Year 12 Chemistry Module 7 · Organic Chemistry ⏱ ~45 min 5 MC · 3 Short Answer Lesson 14 of 23 Core

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.

Today's hook: Ethanoic acid and ethanol both have an O–H bond, so why does the acid boil 40°C higher than the alcohol?

0/5TASKS

Connect: from wine to vinegar

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Wine Turns to Vinegar, A 40°C Mystery

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?

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Key Structures & Equations

General formula

R–COOH  (CₙH₂ₙO₂)

Carboxyl group = C=O + O–H on same carbon

Weak acid ionisation

R–COOH + H₂O ⇌ R–COO⁻ + H₃O⁺

Ka ≪ 1, partially ionises only

With NaHCO₃ (KEY diagnostic)

R–COOH + NaHCO₃ → R–COONa + H₂O + CO₂↑

Only carboxylic acids do this

pKa comparison

Carboxylic acid (~5) < Phenol (~10) < Alcohol (~16)

Lower pKa = stronger acid

Learning Intentions

goals

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

Scan these before reading

vocab

Carboxyl groupThe functional group –COOH, consisting of a carbonyl (C=O) and a hydroxyl (O–H) on the same carbon; defines carboxylic acids.

Carboxylic acidAn organic compound containing the –COOH group; general formula CₙH₂ₙO₂; named with the suffix –oic acid.

DimerA pair of carboxylic acid molecules held together by two simultaneous H-bonds in a cyclic 8-membered ring structure.

Weak acidAn acid that only partially ionises in water; Ka ≪ 1; carboxylic acids have Ka ~ 10⁻⁵.

Carboxylate ionThe conjugate base of a carboxylic acid, R–COO⁻; the negative charge is resonance-stabilised across both oxygen atoms.

Resonance stabilisationThe delocalisation of charge or electrons across multiple bonds or atoms, lowering the energy of a species and making it more stable.

NaHCO₃ diagnostic testAddition of sodium hydrogen carbonate solution; only carboxylic acids (pKa ~5) are strong enough to produce CO₂ gas, phenols and alcohols do not react.

pKaThe negative log of Ka; a lower pKa means a stronger acid. Ranking: carboxylic acid (~5) > phenol (~10) > alcohol (~16).

Core Concepts

Explain: carboxyl structure, naming and dimers

The Carboxyl Group, Structure and Unique Bonding

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.

Two ethanoic acid molecules pair through two simultaneous hydrogen bonds, doubling the effective mass.

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).

Must Do: In any structural formula of a carboxylic acid, draw the –COOH group with both the C=O (two lines from C to O) and the O–H (single bond from the second O to H) explicitly. The two oxygens are not equivalent: one is double-bonded, one carries a hydrogen.

Common Error: "Carboxylic acids have two OH groups and therefore form two H-bonds per molecule." The carboxyl group has only ONE O–H bond. The C=O oxygen does NOT have an H, it accepts H-bonds but cannot donate them. In a dimer, each molecule donates one H-bond and accepts one, the result is two H-bonds per dimer pair.

MICROTASK · +5 XP

Which statement about carboxylic acid dimerisation is correct?

Represent: explain boiling point and solubility trends

Physical Properties, Anomalously High Boiling Points

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

Ethane

−89°C (dispersion) Ethanal

20°C (dipole–dipole) Ethanol

78°C (H-bonding, 1/mol) Ethanoic acid

118°C (dimer, 2/pair)

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.

Must Do: When explaining why a carboxylic acid has a higher boiling point than an alcohol of the same chain length, you MUST mention dimerisation specifically, not just "stronger H-bonds." The key is that the carboxyl group forms TWO H-bonds per dimer pair simultaneously. Name the dimer, explain its formation, and link it to the higher energy required for vaporisation.

Common Error: "Carboxylic acids have higher boiling points because the –COOH group contains more oxygen atoms." The number of oxygen atoms is not a reliable predictor of boiling point. Always reason from IMF type and H-bond quantity, not atom count.

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.

