S5 Chemistry – Unit 8: Esters, Acid Anhydrides, Amides and Nitriles

8.1

Esters

Definition Esters are organic compounds formed by the reaction of a carboxylic acid and an alcohol, with the general structure RCOOR'. They contain the ester functional group –COO–. Named as alkyl alkanoate: the alkyl group from the alcohol comes first, then the acyl part of the acid ending in –oate.

IUPAC Nomenclature of Esters

Identify the alcohol part (R'–O): becomes the alkyl prefix (e.g. methyl–, ethyl–, propyl–).
Identify the acid part (R–CO): name the acid, replace –ic acid with –ate.
Combine: alkyl alkanoate. Example: CH₃COOC₂H₅ = ethanoate (from ethanoic acid) + ethyl (from ethanol) → ethyl ethanoate.

Name	Structure	Acid Part	Alcohol Part	Smell/Use
Methyl methanoate	HCOOCH₃	Methanoic acid	Methanol	Fruity
Ethyl ethanoate	CH₃COOC₂H₅	Ethanoic acid	Ethanol	Nail polish/solvent
Pentyl ethanoate	CH₃COO(CH₂)₄CH₃	Ethanoic acid	Pentan-1-ol	Banana
Octyl ethanoate	CH₃COO(CH₂)₇CH₃	Ethanoic acid	Octan-1-ol	Orange
Ethyl butanoate	C₃H₇COOC₂H₅	Butanoic acid	Ethanol	Pineapple
Methyl salicylate	HOC₆H₄COOCH₃	Salicylic acid	Methanol	Wintergreen (muscle rubs)
Glyceryl trinitrate	Glycerol + 3 HNO₃	Nitric acid	Glycerol	Explosive (dynamite)/angina drug

Physical Properties of Esters

Esters have lower boiling points than carboxylic acids of similar M_r because they cannot form H-bonds with each other (no O–H group). However, they have higher boiling points than alkanes of similar M_r due to dipole–dipole interactions.

Short-chain esters are slightly soluble in water (C=O can accept H-bonds from water). Most esters are insoluble in water but dissolve in organic solvents. They have characteristic pleasant fruity smells and are used as food flavourings and perfumes.

Preparation of Esters

Method 1: Fischer Esterification (Acid + Alcohol)

RCOOH + R'OH <==> RCOOR' + H2O conc. H2SO4 catalyst, heat (reversible) e.g. CH3COOH + C2H5OH <==> CH3COOC2H5 + H2O (ethanoic acid) (ethanol) (ethyl ethanoate)

Method 2: Acyl Chloride + Alcohol (faster, irreversible)

RCOCl + R'OH --> RCOOR' + HCl (no catalyst needed, room temperature, near 100% yield) e.g. CH3COCl + C2H5OH --> CH3COOC2H5 + HCl

Method 3: Acid Anhydride + Alcohol

(RCO)2O + R'OH --> RCOOR' + RCOOH (less vigorous than acyl chloride, no HCl fumes) e.g. (CH3CO)2O + C2H5OH --> CH3COOC2H5 + CH3COOH

Reactions of Esters

Hydrolysis (Acid-catalysed — reversible)

RCOOR' + H2O <==> RCOOH + R'OH H+ catalyst, heat e.g. CH3COOC2H5 + H2O <==> CH3COOH + C2H5OH

Hydrolysis (Base-catalysed — saponification, irreversible)

RCOOR' + NaOH --> RCOONa + R'OH heat (irreversible -- drives equilibrium to completion) e.g. CH3COOC2H5 + NaOH --> CH3COONa + C2H5OH (ethyl ethanoate) (sodium ethanoate) (ethanol)

Base hydrolysis is irreversible because the carboxylate salt (RCOO⁻Na⁺) does not react back with the alcohol. This is the basis of soap making (saponification).

Reduction with LiAlH₄

RCOOR' + 4[H] --LiAlH4, dry ether--> RCH2OH + R'OH (gives two alcohols: primary alcohol from acid part + alcohol from alkyl part) e.g. CH3COOC2H5 + 4[H] --> CH3CH2OH + C2H5OH

Example 1

Naming and Preparing Esters

Name the ester CH₃CH₂COOCH₂CH₃ and write the equation for its preparation from an acid and an alcohol.

