Chemical reactions of acids

A carboxylic acid is not a one-trick functional group. Yes, –COOH is acidic — it fizzes with sodium bicarbonate, dissolves in alkali, releases hydrogen with reactive metals — but that Brønsted chemistry is only the entry point. The real syllabus payoff is the set of named transformations built around the carboxyl carbon: swapping the –OH for chlorine, oxygen or nitrogen to make acid chlorides, anhydrides, esters and amides; stripping away the whole –COOH as $\text{CO}_2$ by two very different routes (soda-lime decarboxylation and Kolbe's electrolysis); attacking the carbon next door in the Hell–Volhard–Zelinsky reaction; and steering an electrophile to the meta position when the acid is aromatic. Each reaction has its own reagents, its own mechanism logic, and its own favourite exam trap. This article threads them into one coherent picture, anchored to one worked mechanism you can reproduce from scratch.

The core insight

Every reaction of a carboxylic acid is governed by two handles on the same functional group — the acidic –OH (neutralisation, salt formation, acid-derivative synthesis) and the destabilised C–COOH bond (decarboxylation) or the acidified α-C–H bond (HVZ halogenation) — with the ring case (aromatic –COOH) adding a third, purely electronic, meta-directing handle.

The core idea

Handle 1 — the acidic –OH. Carboxylic acids are stronger acids than water or alcohols because the conjugate base $\text{RCOO}^-$ is resonance-stabilised across both oxygens. This lets –COOH do ordinary Brønsted chemistry that phenols and alcohols cannot fully match:

2RCOOH + 2Na \rightarrow 2RCOONa + H_2\uparrow

RCOOH + NaOH \rightarrow RCOONa + H_2O

RCOOH + NaHCO_3 \rightarrow RCOONa + H_2O + CO_2\uparrow

where: $R$ is any alkyl or aryl group carried on the carboxyl carbon. The last equation is the standard fizz test: only carboxylic acids (and a few unusually acidic phenols) liberate visible $\text{CO}_2$ from bicarbonate; ordinary phenols do not, because they are too weakly acidic to protonate $\text{HCO}_3^-$ .

Handle 2 — the –OH as a leaving group. Once the carbon is activated, the –OH can be replaced outright, generating the four classical acid derivatives, each with its own reagent:

RCOOH \xrightarrow{PCl_5 \text{ or } SOCl_2} RCOCl \ (\text{acid chloride})

RCOOH \xrightarrow{P_2O_5,\ \Delta} (RCO)_2O \ (\text{anhydride})

RCOOH + R'OH \xrightarrow{H^+} RCOOR' + H_2O \ (\text{ester, Fischer esterification})

RCOOH + NH_3 \rightarrow RCOONH_4 \xrightarrow{\Delta} RCONH_2 \ (\text{amide})

where: each product retains the acyl group ( $\text{RCO}$ –) but carries a different atom on the carbonyl carbon — Cl, O(COR), OR′ or NH₂ respectively. Reactivity of the derivatives toward further substitution runs acid chloride greater than anhydride greater than ester greater than amide, the order you exploit whenever one derivative is converted into another.

Handle 3 — losing the whole carboxyl as CO₂ (decarboxylation). Two routes exist, and examiners love the contrast between them. Simple soda-lime decarboxylation heats the sodium salt with $\text{NaOH/CaO}$ and simply drops off $\text{CO}_2$ , giving the alkane with one fewer carbon:

RCOONa + NaOH \xrightarrow{\text{CaO},\ \Delta} RH + Na_2CO_3

Kolbe's electrolytic decarboxylation instead electrolyses an aqueous solution of the salt; at the anode the carboxylate is oxidised, loses $\text{CO}_2$ as a free radical, and two such alkyl radicals couple:

2RCOO^- \xrightarrow{\text{electrolysis, anode}} 2RCOO\cdot \rightarrow 2R\cdot + 2CO_2 \rightarrow R\text{–}R

At the cathode water is reduced, $2H_2O + 2e^- \rightarrow H_2\uparrow + 2OH^-$ . Kolbe electrolysis gives the symmetric dimer, not the alkane — a mechanistic fork worth memorising on sight.

