Properties and Reactions of Carboxylic Acids

Vinegar smells sharp and dissolves easily in water, yet acetic acid boils at $118^\circ\text{C}$ — hotter than water itself, and far hotter than an alcohol of similar mass. The reason is a molecule that hydrogen-bonds to a copy of itself on two sides at once, locking pairs of molecules into a stable ring. That single structural fact — the cyclic dimer — explains almost everything odd about how carboxylic acids behave physically, and sets up the three signature reactions CBSE Class 12 expects: esterification with alcohols, decarboxylation of their sodium salts, and $\alpha$ -halogenation by the Hell–Volhard–Zelinsky (HVZ) route. This article builds on the mechanisms already covered — protonation, nucleophilic attack, and the Le Chatelier reasoning behind driving esterification to completion — and pushes one step further: turning "excess alcohol shifts the equilibrium" into an actual number, so you can see by how much.

The core insight

Carboxylic acids owe their high boiling points to cyclic hydrogen-bonded dimers, and their three signature reactions — reversible esterification (driven to completion by excess alcohol or water removal), soda-lime decarboxylation (losing one carbon as carbonate), and the red-phosphorus-catalysed HVZ $\alpha$ -halogenation — all follow directly from the reactivity of the $\text{–COOH}$ group.

The core idea

Why the boiling point is so high. A lone $\text{–COOH}$ group has both a hydrogen-bond donor ( $\text{O–H}$ ) and two acceptor sites (the carbonyl and hydroxyl oxygens). Two carboxylic acid molecules pair up so that each donates to the other's carbonyl oxygen, forming a ring closed by two hydrogen bonds simultaneously — stronger, as a pair, than the single hydrogen bond an alcohol can form. Breaking this dimer to vaporise the acid costs extra energy, which is why acetic acid ( $\text{CH}_3\text{COOH}$ , $M = 60$ ) boils higher than ethanol ( $M = 46$ ) despite the acid being heavier by a modest margin only.

Why solubility falls with chain length. The $\text{–COOH}$ group is hydrophilic (it hydrogen-bonds with water), but the alkyl chain attached to it is hydrophobic. Short-chain acids like formic and acetic acid are fully miscible with water; as the chain lengthens (propanoic, butanoic, and beyond), the hydrophobic part dominates and solubility drops steadily.

Esterification — a reversible substitution. Acid-catalysed reaction with an alcohol gives an ester and water:

\text{R–COOH} + \text{R'–OH} \xrightleftharpoons[\Delta]{\text{H}^+} \text{R–COOR'} + \text{H}_2\text{O}

where: $\text{R–COOH}$ is the carboxylic acid, $\text{R'–OH}$ the alcohol, $\text{H}^+$ the catalyst (conc. $\text{H}_2\text{SO}_4$ or dry $\text{HCl}$ ), and the double arrow marks a true equilibrium — the reaction does not go to completion on its own. As covered already, the mechanism is protonation of the carbonyl oxygen, nucleophilic attack by the alcohol, proton transfer, and loss of water from the tetrahedral intermediate.

Decarboxylation — losing one carbon. Heating a sodium carboxylate with soda lime strips the $\text{–COONa}$ group off entirely, replacing it with hydrogen:

\text{R–COONa} + \text{NaOH} \xrightarrow[\Delta]{\text{CaO}} \text{R–H} + \text{Na}_2\text{CO}_3

where: $\text{R–COONa}$ is the sodium salt of the acid, $\text{CaO}$ is an inert dehydrating catalyst that keeps the $\text{NaOH}$ solid and reactive, and $\text{R–H}$ is the resulting alkane, always one carbon shorter than the starting acid because the entire $\text{–COO}$ carbon leaves as carbonate.