Respond: write weak-acid equations and observations

Carboxylic Acids as Weak Acids, Reactions and Ka

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

        R–COOH + H₂O ⇌ R–COO⁻ + H₃O⁺

        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

Reaction 1, With NaOH (neutralisation)

R–COOH + NaOH → R–COONa + H₂O

No gas; salt + water

Reaction 2, With Na₂CO₃

2R–COOH + Na₂CO₃ → 2R–COONa + H₂O + CO₂↑

Effervescence (CO₂ bubbles)

Reaction 3, With NaHCO₃ (KEY DIAGNOSTIC)

R–COOH + NaHCO₃ → R–COONa + H₂O + CO₂↑

ONLY carboxylic acids produce CO₂ with NaHCO₃, phenols and alcohols do NOT

Reaction 4, With reactive metals (e.g. Mg)

2R–COOH + Mg → (R–COO)₂Mg + H₂↑

H₂ gas; metal dissolves; slightly exothermic

KEY Diagnostic: The reaction with NaHCO₃ producing CO₂ is the KEY test distinguishing carboxylic acids from phenols and alcohols. (1) Reacts with NaHCO₃ + CO₂? → carboxylic acid. (2) Reacts with Na₂CO₃ but NOT NaHCO₃? → phenol. (3) Reacts with neither? → alcohol.

Common Error: Students omit CO₂ from the NaHCO₃ equation. The full product set is carboxylate salt + water + CO₂. Without CO₂ your equation is unbalanced, an extra carbon and two extra oxygens are unaccounted for on the right side.

Safety: The carbonate/hydrogencarbonate test releases CO₂ gas, which can spray liquid out of a narrow tube, carry the test out in an open test tube pointed away from your face and others, and add the carbonate slowly to control the effervescence. Dilute carboxylic acids are irritants, and glacial (concentrated) ethanoic acid is corrosive with an irritating vapour, wear goggles, handle small volumes, and work in a ventilated area; rinse any acid splashes on skin immediately with plenty of water.

MICROTASK · +5 XP

A student adds NaHCO₃ solution to three compounds: propanoic acid, propan-1-ol, and phenol. Which gives effervescence?

Feedback: discriminate acids using structure and tests

Acid Strength Comparison, Carboxylic Acid > Phenol > Alcohol

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

Carboxylic acid

pKa ≈ 5 → STRONGEST

Phenol

pKa ≈ 10 → intermediate

Alcohol

pKa ≈ 16 → weakest

Structural Basis for the Ranking

Carboxylic acid

R–COO⁻
Charge on BOTH oxygens equally, high resonance stabilisation

Phenol

C₆H₅–O⁻
Charge spreads into ring carbons, partial delocalisation

Alcohol

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 ✗

Exam tip: For the two in-scope classes, the key discriminating test is NaHCO₃: only a carboxylic acid reacts with NaHCO₃ to release CO₂; an alcohol does not react. That single test separates the acid from the alcohol in most NSW identification questions.

Beyond the syllabus (extension): Aromatic compounds such as phenol (–OH attached to a benzene ring) are not part of NSW HSC Chemistry, so the phenol row of the matrix above is enrichment only. For interest: phenol is a weak acid that reacts with the stronger base Na₂CO₃ but not with NaHCO₃, placing it between carboxylic acids and alcohols in acidity.

Common Error: "Phenol is a type of alcohol because it has an –OH group." Phenol is NOT an alcohol, the –OH is attached to a benzene ring (aromatic carbon), not a tetrahedral carbon with only single bonds. This gives phenol fundamentally different properties: more acidic (pKa 10 vs 16), reacts with NaOH readily, and does not undergo the same oxidation reactions as alcohols.

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.

Apply: identify an unknown and justify each inference

Examples, Writing Equations, Explaining Properties, Identifying Unknowns

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)

(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.

Use the formula: CₙH₂ₙO₂ is consistent with a saturated carboxylic acid or ester.

Use the diagnostic test: CO₂ with NaHCO₃ identifies a carboxylic acid; an ester does not react.

Check the exclusions: negative Tollens' excludes an aldehyde and no bromine-water decolourisation supports a saturated compound.

Conclusion: X is butanoic acid or 2-methylpropanoic acid. The straight-chain isomer has the higher boiling point because its larger contact surface strengthens dispersion forces.

Why does the NaHCO₃ test distinguish a carboxylic acid from aldehydes, ketones and esters?

Practice: identify, compare and explain carboxylic acids

✍️ Fill in the Blanks +4 XP

Complete these statements about carboxylic acids:

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.

Activities

Acid Identification Flowchart

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.

BP Ranking and Explanation

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).

Check Your Understanding

Short Answer

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).

Wine Turns to Vinegar, The Answer

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:

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Review: retrieve the acid evidence

Check what actually stuck

Review, Rapid Recall

DRILL

Five Key Recall Questions

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).

I have completed Lesson 14, Carboxylic Acids