Acid part: CH₃CH₂COO– = propanoate (from propanoic acid).

Alcohol part: –OCH₂CH₃ = ethyl (from ethanol).

Name: ethyl propanoate.

Equation: CH₃CH₂COOH + C₂H₅OH ⇌^{conc. H₂SO₄} CH₃CH₂COOC₂H₅ + H₂O

8.2

Acid Anhydrides

Definition Acid anhydrides are formed by the elimination of water from two carboxylic acid molecules. They contain the group RCO–O–COR'. Named by replacing “acid” with “anhydride”: e.g. ethanoic anhydride = (CH₃CO)₂O. Cyclic anhydrides form from diacids (e.g. maleic anhydride, phthalic anhydride).

Name	Structure	Parent Acid
Ethanoic anhydride (acetic anhydride)	(CH₃CO)₂O	Ethanoic acid
Propanoic anhydride	(C₂H₅CO)₂O	Propanoic acid
Maleic anhydride	Cyclic (C₄H₂O₃)	Maleic acid (cis-butenedioic acid)
Phthalic anhydride	Cyclic (C₈H₄O₃)	Phthalic acid

Reactions of Acid Anhydrides

Acid anhydrides are less reactive than acyl chlorides but more reactive than carboxylic acids. They react with nucleophiles by nucleophilic acyl substitution, giving one mole of ester (or amide) and one mole of carboxylic acid.

With water: (RCO)2O + H2O --> 2 RCOOH With alcohol: (RCO)2O + R'OH --> RCOOR' + RCOOH With ammonia: (RCO)2O + 2NH3 --> RCONH2 + RCOONH4 With amine: (RCO)2O + R'NH2 --> RCONHR' + RCOOH Aspirin synthesis (ethanoic anhydride + salicylic acid): (CH3CO)2O + HOC6H4COOH --> CH3COOC6H4COOH + CH3COOH (aspirin)

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Acid Anhydride vs Acyl Chloride in Industry Ethanoic anhydride is preferred over ethanoyl chloride for manufacturing aspirin because: no corrosive HCl fumes produced; safer to handle; by-product is ethanoic acid (recyclable); more economical at industrial scale.

Example 2

Aspirin Synthesis

Write the equation for the manufacture of aspirin from salicylic acid and ethanoic anhydride. State the advantage over using ethanoyl chloride.

(CH₃CO)₂O + HOC₆H₄COOH → CH₃COOC₆H₄COOH + CH₃COOH

The –OH group of salicylic acid is acylated to form the phenyl ester (aspirin).

✓

Advantage: by-product is ethanoic acid (non-corrosive, recyclable) rather than HCl (corrosive, toxic). Anhydride is also safer to handle.

8.3

Amides

Definition Amides contain the amide functional group (–CONH₂) or –CONHR / –CONR₂. They are formed by the reaction of a carboxylic acid (or acyl chloride) with ammonia or an amine. Named by replacing –oic acid with –amide. N-substituted amides have the prefix N– before the alkyl substituent.

Name	Structure	Class
Methanamide	HCONH₂	Primary amide
Ethanamide	CH₃CONH₂	Primary amide
Propanamide	C₂H₅CONH₂	Primary amide
N-methylethanamide	CH₃CONHCH₃	Secondary amide
N,N-dimethylethanamide	CH₃CON(CH₃)₂	Tertiary amide
Benzamide	C₆H₅CONH₂	Primary amide (aromatic)
Paracetamol (drug)	HOC₆H₄NHCOCH₃	Secondary amide

Physical Properties of Amides

Primary and secondary amides have very high melting and boiling points due to strong N–H···O=C hydrogen bonding between molecules. Ethanamide (M_r=59) has B.P. 221°C — far higher than ethanoic acid (B.P. 118°C) of similar M_r.

Short-chain primary amides are soluble in water due to H-bonding with water. Amides are much weaker bases than amines because the lone pair on N is delocalised into the C=O group, reducing its availability for protonation.

