Edexcel International A Level Chemistry: Organic chemistry: general principles and halogenoalkanes (YCH11) – Revision Notes

For full explanations and longer worked examples, read the study guide for this topic first. These notes condense Topic 10A: General principles and Topic 10B: Halogenoalkanes of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 10.1 to 10.14, with Core Practicals 5 and 6. It is International AS (IAS) content in Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols.

Use them with the printable Edexcel Chemistry checklist, the course hub and the practice questions for this topic. To check which parts of chemistry need the most work, try the free diagnostics.

Key definitions (10.1 to 10.4)

Term	Meaning
Addition	Two reactants combine into one product
Elimination	A small molecule leaves and a C=C forms
Substitution	An atom or group is replaced by another
Oxidation	Organic compound gains O or loses H (includes combustion)
Reduction	Organic compound gains H or loses O
Hydrolysis	Splitting of a bond by water or hydroxide ions
Polymerisation	Many monomers join into one long molecule
Reaction mechanism	The step-by-step route from reactants to products, showing electron movement
Homolytic fission	Bond breaks with one electron to each atom; radicals form
Heterolytic fission	Bond breaks with both electrons to one atom; ions form
Electrophile	Accepts an electron pair to form a covalent bond
Nucleophile	Donates a lone pair of electrons to form a new covalent bond

A reaction can carry two labels. Halogenoalkane + water is substitution and hydrolysis. Alkene + hydrogen is addition and reduction.

Polarity decides the mechanism (10.5)

• Non-polar C–C and C–H in alkanes: homolytic fission, radicals, free-radical substitution.
• Electron-rich C=C in alkenes: attracts electrophiles, electrophilic addition.
• Polar Cδ+–Xδ− in halogenoalkanes: δ+ carbon attracts nucleophiles, nucleophilic substitution.

Naming and classifying (10.6, 10.7)

• Prefix with fluoro-, chloro-, bromo-, iodo-; lowest locants; alphabetical order of substituents.
• Skeletal formulae: lines for C–C bonds, halogen written at the end of its line.
• Classify by the number of carbons bonded to the C–X carbon.

Type	Carbons on the C–X carbon	Example
Primary	1 (or 0 for CH₃X)	CH₃CH₂CH₂CH₂Cl, 1-chlorobutane
Secondary	2	CH₃CHICH₂CH₃, 2-iodobutane
Tertiary	3	(CH₃)₂CBrCH₂CH₃, 2-bromo-2-methylbutane

Reagents table (10.8)

Reagent	Conditions	Reagent acts as	Product	Example from bromoethane
KOH(aq)	Heat under reflux	Nucleophile (OH⁻)	Alcohol	Ethanol
KOH in ethanol	Heat under reflux	Base (OH⁻)	Alkene	Ethene
AgNO₃(aq) in ethanol	Warm	Nucleophile (H₂O)	Alcohol + AgX precipitate	Ethanol + cream AgBr
NH₃ in ethanol	Heat, sealed tube, pressure	Nucleophile (NH₃)	Amine	Ethylamine
KCN in ethanol	Heat under reflux	Nucleophile (CN⁻)	Nitrile, one more carbon	Propanenitrile, CH₃CH₂CN

Key equations (R = alkyl group):

• R–X + OH⁻ → R–OH + X⁻
• R–X + OH⁻ → alkene + H₂O + X⁻ (ethanolic)
• R–X + H₂O → R–OH + H⁺ + X⁻; Ag⁺ + X⁻ → AgX
• R–X + 2NH₃ → R–NH₂ + NH₄X
• R–X + CN⁻ → R–CN + X⁻

Mechanism checklist (10.9)

Method in steps: OH⁻ with a primary halogenoalkane

1. Draw the C–X bond with δ+ on C and δ− on X.
2. Draw OH⁻ with its charge and a lone pair on oxygen.
3. First curly arrow: oxygen lone pair towards the δ+ carbon.
4. Second curly arrow: C–X bonding pair moving onto X.
5. Products: the alcohol and X⁻.

Method in steps: NH₃ with a primary halogenoalkane

1. Same first arrows, but from the lone pair on N of NH₃.
2. Intermediate: R–NH₃⁺ and X⁻.
3. A second NH₃ removes H⁺: arrow from its lone pair to H, arrow from the N–H bond to N.
4. Products: R–NH₂ and NH₄⁺.

SN1 and SN2 labels and mechanisms belong to Unit 4.

Rate trends (10.10, 10.12, 10.13)

Compare	Fastest → slowest	Reason to give
Halogen	R–I > R–Br > R–Cl	C–I is the weakest bond (smallest bond enthalpy), so less energy is needed to break it
Structure	Tertiary > secondary > primary	Learn the trend; the explanation is in Unit 4

C–Cl is the most polar of the three bonds, yet chloroalkanes react most slowly. So bond enthalpy, not polarity, controls this trend.

Relative rate ∝ 1/time. Small worked reminder: if one compound gives a precipitate in 50 s and another in 400 s, 1/t is 0.020 s⁻¹ and 0.0025 s⁻¹, so the first is 400 ÷ 50 = 8 times faster.

