH groups esterified by long-chain carboxylic acids. That gives three ester links in one triglyceride molecule.

The three fatty-acid residues do not have to be identical. When drawing a triglyceride, make sure there are three ester carbonyls and the glycerol carbon skeleton is still visible.

Acid hydrolysis of an ester is reversible and produces the alcohol plus the carboxylic acid. Alkaline hydrolysis gives the alcohol plus a carboxylate salt and is effectively driven to products.

If you want the free carboxylic acid after alkaline hydrolysis, acidify the carboxylate. Do not write RCOOH as the immediate product of hydrolysis in excess NaOH.

For saponification, picture alkaline hydrolysis of all three ester links in a triglyceride. Heating with NaOH or KOH gives glycerol plus the sodium or potassium salts of the long-chain carboxylic acids.

Those long-chain carboxylate salts are soaps. Check the stoichiometry: one triglyceride molecule contains three ester links and can yield three carboxylate ions.

To identify biodiesel, look for long-chain methyl esters, RCOOCH₃, not the original three-armed triglyceride structure. Transesterification converts the glycerol triester into those smaller methyl esters.

A quick structural check is the glycerol backbone: a triglyceride still has the three-arm triester framework, while a biodiesel molecule is a separate long-chain methyl ester such as RCOOCH₃.

Transesterification converts triglycerides in vegetable oil into methyl esters by reaction with methanol. The glycerol part of the triglyceride is displaced and glycerol is formed as a coproduct; the long-chain acyl groups become the methyl esters used as biodiesel.

Keep this distinct from ordinary esterification of a carboxylic acid with an alcohol: here one ester is converted into different ester products by exchanging the alcohol-derived group.

Recognise the functional groups: an acyl chloride contains –COCl, an acid anhydride contains –CO–O–CO–, and an amide contains –CONH₂ or the corresponding N-substituted form.

All contain a carbonyl, so read the atom/group directly attached to the acyl carbon before deciding which derivative you have. A carbonyl alone does not distinguish them.

For acyl chlorides and acid anhydrides, I’d learn one common reaction pattern and then change the nucleophile. Water gives a carboxylic acid, an alcohol gives an ester, ammonia gives an amide and a primary amine gives an N-substituted amide; acyl chlorides also release HCl in the relevant reactions.

The common thread is acyl substitution at the carbonyl carbon. Learn the product pattern first; then the mechanism becomes much easier to reconstruct.

Compare ethanoic anhydride with ethanoyl chloride for aspirin manufacture: think process chemistry as well as reactivity. Ethanoic anhydride is easier and safer to handle on scale and gives ethanoic acid as the by-product; ethanoyl chloride gives corrosive HCl.

The choice is a process decision involving reagent reactivity, hazard, by-products and handling — not a claim that the anhydride is chemically incapable of acylation.

For an acyl-chloride nucleophilic addition–elimination mechanism, attack the carbonyl carbon with the nucleophile, move the C=O π pair to oxygen, then reform C=O as Cl⁻ leaves. Add the required proton-transfer step for water, alcohol, ammonia or amine nucleophiles.

Locate the δ⁺ acyl carbon first. The nucleophile attacks there, the tetrahedral intermediate forms, and C=O then reforms as the leaving group departs. Once you can see that sequence, changing water to an alcohol or amine mainly changes the nucleophile and product.