Preparation of Amides

From acyl chloride + ammonia (best method): RCOCl + 2NH3 --> RCONH2 + NH4Cl From acid anhydride + ammonia: (RCO)2O + 2NH3 --> RCONH2 + RCOONH4 From carboxylic acid + ammonia (indirect -- via ammonium salt, then heat): RCOOH + NH3 --> RCOONH4 (ammonium salt) RCOONH4 --heat--> RCONH2 + H2O

Reactions of Amides

Hydrolysis (Acid or Base)

Acid hydrolysis: RCONH2 + H2O --dil. HCl, reflux--> RCOOH + NH4Cl Base hydrolysis: RCONH2 + NaOH --reflux--> RCOONa + NH3(g)

Dehydration → Nitrile

RCONH2 --P2O5, heat (dehydrating agent)--> RCN + H2O e.g. CH3CONH2 --P2O5--> CH3CN + H2O (ethanamide) (ethanenitrile)

Reduction with LiAlH₄ → Amine

RCONH2 + 4[H] --LiAlH4, dry ether--> RCH2NH2 e.g. CH3CONH2 + 4[H] --> CH3CH2NH2 (ethylamine)

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Amide Bond in Biology — The Peptide Bond The amide bond (–CO–NH–) is the same as the peptide bond that links amino acids in proteins. The C–N bond in amides has partial double-bond character due to resonance delocalisation, making it planar and rigid — this is fundamental to protein secondary structure (alpha helices, beta sheets).

8.4

Nitriles

Definition Nitriles contain the cyano group (–C≡N). General formula: R–C≡N. Named by counting the CN carbon as part of the chain and adding the suffix –nitrile, or as alkanenitriles. Example: CH₃CN = ethanenitrile (acetonitrile); CH₃CH₂CN = propanenitrile.

Physical Properties

Nitriles are polar (C≡N is very polar) and have higher boiling points than alkanes but lower than amides of similar M_r. Lower nitriles are miscible with water; they are also good polar aprotic solvents for organic reactions.

Preparation of Nitriles

From halogenoalkanes + KCN (nucleophilic substitution): R-X + KCN(alc) --heat--> R-CN + KX (chain extended by 1C) From amides (dehydration): RCONH2 --P2O5, heat--> RCN + H2O Industrial: from alkenes + HCN (hydrocyanation): CH2=CH2 + HCN --catalyst--> CH3CH2CN (propanenitrile)

Reactions of Nitriles

Hydrolysis to carboxylic acid: R-CN + H2O --dil. HCl or NaOH, reflux--> RCOOH + NH3 Reduction to primary amine (LiAlH4): R-CN + 4[H] --LiAlH4, dry ether--> RCH2NH2 (chain extended by 1C) e.g. CH3CN + 4[H] --> CH3CH2NH2 (ethylamine) Addition of HCN to carbonyl (Unit 6): RCHO + HCN --> RCH(OH)CN (hydroxynitrile)

Example 3

Synthetic Route: Halogenoalkane → Amine via Nitrile

Show how 1-bromopropane can be converted to butan-1-amine (a primary amine with 4 carbons) in two steps.

Step 1: CH₃CH₂CH₂Br + KCN(alc) → CH₃CH₂CH₂CN + KBr
Product: butanenitrile (4C — chain extended by 1C).

Step 2: CH₃CH₂CH₂CN + 4[H] →^{LiAlH₄, dry ether} CH₃CH₂CH₂CH₂NH₂
Product: butan-1-amine (primary amine, 4C).

✓

Net: 3C halogenoalkane → 4C primary amine. Both steps extend/use the chain effectively.

8.5

Saponification and Fats

Saponification Saponification is the base hydrolysis of an ester (especially a fat or oil) to give a carboxylate salt (soap) and glycerol. The word comes from Latin sapo (soap). The reaction is irreversible because the carboxylate salt does not react back with the alcohol.

Fats and Oils — Triglycerides

Fats (solid) and oils (liquid) are triglycerides — triesters of glycerol (propane-1,2,3-triol) with three long-chain fatty acids (C₁₂–C₂₀). Fats have mainly saturated fatty acid chains (all single C–C bonds → can pack closely → solid at room temperature). Oils have unsaturated chains (C=C bonds → cannot pack as closely → liquid at room temperature).

