Carboxylic acids are everywhere you can smell or taste sourness — vinegar, citrus fruit, rancid butter, and the lactic acid in sore muscles — and their chemistry connects the physical properties of IMF to the equilibrium chemistry of Module 6 in one of the most integrated lessons in the HSC course.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Ethanoic acid (acetic acid) and ethanol are both two-carbon compounds with oxygen, and both are found in wine — ethanol as the intended product of fermentation, ethanoic acid as the unwanted product when wine sours on exposure to air. Vinegar is 5% ethanoic acid in water.
Despite having similar molecular masses (ethanol 46 g/mol, ethanoic acid 60 g/mol), ethanoic acid boils at 118°C while ethanol boils at 78°C — a difference of 40°C despite the acid being heavier. Both compounds have an O–H bond and can form hydrogen bonds.
Before you read on: Write down your explanation for why ethanoic acid has such a dramatically higher boiling point than ethanol. What do you think the carboxyl group (–COOH) can do that the hydroxyl group (–OH) cannot? Record your best prediction below.
The carboxyl group (–COOH) is not simply a carbonyl plus a hydroxyl side by side — the two parts interact electronically in ways that give carboxylic acids properties that neither aldehydes, ketones, nor alcohols possess individually.
The carboxyl group consists of a carbonyl (C=O) and a hydroxyl (–OH) bonded to the same carbon. That carbon is sp²-hybridised (trigonal planar, ~120° bond angles). The functional group is written –COOH in condensed formulae. In a full structural formula, both oxygens must be drawn explicitly — one with a double bond, one with a single bond to H. The two oxygens are not equivalent.
Each carboxylic acid molecule can both donate and accept H-bonds:
Two R–COOH molecules can align face-to-face and form two simultaneous hydrogen bonds in a cyclic, 8-membered ring structure called a dimer:
The dimer is highly stable. In liquid carboxylic acids, a significant proportion of molecules exist as dimers rather than free monomers. Near the boiling point, carboxylic acid vapour density measurements show an effective molecular mass approximately double the monomer mass — confirming molecules predominantly leave the liquid surface as pairs.
Ethanoic acid hydrogen-bonded dimer. Two carboxyl groups lock together via two simultaneous H-bonds (dashed blue lines, crossing to form the 8-membered ring). Both H-bonds must break to vaporise — this is why carboxylic acid BPs are anomalously high.
Suffix: –oic acid. The –COOH carbon is always C1 (no locant needed). The chain length includes the carbonyl carbon.
Carboxylic acids consistently have the highest boiling points among organic compounds of comparable chain length — higher than even alcohols — and the explanation is dimerisation, not simply "strong H-bonds."
Alkane < Aldehyde ≈ Ketone < Alcohol < Carboxylic acid
Both ethanoic acid and ethanol have one O–H bond and can form hydrogen bonds. The difference is dimerisation:
Short-chain carboxylic acids (C1–C4) are fully miscible with water — the –COOH group forms strong H-bonds with water and the carboxylate anion produced by partial ionisation is fully solvated. From C5 onward, solubility decreases as the non-polar alkyl chain increasingly dominates. Long-chain fatty acids (C12–C22) are essentially insoluble — a property central to biological membranes and soaps (Lesson 17).
Carboxylic acids are weak acids — they partially ionise in water rather than fully dissociating — and this equilibrium behaviour, learned in Module 6, now applies directly to predicting which reagents they react with and at what rate.
In water, carboxylic acids establish a partial ionisation equilibrium:
For ethanoic acid, Ka = 1.8 × 10⁻⁵ (pKa = 4.74). At equilibrium, only ~1% of CH₃COOH molecules are ionised in a 0.1 mol/L solution. Despite being "weak," carboxylic acids are significantly more acidic than alcohols (Ka ~ 10⁻¹⁶) or water.
After a carboxylic acid donates a proton, the carboxylate anion (R–COO⁻) is stabilised by resonance — the negative charge is delocalised across both oxygens equally. This makes the carboxylate a weaker conjugate base (less tendency to re-accept H⁺), shifting equilibrium further right.
In an alcohol, after proton donation, the alkoxide (R–O⁻) has the negative charge localised on one oxygen — no resonance stabilisation. Stronger base → equilibrium lies far left → much weaker acid.
The acid strength ranking of organic compounds directly reflects the structural stability of their conjugate bases — the more stable the conjugate base, the stronger the acid, and this can be predicted from structure without memorising a list.
R–COO⁻
R–C(=O)–O⁻ ↔ R–C(–O⁻)=O
Charge on BOTH oxygens equally — high resonance stabilisation
pKa ≈ 5
C₆H₅–O⁻
Charge spreads into ring carbons
Partial delocalisation — moderate stabilisation
pKa ≈ 10
R–O⁻
Charge on ONE oxygen only
No resonance — unstabilised, very strong base
pKa ≈ 16
| Compound class | NaOH | Na₂CO₃ | NaHCO₃ |
|---|---|---|---|
| Carboxylic acid (pKa ~5) | Reacts ✓ | CO₂ ✓ | CO₂ ✓ |
| Phenol (pKa ~10) | Reacts ✓ | Reacts ✓ | No reaction ✗ |
| Alcohol (pKa ~16) | No reaction ✗ | No reaction ✗ | No reaction ✗ |
"The –COOH group has two O–H bonds." It has ONE O–H bond (the hydroxyl part). The C=O oxygen has no attached hydrogen — it only accepts H-bonds, it cannot donate them.
"Carboxylic acids form stronger individual H-bonds than alcohols." The O–H hydrogen bond type is essentially the same in both. The difference in boiling point comes from dimerisation (two simultaneous H-bonds per pair), not from individual bond strength.
"Carboxylic acids fully ionise in water because they react with NaHCO₃." The reaction with NaHCO₃ reflects acid strength (pKa ~5 vs ~10), not complete ionisation. Carboxylic acids remain weak acids — Ka ≪ 1, only ~1% ionised at typical concentrations.
Propanoic acid (CH₃CH₂COOH). Write balanced equations for reactions with: (a) NaOH, (b) Na₂CO₃, (c) NaHCO₃, (d) Mg.
Balanced equations and names of products for each reaction.
For each reaction, identify the base/metal, apply the general carboxylic acid reaction pattern, and verify atom balance.
(a) With NaOH — acid-base neutralisation; H from –COOH transfers to OH⁻:
(b) With Na₂CO₃ — two –COOH groups donate H⁺ to CO₃²⁻; forms H₂CO₃ → H₂O + CO₂:
(c) With NaHCO₃ — one –COOH donates H⁺ to HCO₃⁻ → H₂CO₃ → H₂O + CO₂:
(d) With Mg — two –COOH groups donate H⁺ to Mg; Mg oxidised; H₂ gas produced:
(a) CH₃CH₂COOH + NaOH → CH₃CH₂COONa + H₂O | (b) 2CH₃CH₂COOH + Na₂CO₃ → 2CH₃CH₂COONa + H₂O + CO₂ | (c) CH₃CH₂COOH + NaHCO₃ → CH₃CH₂COONa + H₂O + CO₂ | (d) 2CH₃CH₂COOH + Mg → (CH₃CH₂COO)₂Mg + H₂
(a) Ethanoic acid (M = 60 g/mol, BP 118°C) has a significantly higher boiling point than ethanol (M = 46 g/mol, BP 78°C). Explain using IMF reasoning, noting ethanol has a lower molecular mass. (b) Explain why ethanoic acid is a stronger acid than ethanol using structural reasoning.
Both ethanoic acid and ethanol have O–H bonds and can form hydrogen bonds. The BP difference (40°C) cannot be explained by molecular mass (acid is heavier) or by one O–H bond alone (each has one O–H).
The key difference is dimerisation. Two ethanoic acid molecules form a cyclic dimer with two simultaneous H-bonds: O–H (mol 1) ··· O=C (mol 2) and O=C (mol 1) ··· H–O (mol 2). To vaporise ethanoic acid, both H-bonds of a dimer must be broken simultaneously — greater energy than breaking the single H-bond between two ethanol molecules. The effective unit leaving the liquid is the dimer (~120 g/mol effective mass).
Both compounds lose a proton from an O–H bond. The difference is in conjugate base stability:
(a) Ethanoic acid forms H-bonded dimers (two simultaneous H-bonds per pair). The effective vaporising unit is the dimer (~120 g/mol), requiring more energy than breaking ethanol's single H-bond → higher BP despite lower monomer mass. (b) Ethanoate ion is resonance-stabilised (charge on both O equally) → weaker conjugate base → stronger acid (pKa ~5). Ethoxide has charge on one O only — no resonance → stronger base → ethanol is a much weaker acid (pKa ~16).
Unknown compound X has molecular formula C₄H₈O₂. Tests performed:
(a) Identify the functional group class and explain each test. (b) Write TWO structural isomers of X with IUPAC names. (c) Explain how NaHCO₃ distinguishes a carboxylic acid from an ester (both C₄H₈O₂). (d) Which isomer has the higher boiling point and why?
C₄H₈O₂ fits CₙH₂ₙO₂ — could be a carboxylic acid or an ester (both have two oxygens).
C₄H₈O₂ carboxylic acid: –COOH group accounts for 1C + 2O. Remaining: 3C, 7H → propyl group (C₃H₇–).
Esters (R–COO–R') have no O–H bond — no acidic hydrogen to donate to HCO₃⁻. The ester linkage (–COO–) does not react with NaHCO₃ under ordinary conditions. Carboxylic acids (pKa ~5) are acidic enough to donate H⁺ to HCO₃⁻, producing H₂CO₃ → H₂O + CO₂ gas (effervescence). The NaHCO₃ test therefore distinguishes: carboxylic acid → CO₂; ester → no reaction.
Both are carboxylic acids with one –COOH — same H-bonding capability. The difference is dispersion forces: branching in 2-methylpropanoic acid creates a more compact shape → less surface area → weaker dispersion forces. Butanoic acid (straight chain, greater surface area, stronger dispersion) has the higher boiling point (164°C vs 154°C).
(a) X is a saturated carboxylic acid. NaHCO₃ + CO₂ confirms. Tollens'/Fehling's negative rules out aldehyde. No Br₂ decolourisation → saturated. K₂Cr₂O₇ unchanged → already at maximum oxidation state. (b) Butanoic acid (CH₃CH₂CH₂COOH) and 2-methylpropanoic acid ((CH₃)₂CHCOOH). (c) Esters lack acidic O–H; carboxylic acids (pKa ~5) donate H⁺ to HCO₃⁻ → CO₂. (d) Butanoic acid (straight chain, greater surface area, stronger dispersion forces, BP 164°C) > 2-methylpropanoic acid (branched, BP 154°C).
Structure: Carboxyl group = –COOH (carbonyl C=O + hydroxyl –OH on same C). General formula: CₙH₂ₙO₂. Suffix: –oic acid; –COOH carbon always C1.
Dimerisation: Two –COOH molecules form a cyclic dimer via two simultaneous H-bonds. To vaporise: both H-bonds must break simultaneously → highest BPs among organic compounds of same chain length (higher than alcohols).
BP trend (same chain length): Alkane < Aldehyde ≈ Ketone < Alcohol < Carboxylic acid
Weak acid: R–COOH + H₂O ⇌ R–COO⁻ + H₃O⁺ Ka ≪ 1. Ethanoic acid Ka = 1.8 × 10⁻⁵.
Reactions:
+ NaOH → salt + H₂O (no gas)
+ Na₂CO₃ → salt + H₂O + CO₂↑
+ NaHCO₃ → salt + H₂O + CO₂↑ (KEY diagnostic)
+ Na/Mg → salt + H₂↑
Acid strength: Carboxylic acid (pKa ~5) > Phenol (~10) > Alcohol (~16). Carboxylate R–COO⁻ is resonance-stabilised (charge on 2 O) → weaker base → stronger acid.
NaHCO₃ test: Carboxylic acid → CO₂; Phenol → no reaction; Alcohol → no reaction.
Three unknown compounds (P, Q, R) each have molecular formula C₃H₆O₂. The following tests are carried out. Complete the table and identify each compound.
| Test | P | Q | R |
|---|---|---|---|
| NaHCO₃ solution | Effervescence | No reaction | No reaction |
| Tollens' reagent | No reaction | Silver mirror | No reaction |
| K₂Cr₂O₇/H⁺ | Orange → stays orange | Orange → green | Orange → stays orange |
Identify P, Q, and R. For each, write the IUPAC name and a short justification (1–2 sentences per compound). Note: C₃H₆O₂ compounds can include carboxylic acids, esters, and aldehydes.
Rank the following five compounds from lowest to highest boiling point. Then explain the position of each compound using IMF reasoning.
1. Which of the following correctly describes why ethanoic acid (BP 118°C) has a higher boiling point than ethanol (BP 78°C), despite ethanol having a lower molecular mass?
2. A student adds NaHCO₃ solution to three unlabelled compounds: propanoic acid (P), propan-1-ol (Q), and propyl propanoate (R, an ester). Which observation correctly distinguishes them?
3. Which structural feature of the carboxylate ion (R–COO⁻) makes carboxylic acids stronger acids than alcohols?
4. A compound with molecular formula C₃H₆O₂ produces CO₂ when added to NaHCO₃ solution and gives no reaction with Tollens' reagent. Which compound is it?
5. A 0.1 mol/L solution of ethanoic acid has a pH of approximately 2.9, while a 0.1 mol/L solution of HCl has a pH of 1.0. Which statement best explains this difference?
Question 6 (4 marks) — Butanoic acid (CH₃CH₂CH₂COOH) is the compound responsible for the smell of rancid butter. Write balanced equations for the reactions of butanoic acid with each of the following reagents. Name all products and state one observable for each reaction.
(a) NaOH(aq) (b) NaHCO₃(aq) (c) Magnesium metal (Mg)
Question 7 (5 marks) — Explain why propanoic acid (C₂H₅COOH, BP 141°C) has a significantly higher boiling point than propan-1-ol (C₃H₇OH, BP 97°C), given that both have an O–H bond and similar molecular masses (propanoic acid 74 g/mol, propan-1-ol 60 g/mol). In your response, refer to the structure of the carboxylic acid dimer and explain what must occur during vaporisation.
Question 8 (6 marks) — "Carboxylic acids are significantly more acidic than alcohols despite both containing an O–H bond." Using structural arguments, explain this statement by comparing the stability of the carboxylate anion (R–COO⁻) with the alkoxide anion (R–O⁻). Then evaluate the following student claim: "We could increase the acid strength of ethanoic acid by substituting electronegative atoms into the alkyl chain, such as forming chloroethanoic acid (CH₂ClCOOH, pKa 2.86)." Is the student correct? Justify your answer.
Q1 — B. Dimerisation is the key. Two ethanoic acid molecules form a cyclic pair with two simultaneous H-bonds; breaking both requires more energy than breaking ethanol's single H-bond. Option A counts O atoms — not valid. Option C confuses intramolecular bond strength with intermolecular forces. Option D confuses viscosity (a consequence) with the cause.
Q2 — A. Only carboxylic acids react with NaHCO₃ to produce CO₂. Propan-1-ol (pKa ~16) is far too weak an acid. Propyl propanoate is an ester — no acidic O–H. The NaHCO₃ test uniquely identifies the carboxylic acid.
Q3 — B. Resonance delocalises the negative charge across both O atoms in R–COO⁻, stabilising it. A more stable conjugate base = weaker base = stronger acid. Option A (more lone pairs) is not a valid stability argument. Option C (size) is a secondary effect, not the primary mechanism at HSC level.
Q4 — B. C₃H₆O₂ with NaHCO₃ → CO₂ means carboxylic acid. Propanoic acid (CH₃CH₂COOH) fits. Methyl ethanoate and ethyl methanoate are esters — no reaction with NaHCO₃. Propanal has formula C₃H₆O (only one oxygen).
Q5 — B. HCl fully dissociates → [H₃O⁺] = 0.1 mol/L → pH 1.0. Ethanoic acid partially ionises (Ka = 1.8 × 10⁻⁵) → only ~1% ionised → [H₃O⁺] ≪ 0.1 mol/L → pH ~2.9. Molecular mass (A) is irrelevant. There is no temperature requirement for the ionisation (C). CH₃ groups do not neutralise H₃O⁺ (D).
