Cambridge AS & A Level9701

Carboxylic acids (33)

Chemistry 9701 Chapter Notes

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Carboxylic acidsEstersAcyl chlorides
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1. What Makes Carboxylic Acids Acidic?

Carboxylic acids contain the carboxyl functional group, -COOH. They are classified as weak acids because they only partially dissociate (ionise) when dissolved in water. An equilibrium is established where most molecules remain undissociated. For example, with ethanoic acid: CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq). The acidity arises from two main factors: 1) The highly electronegative oxygen atom of the carbonyl group (C=O) withdraws electron density from the O-H bond, weakening it and making the proton (H⁺) easier to release. 2) The resulting carboxylate anion (R-COO⁻) is stabilised by resonance. The negative charge is delocalised (spread out) over the two oxygen atoms, which makes the anion more stable and less likely to re-join with a proton.

R-COOH(aq) ⇌ R-COO⁻(aq) + H⁺(aq)

Key term

Weak Acid: An acid that only partially dissociates or ionises in an aqueous solution.

Examiner insight

Examiners expect students to clearly explain *both* the weakening of the O-H bond and the stabilisation of the carboxylate ion to gain full marks for explaining acidity.

Worked example 13 marks

Explain, with reference to its structure, why ethanoic acid (CH₃COOH) behaves as an acid in aqueous solution.

  1. 1

    Step 1: State that ethanoic acid is a weak acid and partially dissociates in water, releasing H⁺ ions: CH₃COOH ⇌ CH₃COO⁻ + H⁺.

  2. 2

    Step 2: Explain the first reason for acidity. The C=O group is electron-withdrawing. This pulls electron density away from the O-H bond, weakening it.

  3. 3

    Step 3: Explain the second reason. The negative charge on the resulting ethanoate ion (CH₃COO⁻) is delocalised across the two oxygen atoms.

  4. 4

    Step 4: Conclude that this delocalisation stabilises the ethanoate ion, making the dissociation more favourable and allowing the H⁺ to be released.

Recap

  • Carboxylic acids contain the -COOH functional group.
  • They are weak acids, meaning they only partially ionise in water.
  • The O-H bond is weakened by the electron-withdrawing C=O group.
  • The carboxylate anion is stabilised by the delocalisation of the negative charge.
  • This stability is the primary reason for their acidic properties.

Quick check

  1. Draw the displayed formula of the ethanoate ion and show the delocalisation of the negative charge.2 marks

2. How Substituents Affect Acid Strength

The strength of a carboxylic acid is significantly influenced by other atoms or groups attached to the carbon chain, particularly on the carbon atom adjacent to the -COOH group (the alpha-carbon). Electron-withdrawing groups, such as chlorine atoms, pull electron density away from the carboxylate group through the carbon chain (a negative inductive effect). This further spreads out the negative charge on the carboxylate ion, increasing its stability. A more stable anion means the acid is more likely to dissociate, so the acid becomes stronger. The more electron-withdrawing groups present, or the more electronegative they are, the stronger the acid. For example, trichloroethanoic acid (CCl₃COOH) is a much stronger acid than ethanoic acid (CH₃COOH). Conversely, electron-donating groups, like alkyl groups, push electron density towards the carboxylate group, destabilising the anion and making the acid weaker.

Ka = ([RCOO⁻][H⁺]) / [RCOOH]

Order of increasing acidity: CH₃COOH < CH₂ClCOOH < CHCl₂COOH < CCl₃COOH

Key term

Inductive Effect: The transmission of charge through a chain of atoms in a molecule, resulting in a permanent dipole in a bond.

Common pitfall

Confusing electron-withdrawing groups with electron-donating groups and their respective effects on acid strength. Remember: withdrawing = stronger acid.

Worked example 14 marks

Place the following acids in order of increasing strength and explain your reasoning: ethanoic acid, chloroethanoic acid, dichloroethanoic acid.

  1. 1

    Step 1: State the correct order of increasing acid strength: ethanoic acid < chloroethanoic acid < dichloroethanoic acid.

  2. 2

    Step 2: Explain the effect of the chlorine atom. Chlorine is an electronegative atom that acts as an electron-withdrawing group.