Structure of a triglyceride (fat/oil): CH2-OOC-R1 R1, R2, R3 = long-chain fatty acid chains | (e.g. C15H31- for palmitic acid) CH-OOC-R2 | CH2-OOC-R3

Saponification Reaction (Soap Making)

Triglyceride + 3 NaOH --heat--> Glycerol + 3 RCOONa (fat/oil) (propane-1,2,3-triol) (soap = sodium salt of fatty acid) Example with glyceryl tristearate (hard fat): (C17H35COO)3C3H5 + 3NaOH --> C3H5(OH)3 + 3 C17H35COONa (tristearate) (glycerol) (sodium stearate = soap)

The soap (sodium salt) is precipitated by adding NaCl to the reaction mixture (salting out), then separated and processed. Potassium salts (made with KOH) give softer soaps (e.g. shaving cream).

Common Fatty Acids

Name	Formula	Saturated?	Source
Palmitic acid	C₁₅H₃₁COOH (C16:0)	Yes	Palm oil, animal fat
Stearic acid	C₁₇H₃₅COOH (C18:0)	Yes	Animal fat, cocoa butter
Oleic acid	C₁₇H₃₃COOH (C18:1)	No (1 C=C)	Olive oil
Linoleic acid	C₁₇H₃₁COOH (C18:2)	No (2 C=C)	Sunflower oil
Linolenic acid	C₁₇H₂₉COOH (C18:3)	No (3 C=C)	Linseed oil

8.6

Detergents and Soaps

Structure of Soap Molecules

A soap molecule (e.g. sodium stearate, C₁₇H₃₅COONa) has two distinct parts:

Hydrophilic (water-loving) head: the charged carboxylate group (–COO⁻Na⁺) — ionic, interacts with water.
Hydrophobic (water-hating) tail: the long non-polar hydrocarbon chain (C₁₇H₃₅–) — interacts with grease and oils.

This dual character makes soaps surfactants (surface-active agents) that can emulsify grease in water.

How Soap Cleans — Micelle Formation

Soap molecules arrange around grease droplets with hydrophobic tails pointing inward (into the grease) and hydrophilic heads pointing outward (into the water).
This spherical arrangement is called a micelle.
The charged outer surface of the micelle keeps it dispersed in water (repulsion between like charges).
Grease is thus emulsified and can be rinsed away with water.

Hard Water Problem with Soaps

In hard water (containing Ca²⁺ and Mg²⁺ ions), soap molecules form insoluble precipitates called scum:

2 RCOONa + CaCl2 --> (RCOO)2Ca(s) + 2 NaCl (calcium soap = scum, insoluble)

This wastes soap, leaves deposits, and reduces cleaning efficiency.

Soapless (Synthetic) Detergents

Synthetic detergents are also surfactants but their calcium and magnesium salts are soluble — they work in hard water without forming scum. They are made from petroleum-derived alkylbenzene sulfonates or alkyl sulfates:

R-C6H4-SO3-Na+ (sodium alkylbenzenesulfonate -- anionic detergent) R-OSO3-Na+ (sodium alkyl sulfate -- e.g. sodium lauryl sulfate in shampoos)

Types of synthetic detergents: anionic (sulfonate/sulfate head, most common), cationic (ammonium head, fabric softeners, disinfectants), non-ionic (polyether head, gentle, used in washing-up liquid).

Property	Soaps	Synthetic Detergents
Raw material	Natural fats/oils (renewable)	Petroleum products (non-renewable)
Hard water	Forms scum (insoluble Ca/Mg salts)	No scum (soluble Ca/Mg salts)
Biodegradability	Fully biodegradable	Some are biodegradable; early ones were not
Skin sensitivity	Generally mild; slightly alkaline	Some can irritate skin (strong anionic types)
Cost	Relatively cheap	Variable; specialty types more expensive
pH	Slightly alkaline (~9–10)	Near neutral (some formulations)

Example 4

Saponification Calculation

Calculate the mass of NaOH needed to completely saponify 89 g of glyceryl tripalmitate (M_r = 807 g/mol, formula (C₁₅H₃₁COO)₃C₃H₅).

