Edexcel International A Level Chemistry: Carboxylic acids and their derivatives (YCH11)

This guide teaches parts 15C (Carboxylic acids) and 15D (Carboxylic acid derivatives) of Topic 15 in the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. Learning outcomes 15.9 to 15.16 are all here. The content is International A2 (IA2); Pearson places it in Unit 4: Rates, Equilibria and Further Organic Chemistry. Alcohol oxidation from Unit 2 comes back here.

Other units of this course are listed on the YCH11 course hub, and every outcome can be ticked off on the YCH11 checklist. For a shorter version, read the carboxylic acids revision notes; to test yourself, try the carboxylic acids practice questions. A free diagnostic will point out your weak spots across the course.

15C and 15D at a glance

Spec outcome	You should be able to
15.9	Name carboxylic acids; draw structural, displayed and skeletal formulae
15.10	Explain how hydrogen bonding affects boiling temperatures and solubility
15.11	Describe preparation by oxidising alcohols or aldehydes, and by hydrolysing nitriles
15.12	Give the reactions with LiAlH₄ in dry ether, bases, PCl₅, and alcohols with an acid catalyst
15.13	Name acyl chlorides and esters; draw their structural, displayed and skeletal formulae
15.14	Give the reactions of acyl chlorides with water, alcohols, concentrated ammonia and amines
15.15	Describe ester hydrolysis in acidic and in alkaline solution
15.16	Explain how polyesters such as Terylene form by condensation polymerisation

The specification suggests practicals for 15C and 15D (solubility, oxidation, ester preparation and hydrolysis, ethanoyl chloride reactions), but none is a numbered Core Practical.

15.9 Naming and drawing carboxylic acids

The functional group is the carboxyl group, –COOH. Its carbon holds a double-bonded oxygen and an –OH group. It is always carbon 1 of the chain, so it never needs a locant.

• Count every carbon in the longest chain that includes the COOH carbon.
• Replace the final -e of the alkane with -oic acid: methanoic, ethanoic, propanoic, butanoic, pentanoic, hexanoic acid.
• Number side chains from the COOH carbon.
• An acid with a COOH at both ends is a -dioic acid, for example hexanedioic acid, HOOC(CH₂)₄COOH.

Worked example 1: one acid, three formulae

Name (CH₃)₂CHCH₂COOH and describe its three formulae.

1. Longest chain through the COOH carbon: four carbons, so the stem is butanoic acid.
2. Counting from the COOH carbon, the methyl branch sits on carbon 3.
3. Name: 3-methylbutanoic acid.

Structural formula: (CH₃)₂CHCH₂COOH. Displayed formula (every bond shown):

            H
            |
        H - C - H
            |
    H       |       H       O
    |       |       |       ||
H - C ----- C ----- C ----- C - O - H
    |       |       |
    H       H       H

In the skeletal formula draw a zigzag of three lines, with a short branch line from the second carbon from the left. At the right-hand end, draw a double line up to an O and a single line to OH. Write the O and OH as atoms; leave out hydrogens on carbon.

15.10 Hydrogen bonding and physical properties

The –COOH group has a very polar O–H bond and lone pairs on both oxygens. So carboxylic acid molecules form hydrogen bonds with each other. In the pure liquid, two molecules often pair up through two hydrogen bonds to form a dimer. Dimers are larger particles, so London forces between them are also stronger.

Worked example 2: comparing boiling temperatures

Use these values: propanoic acid (Mr 74.0) boils at 141 °C; butan-1-ol (Mr 74.0) at 118 °C; butanal (Mr 72.0) at 75 °C; pentane (Mr 72.0) at 36 °C. Explain the order.

1. The Mr values are almost equal, so London forces are similar.
2. Pentane has only London forces: lowest.
3. Butanal adds permanent dipole–dipole forces from its polar C=O, but has no O–H to hydrogen bond to itself.
4. Butan-1-ol forms hydrogen bonds between molecules through its O–H group.
5. Propanoic acid forms hydrogen-bonded dimers using its O–H and its C=O oxygen, so it has the highest boiling temperature.

Solubility. Carboxylic acids hydrogen bond with water through both oxygens and the O–H hydrogen. Methanoic, ethanoic, propanoic and butanoic acids are miscible with water. Each extra CH₂ adds a non-polar section that water cannot hydrogen bond to, so solubility falls as the chain grows.

15.11 Preparing carboxylic acids

Oxidation. Heat a primary alcohol or an aldehyde under reflux with excess acidified potassium dichromate(VI). Reflux stops the aldehyde escaping. The colour changes from orange to green as Cr₂O₇²⁻ is reduced to Cr³⁺. In equations, write the oxidising agent as [O]:

• primary alcohol: RCH₂OH + 2[O] → RCOOH + H₂O
• aldehyde: RCHO + [O] → RCOOH

Hydrolysis of a nitrile. Heat the nitrile under reflux with dilute hydrochloric acid. The –C≡N carbon becomes the –COOH carbon, so the carbon count does not change:

CH₃CH₂CH₂CN + 2H₂O + HCl → CH₃CH₂CH₂COOH + NH₄Cl

Butanenitrile gives butanoic acid. Heating with NaOH(aq) instead gives the sodium salt and ammonia; acidify to release the acid.

Worked example 3: yield of an oxidation

A student refluxes 3.57 g of 3-methylpentan-1-ol, CH₃CH₂CH(CH₃)CH₂CH₂OH, with excess acidified potassium dichromate(VI). After purification, 2.84 g of 3-methylpentanoic acid is obtained. Work out the percentage yield. Use relative atomic masses H = 1.0, C = 12.0, O = 16.0.

CH₃CH₂CH(CH₃)CH₂CH₂OH + 2[O] → CH₃CH₂CH(CH₃)CH₂COOH + H₂O
Mr(alcohol, C₆H₁₄O) = 102.0    n = 3.57 ÷ 102.0 = 0.0350 mol
Mr(acid, C₆H₁₂O₂) = 116.0      maximum mass = 0.0350 × 116.0 = 4.06 g
yield = 2.84 ÷ 4.06 × 100 = 70.0%

15.12 Reactions of carboxylic acids

Reagent	Conditions	Product	Equation (R = alkyl)
LiAlH₄	dry ether (ethoxyethane), room temperature, then dilute acid	primary alcohol	RCOOH + 4[H] → RCH₂OH + H₂O
NaOH(aq)	room temperature	salt + water	RCOOH + NaOH → RCOONa + H₂O
Na₂CO₃ or NaHCO₃	room temperature	salt + water + CO₂	RCOOH + NaHCO₃ → RCOONa + H₂O + CO₂
PCl₅	dry, room temperature	acyl chloride	RCOOH + PCl₅ → RCOCl + POCl₃ + HCl
alcohol R′OH	warm or reflux, concentrated H₂SO₄ catalyst	ester + water	RCOOH + R′OH ⇌ RCOOR′ + H₂O

Points to note:

• LiAlH₄ reacts violently with water, so the ether must be dry. The acid needs 4[H]; an aldehyde needs 2[H].
• Bases. Fizzing with a carbonate or hydrogencarbonate tells a carboxylic acid from an alcohol. Propanoic acid gives sodium propanoate.
• PCl₅ gives steamy fumes of HCl. Alcohols do this too, so it is a test for an –OH group, not for –COOH alone.
• Esterification is slow and reversible, so the equilibrium limits the yield.

Worked example 4: identifying an acid by titration

0.370 g of a carboxylic acid with one COOH group needs 25.00 cm³ of 0.200 mol dm⁻³ NaOH for neutralisation. Identify the acid.

1. n(NaOH) = 25.00 ÷ 1000 × 0.200 = 0.00500 mol.
2. The ratio is 1 : 1, so n(acid) = 0.00500 mol.
3. Mr = 0.370 ÷ 0.00500 = 74.0.
4. CₙH₂ₙ₊₁COOH with Mr 74.0: 14n + 46 = 74, so n = 2. The acid is C₂H₅COOH, propanoic acid.

15.13 Naming acyl chlorides and esters

Acyl chlorides contain –COCl. Replace -oic acid with -oyl chloride: ethanoyl chloride CH₃COCl, butanoyl chloride CH₃CH₂CH₂COCl.

Esters contain –COO–. The name has two words:

1. Find the C=O carbon. The chain that includes it comes from the acid and gives the second word, ending -oate.
2. The group on the other side of the single-bonded oxygen comes from the alcohol and gives the first word, ending -yl.

So CH₃CH₂COOCH₂CH(CH₃)₂ is 2-methylpropyl propanoate: the CH₃CH₂CO– part is propanoate and the four-carbon branched group on the oxygen is 2-methylpropyl.

15.14 Reactions of acyl chlorides

Acyl chlorides react much faster than carboxylic acids. Oxygen and chlorine both pull electron density away from the carbonyl carbon, which makes it strongly δ+, and Cl⁻ leaves easily. Each reaction gives HCl (steamy fumes) unless a base absorbs it.

Reagent	Organic product	Equation for butanoyl chloride
water	carboxylic acid	C₃H₇COCl + H₂O → C₃H₇COOH + HCl
alcohol	ester	C₃H₇COCl + CH₃CH₂CH₂OH → C₃H₇COOCH₂CH₂CH₃ + HCl
concentrated ammonia	amide	C₃H₇COCl + 2NH₃ → C₃H₇CONH₂ + NH₄Cl
amine (propylamine)	N-substituted amide	C₃H₇COCl + 2C₃H₇NH₂ → C₃H₇CONHC₃H₇ + C₃H₇NH₃⁺Cl⁻

With ammonia or an amine, one molecule forms the amide and a second one neutralises the HCl, giving a salt. The amine row gives N-propylbutanamide. With an alcohol the reaction is fast and not reversible, so it gives a better ester yield than the acid does.

Worked example 5: an amide from an acyl chloride

4.26 g of 2-methylpropanoyl chloride, (CH₃)₂CHCOCl, is added to excess concentrated ammonia. Find the maximum mass of 2-methylpropanamide that can form. Use H = 1.0, C = 12.0, N = 14.0, O = 16.0, Cl = 35.5.

(CH₃)₂CHCOCl + 2NH₃ → (CH₃)₂CHCONH₂ + NH₄Cl
Mr((CH₃)₂CHCOCl) = 106.5     n = 4.26 ÷ 106.5 = 0.0400 mol
1 mol acyl chloride → 1 mol amide
Mr((CH₃)₂CHCONH₂) = 87.0     mass = 0.0400 × 87.0 = 3.48 g

15.15 Hydrolysis of esters

• Acidic solution (dilute HCl or H₂SO₄): the reverse of esterification. It is reversible, so an equilibrium mixture of ester, acid, alcohol and water forms. RCOOR′ + H₂O ⇌ RCOOH + R′OH.
• Alkaline solution (NaOH(aq)): gives the carboxylate salt and the alcohol. RCOOR′ + NaOH → RCOONa + R′OH. It goes to completion: the negatively charged carboxylate ion has no tendency to re-form an ester, so nothing reverses. Add a strong acid afterwards to get the free carboxylic acid.

Worked example 6: alkaline hydrolysis

2.32 g of ethyl 2-methylpropanoate, (CH₃)₂CHCOOCH₂CH₃, is heated under reflux with excess NaOH(aq). Name the products and find the mass of salt formed. Use H = 1.0, C = 12.0, O = 16.0, Na = 23.0.

1. Products: sodium 2-methylpropanoate, (CH₃)₂CHCOONa, and ethanol.
2. Mr(C₆H₁₂O₂) = 116.0, so n = 2.32 ÷ 116.0 = 0.0200 mol.
3. Mr((CH₃)₂CHCOONa) = 110.0, so mass = 0.0200 × 110.0 = 2.20 g.

15.16 Polyesters and condensation polymerisation

In condensation polymerisation, monomers with two functional groups join, and a small molecule is lost at each link. For a polyester, a diol reacts with a dicarboxylic acid (losing H₂O) or with a diacyl dichloride (losing HCl). Each new link is an ester group, –COO–.

Terylene (PET) is made by condensing ethane-1,2-diol with benzene-1,4-dicarboxylic acid. Its repeat unit is –OC–C₆H₄–COO–CH₂CH₂–O–. It is used for clothing fibres and drinks bottles.

Worked example 7: a polyester from two monomers

Hexanedioic acid, HOOC(CH₂)₄COOH, reacts with butane-1,4-diol, HO(CH₂)₄OH. Draw the repeat unit and find its Mr.

1. Remove –OH from each acid end and –H from each alcohol end. Join the C=O carbon to the diol oxygen.
2. Repeat unit: –OC(CH₂)₄COO(CH₂)₄O–, which is C₁₀H₁₆O₄.
3. Mr = 10(12.0) + 16(1.0) + 4(16.0) = 200.0. Two water molecules are lost per repeat unit.

To find monomers from a given polyester, split each ester link at the C–O single bond. Add –OH to the C=O carbon and –H to the oxygen.

Common errors

• Giving the COOH carbon a number in the name, as in “butan-1-oic acid”.
• Naming esters backwards: the alcohol word comes first.
• Distilling instead of refluxing when the aim is the acid, which collects the aldehyde instead.
• Writing 2[H] for reducing a carboxylic acid with LiAlH₄; it needs 4[H].
• Using one NH₃ for each acyl chloride; two are needed.
• Naming the free acid as a product of alkaline hydrolysis.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 4, Topic 15, parts 15C Carboxylic acids and 15D Carboxylic acid derivatives, outcomes 15.9 to 15.16.