Q6 (4 marks):
(a) CH₃CH₂CH₂COOH + NaOH → CH₃CH₂CH₂COONa + H₂O | Products: sodium butanoate + water | Observable: no gas produced; pH rises (1 mark)
(b) CH₃CH₂CH₂COOH + NaHCO₃ → CH₃CH₂CH₂COONa + H₂O + CO₂(g) | Products: sodium butanoate + water + carbon dioxide | Observable: effervescence/CO₂ bubbles (1 mark)
(c) 2CH₃CH₂CH₂COOH + Mg → (CH₃CH₂CH₂COO)₂Mg + H₂(g) | Products: magnesium butanoate + hydrogen gas | Observable: Mg dissolves; H₂ gas bubbles (1 mark); overall balance (1 mark)
Q7 (5 marks) — sample: Both propanoic acid and propan-1-ol possess an O–H bond capable of hydrogen bonding (1 mark). However, propanoic acid molecules form hydrogen-bonded dimers — two molecules align and simultaneously form two H-bonds: O–H (mol 1) ··· O=C (mol 2) and O=C (mol 1) ··· H–O (mol 2), creating a stable cyclic structure (1 mark). To vaporise propanoic acid, both H-bonds of the dimer must be broken simultaneously — the effective unit leaving the liquid is the dimer (effective mass ~148 g/mol), not the monomer (74 g/mol) (1 mark). Propan-1-ol forms only one H-bond per molecular interaction; breaking this single H-bond requires less energy (1 mark). Therefore, despite propanoic acid having a slightly higher molecular mass, the energy required to disrupt its dimer is far greater, giving it a 44°C higher boiling point than propan-1-ol (1 mark).
Q8 (6 marks) — sample:
Carboxylate vs alkoxide: When ethanoic acid loses H⁺, it forms the ethanoate ion (CH₃COO⁻). The negative charge is delocalised across both oxygen atoms by resonance: CH₃–C(=O)–O⁻ ↔ CH₃–C(–O⁻)=O — each oxygen carries approximately half a negative charge (2 marks). This resonance stabilisation makes the ethanoate ion a relatively weak base (less tendency to re-accept H⁺), so the equilibrium lies further toward ionisation — ethanoic acid is a stronger acid (pKa ~5) (1 mark). When ethanol loses H⁺, the ethoxide (CH₃CH₂O⁻) has the full negative charge localised on a single oxygen — no resonance stabilisation. This strong base readily re-accepts H⁺, so the equilibrium lies far left — ethanol is a very weak acid (pKa ~16) (1 mark).
Student claim evaluation: The student is correct (1 mark). Electronegative atoms like Cl attached to the alkyl chain exert an inductive effect — they withdraw electron density through the C–C bonds toward themselves, which partially stabilises the negative charge on the carboxylate oxygens beyond what resonance alone achieves (it reduces the electron density of the carboxyl group, making the O–H bond more polar and the conjugate base more stable). Chloroethanoic acid (CH₂ClCOOH) has pKa 2.86, significantly lower than ethanoic acid's 4.74, confirming it is a stronger acid (1 mark).
Now that you have studied carboxylic acids, revisit your Think First prediction. The reason ethanoic acid (118°C) has a dramatically higher boiling point than ethanol (78°C) — despite the acid being heavier — is dimerisation. The carboxyl group can simultaneously donate AND accept a H-bond, allowing two carboxylic acid molecules to lock together in a cyclic dimer via two H-bonds. Breaking this pair requires far more energy than breaking ethanol's single H-bond. This is something the simple –OH group in ethanol cannot do.
The acid strength difference (ethanoic acid pKa 4.74 vs ethanol pKa ~16) traces to the same structural feature: the adjacent C=O allows resonance in the carboxylate anion (charge on both O), making it a much weaker conjugate base than the ethoxide (charge on one O). One group — –COOH — gives carboxylic acids two major, interconnected advantages over alcohols.
Carboxylic Acids