Carboxylic acids

1. Overview

Carboxylic acids contain the carboxyl group, -COOH. This section covers three ways to make them (oxidising primary alcohols or aldehydes, hydrolysing nitriles, hydrolysing esters) and five of their reactions: with reactive metals, alkalis, carbonates, alcohols and LiAlH₄. Carboxylic acids link many other functional groups, so these reagents and equations recur across organic chemistry.

Key Definitions

• Carboxylic acid: an organic compound containing the carboxyl group, -COOH, in which one carbon atom is bonded both to an oxygen atom by a C=O double bond and to an -OH group.
• Carboxylate ion: the ion RCOO⁻ formed when a carboxylic acid loses the hydrogen ion from its -COOH group; its salts are named with the ending -oate, for example sodium ethanoate, CH₃COONa.
• Weak acid: an acid that is only partially dissociated into ions in aqueous solution; carboxylic acids are weak acids.
• Heating under reflux: heating a reaction mixture in a flask fitted with a vertical condenser, so that vapours condense and run back into the flask; the mixture can be heated for a long time without losing volatile reactants or products.
• Hydrolysis: a reaction in which a compound is broken down by reaction with water; nitriles and esters are hydrolysed by heating with dilute acid or dilute alkali.
• Acidification: adding a dilute strong acid, such as dilute hydrochloric acid, to a solution of a carboxylate salt so that the carboxylate ions gain hydrogen ions and the carboxylic acid forms.
• Esterification: the reversible reaction of a carboxylic acid with an alcohol, with concentrated sulfuric acid as a catalyst, to form an ester and water; it is a condensation reaction.

Content

The carboxyl group

In the carboxyl group one carbon atom carries both a C=O double bond and an -OH group. Write it as -COOH or -CO₂H, so a general carboxylic acid is RCOOH. Names end in -oic acid, and the carbon of -COOH counts in the chain: CH₃COOH is ethanoic acid and CH₃CH₂COOH is propanoic acid.

Carboxylic acids are weak acids (7.2): only a small fraction of the molecules dissociate in water:

CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq)

The ion formed is a carboxylate ion. Its salts end in -oate, such as sodium ethanoate.

Making carboxylic acids: the three routes

Start from	Reagents and conditions	Product
primary alcohol or aldehyde	acidified K₂Cr₂O₇ or KMnO₄, heat under reflux	RCOOH
nitrile, RCN	heat under reflux with dilute acid; or with dilute alkali, then acidify	RCOOH
ester, RCOOR′	heat under reflux with dilute acid; or with dilute alkali, then acidify	RCOOH and R′OH

Route 1: oxidising primary alcohols and aldehydes

Heat the alcohol or aldehyde under reflux with an excess of acidified potassium dichromate(VI), K₂Cr₂O₇, or acidified potassium manganate(VII), KMnO₄. "Acidified" means dilute sulfuric acid is added. Dichromate turns from orange to green; manganate(VII) turns from purple to colourless.

The alcohol is oxidised first to an aldehyde, then to the acid. In equations, [O] stands for one oxygen atom from the oxidising agent; balance it with a number.

CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O

CH₃CH₂CHO + [O] → CH₃CH₂COOH

The acid has the same number of carbon atoms as the alcohol. Secondary alcohols give ketones and tertiary alcohols are not oxidised (16.1).

Heating under reflux

The aldehyde formed first boils at a lower temperature than the alcohol or the acid: its molecules have no O-H group, so they cannot hydrogen bond to each other. In an open flask much of it would escape before being oxidised (distilling it off on purpose is how aldehydes are made, 17.1). Under reflux:

1. A vertical condenser, open at the top, is fitted to the flask.
2. Vapour rises, condenses and drips back into the flask.
3. The aldehyde stays with the oxidising agent until it becomes the acid.

Cold water enters the condenser at the bottom and leaves at the top. Reflux keeps the volatile organic substances in the flask; the dissolved oxidising agent is not volatile.

Route 2: hydrolysing nitriles

A nitrile contains the -C≡N group. Heat it under reflux with dilute acid or dilute alkali. The carbon of -C≡N becomes the carbon of -COOH.

