羧酸及其衍生物

A carboxylic acid 羧酸 has the $\text{–COOH}$ (carboxyl) functional group. It is a weak acid.

Making carboxylic acids

• oxidation 氧化 of a primary alcohol 醇 or an aldehyde 醛 with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ or $\text{KMnO}_4$, with reflux 回流 (so it is fully oxidised).
• hydrolysis 水解 of a nitrile 腈 with dilute acid or alkali, then acidifying.
• hydrolysis of an ester 酯 with dilute acid or alkali and heat, then acidifying.

Reactions of carboxylic acids

These reactions all show that carboxylic acids are acids:

• with a reactive metal (a redox 氧化还原 reaction): gives a salt 盐 and hydrogen.

• with an alkali (a neutralisation 中和): gives a salt and water.

• with a carbonate 碳酸盐: gives a salt, water and carbon dioxide. The fizzing of $\text{CO}_2$ is a test for a carboxylic acid.

Two reactions change the functional group:

• esterification 酯化 with an alcohol (concentrated $\text{H}_2\text{SO}_4$ catalyst) gives an ester.
• reduction 还原 by $\text{LiAlH}_4$ gives a primary alcohol.

An ester has the $\text{–COO–}$ group. It often smells sweet or fruity.

Making esters

An ester forms in a condensation 缩合 reaction between an alcohol and a carboxylic acid, with concentrated $\text{H}_2\text{SO}_4$ as catalyst. A water molecule is lost, and the reaction is reversible:

Hydrolysis of esters

Hydrolysis splits the ester back apart. The conditions change the products:

• dilute acid and heat: reversible. Gives back the carboxylic acid and the alcohol.
• dilute alkali and heat: not reversible. Gives the alcohol and the salt of the carboxylic acid (the carboxylate ion).

Why alkali takes it to completion. With alkali the acid formed is at once neutralised to the carboxylate ion, $\text{RCOO}^-$. A negative ion cannot react with the alcohol, so the reverse reaction stops and all the ester is hydrolysed. To get the free carboxylic acid afterwards, add a strong acid.

Worked example. Name the ester made from ethanol and propanoic acid, and say which part comes from which. In an ester name the alcohol gives the first word (the alkyl part) and the acid gives the second (the -oate part). Ethanol supplies $\text{C}_2\text{H}_5-$ and propanoic acid supplies $\text{CH}_3\text{CH}_2\text{COO}-$, so the ester is ethyl propanoate, $\text{CH}_3\text{CH}_2\text{COOC}_2\text{H}_5$. It is made by refluxing the two with a concentrated $\text{H}_2\text{SO}_4$ catalyst, and the reaction is reversible, so the yield is never complete. The trap is naming it backwards: propyl ethanoate is a completely different ester (made from propan-1-ol and ethanoic acid). Alcohol first, acid second.

Worked example (working backwards from a formula). An ester is $\text{HCOOCH}_2\text{CH}_2\text{CH}_3$. Name it and give the products of heating it with $\text{NaOH(aq)}$.

1. Find the $-\text{COO}-$ and split the formula after its single-bonded O. The part with the C=O, $\text{HCOO}-$, comes from methanoic acid. The part on the other O, $-\text{CH}_2\text{CH}_2\text{CH}_3$, comes from propan-1-ol.
2. Name: alcohol part first, acid part second: propyl methanoate.
3. With NaOH the products are propan-1-ol and sodium methanoate, $\text{HCOONa}$:

Common mistakes.

• Reading an ester formula from left to right as "alcohol, then acid". In $\text{CH}_3\text{COOC}_2\text{H}_5$ the acid part is written first ($\text{CH}_3\text{COO}-$), but the name puts the alcohol part first: ethyl ethanoate.
• Counting the C of the C=O in the alcohol part. It always belongs to the acid part.
• Giving the carboxylic acid as a product of alkaline hydrolysis. Alkali gives its salt; you get the acid only after adding a strong acid.
• Saying concentrated $\text{H}_2\text{SO}_4$ is a reactant. It is a catalyst (it also removes some of the water, which helps the yield).

• Carboxylic acids react with carbonates to give $\text{CO}_2$ (fizzing) — this distinguishes them from phenols.
• Explain acid strength via the delocalised carboxylate ion; electron-withdrawing groups (e.g. Cl) increase strength.
• Esterification (with concentrated $\text{H}_2\text{SO}_4$) is reversible; name esters as alkyl alkanoate.
• Know the products of acid vs alkaline hydrolysis of an ester (they differ).