  3. 3

    Step 3: Describe the mechanism. The chlorine atom(s) pull electron density away from the -COO⁻ group of the carboxylate ion via a negative inductive effect.

  4. 4

    Step 4: Link this effect to stability and strength. This delocalises the negative charge, stabilising the anion. Dichloroethanoic acid has two chlorine atoms, so the effect is greater, the anion is more stable, and the acid is stronger than chloroethanoic acid, which is stronger than ethanoic acid (which has no chlorine atoms).

Recap

  • Acid strength depends on the stability of the carboxylate anion.
  • Electron-withdrawing groups (e.g., Cl, F) increase acid strength.
  • These groups stabilise the anion by pulling electron density away from it (negative inductive effect).
  • Electron-donating groups (e.g., CH₃) decrease acid strength.
  • The more electron-withdrawing groups there are, the stronger the acid.

Quick check

  1. Which is the stronger acid: 2-chlorobutanoic acid or 3-chlorobutanoic acid? Explain your answer.2 marks

3. Reactions as Typical Acids

Because they release H⁺ ions in solution, carboxylic acids undergo the typical reactions of any acid, although the reactions may be slower than with strong acids like HCl. They react with reactive metals, bases (alkalis), and carbonates to form a carboxylate salt. The name of the salt is derived from the acid, for example, ethanoic acid forms ethanoate salts.

  1. With reactive metals (e.g., Mg, Zn, Na): Produces a salt and hydrogen gas.
  2. With bases/alkalis (e.g., NaOH, CaO): A neutralisation reaction occurs, producing a salt and water.
  3. With carbonates (e.g., Na₂CO₃, K₂CO₃): Produces a salt, water, and carbon dioxide gas. The production of CO₂ (effervescence) is a useful test for the carboxylic acid group, as alcohols do not react this way.

2RCOOH + Mg → (RCOO)₂Mg + H₂(g)

RCOOH + NaOH → RCOONa + H₂O(l)

2RCOOH + Na₂CO₃ → 2RCOONa + H₂O(l) + CO₂(g)

Key term

Carboxylate Salt: The salt formed when a carboxylic acid reacts with a base, metal, or carbonate, where the acidic proton is replaced by a metal ion.

Fun fact

Vinegar is a dilute solution of ethanoic acid. The reaction of vinegar with sodium bicarbonate (a carbonate) is the basis for the classic 'volcano' science experiment.

Worked example 12 marks

Write a balanced chemical equation for the reaction between propanoic acid (CH₃CH₂COOH) and solid potassium carbonate (K₂CO₃).

  1. 1

    Step 1: Identify the reactants: propanoic acid (an acid) and potassium carbonate (a carbonate).

  2. 2

    Step 2: Recall the general products for an acid + carbonate reaction: salt + water + carbon dioxide.

  3. 3

    Step 3: Determine the formula of the salt. Propanoic acid (CH₃CH₂COOH) forms the propanoate ion (CH₃CH₂COO⁻). Potassium is K⁺. The salt is potassium propanoate, CH₃CH₂COOK.

  4. 4

    Step 4: Write the unbalanced equation: CH₃CH₂COOH + K₂CO₃ → CH₃CH₂COOK + H₂O + CO₂.

  5. 5

    Step 5: Balance the equation. There are 2 K atoms on the left, so we need 2 CH₃CH₂COOK on the right. This requires 2 CH₃CH₂COOH on the left. The final balanced equation is: 2CH₃CH₂COOH + K₂CO₃ → 2CH₃CH₂COOK + H₂O + CO₂.

Recap

  • Carboxylic acids react with reactive metals to form a salt and hydrogen.
  • They neutralise bases and alkalis to form a salt and water.
  • They react with carbonates to form a salt, water, and carbon dioxide.
  • The reaction with carbonates (producing fizzing) is a key test for a carboxylic acid.

Quick check

  1. What are the names of the three products formed when ethanoic acid reacts with magnesium carbonate?3 marks

4. Synthesising Carboxylic Acids

There are two principal methods for preparing carboxylic acids in the laboratory.

