Carboxylic Acids, Structure, Properties & Reactions
Discover how a single functional group, the carboxyl, unlocks acid chemistry, anomalous boiling points, and the diagnostic tests that identify organic acids in the lab.
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Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
Ethanoic acid (acetic acid) and ethanol are both two-carbon compounds with oxygen. 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?
Know
- The carboxyl group structure (C=O + O–H on same C)
- General formula CₙH₂ₙO₂; suffix –oic acid
- Carboxylic acids form H-bonded dimers
- Ka ≪ 1, weak acid, partially ionises
- Acid strength: carboxylic acid > phenol > alcohol
Understand
- Why dimerisation raises BP above alcohols of same chain
- Why R–COO⁻ is resonance-stabilised and more stable than R–O⁻
- Why NaHCO₃ distinguishes carboxylic acids from phenols and alcohols
Can Do
- Write IUPAC names and formulae for C1–C5 carboxylic acids
- Write balanced equations for reactions with NaOH, Na₂CO₃, NaHCO₃, Mg
- Explain dimerisation and link to anomalous BPs
- Apply the NaHCO₃ diagnostic in identification questions
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.
Structure of the Carboxyl Group
The carboxyl group consists of a carbonyl (C=O) and a hydroxyl (–OH) bonded to the same carbon. That carbon has a double bond, trigonal planar geometry, ~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.
Two H-Bonding Roles in One Group
Each carboxylic acid molecule can both donate and accept H-bonds:
- O–H bond → H-bond donor: the O–H hydrogen is δ⁺. The adjacent C=O withdraws electron density, making the O–H bond more polar and the H more acidic than in a simple alcohol.
- C=O group → H-bond acceptor: lone pairs on the carbonyl oxygen accept H-bonds (same as in aldehydes and ketones).
Dimerisation
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, 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.
IUPAC Naming
Suffix: –oic acid. The –COOH carbon is always C1 (no locant needed). The chain length includes the carbonyl carbon. Examples: methanoic acid (HCOOH), ethanoic acid (CH₃COOH), propanoic acid (CH₃CH₂COOH).
MICROTASK · +5 XP
Which statement about carboxylic acid dimerisation is correct?
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."
Boiling Point Trend (same carbon number)
Alkane < Aldehyde ≈ Ketone < Alcohol < Carboxylic acid
Why Carboxylic Acid BP > Alcohol BP (same chain length)
Both ethanoic acid and ethanol have one O–H bond and can form hydrogen bonds. The difference is dimerisation:
- Two carboxylic acid molecules form a cyclic dimer with two simultaneous H-bonds. To vaporise ethanoic acid, both H-bonds of a dimer must be broken simultaneously.
- The energy required to disrupt this doubly-bonded pair is significantly greater than the energy to break the single O–H H-bond between two ethanol molecules.
- The effective unit leaving the liquid surface is the dimer (~120 g/mol for ethanoic acid), not the monomer (60 g/mol). This is confirmed by vapour density measurements near the boiling point.
Water Solubility
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.
MICROTASK · +5 XP
True or False: Ethanoic acid (BP 118°C) has a higher boiling point than ethanol (BP 78°C) because the individual O–H hydrogen bonds in ethanoic acid are stronger than those in ethanol.
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.
Weak Acid Ionisation
Ka = [R–COO⁻][H₃O⁺] / [R–COOH] ≪ 1
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.
Why Carboxylic Acids Are Stronger Acids Than Alcohols
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, the alkoxide (R–O⁻) has the negative charge localised on one oxygen, no resonance stabilisation.
The Four Reactions
MICROTASK · +5 XP
A student adds NaHCO₃ solution to three compounds: propanoic acid, propan-1-ol, and phenol. Which gives effervescence?
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.
The pKa Ranking
Structural Basis for the Ranking
R–COO⁻Charge on BOTH oxygens equally, high resonance stabilisation
C₆H₅–O⁻Charge spreads into ring carbons, partial delocalisation
R–O⁻Charge on ONE oxygen only, no resonance, very unstable
The NaHCO₃/Na₂CO₃ Discrimination Ladder
| 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 ✗ |
MICROTASK · +5 XP
Fill in the blank: The carboxylate anion (R–COO⁻) is a stronger acid's conjugate base than the alkoxide (R–O⁻) because the negative charge in R–COO⁻ is _______ across both oxygen atoms by resonance, making it a more stable, weaker base.
Example 1, Equations for Propanoic Acid Reactions
Given: Propanoic acid (CH₃CH₂COOH). Write balanced equations for reactions with NaOH, Na₂CO₃, NaHCO₃, and Mg.
(a) CH₃CH₂COOH + NaOH → CH₃CH₂COONa + H₂O (no gas)
(b) 2CH₃CH₂COOH + Na₂CO₃ → 2CH₃CH₂COONa + H₂O + CO₂(g)
(c) CH₃CH₂COOH + NaHCO₃ → CH₃CH₂COONa + H₂O + CO₂(g)
(d) 2CH₃CH₂COOH + Mg → (CH₃CH₂COO)₂Mg + H₂(g)
Example 2, BP Comparison and Acid Strength
BP: 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.