With dilute acid, such as dilute HCl, the nitrogen ends up as ammonium ions:

CH₃CH₂CN + 2H₂O + H⁺ → CH₃CH₂COOH + NH₄⁺

With dilute alkali, such as NaOH(aq), the carboxylate ion forms and ammonia is given off (it turns damp red litmus blue):

CH₃CH₂CN + H₂O + OH⁻ → CH₃CH₂COO⁻ + NH₃

Nitriles are made by heating halogenoalkanes with KCN in ethanol (15.1, 19.2), which adds one carbon atom: 1-bromopentane gives hexanenitrile, then hexanoic acid.

Route 3: hydrolysing esters

Heating an ester under reflux with dilute acid or dilute alkali gives a carboxylic acid (or its salt) and an alcohol. With acid the reaction is reversible; with alkali it goes to completion. For methyl propanoate:

CH₃CH₂COOCH₃ + H₂O ⇌ CH₃CH₂COOH + CH₃OH (dilute acid)

CH₃CH₂COOCH₃ + NaOH → CH₃CH₂COONa + CH₃OH (dilute alkali)

More on esters is in 18.2.

Acidifying after alkaline hydrolysis

With an alkali, Routes 2 and 3 give the salt. To obtain the acid, add a dilute strong acid such as dilute HCl; the carboxylate ions take up H⁺ ions because the carboxylic acid is weak:

CH₃CH₂COO⁻ + H⁺ → CH₃CH₂COOH

Worked example 1: two routes to hexanoic acid

Question: Hexanoic acid, CH₃(CH₂)₄COOH, can be made from hexan-1-ol or from hexanenitrile, CH₃(CH₂)₄CN. (a) For each route, give the reagents and conditions and a balanced equation. (b) If NaOH(aq) is used for the nitrile instead, name the organic product before acidification and the gas given off.

(a) From hexan-1-ol: acidified potassium dichromate(VI), heat under reflux. The alcohol gains one O atom and loses two H atoms (as water), so 2[O] are needed.

CH₃(CH₂)₄CH₂OH + 2[O] → CH₃(CH₂)₄COOH + H₂O

(a) From hexanenitrile: dilute hydrochloric acid, heat under reflux. Two water molecules supply the two O atoms; the nitrogen leaves as ammonium chloride.

CH₃(CH₂)₄CN + 2H₂O + HCl → CH₃(CH₂)₄COOH + NH₄Cl

Check: 6 C, 16 H, 1 N, 2 O and 1 Cl on each side.

(b) With NaOH(aq):

CH₃(CH₂)₄CN + H₂O + NaOH → CH₃(CH₂)₄COONa + NH₃

Answer: (a) Acidified K₂Cr₂O₇, heat under reflux; dilute HCl, heat under reflux (equations above). (b) Sodium hexanoate and ammonia; dilute HCl then gives hexanoic acid.

Worked example 2: percentage yield of an oxidation

Question: 7.50 cm³ of butan-1-ol (density 0.810 g cm⁻³) is heated under reflux with an excess of acidified potassium dichromate(VI). 5.06 g of butanoic acid is collected. Calculate the percentage yield, and explain why the mixture is refluxed rather than distilled. ($A_{\text{r}}$: H 1.0, C 12.0, O 16.0)

CH₃CH₂CH₂CH₂OH + 2[O] → CH₃CH₂CH₂COOH + H₂O

Step 1: mass of butan-1-ol. $$m = 7.50 \times 0.810 = 6.075\ \text{g}$$

Step 2: moles of butan-1-ol. $M$(C₄H₁₀O) = 74.0 g mol⁻¹. $$n = \frac{6.075}{74.0} = 0.08209\ \text{mol}$$

Step 3: theoretical mass of acid. The ratio is 1 : 1 and $M$(C₄H₈O₂) = 88.0 g mol⁻¹. $$0.08209 \times 88.0 = 7.224\ \text{g}$$

Step 4: percentage yield. $$\frac{5.06}{7.224} \times 100 = 70.0\ \%$$

Why reflux: butanal forms first and is volatile. Under reflux its vapour condenses and returns to the flask, so it is oxidised to butanoic acid. Distilling would remove the butanal as it forms.

Answer: percentage yield = 70.0 % (3 s.f.). Keep unrounded values to the end: rounding the theoretical mass to 7.22 g first gives 70.1 %.

Reaction with reactive metals

This is a redox reaction: the metal is oxidised (Mg from 0 to +2) and hydrogen is reduced (from +1 to 0).

2CH₃COOH(aq) + Mg(s) → (CH₃COO)₂Mg(aq) + H₂(g)

You see effervescence as the magnesium dissolves; the gas 'pops' with a lighted splint. Two moles of acid give one mole of H₂. Fizzing is slower than with hydrochloric acid of the same concentration, because the weak acid gives a lower concentration of H⁺ ions (7.2).

Sodium reacts in the same way, but it also reacts with alcohols and water, so fizzing with sodium does not prove -COOH is present.

Neutralisation by alkalis

CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)

The ionic equation is usually written with the acid as molecules, because most of it starts undissociated (H⁺ + OH⁻ → H₂O is also correct, 7.2):

CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l)

The mixture warms slightly but looks unchanged, so titrations need an indicator (7.2). Write the salt as CH₃COONa or CH₃COO⁻Na⁺, never a mixture such as CH₃COO⁻Na.

Reaction with carbonates: the test for -COOH

2CH₃COOH(aq) + Na₂CO₃(aq) → 2CH₃COONa(aq) + H₂O(l) + CO₂(g)

You see effervescence, and a solid carbonate such as CaCO₃ dissolves. The gas gives a white precipitate with limewater, so it is CO₂. Two moles of acid give one mole of CO₂.

Alcohols do not react with carbonates or aqueous alkalis (16.1), so fizzing with Na₂CO₃(aq) tells a carboxylic acid apart from an alcohol.

Esterification with alcohols

Heat the acid with an alcohol and a few drops of concentrated sulfuric acid as a catalyst. An ester and water form, reversibly. It is a condensation reaction, because water is lost as the two molecules join.

CH₃CH₂COOH + CH₃OH ⇌ CH₃CH₂COOCH₃ + H₂O

The alcohol gives the first part of the name and the acid the -oate part: methyl propanoate. This is the reverse of Route 3 with acid (more in 18.2).

Reduction by LiAlH₄

Lithium aluminium hydride, LiAlH₄, reduces -COOH to -CH₂OH, giving a primary alcohol with the same number of carbon atoms. It is used in dry ether, because it reacts violently with water. [H] stands for one hydrogen atom from the reducing agent; four are needed.

CH₃CH₂COOH + 4[H] → CH₃CH₂CH₂OH + H₂O

NaBH₄ cannot reduce a carboxylic acid. It is a weaker reducing agent than LiAlH₄: it reduces aldehydes and ketones (17.1), but not -COOH. Neither reagent reduces a C=C bond, so an alkene group in the acid is left unchanged.

Worked example 3: an alcohol group and an acid group in one molecule

Question: 6-hydroxyhexanoic acid, HO(CH₂)₅COOH, has an alcohol group and a carboxylic acid group. Give the organic product with each reagent and say which group reacts: (a) NaOH(aq); (b) Na₂CO₃(aq); (c) an excess of sodium; (d) LiAlH₄ in dry ether.

(a) and (b): only -COOH reacts. The alcohol group is too weakly acidic to react with an alkali or a carbonate. Both give HO(CH₂)₅COONa; the carbonate also gives CO₂.

2HO(CH₂)₅COOH + Na₂CO₃ → 2HO(CH₂)₅COONa + H₂O + CO₂

(c): both groups react. Sodium replaces the H of every O-H group, in acids and alcohols alike.

HO(CH₂)₅COOH + 2Na → NaO(CH₂)₅COONa + H₂

(d): only -COOH is reduced. LiAlH₄ does not reduce the alcohol group; -COOH becomes -CH₂OH.

HO(CH₂)₅COOH + 4[H] → HO(CH₂)₆OH + H₂O

Answer: (a) and (b) -COOH only: HO(CH₂)₅COONa, sodium 6-hydroxyhexanoate. (c) Both groups: NaO(CH₂)₅COONa. (d) -COOH only: hexane-1,6-diol, HO(CH₂)₆OH.

Key Equations

Oxidation (acidified K₂Cr₂O₇ or KMnO₄, heat under reflux):

RCH₂OH + 2[O] → RCOOH + H₂O

RCHO + [O] → RCOOH

Nitrile hydrolysis (heat under reflux with dilute acid, or with dilute alkali and then acidify):

RCN + 2H₂O + H⁺ → RCOOH + NH₄⁺

RCN + H₂O + OH⁻ → RCOO⁻ + NH₃

Ester hydrolysis (heat under reflux; R′ comes from the alcohol):

RCOOR′ + H₂O ⇌ RCOOH + R′OH

RCOOR′ + OH⁻ → RCOO⁻ + R′OH

Acidification (dilute strong acid): RCOO⁻ + H⁺ → RCOOH

Reactions as an acid (with Mg or Na₂CO₃, 2 mol of acid give 1 mol of gas):

2RCOOH + Mg → (RCOO)₂Mg + H₂

RCOOH + NaOH → RCOONa + H₂O

2RCOOH + Na₂CO₃ → 2RCOONa + H₂O + CO₂

Esterification (concentrated H₂SO₄ catalyst, heat): RCOOH + R′OH ⇌ RCOOR′ + H₂O

Reduction (LiAlH₄ in dry ether): RCOOH + 4[H] → RCH₂OH + H₂O

Percentage yield (both masses in g): $$\%\ \text{yield} = \frac{\text{actual}}{\text{theoretical}} \times 100$$

Common Mistakes to Avoid

• Mistake: "heat" or "distil" as the condition for oxidising an alcohol to an acid. Fix: write "heat under reflux". Distilling butan-1-ol with acidified K₂Cr₂O₇ collects mainly butanal.
• Mistake: unbalanced [O] or [H] equations, [H⁺] in place of [H], or O₂ as a by-product. Fix: alcohol to acid needs 2[O]; acid to alcohol needs 4[H]; both form one H₂O, as in CH₃CH₂COOH + 4[H] → CH₃CH₂CH₂OH + H₂O.
• Mistake: choosing NaBH₄ to reduce a carboxylic acid, or losing a carbon atom. Fix: only LiAlH₄ works, and the chain stays the same length: butanoic acid gives butan-1-ol, not propan-1-ol.
• Mistake: leaving water or NH₄⁺ out of a nitrile hydrolysis, or giving the free acid as the product with alkali. Fix: RCN + 2H₂O + H⁺ → RCOOH + NH₄⁺. With NaOH(aq), hexanenitrile gives sodium hexanoate until dilute HCl is added.
• Mistake: a 1 : 1 ratio of acid to gas with a metal or a carbonate. Fix: 2 mol of acid give 1 mol of H₂ or CO₂. 0.0400 mol of ethanoic acid with excess magnesium gives 0.0200 mol of H₂, which is 0.480 dm³ at room conditions, not 0.960 dm³. Only a hydrogencarbonate, such as NaHCO₃, reacts 1 : 1.
• Mistake: "CO₂ is produced" or "the limewater goes milky" as an observation. Fix: write "effervescence" and "the gas gives a white precipitate with limewater".

Exam Tips

• State reagents and conditions in full, for example "acidified potassium dichromate(VI), heat under reflux" or "LiAlH₄ in dry ether". Do not add reflux to room-temperature reactions with Na₂CO₃(aq), NaOH(aq) or Mg.
• Construct an equation with [O] or [H] and balancing numbers, then count each atom on both sides. Write an aldehyde as -CHO, not -COH.
• Describe observations you can see (effervescence, a solid dissolving, orange to green), then the gas test: white precipitate with limewater for CO₂, a 'pop' with a lighted splint for H₂. To explain reflux, link three steps: the aldehyde is volatile, its vapour condenses and returns, so it is oxidised to the acid.
• Deduce groups from tests: fizzing with Na₂CO₃(aq) means -COOH (an alcohol -OH does not react); fizzing with sodium alone could be -COOH or an alcohol. Name the group "carboxylic acid" (or carboxyl), never just "carboxylic", and check whether the question wants the starting compound or the product.