  1. Oxidation of Primary Alcohols and Aldehydes: Primary alcohols can be oxidised to aldehydes, which can then be further oxidised to carboxylic acids. To ensure the final product is the carboxylic acid, a strong oxidising agent (like acidified potassium dichromate(VI), K₂Cr₂O₇/H₂SO₄) is used, and the mixture is heated under reflux. Refluxing prevents any volatile aldehyde from escaping, ensuring it is fully oxidised. The colour change observed is from orange (Cr₂O₇²⁻) to green (Cr³⁺).
  2. Hydrolysis of Nitriles: Nitriles (compounds containing the -C≡N group) can be converted to carboxylic acids by heating them under reflux with either dilute acid (e.g., HCl) or dilute alkali (e.g., NaOH). This reaction is called hydrolysis. Acid hydrolysis produces the carboxylic acid directly, along with an ammonium salt. Alkaline hydrolysis produces a carboxylate salt, which must then be acidified in a separate step to form the free carboxylic acid.

RCH₂OH + 2[O] --(reflux)--> RCOOH + H₂O

RCN + 2H₂O + H⁺ --(reflux)--> RCOOH + NH₄⁺

RCN + OH⁻ + H₂O --(reflux)--> RCOO⁻ + NH₃ then RCOO⁻ + H⁺ → RCOOH

Key term

Reflux: The process of heating a chemical reaction for a specific amount of time, while continually cooling the vapour produced back into liquid form, using a condenser, so it returns to the reaction.

Examiner insight

For the oxidation route, specifying the conditions (reflux, excess oxidising agent) is crucial for distinguishing it from the formation of an aldehyde, which requires distillation.

Worked example 14 marks

Describe how you would prepare a sample of propanoic acid starting from propan-1-ol. Include essential reagents, conditions and a balanced equation.

  1. 1

    Step 1: Identify the transformation: Primary alcohol → Carboxylic acid. This requires oxidation.

  2. 2

    Step 2: State the reagents: Propan-1-ol and a suitable oxidising agent, such as potassium dichromate(VI) acidified with dilute sulfuric acid.

  3. 3

    Step 3: State the conditions: The mixture must be heated under reflux to ensure complete oxidation and prevent the intermediate aldehyde from being lost.

  4. 4

    Step 4: Write the balanced equation using [O] to represent the oxidising agent: CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O.

  5. 5

    Step 5: Note the observation: The colour of the solution will change from orange to green.

Recap

  • Carboxylic acids are formed by oxidising primary alcohols or aldehydes.
  • This oxidation requires a strong oxidising agent (e.g., K₂Cr₂O₇/H⁺) and heating under reflux.
  • Carboxylic acids can also be formed by the acid or alkaline hydrolysis of nitriles.
  • Hydrolysis of nitriles increases the carbon chain length by one.

Quick check

  1. What is the organic product of the acid hydrolysis of butanenitrile?1 mark

5. Reduction to Alcohols and Esterification

Carboxylic acids can undergo two other important transformations: reduction and esterification.

  1. Reduction: Carboxylic acids are resistant to reduction by common reducing agents. They can only be reduced to primary alcohols using a very powerful reducing agent, lithium tetrahydridoaluminate (LiAlH₄). The reaction is typically carried out in a dry ether solvent, as LiAlH₄ reacts violently with water. The overall reaction can be represented as adding four hydrogen atoms. Note that sodium borohydride (NaBH₄) is not strong enough to reduce carboxylic acids.
  2. Esterification: This is a condensation reaction between a carboxylic acid and an alcohol to form an ester and water. The reaction is reversible and has a slow rate at room temperature. To improve the yield and rate, the reaction is heated under reflux with a strong acid catalyst, typically concentrated sulfuric acid. The catalyst both speeds up the reaction and absorbs the water produced, helping to shift the equilibrium to the right.

RCOOH + 4[H] --(1. LiAlH₄ in dry ether, 2. H₂O)--> RCH₂OH + H₂O

RCOOH + R'OH ⇌ RCOOR' + H₂O (with H⁺ catalyst and heat)

Key term

Esterification: The reaction between an alcohol and a carboxylic acid (or its derivative) to form an ester and water.

Common pitfall

Using NaBH₄ to reduce a carboxylic acid – it is not a strong enough reducing agent. Only LiAlH₄ is suitable for this transformation.

Worked example 12 marks

Butanoic acid reacts with ethanol in the presence of an acid catalyst. Draw the structure of the organic product and name it.

  1. 1

    Step 1: Identify the reactants: Butanoic acid (CH₃CH₂CH₂COOH) and ethanol (CH₃CH₂OH).

  2. 2

    Step 2: Understand the reaction: This is esterification. The -OH from the carboxylic acid and the -H from the alcohol's -OH group are removed to form water.

  3. 3

    Step 3: Join the remaining fragments. The C=O of the butanoic acid fragment joins to the oxygen of the ethanol fragment.

  4. 4

    Step 4: Draw the structure: CH₃CH₂CH₂COOCH₂CH₃.

  5. 5

    Step 5: Name the ester. The first part of the name comes from the alcohol ('ethyl' from ethanol). The second part comes from the carboxylic acid ('butanoate' from butanoic acid). The name is ethyl butanoate.

Worked example 22 marks

What reagent is needed to convert propanoic acid to propan-1-ol, and what conditions are required?

  1. 1

    Step 1: Identify the reaction type: This is a reduction of a carboxylic acid to a primary alcohol.

  2. 2

    Step 2: State the reagent: A powerful reducing agent is needed, which is lithium tetrahydridoaluminate (LiAlH₄).

  3. 3

    Step 3: State the conditions: The reaction must be carried out in a non-aqueous solvent, such as dry ether, followed by an aqueous workup.

Recap

  • Carboxylic acids are reduced to primary alcohols by LiAlH₄ in dry ether.
  • Weaker reducing agents like NaBH₄ do not work.
  • Esterification is the reversible reaction of a carboxylic acid and an alcohol.
  • Esterification requires heat (reflux) and a strong acid catalyst (e.g., conc. H₂SO₄).

Quick check

  1. Name the two organic compounds that react to form the ester methyl propanoate.2 marks

6. Formation of Acyl Chlorides

Acyl chlorides (or acid chlorides) are highly reactive derivatives of carboxylic acids, with the general formula R-COCl. They are formed by replacing the -OH group of the carboxylic acid with a -Cl atom. This is a key step in synthesis as acyl chlorides are much more reactive than the parent carboxylic acids. There are three common reagents used for this conversion:

  1. Thionyl chloride (SOCl₂): This is often the preferred method as the other products, sulfur dioxide (SO₂) and hydrogen chloride (HCl), are gases and can be easily removed from the reaction mixture, leaving a purer product.
  2. Phosphorus(V) chloride (PCl₅): This solid reagent reacts vigorously at room temperature, producing the acyl chloride, phosphorus oxychloride (POCl₃) and hydrogen chloride gas.
  3. Phosphorus(III) chloride (PCl₃): This liquid reagent also works, but requires gentle heating. The stoichiometry is different, with 3 moles of carboxylic acid reacting per mole of PCl₃.

RCOOH + SOCl₂ → RCOCl + SO₂(g) + HCl(g)

RCOOH + PCl₅ → RCOCl + POCl₃(l) + HCl(g)

3RCOOH + PCl₃ → 3RCOCl + H₃PO₃(aq)

Key term

Acyl Chloride: A reactive organic compound containing the functional group -COCl, derived from a carboxylic acid by replacing the hydroxyl group with a chlorine atom.

Examiner insight

Knowing the balanced equations for all three reagents is important, as any could appear on an exam. Pay particular attention to the different by-products and stoichiometries.

Worked example 12 marks

Write a balanced equation for the preparation of propanoyl chloride from propanoic acid using thionyl chloride.

  1. 1

    Step 1: Write the formula for propanoic acid: CH₃CH₂COOH.

  2. 2

    Step 2: Write the formula for propanoyl chloride: CH₃CH₂COCl.

  3. 3

    Step 3: Recall the general equation for the reaction with thionyl chloride: RCOOH + SOCl₂ → RCOCl + SO₂ + HCl.

  4. 4

    Step 4: Substitute the specific 'R' group (CH₃CH₂-) into the equation: CH₃CH₂COOH + SOCl₂ → CH₃CH₂COCl + SO₂ + HCl.

Recap

  • Acyl chlorides are reactive derivatives of carboxylic acids.
  • They are formed by replacing the -OH group with a -Cl atom.
  • Common reagents are thionyl chloride (SOCl₂), phosphorus(V) chloride (PCl₅), and phosphorus(III) chloride (PCl₃).
  • SOCl₂ is often preferred because its by-products are gaseous and easy to remove.

Quick check

  1. Name the liquid by-product formed when ethanoyl chloride is made using PCl₅.1 mark

7. The Vigorous Reactions of Acyl Chlorides

Acyl chlorides are among the most reactive organic functional groups. Their high reactivity is due to the large partial positive charge (δ+) on the carbonyl carbon, which is made highly electron-deficient by both the electronegative oxygen and chlorine atoms. This makes it an excellent target for nucleophiles. The reactions are typically nucleophilic addition-elimination reactions and are much faster and more vigorous than with carboxylic acids. Key reactions include:

  1. Hydrolysis: A violent reaction with cold water to form the parent carboxylic acid and misty fumes of hydrogen chloride gas. RCOCl + H₂O → RCOOH + HCl.
  2. Reaction with Alcohols: A vigorous, irreversible reaction to form an ester and HCl. This is a more efficient way to make esters than direct esterification, giving a higher yield. RCOCl + R'OH → RCOOR' + HCl.
  3. Reaction with Phenols: Similar to alcohols, phenols react with acyl chlorides to form phenyl esters and HCl.
  4. Reaction with Amines and Ammonia: A violent reaction with ammonia or primary amines to form amides and HCl. The HCl produced will then react with any excess amine to form a salt. RCOCl + 2NH₂R' → RCONHR' + R'NH₃⁺Cl⁻.

RCOCl + H₂O → RCOOH + HCl

RCOCl + R'OH → RCOOR' + HCl

RCOCl + C₆H₅OH → RCOOC₆H₅ + HCl

RCOCl + 2R'NH₂ → RCONHR' + R'NH₃⁺Cl⁻

Key term

Nucleophilic Addition-Elimination: A two-step reaction mechanism where a nucleophile adds to an unsaturated electrophile, followed by the elimination of a leaving group.

Common pitfall

Forgetting that the reactions of acyl chlorides produce HCl as a by-product. In the case of reaction with amines, this HCl reacts with a second molecule of the amine.

Worked example 13 marks

Ethanoyl chloride is added to methylamine (CH₃NH₂). Identify the products and write an equation for the overall reaction.

  1. 1

    Step 1: Identify the reactants: an acyl chloride (ethanoyl chloride, CH₃COCl) and a primary amine (methylamine, CH₃NH₂).

  2. 2

    Step 2: Predict the initial products. The nucleophilic amine attacks the carbonyl carbon, and HCl is eliminated. This forms an N-substituted amide, N-methylethanamide (CH₃CONHCH₃), and hydrogen chloride (HCl).

  3. 3

    Step 3: Consider the side reaction. Methylamine is a base, and the HCl produced is an acid. They will react together: CH₃NH₂ + HCl → CH₃NH₃⁺Cl⁻.

  4. 4

    Step 4: Write the overall equation. One mole of methylamine is used to form the amide, and a second mole is used to neutralise the HCl. The overall equation is: CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃⁺Cl⁻.

Recap

  • Acyl chlorides are very reactive due to the highly electron-deficient carbonyl carbon.
  • They react vigorously with nucleophiles like water, alcohols, phenols, and amines.
  • All these reactions produce misty fumes of hydrogen chloride gas.
  • Reactions with acyl chlorides are faster and less reversible than with carboxylic acids.

Quick check

  1. What two products are formed when propanoyl chloride reacts with ethanol?2 marks

8. Why Acyl Chlorides Hydrolyse Easily

The rate at which different types of organic chlorides hydrolyse (react with water) varies dramatically. The order of reactivity is: Acyl chloride >> Alkyl chloride > Aryl chloride.

  • Acyl Chlorides (e.g., ethanoyl chloride, CH₃COCl): These hydrolyse extremely rapidly, even in cold water. The carbonyl carbon is strongly δ+ because it is bonded to two highly electronegative atoms (O and Cl). This makes it very attractive to the lone pair on the oxygen of a water molecule (a nucleophile). The reaction proceeds via a fast nucleophilic addition-elimination mechanism.
  • Alkyl Chlorides (e.g., chloroethane, CH₃CH₂Cl): These hydrolyse much more slowly, usually requiring heat and often a catalyst (like NaOH(aq) in nucleophilic substitution). The carbon bonded to chlorine is δ+, but less so than in an acyl chloride as it's only attached to one electronegative atom.
  • Aryl Chlorides (e.g., chlorobenzene, C₆H₅Cl): These are extremely resistant to hydrolysis and do not react with water under normal conditions. This is because the lone pair of electrons on the chlorine atom is delocalised into the pi-system of the benzene ring. This gives the C-Cl bond partial double bond character, making it much stronger and harder to break.

Key term

Delocalisation: The phenomenon where electrons are shared among three or more atoms in a molecule, rather than being localised between two atoms.

Examiner insight

Marks are often awarded for comparing the magnitude of the partial positive charge (δ+) on the carbon atom being attacked in each type of chloride, and for explaining the stability of the C-Cl bond in aryl chlorides.

Worked example 14 marks

Explain the difference in reactivity towards water for ethanoyl chloride and chloroethane.

  1. 1

    Step 1: State the relative reactivity. Ethanoyl chloride reacts vigorously with cold water, while chloroethane only hydrolyses slowly on heating.

  2. 2

    Step 2: Compare the electrophilicity of the carbon atom. In ethanoyl chloride, the carbonyl carbon is attached to two electronegative atoms (O and Cl), making it highly electron-deficient (very δ+).

  3. 3

    Step 3: In chloroethane, the carbon is attached to only one electronegative atom (Cl), so it is less electron-deficient (less δ+).

  4. 4

    Step 4: Relate electrophilicity to the reaction. The highly δ+ carbon in ethanoyl chloride is much more readily attacked by the lone pair on the nucleophile (water) than the carbon in chloroethane, leading to a much faster reaction.

Recap

  • Reactivity towards hydrolysis: Acyl chloride >> Alkyl chloride > Aryl chloride.
  • Acyl chlorides are most reactive because the carbonyl carbon is highly electron-deficient (δ+).
  • Aryl chlorides are least reactive because delocalisation gives the C-Cl bond partial double bond character.
  • The mechanism for acyl chloride hydrolysis is nucleophilic addition-elimination.

Quick check

  1. Place the following in order of increasing rate of hydrolysis: chlorobenzene, propanoyl chloride, 1-chloropropane.1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Draw the structure of 2,2-dichloropropanoic acid and explain why it is a stronger acid than propanoic acid.4 marks
  2. Write a balanced chemical equation for the reaction of butanoic acid with magnesium carbonate.2 marks
  3. Describe how you would convert propan-1-ol into propanoic acid in a laboratory. Include reagents, conditions, and the type of reaction.4 marks
  4. Ethanoyl chloride reacts vigorously with water. Name the organic product and the other product, and outline the mechanism of this reaction using curly arrows.5 marks
  5. Give the structural formula of the ester formed when methanol reacts with propanoic acid. Name the ester and state the necessary conditions for the reaction.3 marks
  6. A student suggests reducing pentanoic acid using sodium borohydride (NaBH₄). Explain why this reaction will not be successful and suggest a suitable reagent and condition.3 marks
  7. Starting from ethanenitrile, CH₃CN, outline a two-step synthesis to produce the ester ethyl ethanoate. You should state the reagents and conditions for each step.5 marks
  8. Write an equation for the formation of butanoyl chloride from butanoic acid using phosphorus(V) chloride.2 marks
  9. Explain in terms of electronic structure why ethanoyl chloride is readily hydrolysed by water but chlorobenzene is not.4 marks
  10. A carboxylic acid X has the empirical formula CH₂O and a relative molecular mass of 90.0. Deduce the molecular formula of X and draw the structure of a possible isomer of X.3 marks

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