Core Practical 5 summary (10.11)

• Mix each halogenoalkane with ethanol and aqueous silver nitrate at the same temperature (water bath).
• Ethanol is there so the halogenoalkane and the aqueous solution mix.
• Time how long until a precipitate appears. Observe colour: white (Cl), cream (Br), yellow (I).
• Control: temperature, volumes, AgNO₃ concentration, amount of halogenoalkane.
• Use structural isomers to compare 1°, 2°, 3°; use primary chloro-, bromo-, iodoalkanes to compare halogens.

Core Practical 6 summary (10.14)

(CH₃)₃COH + HCl → (CH₃)₃CCl + H₂O

Step	Purpose
Shake alcohol with concentrated HCl in a separating funnel	Carries out the substitution at room temperature
Open the tap often	Releases pressure
Discard the lower aqueous layer	The product is less dense than water and sits on top
Wash with NaHCO₃(aq)	Neutralises remaining acid (CO₂ is given off)
Wash with water	Removes soluble salts
Add anhydrous Na₂SO₄	Removes water; liquid turns from cloudy to clear
Distil, collect near 51 °C	Separates pure product from unreacted alcohol

Further suggested practicals

• Halogen test: warm with NaOH(aq), acidify with dilute HNO₃, add AgNO₃(aq). Then ammonia: AgCl dissolves in dilute, AgBr in concentrated, AgI in neither.
• 1-bromobutane: heat butan-1-ol with KBr and sulfuric acid, which together form HBr.

Must-know distinctions

• Hydroxide in water vs hydroxide in ethanol: in water it substitutes (alcohol); in ethanol it eliminates (alkene).
• Water vs hydroxide as nucleophile: silver nitrate in ethanol relies on H₂O, a weaker nucleophile, so reactions are slow enough to time.
• Homolytic vs heterolytic: radicals vs ions.
• Electrophile vs nucleophile: accepts vs donates an electron pair.
• Polarity vs bond enthalpy: polarity explains why attack happens at carbon; bond enthalpy explains the I > Br > Cl rate order.
• Ammonia vs cyanide: both are nucleophiles; only cyanide adds a carbon atom.
• Primary vs tertiary classification: count carbons bonded to the C–X carbon, not the total number of carbons in the molecule.
• Halogen test vs rate test: the halogen test hydrolyses with NaOH(aq) first and must be acidified; the rate test uses water in ethanol, so no acidifying is needed.

Quick self-test

1. Define the term nucleophile.
2. Classify: CH₂=CH₂ + Br₂ → CH₂BrCH₂Br.
3. Classify: CH₃CHClCH₃ + KOH (ethanol) → CH₃CH=CH₂ + KCl + H₂O.
4. Is 2-iodo-2-methylpentane primary, secondary or tertiary?
5. Name CH₃CH₂CHBrCH₂CH₃.
6. Give the reagent and conditions to turn 1-bromopropane into butanenitrile.
7. 2.46 g of 1-bromopropane is converted fully into butanenitrile. What mass of nitrile forms? (H = 1.0, C = 12.0, N = 14.0, Br = 79.9)
8. What colour precipitate forms when 1-iodobutane is warmed with silver nitrate in ethanol?
9. Which reacts fastest with aqueous silver nitrate in ethanol: 1-chlorobutane, 1-bromobutane or 1-iodobutane? Give the reason.
10. Why are two moles of ammonia needed per mole of halogenoalkane?
11. In Core Practical 6, why is sodium hydrogencarbonate used rather than sodium hydroxide to wash the product?
12. Which bond type in a molecule leads to free-radical substitution?

Answers

1. A species that donates a lone pair of electrons to form a new covalent bond.
2. Addition.
3. Elimination (HCl is removed).
4. Tertiary: the C–I carbon is bonded to three carbons.
5. 3-bromopentane.
6. KCN dissolved in ethanol, heated under reflux.
7. M(C₃H₇Br) = 122.9, so n = 2.46 ÷ 122.9 = 0.0200 mol. M(C₄H₇N) = 69.0, so mass = 0.0200 × 69.0 = 1.38 g.
8. Yellow (AgI).
9. 1-iodobutane. C–I is the weakest of the three carbon–halogen bonds, so it needs the least energy to break.
10. One NH₃ is the nucleophile; the second removes H⁺ from the RNH₃⁺ ion, forming NH₄⁺.
11. Hydroxide ions would hydrolyse some of the 2-chloro-2-methylpropane back to the alcohol. Hydrogencarbonate is a weak base that removes acid without doing this.
12. Non-polar C–H or C–C bonds, as in alkanes.

Where marks are usually lost

• Writing “KOH” or “alkali” with no solvent, so the examiner cannot tell substitution from elimination.
• Calling ethanolic hydroxide a nucleophile instead of a base.
• Curly arrows that start on an atom or a minus sign rather than a lone pair or a bond.
• Missing the δ+ and δ− on the C–X bond in a mechanism.
• Stopping the ammonia mechanism at RNH₃⁺ without the second ammonia.
• Explaining the I > Br > Cl order by polarity or electronegativity.
• Getting nitrile names wrong by not counting the CN carbon.
• Describing a “white” precipitate for a bromoalkane. AgBr is cream.
• Forgetting to acidify with dilute nitric acid before adding silver nitrate in the halogen test.

Next, test yourself with the practice questions, or go back to the study guide for any section that felt weak.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 2, Topic 10: Organic Chemistry: Halogenoalkanes, Alcohols and Spectra, parts 10A General principles and 10B Halogenoalkanes, outcomes 10.1 to 10.14.