Carboxylic Acids: Nomenclature, Structure & Preparation

Carboxylic Acids — Structure and Nomenclature

A carboxylic acid contains the carboxyl group, -COOH (a carbonyl C=O joined to a hydroxyl -OH). The two groups interact: the lone pair on the -OH oxygen is delocalised toward the C=O, which is why -COOH behaves very differently from a simple alcohol or ketone.

Common names: HCOOH = formic acid, CH$_3$COOH = acetic acid, C$_6$H$_5$COOH = benzoic acid, (COOH)$_2$ = oxalic acid.

IUPAC names (alkanoic acid): replace the -e of the alkane with -oic acid; the carboxyl carbon is always C-1.

• HCOOH → methanoic acid
• CH$_3$COOH → ethanoic acid
• CH$_3$CH$_2$COOH → propanoic acid
• HOOC-COOH → ethanedioic acid (oxalic acid)

For -COOH on a ring, use -carboxylic acid (e.g. benzenecarboxylic acid = benzoic acid).

Key Point: -COOH = carboxyl. IUPAC = alkane → alkanoic acid (carboxyl is C-1). Ring-COOH = carboxylic acid suffix.

Preparation of Carboxylic Acids — From Oxidation

From primary alcohols and aldehydes (oxidation): 1° alcohols and aldehydes are oxidised by strong oxidants (KMnO$_4$, K$_2$Cr$_2$O$_7$, conc. HNO$_3$) to carboxylic acids with the same number of carbons.

CH$_3$CH$_2$OH →(KMnO$_4$)→ CH$_3$COOH

From alkylbenzenes (side-chain oxidation): an alkylbenzene is oxidised by hot KMnO$_4$ to benzoic acid — the whole side chain is oxidised to a single -COOH regardless of its length (as long as there is a benzylic H). So toluene, ethylbenzene and propylbenzene all give benzoic acid.

Key Point: 1° alcohol or aldehyde + strong oxidant → carboxylic acid (same C count); any alkylbenzene with a benzylic H → benzoic acid (the side chain is cut to -COOH).

Preparation of Carboxylic Acids — From Nitriles, Grignard and Derivatives

From nitriles (hydrolysis): a nitrile is hydrolysed (acid or base) to a carboxylic acid (via the amide). R-C≡N → R-COOH. This adds one carbon (the nitrile is usually made from an alkyl halide + KCN), so it is a way to ascend the series.

From Grignard reagents (with CO$_2$): a Grignard reagent adds to carbon dioxide (dry ice) and the product is hydrolysed to a carboxylic acid with one more carbon than the Grignard:

R-MgX + CO$_2$ → R-COO-MgX →(H$_3$O⁺)→ R-COOH

From acyl halides, anhydrides and esters (hydrolysis): all are hydrolysed to the parent carboxylic acid (esters need acid or base; base gives the salt — saponification).

Key Point: nitrile hydrolysis and Grignard + CO$_2$ both give an acid with one extra carbon; acyl halides, anhydrides and esters hydrolyse back to the acid.

Solved Examples

Example 1: IUPAC naming

Give the IUPAC name of (CH$_3$)$_2$CHCOOH.

Solution: A 3-carbon acid (-COOH is C-1) with a methyl branch on C-2: 2-methylpropanoic acid.

Example 2: Oxidation route

What acid forms when propan-1-ol is oxidised by acidified KMnO$_4$?

Solution: A 1° alcohol is oxidised to the acid with the same carbons: propanoic acid (CH$_3$CH$_2$COOH).

Example 3: Side-chain oxidation

What is the product when ethylbenzene is heated with KMnO$_4$?

Solution: The whole side chain is oxidised to -COOH, giving benzoic acid (C$_6$H$_5$COOH) (plus CO$_2$ from the lost carbons).

Example 4: Grignard + CO$_2$

How would you make propanoic acid from ethylmagnesium bromide?

Solution: C$_2$H$_5$MgBr + CO$_2$ (dry ice), then hydrolyse with dilute acid → propanoic acid (CH$_3$CH$_2$COOH) (one carbon more than the ethyl group).

Example 5: Nitrile hydrolysis (ascending the series)

How is acetic acid prepared from methyl bromide?

Solution: CH$_3$Br + KCN → CH$_3$CN (acetonitrile), then hydrolyse (acid/base) → CH$_3$COOH (acetic acid). The CN adds one carbon.

Example 6: From an ester

What acid (and other product) forms when ethyl ethanoate is hydrolysed with dilute acid?

Solution: Ethanoic acid (CH$_3$COOH) + ethanol (C$_2$H$_5$OH) (acid-catalysed hydrolysis is the reverse of esterification).

Example 7: Choosing a method to add a carbon

You want to convert CH$_3$CH$_2$Br to CH$_3$CH$_2$COOH. Give two ways.

Solution: (a) CH$_3$CH$_2$Br + KCN → CH$_3$CH$_2$CN → (hydrolysis) → CH$_3$CH$_2$COOH. (b) CH$_3$CH$_2$MgBr + CO$_2$, then H$_3$O⁺ → CH$_3$CH$_2$COOH. Both add one carbon.

Example 8: Oxalic acid name

Give the IUPAC name of HOOC-COOH.

Solution: Ethanedioic acid (common name oxalic acid) — a two-carbon diacid.

Example 9: All alkylbenzenes → one acid

Why do toluene and propylbenzene both give benzoic acid on oxidation?

Solution: KMnO$_4$ oxidises the benzylic carbon and cleaves the rest of the side chain, leaving a single -COOH on the ring regardless of chain length — so both give benzoic acid.

Example 10: Hydrolysis of a nitrile

Write the product of hydrolysis of benzonitrile (C$_6$H$_5$CN).

Solution: Benzoic acid (C$_6$H$_5$COOH) (via benzamide as the intermediate).