Acid strength: Ethanoate ion (CH₃COO⁻) is resonance-stabilised, charge delocalised across both oxygens → weaker conjugate base → stronger acid (pKa ~5). Ethoxide (CH₃CH₂O⁻) has charge on one O only, no resonance → stronger base → ethanol is much weaker acid (pKa ~16).
Example 3, Identifying an Unknown from Test Results
Given: Unknown X: C₄H₈O₂, NaHCO₃ → CO₂ (effervescence), Tollens' → negative, Br₂ water → no decolourisation.
Identify: CₙH₂ₙO₂ = carboxylic acid or ester. NaHCO₃ + CO₂ → must be a carboxylic acid (esters don't react). Tollens' negative → not an aldehyde. No Br₂ decolourisation → saturated. → X is a saturated carboxylic acid.
Structural isomers: Butanoic acid (CH₃CH₂CH₂COOH) and 2-methylpropanoic acid ((CH₃)₂CHCOOH). Butanoic acid has higher BP (straight chain, greater dispersion forces).
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Carboxylic acids contain the functional group ____. They are weak acids because they only ____ ionise in water. Carboxylic acids react with alcohols in the presence of an acid catalyst to form ____ and water in a condensation reaction. With Na₂CO₃, carboxylic acids produce a salt, water, and ____ gas.
Three unknown compounds (P, Q, R) each have molecular formula C₃H₆O₂. The following tests are carried out:
| 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 compounds in order of increasing boiling point and provide a brief justification for each ranking step: propane (C₃H₈, M = 44), propanal (C₃H₆O, M = 58), propan-1-ol (C₃H₈O, M = 60), and propanoic acid (C₃H₆O₂, M = 74).
Q1. 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?
Q2. 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?
Q3. Which structural feature of the carboxylate ion (R–COO⁻) makes carboxylic acids stronger acids than alcohols?
Q4. A compound with molecular formula C₃H₆O₂ produces CO₂ when added to NaHCO₃ solution and gives no reaction with Tollens' reagent. Which compound is it?
Q5. 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. 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.
Reveal Answers
Multiple Choice Answers
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.
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.
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).
Short Answer Sample Answers
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 + CO₂ | Observable: effervescence/CO₂ bubbles (1 mark)
(c) 2CH₃CH₂CH₂COOH + Mg → (CH₃CH₂CH₂COO)₂Mg + H₂(g) | Products: magnesium butanoate + H₂ gas | Observable: Mg dissolves; H₂ gas bubbles (1 mark); overall balance (1 mark)
Q7 (5 marks): 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 (~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 (1 mark).
Q8 (6 marks): 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, 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 → strong base → equilibrium far left → ethanol is a very weak acid (pKa ~16) (1 mark). The student is correct (1 mark). Electronegative Cl exerts an inductive effect, withdrawing electron density through C–C bonds partially stabilises the carboxylate anion beyond resonance alone. Chloroethanoic acid (pKa 2.86) confirms this is a stronger acid than ethanoic acid (pKa 4.74) (1 mark).
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).
Now review your Think First prediction below:
1. What does the NaHCO₃ test uniquely identify, and what is the observable result?
Show answer
The NaHCO₃ test uniquely identifies carboxylic acids. Adding NaHCO₃ to a carboxylic acid produces effervescence (CO₂ bubbles), the only organic functional group class acidic enough (pKa ~5) to react with HCO₃⁻ and release CO₂. Phenols and alcohols do not react.
2. Why does carboxylic acid have a higher boiling point than the corresponding alcohol (same carbon number)?
Show answer
Carboxylic acids form hydrogen-bonded dimers, two molecules lock together via two simultaneous H-bonds in a cyclic structure. To vaporise, both H-bonds of the dimer must break simultaneously. Alcohols form only one H-bond per interaction. The extra energy required to break two bonds elevates the carboxylic acid's boiling point.
3. Why is the carboxylate anion (R–COO⁻) more stable than the alkoxide anion (R–O⁻)?
Show answer
In R–COO⁻, the negative charge is delocalised across both oxygen atoms by resonance (R–C(=O)–O⁻ ↔ R–C(–O⁻)=O). Each oxygen bears ~½ negative charge, a lower energy arrangement. In R–O⁻, the full negative charge is localised on one oxygen, higher energy, less stable, stronger base.
4. Write the balanced equation for propanoic acid reacting with Na₂CO₃.
Show answer
2CH₃CH₂COOH + Na₂CO₃ → 2CH₃CH₂COONa + H₂O + CO₂(g). Products: sodium propanoate + water + carbon dioxide gas. Observable: effervescence as CO₂ is produced.
5. Give the acid strength ranking of carboxylic acid, phenol, and alcohol with approximate pKa values.
Show answer
Carboxylic acid (pKa ~5) > Phenol (pKa ~10) > Alcohol (pKa ~16). Carboxylic acid is strongest (resonance in carboxylate across 2 O), phenol intermediate (partial delocalisation into benzene ring), alcohol weakest (charge localised on 1 O, no resonance).