Handle 4 — the α-C–H (HVZ reaction). A carboxylic acid with at least one α-hydrogen can be halogenated next to the carbonyl using $\text{X}_2$ and a catalytic trace of red phosphorus:

RCH_2COOH + X_2 \xrightarrow{\text{red P (catalytic)}} RCHXCOOH + HX

Red phosphorus is not a stoichiometric reagent — it converts a small amount of the acid to the far more enolisable acid halide, which is halogenated first and then hands the halogen back to fresh acid, regenerating the catalytic cycle (worked below).

Handle 5 — the ring, if the acid is aromatic. The –COOH group withdraws electron density from the benzene ring by both induction and resonance, so it is ring-deactivating and, crucially, meta-directing — behaving exactly like –CHO or –NO₂, not like an electron-donating group:

C_6H_5COOH + HNO_3 \xrightarrow{\text{conc. } H_2SO_4} m\text{-O}_2N\text{–}C_6H_4\text{–}COOH + H_2O

Figure

Five reaction arrows radiating from a central carboxylic acid box to acid chloride anhydride ester amide and decarboxylation product boxes

Derivation AWhy Kolbe electrolysis gives the dimer R–R, not the alkane

The goal is to build up the net Kolbe equation $2RCOO^- \xrightarrow{\text{electrolysis}} R\text{–}R + 2CO_2\uparrow$ from the two separate electrode processes and the radical-coupling step, so that the symmetric-dimer outcome falls out of the mechanism rather than being memorised as a fact — and so that it is visibly a different product from the soda-lime alkane.

Step 1. Identify the species at the anode. In the aqueous salt solution, the carboxylate anion $\text{RCOO}^-$ migrates to the positive electrode (the anode), because that electrode is where oxidation — loss of electrons — happens.

Step 2. Oxidise the carboxylate. At the anode surface, $\text{RCOO}^-$ loses one electron to the electrode: $RCOO^- \rightarrow RCOO\cdot + e^-$ . The product is a carboxyl radical, an unstable species with an odd electron on oxygen.

Step 3. Let the radical fragment. The carboxyl radical is not isolable — it instantly expels carbon dioxide, the most stable small molecule it can shed, leaving an alkyl radical: $RCOO\cdot \rightarrow R\cdot + CO_2$ . This is the decarboxylation step, and it happens for every carboxylate ion that reaches the anode.

Step 4. Couple the radicals. Two alkyl radicals formed side by side at the electrode surface combine by simple radical–radical bond formation, since each has one unpaired electron looking for a partner: $R\cdot + R\cdot \rightarrow R\text{–}R$ . Because both radicals came from the same starting acid, the product is necessarily the symmetric dimer, not a mixture.

Step 5. Add the two anode half-processes for two carboxylate ions and combine with the coupling step to get the net anode result: $2RCOO^- \rightarrow 2RCOO\cdot + 2e^- \rightarrow 2R\cdot + 2CO_2 \rightarrow R\text{–}R + 2CO_2$ . The two electrons released at the anode are consumed at the cathode, where water is reduced, $2H_2O + 2e^- \rightarrow H_2\uparrow + 2OH^-$ , closing the electrical circuit. The dimer $R$ – $R$ has twice the carbon count of $R$ (minus the two carboxyl carbons, which left as $\text{CO}_2$ ) — the signature that distinguishes Kolbe's product from the single-alkane product $RH$ of soda-lime decarboxylation.

Worked example

Problem: Sodium propanoate ( $CH_3CH_2COONa$ ) is subjected to Kolbe's electrolytic decarboxylation. Identify the anode process, the cathode process, and the final organic product, then contrast it with what soda-lime decarboxylation of the same salt would give.

Step 1. Anode: the propanoate ion loses an electron and immediately loses $\text{CO}_2$ , giving an ethyl radical: $CH_3CH_2COO^- \rightarrow CH_3CH_2COO\cdot + e^- \rightarrow CH_3CH_2\cdot + CO_2$ .