Equation: (C₁₅H₃₁COO)₃C₃H₅ + 3 NaOH → C₃H₅(OH)₃ + 3 C₁₅H₃₁COONa

Moles of fat = 89 ÷ 807 = 0.110 mol

Moles of NaOH = 3 × 0.110 = 0.330 mol

Mass of NaOH = 0.330 × 40 = 13.2 g

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✏️

Exercises

Name the following esters and give the acid and alcohol from which each is made:
(a) CH₃COOCH₃ (b) C₂H₅COOC₃H₇ (c) HCOOC₂H₅

(a) Methyl ethanoate; from ethanoic acid + methanol
(b) Propyl propanoate; from propanoic acid + propan-1-ol
(c) Ethyl methanoate; from methanoic acid + ethanol
Compare the reactions of ethanoic anhydride and ethanoyl chloride with ethanol. Write equations for both and state which is preferred industrially and why.

Ethanoyl chloride:
CH₃COCl + C₂H₅OH → CH₃COOC₂H₅ + HCl (fumes, very fast)

Ethanoic anhydride:
(CH₃CO)₂O + C₂H₅OH → CH₃COOC₂H₅ + CH₃COOH (milder, no HCl)

Industrial preference: ethanoic anhydride — no corrosive HCl produced; by-product (ethanoic acid) is recyclable; safer handling; used for aspirin manufacture.
Write equations for the acid and base hydrolysis of ethanamide. What gas is produced in the base hydrolysis?

Acid hydrolysis:
CH₃CONH₂ + H₂O →^{dil. HCl, reflux} CH₃COOH + NH₄Cl

Base hydrolysis:
CH₃CONH₂ + NaOH →^reflux CH₃COONa + NH₃(g)

Gas produced: ammonia (NH₃) — recognisable by its pungent smell and turning damp red litmus blue.
Explain, using a diagram description, how soap cleans greasy dishes. Why does soap not work well in hard water?

Soap molecules have a hydrophilic head (–COO⁻Na⁺) and a hydrophobic tail (long alkyl chain). When soap is added to greasy water, the hydrophobic tails bury into the grease while the hydrophilic heads point outward into the water, forming micelles (spherical clusters). The negatively charged outer surface keeps micelles dispersed in water and prevents them from re-depositing. The emulsified grease is then rinsed away.

Hard water problem: Ca²⁺ and Mg²⁺ ions in hard water react with soap to form insoluble scum: 2 RCOONa + CaCl₂ → (RCOO)₂Ca ↓ + 2 NaCl. This wastes soap and leaves grey deposits on surfaces.
Starting from ethanenitrile (CH₃CN), show how you would prepare: (a) ethanoic acid; (b) ethylamine (CH₃CH₂NH₂); (c) ethanamide.

(a) CH₃CN + H₂O →^{dil. HCl, reflux} CH₃COOH + NH₃ — ethanoic acid
(b) CH₃CN + 4[H] →^{LiAlH₄, dry ether} CH₃CH₂NH₂ — ethylamine
(c) CH₃CN + H₂O →^{partial hydrolysis, controlled} CH₃CONH₂ — ethanamide (stop before full hydrolysis to acid)
A fat (triglyceride) has M_r = 890. Calculate the mass of NaOH (M_r = 40) needed to saponify 44.5 g of this fat completely.

Moles of fat = 44.5 ÷ 890 = 0.050 mol
Each mole of triglyceride requires 3 moles of NaOH.
Moles of NaOH = 3 × 0.050 = 0.150 mol
Mass of NaOH = 0.150 × 40 = 6.0 g

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Unit Test

ℹ️

Instructions Total: 50 marks | Time: 50 minutes | Attempt all questions | Show all working.

Section A — Short Answer

30 marks

Q1 [4 marks]

Name the following compounds:
(a) CH₃COOCH₂CH₃ (b) C₃H₇COOCH₃ (c) CH₃CH₂CONH₂ (d) CH₃CH₂C≡N

(a) Ethyl ethanoate
(b) Methyl butanoate
(c) Propanamide
(d) Propanenitrile

Q2 [5 marks]

Write equations for:
(a) Preparation of ethyl propanoate from propanoic acid and ethanol
(b) Preparation of ethyl propanoate from propanoyl chloride and ethanol
(c) Base hydrolysis (saponification) of ethyl propanoate
(d) Reduction of ethyl ethanoate with LiAlH₄
(e) Acid hydrolysis of propanamide

(a) C₂H₅COOH + C₂H₅OH ⇌^{conc. H₂SO₄} C₂H₅COOC₂H₅ + H₂O
(b) C₂H₅COCl + C₂H₅OH → C₂H₅COOC₂H₅ + HCl
(c) C₂H₅COOC₂H₅ + NaOH → C₂H₅COONa + C₂H₅OH
(d) CH₃COOC₂H₅ + 4[H] →^LiAlH₄ CH₃CH₂OH + C₂H₅OH
(e) C₂H₅CONH₂ + H₂O →^{dil. HCl, reflux} C₂H₅COOH + NH₄Cl

Q3 [5 marks]

Describe the industrial manufacture of soap by saponification. Include: the raw materials, the reaction equation (use general formula), conditions, and how the soap is separated from the reaction mixture.

Raw materials: Animal fat or vegetable oil (triglyceride) + concentrated NaOH solution (lye).

Equation:
(RCOO)₃C₃H₅ + 3 NaOH →^heat C₃H₅(OH)₃ + 3 RCOONa
(triglyceride) (glycerol) (soap)

Conditions: Heat with concentrated NaOH; boil for several hours; reaction is irreversible.

Separation: Add sodium chloride (salting out) — the increased ionic strength reduces soap solubility and it precipitates. The soap floats to the top and is skimmed off. Glycerol remains in the aqueous layer and is recovered as a valuable by-product.

Q4 [5 marks]

Explain how soap cleans greasy surfaces. Why does soap not work well in hard water, and how do synthetic detergents overcome this problem?

Cleaning action: Soap molecules are amphiphilic — they have a hydrophobic tail (long alkyl chain) and a hydrophilic head (–COO⁻Na⁺). In water, the hydrophobic tails embed into grease particles while the hydrophilic heads project into the water, forming spherical micelles. The negatively charged micelle surface prevents aggregation; grease is emulsified and rinsed away.

Hard water problem: Ca²⁺/Mg²⁺ ions react with soap: 2 RCOONa + Ca²⁺ → (RCOO)₂Ca↓ (scum, insoluble). This wastes soap and leaves deposits.

Synthetic detergents: Have the same amphiphilic structure but the ionic head is a sulfonate (–SO₃⁻) or sulfate (–OSO₃⁻). Their calcium/magnesium salts are soluble → no scum forms in hard water.

Q5 [5 marks]

Show how butanenitrile (CH₃CH₂CH₂CN) can be prepared from 1-bromopropane, and then converted into: (a) pentanoic acid; (b) butan-1-amine; (c) butanamide. Write equations and conditions for each step.

Preparation of butanenitrile:
CH₃CH₂CH₂Br + KCN(alc) →^heat/reflux CH₃CH₂CH₂CN + KBr

(a) Pentanoic acid:
CH₃CH₂CH₂CN + H₂O →^{dil. HCl, reflux} CH₃CH₂CH₂COOH + NH₃
(4C nitrile → 4C acid = butanoic acid, not pentanoic. For pentanoic acid, start from 1-bromobutane.)

Corrected (a): Butanenitrile → butanoic acid (C₄): CH₃(CH₂)₂CN + H₂O → CH₃(CH₂)₂COOH + NH₃

(b) Butan-1-amine:
CH₃CH₂CH₂CN + 4[H] →^{LiAlH₄, dry ether} CH₃CH₂CH₂CH₂NH₂ (butan-1-amine, 4C)

(c) Butanamide:
CH₃CH₂CH₂CN + H₂O →^{controlled partial hydrolysis} CH₃CH₂CH₂CONH₂ (butanamide)

Q6 [6 marks]

A student prepares a sample of ethanamide by three different routes from ethanoyl chloride, ethanoic anhydride, and ethanoic acid respectively. Write equations for all three preparations and compare the advantages and disadvantages of each route.

Route 1: Ethanoyl chloride + ammonia
CH₃COCl + 2NH₃ → CH₃CONH₂ + NH₄Cl
Fast, room temperature, near-quantitative yield. Disadvantage: HCl/NH₄Cl by-products; ethanoyl chloride is corrosive and moisture-sensitive.

Route 2: Ethanoic anhydride + ammonia
(CH₃CO)₂O + 2NH₃ → CH₃CONH₂ + CH₃COONH₄
Safer, no HCl fumes. By-product is ammonium ethanoate (can recycle acid). Slightly slower than acyl chloride.

Route 3: Ethanoic acid + ammonia (two steps)
Step 1: CH₃COOH + NH₃ → CH₃COONH₄ (ammonium ethanoate)
Step 2: CH₃COONH₄ →^{heat strongly} CH₃CONH₂ + H₂O
Cheapest reagents; no corrosive intermediates. Disadvantage: requires high temperature; partial decomposition; lowest yield; two steps needed.

Section B — Extended Response

20 marks

Q7 [10 marks]

(a) Describe the structure of fats and oils as triglycerides. Explain the difference between saturated and unsaturated fatty acids and how this explains why fats are solid and oils are liquid at room temperature. [4 marks]

(b) Write the equation for the saponification of glyceryl tristearate (C₁₇H₃₅COO)₃C₃H₅ with NaOH. Calculate the mass of soap produced from 100 g of this fat. (M_r: glyceryl tristearate = 890; sodium stearate = 306.) [4 marks]

(c) Explain the difference between soaps and synthetic detergents in terms of structure and behaviour in hard water. [2 marks]

(a) Triglycerides are triesters of glycerol (propane-1,2,3-triol) with three fatty acid molecules (R₁COOH, R₂COOH, R₃COOH). The three –OH groups of glycerol form ester linkages with the carboxyl groups of the fatty acids.

Saturated fatty acids: all C–C single bonds → straight, flexible chains → can pack closely together → stronger van der Waals forces → higher melting point → solid (fats). Unsaturated fatty acids: contain C=C double bonds → introduces kinks/bends in chain → cannot pack as tightly → weaker forces → lower melting point → liquid (oils).

(b) Equation:
(C₁₇H₃₅COO)₃C₃H₅ + 3NaOH → C₃H₅(OH)₃ + 3C₁₇H₃₅COONa
Moles of fat = 100 ÷ 890 = 0.1124 mol
Moles of soap = 3 × 0.1124 = 0.3371 mol
Mass of soap = 0.3371 × 306 = 103.1 g ≈ 103 g

(c) Both soaps and synthetic detergents have an amphiphilic structure (hydrophobic tail + hydrophilic head). In soaps, the head is –COO⁻Na⁺; its Ca/Mg salt is insoluble → forms scum in hard water. In synthetic detergents, the head is –SO₃⁻Na⁺ or –OSO₃⁻Na⁺; their Ca/Mg salts are soluble → no scum in hard water.

Q8 [10 marks]

(a) Complete the following reaction summary for ethanoic anhydride, writing equations for its reactions with: water, ethanol, ammonia, and methylamine. Name all organic products. [4 marks]

(b) Starting from propan-1-ol only (and any inorganic reagents), design a four-step synthesis of pentanenitrile (C₄H₉CN). Write an equation for each step and name all organic intermediates. [6 marks]

(a)
Water: (CH₃CO)₂O + H₂O → 2 CH₃COOH — ethanoic acid
Ethanol: (CH₃CO)₂O + C₂H₅OH → CH₃COOC₂H₅ + CH₃COOH — ethyl ethanoate + ethanoic acid
Ammonia: (CH₃CO)₂O + 2NH₃ → CH₃CONH₂ + CH₃COONH₄ — ethanamide
Methylamine: (CH₃CO)₂O + CH₃NH₂ → CH₃CONHCH₃ + CH₃COOH — N-methylethanamide

(b) Synthesis of pentanenitrile from propan-1-ol (3C → 5C):
Target: C₄H₉CN = pentanenitrile (5C). Need to add 2 carbons. Use two chain-extension steps via nitrile.

Step 1: C₃H₇OH + HBr → C₃H₇Br + H₂O — 1-bromopropane
Step 2: C₃H₇Br + KCN(alc) → C₃H₇CN + KBr — butanenitrile (4C)
Step 3: C₃H₇CN + 4[H] →^LiAlH₄ C₃H₇CH₂NH₂ — butan-1-amine (4C)
Step 4: C₃H₇CH₂NH₂ → convert to bromide then CN: C₄H₉Br + KCN → C₄H₉CN — pentanenitrile (5C)

Alternative route (2 chain extensions via nitrile each time): propan-1-ol → 1-bromopropane → butanenitrile → butanoic acid → 1-bromobutane → pentanenitrile.

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