HVZ $\alpha$ -halogenation. Acids bearing an $\alpha$ -hydrogen react with $\text{Cl}_2$ or $\text{Br}_2$ in the presence of red phosphorus:

\text{R–CH}_2\text{COOH} \xrightarrow[\text{red P}]{\text{X}_2} \text{R–CHX–COOH}

where: $\text{X}_2$ is the halogen, red phosphorus first converts a trace of the acid into the far more enolisable acid halide, and the halogen then substitutes at the $\alpha$ -carbon before the halide exchanges back to regenerate more acid halide, propagating the reaction through the whole batch.

Figure

Bar chart comparing ester yield at equilibrium for equimolar reactants against a three fold excess of alcohol, showing the yield rising from about sixty seven percent to about ninety percent

Derivation AHow excess alcohol raises the equilibrium yield of an ester

The goal is to turn the qualitative claim "excess alcohol shifts esterification toward the product side" into an actual equilibrium-yield number, for a fixed equilibrium constant $K_c = 4$ , comparing an equimolar mixture against a threefold excess of alcohol, using nothing beyond the equilibrium-constant expression and the quadratic formula a Class 12 student already owns.

Step 1. Write $K_c$ for the reaction $\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$ as $K_c = \dfrac{[\text{ester}][\text{H}_2\text{O}]}{[\text{acid}][\text{alcohol}]}$ . Because one mole of each product forms for every mole of each reactant consumed, the mole-ratio form works without needing actual concentrations, provided we track moles reacted, $x$ , out of a fixed total volume that cancels out of the expression.

Step 2. Equimolar case. Start with $1$ mol acid and $1$ mol alcohol, no ester or water present. At equilibrium, $x$ mol of each reactant has converted, leaving $(1-x)$ mol of acid and alcohol and forming $x$ mol each of ester and water. Substituting into $K_c$ : $4 = \dfrac{x \cdot x}{(1-x)(1-x)} = \left(\dfrac{x}{1-x}\right)^2$ , so $\dfrac{x}{1-x} = 2$ , giving $x = 2 - 2x$ , hence $3x = 2$ and $x = 0.667$ . Two-thirds of the acid converts — a $66.7\%$ yield, exactly matching the "reversible, does not go to completion" behaviour already established.

Step 3. Excess-alcohol case. Now start with $1$ mol acid but $3$ mol alcohol. At equilibrium $x$ mol acid has reacted, leaving $(1-x)$ mol acid and $(3-x)$ mol alcohol, with $x$ mol each of ester and water formed. The same $K_c$ gives $4 = \dfrac{x^2}{(1-x)(3-x)}$ , so $x^2 = 4(1-x)(3-x) = 4(3 - 4x + x^2)$ , which rearranges to $3x^2 - 16x + 12 = 0$ .

Step 4. Solve the quadratic. $x = \dfrac{16 \pm \sqrt{256 - 144}}{6} = \dfrac{16 \pm \sqrt{112}}{6} = \dfrac{16 \pm 10.58}{6}$ . The root $x = 4.43$ exceeds the $1$ mol of acid available and is physically impossible (the acid cannot convert more than $100\%$ ), so the valid root is $x = \dfrac{16 - 10.58}{6} = 0.903$ .

Step 5. Compare and read the physics. With a threefold excess of alcohol, the acid's conversion rises from $66.7\%$ to $90.3\%$ — a large, quantitative confirmation of Le Chatelier's principle: adding more of a reactant shifts the equilibrium toward products, consuming a larger fraction of the limiting reagent even though $K_c$ itself never changes. The same reasoning, run the other way, is why continuously removing water (e.g. by distillation, or with a drying agent) also pushes the equilibrium toward the ester.

Worked example

Using $K_c = 4$ from above, find the mass of ethyl acetate formed at equilibrium when $1$ mol acetic acid ( $60\ \text{g/mol}$ ) is heated with $3$ mol ethanol and a trace of $\text{H}_2\text{SO}_4$ .

Step 1. From the derivation, the equilibrium extent of reaction for a $1{:}3$ acid-to-alcohol ratio with $K_c = 4$ is $x = 0.903$ mol.

Step 2. This means $0.903$ mol of the ester, ethyl acetate ( $\text{CH}_3\text{COOC}_2\text{H}_5$ , $M = 88\ \text{g/mol}$ ), forms at equilibrium, since ester and acid convert in a $1{:}1$ mole ratio.

Step 3. Convert to mass: mass $= n \times M = 0.903 \times 88 = 79.5\ \text{g}$ .

The equilibrium mixture contains 79.5 g of ethyl acetate, corresponding to 90.3% conversion of the starting acetic acid.

Competitive stretch

If the same $1$ mol acid were instead reacted with only $1$ mol alcohol (no excess), Step 2 of the derivation gives $x = 0.667$ mol, so the ester mass would be $0.667 \times 88 = 58.7\ \text{g}$ — a full $20.8\ \text{g}$ less product from the same starting acid, purely because the alcohol was not in excess. This is the quantitative reason textbooks always recommend excess alcohol (or water removal) for a good esterification yield.

Common traps

Treating esterification as irreversible and assuming $100\%$ conversion by default — it is an equilibrium, and the actual yield depends on the mole ratio used, as the derivation above shows numerically.
Forgetting that decarboxylation removes the entire carboxylate carbon as carbonate, so the alkane product always has one fewer carbon than the starting acid — a propanoate salt ( $\text{C}_3$ ) gives ethane ( $\text{C}_2$ ), not propane.
Assuming $\text{Br}_2$ alone will $\alpha$ -halogenate a carboxylic acid — the acid itself enolises too slowly; red phosphorus is essential to generate the reactive acid halide intermediate first.
Mixing up which oxygen ends up in the water molecule during esterification — it is the $\text{–OH}$ oxygen of the acid, not the alcohol's oxygen, confirmed by isotope labelling (the labelling experiment itself is beyond board-level detail, but the fact is sometimes tested conceptually).

Ask the tutor: want to try the $K_c$ calculation with a different alcohol excess, or work through the full curved-arrow mechanism for the HVZ reaction step by step?

If esterification never fully finishes, how do labs actually get a good yield of ester?

By exploiting Le Chatelier's principle rather than fighting the equilibrium: use a large excess of the cheaper reactant, usually the alcohol, and continuously remove one of the products, typically water, by distillation or with a drying agent such as concentrated $\text{H}_2\text{SO}_4$ . Both tricks pull the equilibrium toward the ester side. The worked example above shows this numerically — moving from a $1{:}1$ to a $1{:}3$ acid-to-alcohol ratio raised the acid's conversion from $66.7\%$ to $90.3\%$ , with $K_c$ never changing.

Why does soda-lime decarboxylation remove exactly one carbon, never more or less

Because the reaction only ever breaks the bond between the carboxylate carbon and the rest of the chain, replacing the whole $\text{–COONa}$ group with a single hydrogen atom. The carboxylate carbon leaves bound to two of its original oxygens, as part of $\text{Na}_2\text{CO}_3$ ; it never survives in the alkane product. No other carbon-carbon bond in the molecule is touched, so exactly one carbon is always lost, regardless of how long the original chain is.

Why is red phosphorus needed for the Hell-Volhard-Zelinsky reaction instead of just adding bromine directly to the acid

A carboxylic acid enolises extremely slowly because the $\text{–COOH}$ group is a poor electrophile once ionised or hydrogen-bonded, so its $\alpha$ -carbon is barely reactive toward $\text{Br}_2$ on its own. Red phosphorus reacts with a small amount of the halogen and the acid to generate the corresponding acid halide in situ, which enolises far more readily and gets halogenated at the $\alpha$ -carbon. That $\alpha$ -halo acid halide then exchanges with fresh starting acid, regenerating acid halide and propagating the cycle, so only a catalytic amount of phosphorus is needed to convert the whole batch.

Properties and Reactions of Carboxylic Acids

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads