Carboxylic Acids: Nomenclature and Structure

Naming the carboxyl group. The carboxyl carbon, $\text{–COOH}$, is always the highest-priority group in these molecules, so it is always numbered C-1 — it has to sit at a chain terminus, exactly like an aldehyde carbon does. The parent alkane's final $-e$ is dropped and replaced with $-oic\ acid$: propane becomes propanoic acid, butane becomes butanoic acid. For a benzene ring bearing $\text{–COOH}$, the ring itself is named as the parent, giving benzoic acid, $\text{C}_6\text{H}_5\text{COOH}$.

Why the acid is acidic at all. An alcohol's $\text{–OH}$ barely ionises because losing that proton dumps the whole negative charge onto one lone oxygen — an unstable, high-energy alkoxide ion. A carboxylic acid does something an alcohol cannot: on losing its $\text{O–H}$ proton it forms the carboxylate ion, $\text{RCOO}^-$, in which the negative charge is shared, by resonance, across both oxygens rather than being stuck on one.

where: the two structures are resonance contributors of the same ion, not two different species — the true carboxylate has both C–O bonds equal in length, intermediate between a single and a double bond, and the negative charge is spread half onto each oxygen.

Tuning acid strength with substituents. Once the resonance baseline is set, nearby atoms shift $K_a$ up or down through the inductive effect. Electron-withdrawing groups (halogens, $\text{–NO}_2$) pull electron density away from the carboxylate, spreading its negative charge even further and stabilising it more — this raises $K_a$, making the acid stronger. Electron-donating groups (alkyl chains) push electron density toward the carboxylate, concentrating the negative charge and destabilising it — this lowers $K_a$. This inductive pull weakens sharply with distance, so a substituent on the carbon right next to $\text{–COOH}$ (the $\alpha$-carbon) matters far more than the same substituent three or four carbons away.

Step 1. Compare the conjugate bases directly. Deprotonating an alcohol gives $\text{RO}^-$: a single oxygen carrying the full negative charge, with no neighbouring atom able to share it. Deprotonating a carboxylic acid gives $\text{RCOO}^-$: two chemically equivalent oxygens, each carrying half the charge through resonance delocalisation.

Step 2. Connect stability of the conjugate base to acid strength. The more stable (lower-energy) the conjugate base, the more the equilibrium $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$ favours the products, so the larger $K_a$ and the stronger the acid. Resonance delocalisation over two oxygens is a large stabilisation that an alkoxide simply cannot access — this is the single biggest reason carboxylic acids ($K_a \sim 10^{-4}$–$10^{-5}$) beat alcohols ($K_a \sim 10^{-16}$) by roughly eleven orders of magnitude.

Step 3. Layer on the inductive effect as a second, smaller correction. An electron-withdrawing substituent near the carboxyl carbon pulls electron density toward itself and away from the carboxylate oxygens, which lets the negative charge spread even more comfortably — a further stabilisation on top of resonance. This is why a chlorine on the $\alpha$-carbon raises $K_a$ noticeably, while the same chlorine four carbons away barely moves it: inductive pull is transmitted through bonds and fades fast with distance.

Step 4. Note the phenol comparison for completeness. Phenol also ionises through a resonance-stabilised conjugate base (phenoxide, spread over the ring), which is why phenol is more acidic than an alcohol — but a carboxylate's two-oxygen resonance is a tighter, more effective way to share charge than phenoxide's ring delocalisation, so carboxylic acids remain roughly $10^5$–$10^6$ times more acidic than phenol.

Name $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ and explain, using resonance, why it is a far stronger acid than butan-1-ol; then rank it against chloroacetic and dichloroacetic acid.

Step 1. Nomenclature. Four carbons total, carboxyl carbon fixed as C1: drop butane's $-e$ and add $-oic\ acid$, giving butanoic acid.

Step 2. Resonance in the conjugate base. Removing the $\text{–OH}$ proton gives $\text{CH}_3\text{CH}_2\text{CH}_2\text{COO}^-$, whose negative charge is shared equally between its two oxygens — confirmed experimentally by both C–O bonds having identical, intermediate bond lengths, unlike the distinct C=O and C–OH bonds of the neutral acid.

Step 3. Compare with the alcohol's conjugate base. Deprotonating butan-1-ol gives $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{O}^-$, with no second oxygen to share the charge — no resonance stabilisation at all. So butanoic acid ($K_a \approx 1.5\times10^{-5}$) is enormously stronger than butan-1-ol ($K_a \sim 10^{-16}$).

Step 4. Bring in inductive substituents for the ranking. Chloroacetic acid ($\text{ClCH}_2\text{COOH}$) has one electron-withdrawing chlorine on the $\alpha$-carbon; dichloroacetic acid ($\text{Cl}_2\text{CHCOOH}$) has two. Each extra chlorine further stabilises the carboxylate beyond resonance alone, so the final order of increasing acid strength is: butanoic acid (no EWG) < chloroacetic acid (one Cl) < dichloroacetic acid (two Cl).

Competitive stretch

If asked to place acetic acid ($\text{CH}_3\text{COOH}$) into this same ranking, reason from its substituent alone: it carries only the electron-donating $-\text{CH}_3$ group and no electron-withdrawing chlorine, so it sits below both chlorinated acids: acetic acid < chloroacetic acid < dichloroacetic acid.

• Treating the two C–O bonds of the neutral acid (one C=O, one C–OH) as equivalent — they are only equal in length in the deprotonated carboxylate ion, never in the neutral acid.
• Forgetting that inductive effects fade quickly with distance — a chlorine on the $\alpha$-carbon changes $K_a$ far more than the same chlorine three or four carbons down the chain.
• Mixing up electron-withdrawing and electron-donating substituents — EWGs (halogens, $\text{–NO}_2$) raise acid strength; EDGs (alkyl groups) lower it.
• Assuming carboxylic acids and phenol are similarly acidic — carboxylic acids are typically about $10^5$–$10^6$ times stronger, because two-oxygen resonance beats ring delocalisation at spreading charge.

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