Step 2. Two ethyl radicals formed at the anode couple: $2CH_3CH_2\cdot \rightarrow CH_3CH_2\text{–}CH_2CH_3$ , which is n-butane, a four-carbon chain.

Step 3. Cathode (accounting for the electrons released above): $2H_2O + 2e^- \rightarrow H_2\uparrow + 2OH^-$ .

Step 4. Contrast with soda-lime: heating sodium propanoate with $\text{NaOH/CaO}$ instead gives $CH_3CH_2COONa + NaOH \xrightarrow{\text{CaO},\Delta} CH_3CH_3 + Na_2CO_3$ — ethane, the three-carbon acid's alkane with one carbon removed. Kolbe's electrolysis of sodium propanoate gives n-butane (C4), while soda-lime decarboxylation of the same salt gives ethane (C2) — same starting salt, two different mechanisms, two different products.

Competitive stretch

For propanoic acid treated with $\text{Br}_2$ and catalytic red phosphorus (HVZ), the same "different mechanism, different site" logic applies: the reaction happens at the α-carbon, not the carboxyl carbon, giving 2-bromopropanoic acid, $CH_3CHBrCOOH$ , via a transient acid-bromide intermediate that is continuously regenerated — never consumed — through the catalytic cycle.

Common traps

Treating soda-lime decarboxylation and Kolbe's electrolytic decarboxylation as interchangeable — they need different conditions (dry heating with NaOH/CaO versus electrolysis of an aqueous salt solution) and give structurally different products (alkane $RH$ versus symmetric dimer $R$ – $R$ ).
Forgetting the HVZ reaction needs at least one α-hydrogen and uses red phosphorus as a catalyst, not a stoichiometric reagent — quoting "excess red P" as a requirement is wrong.
Predicting ortho/para attack on an aromatic carboxylic acid; –COOH, like –CHO, is electron-withdrawing and meta-directing, never an activating, ring-donating group.
Missing that the $\text{NaHCO}_3$ fizz test distinguishes carboxylic acids (which react, releasing visible $\text{CO}_2$ ) from phenols (which do not, despite both being acidic) — a favourite short-answer qualitative test question.

Ask the tutor: if the electron flow in Kolbe's anode mechanism or the catalytic cycle of the HVZ reaction still feels fuzzy, ask the AI tutor to redraw each step with you.

Why does Kolbe electrolysis give a dimer while soda-lime decarboxylation gives an alkane?

The mechanisms are completely different. Soda-lime decarboxylation is a thermal, ionic process — the carboxylate salt is heated with a strong base and simply loses $\text{CO}_2$ as one molecule, leaving one $\text{H}$ from the base to cap the new C–H bond, so you get $RH$ . Kolbe electrolysis is a radical process at an anode: two separate carboxylate ions each lose $\text{CO}_2$ to form two alkyl radicals, and those two radicals then bond to each other, so the carbon skeleton doubles into $R$ – $R$ . Same starting material, two unrelated reaction pathways.

Does the HVZ reaction work on any carboxylic acid?

Only on acids that have at least one hydrogen on the carbon next to –COOH (the α-carbon), because that is the position being halogenated. Formic acid ( $HCOOH$ ) and acids with a fully substituted α-carbon cannot undergo HVZ in the usual sense, since there is no α-C–H to replace. Also remember red phosphorus is catalytic — it works by converting a little of the acid into the far more reactive acid halide, which enolises and is halogenated first, then passes the halogen on to more of the parent acid.

Why is -COOH meta-directing when it looks like it could donate electrons through its oxygen lone pairs?

The oxygens in –COOH are also attached to a carbonyl carbon that is strongly electron-withdrawing by resonance (pulling electron density toward itself, not releasing it into the ring the way –OH or –NH2 would). Overall, the whole –COOH group withdraws electron density from the ring by both the inductive effect and resonance, which destabilises positive charge developing at the ortho and para positions during electrophilic attack. The meta position is left comparatively less destabilised, so that is where substitution occurs — the same logic as –CHO and –NO2, both carbonyl-type or strongly electron-withdrawing groups.

Chemical reactions of acids

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads