S5 Chemistry – Unit 4: Halogenoalkanes

4.1

Definition and Classification

Definition Halogenoalkanes (also called haloalkanes or alkyl halides) are organic compounds derived from alkanes in which one or more hydrogen atoms have been replaced by a halogen atom (F, Cl, Br, or I). They contain the functional group C–X where X = F, Cl, Br, or I.

General Formula

For a monohalogenoalkane: C_nH_2n+1X or written as R–X where R is an alkyl group and X is a halogen.

Classification by Degree of Substitution

Halogenoalkanes are classified according to how many carbon atoms are directly bonded to the carbon carrying the halogen:

Class	Definition	Carbon bearing X bonded to:	Example	IUPAC Name
Primary (1°)	C bearing X is attached to 1 other C	1 carbon + 2 H	CH₃CH₂Br	bromoethane
Secondary (2°)	C bearing X is attached to 2 other C	2 carbons + 1 H	CH₃CHBrCH₃	2-bromopropane
Tertiary (3°)	C bearing X is attached to 3 other C	3 carbons + 0 H	(CH₃)₃CBr	2-bromo-2-methylpropane

Classification by Halogen

Halogen	Prefix	Example	Name
Fluorine (F)	fluoro–	CH₃F	fluoromethane
Chlorine (Cl)	chloro–	CH₃Cl	chloromethane
Bromine (Br)	bromo–	CH₃Br	bromomethane
Iodine (I)	iodo–	CH₃I	iodomethane

IUPAC Nomenclature

Find the longest carbon chain containing the carbon with the halogen → parent alkane name.
Number the chain to give the halogen the lowest locant.
Name the halogen as a prefix: fluoro–, chloro–, bromo–, iodo–.
Multiple halogens: use di–, tri– etc. List halogens alphabetically.
If other substituents present, list all alphabetically before the chain name.

Example 1

Naming Halogenoalkanes

Name: (a) CH₃CHClCH₂CH₃ (b) CH₃CBr₂CH₃ (c) CHCl₂CH₂Br

Longest chain = 4 C (butane). Cl on C2 from left (or C3 from right; C2 is lower). → 2-chlorobutane. Classification: secondary (C2 bonded to C1 and C3).

Longest chain = 3 C (propane). Two Br on C2. → 2,2-dibromopropane. Classification: tertiary (C2 bonded to C1 and C3 and C3... wait, two Br on C2 = C2 bonded to C1, C3, Br, Br → C bearing X attached to 2 C = secondary).

Longest chain = 2 C (ethane). C1 has 2 Cl; C2 has 1 Br. Number to give lower set. → 2-bromo-1,1-dichloroethane.

4.2

Physical Properties

Polarity of the C–X Bond

Halogens are more electronegative than carbon, so the C–X bond is polar (C^δ+–X^δ−). This polarity makes halogenoalkanes more reactive than alkanes and gives them slightly higher boiling points than alkanes of similar M_r.

Electronegativity order: F > Cl > Br > I. Bond polarity: C–F > C–Cl > C–Br > C–I.

Boiling Points

Boiling points increase with:

Chain length (more surface area → stronger van der Waals forces)
Atomic mass of halogen: R–F < R–Cl < R–Br < R–I (larger halogens have more electrons → stronger dispersion forces)
Degree of substitution: 1° < 2° < 3° (for same chain length, more branching lowers B.P.)

Compound	Formula	M_r	B.P. (°C)	State
Chloromethane	CH₃Cl	50.5	−24	Gas
Bromomethane	CH₃Br	95	+4	Gas/Liquid
Iodomethane	CH₃I	142	+43	Liquid
Chloroethane	C₂H₅Cl	64.5	+12	Gas/Liquid
Bromoethane	C₂H₅Br	109	+38	Liquid
Iodoethane	C₂H₅I	156	+72	Liquid
1-chloropropane	C₃H₇Cl	78.5	+47	Liquid
1-bromopropane	C₃H₇Br	123	+71	Liquid

Solubility and Density

Halogenoalkanes are immiscible with water (they cannot form hydrogen bonds as acceptors but their polarity is not sufficient to overcome the strong H-bonds in water). They dissolve readily in organic solvents.

Most liquid halogenoalkanes are denser than water (especially bromo- and iodo-compounds). They form the lower layer when shaken with water.

Bond Strength and Reactivity Trend

Bond dissociation energies: C–F (484) > C–Cl (338) > C–Br (276) > C–I (238) kJ/mol.

Reactivity in nucleophilic substitution: R–I > R–Br > R–Cl > R–F (weaker bond = easier to break = faster reaction, despite F being most electronegative).

4.3

Preparation of Halogenoalkanes

Method 1: Free-Radical Halogenation of Alkanes

Alkanes react with Cl₂ or Br₂ under UV light (free-radical substitution, see Unit 2). Gives a mixture of products.

CH4 + Cl2 --(hv)--> CH3Cl + HCl (chloromethane)

Method 2: Addition of HX to Alkenes

HCl, HBr, or HI add across the C=C bond. Markovnikov's rule gives the major product.

CH2=CH2 + HBr --> CH3CH2Br (bromoethane) CH3CH=CH2 + HBr --> CH3CHBrCH3 (2-bromopropane, major)

Method 3: Addition of X₂ to Alkenes

CH2=CH2 + Cl2 --> CH2ClCH2Cl (1,2-dichloroethane)

Method 4: Reaction of Alcohols with Halogenating Agents

This is the most reliable laboratory method for preparing a specific halogenoalkane:

With PCl5 (phosphorus pentachloride): R-OH + PCl5 --> R-Cl + POCl3 + HCl e.g. CH3CH2OH + PCl5 --> CH3CH2Cl + POCl3 + HCl With PCl3 (phosphorus trichloride): 3 R-OH + PCl3 --> 3 R-Cl + H3PO3 With SOCl2 (thionyl chloride) -- best for chloroalkanes: R-OH + SOCl2 --> R-Cl + SO2 + HCl With HBr (conc. H2SO4 + NaBr, or direct HBr): R-OH + HBr --> R-Br + H2O With red phosphorus + I2 (for iodoalkanes): 3 R-OH + PI3 (formed in situ) --> 3 R-I + H3PO3

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Which Reagent to Use? For chloroalkanes: PCl₅, PCl₃, or SOCl₂ (SOCl₂ is cleanest — only gaseous by-products).
For bromoalkanes: NaBr + conc. H₂SO₄, or direct HBr.
For iodoalkanes: red P + I₂ (NOT HI + H₂SO₄ — H₂SO₄ oxidises HI).
For fluoroalkanes: special methods (e.g. Balz-Schiemann reaction) — not simple lab prep.

Example 2

Preparation of 1-bromopropane from propan-1-ol

Write the equation for preparing 1-bromopropane from propan-1-ol. State the reagent and any conditions.

React propan-1-ol with NaBr and concentrated H₂SO₄ (which generates HBr in situ) or directly with conc. HBr.

CH₃CH₂CH₂OH + HBr → CH₃CH₂CH₂Br + H₂O

Conditions: heat under reflux. Product is separated by distillation.

✓

Product: 1-bromopropane (a primary halogenoalkane) ✔

4.4

Chemical Reactions

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Why Halogenoalkanes are Reactive The C^δ+–X^δ− bond is polar. The δ+ carbon is susceptible to attack by nucleophiles (electron-pair donors: OH⁻, CN⁻, NH₃, RO⁻) and the δ− halogen acts as a leaving group. The two main reaction pathways are nucleophilic substitution (S_N) and elimination (E).

Reaction 1: Nucleophilic Substitution with Aqueous NaOH / KOH → Alcohol

Conditions: aqueous NaOH or KOH, heat under reflux. The OH⁻ nucleophile replaces X.

R-X + NaOH(aq) --heat/reflux--> R-OH + NaX e.g. CH3CH2Br + NaOH(aq) --> CH3CH2OH + NaBr (bromoethane) (ethanol)

Reaction 2: Nucleophilic Substitution with KCN → Nitrile (chain lengthening)

Conditions: alcoholic KCN, heat under reflux. CN⁻ is the nucleophile. The product nitrile has one more carbon than the halogenoalkane — this reaction lengthens the carbon chain.

R-X + KCN(alc) --heat/reflux--> R-CN + KX e.g. CH3CH2Br + KCN --> CH3CH2CN + KBr (bromoethane) (propanenitrile) -- chain extended by 1 C

Reaction 3: Nucleophilic Substitution with NH₃ → Amine

Conditions: excess concentrated NH₃ in ethanol (sealed tube), heat. NH₃ acts as nucleophile. A mixture of primary, secondary, tertiary amines and quaternary ammonium salts may form.

R-X + NH3(excess, alc) --> R-NH2 + HX e.g. CH3Br + NH3 --> CH3NH2 + HBr (bromomethane) (methylamine, primary amine)

Reaction 4: Elimination with Alcoholic KOH → Alkene

Conditions: alcoholic KOH (KOH dissolved in ethanol), heat under reflux. HX is eliminated to form an alkene. Zaitsev's rule: the more substituted alkene is the major product.

R-CHX-CH2-R' + KOH(alc) --heat--> R-CH=CH-R' + KX + H2O e.g. CH3CH2Br + KOH(alc) --> CH2=CH2 + KBr + H2O (bromoethane) (ethene)

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Aqueous vs Alcoholic KOH Aqueous KOH → substitution (gives alcohol). The water acts as solvent and OH⁻ is a good nucleophile in polar protic solvent.
Alcoholic KOH → elimination (gives alkene). The ethanol solvent reduces nucleophilicity of OH⁻ but favours its role as a base, abstracting a proton from the β-carbon.

Reaction 5: Substitution with AgNO₃ in Ethanol → Silver Halide Precipitate

This reaction is used as a chemical test (see Section 4.5). AgNO₃ in ethanol reacts with halogenoalkanes on warming:

R-X + AgNO3(ethanolic) --warm--> R-ONO2 + AgX(s) AgCl: white precipitate AgBr: cream precipitate AgI: yellow precipitate

Example 3

Substitution vs Elimination — Choosing Conditions

2-bromopropane is treated with KOH. (a) What product forms with aqueous KOH? (b) What product forms with alcoholic KOH? Write equations for both.

Aqueous KOH → substitution:
CH₃CHBrCH₃ + KOH(aq) → CH₃CH(OH)CH₃ + KBr
Product: propan-2-ol

Alcoholic KOH → elimination:
CH₃CHBrCH₃ + KOH(alc) → CH₃CH=CH₂ + KBr + H₂O
Product: propene

S_N1 vs S_N2 Mechanisms

S_N2 (Bimolecular)

One-step mechanism. Nucleophile attacks the back of the C–X bond simultaneously as X leaves. Favoured by primary halogenoalkanes (less steric hindrance).

Nu:- + R-X --> [Nu...C...X] --> Nu-R + X- (backside attack, inversion of config)

Rate = k[R–X][Nu] — depends on both reactant concentrations.

S_N1 (Unimolecular)

Two-step mechanism. First, C–X ionises to give a carbocation; then nucleophile attacks. Favoured by tertiary halogenoalkanes (more stable carbocation).

Step 1: R-X --> R+ + X- (slow, rate-determining) Step 2: R+ + Nu- --> R-Nu (fast)

Rate = k[R–X] — depends only on halogenoalkane concentration.

4.5

Chemical Tests for Halogenoalkanes

Test: Identifying the Halogen Present

The identity of the halogen (Cl, Br, or I) is determined using ethanolic silver nitrate solution after first hydrolysing the halogenoalkane:

Add the halogenoalkane to ethanolic AgNO₃ solution in a test tube.
Warm the mixture in a water bath.
Observe the precipitate formed.
To confirm, add dilute HNO₃ then dilute ammonia solution.

Halogen	Precipitate with AgNO₃/EtOH	Colour	Solubility in dilute NH₃	Solubility in conc. NH₃
Chlorine (Cl)	AgCl	White	Soluble (dissolves)	Soluble
Bromine (Br)	AgBr	Cream	Insoluble	Soluble
Iodine (I)	AgI	Yellow	Insoluble	Insoluble

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Relative Rates of Reaction The rate of precipitate formation also reveals the C–X bond strength. Iodoalkanes react fastest (weakest C–I bond, easiest to hydrolyse); chloroalkanes react slowest (strongest C–Cl bond). This gives the reactivity order: R–I > R–Br > R–Cl.

Example 4

Identifying an Unknown Halogenoalkane

Three test tubes contain 1-chlorobutane, 1-bromobutane, and 1-iodobutane. Describe how you would identify which is which using one reagent.

Add a few drops of each compound to separate test tubes containing ethanolic AgNO₃ solution.

Warm each test tube in a water bath at ~60°C and observe.

Results:
• White precipitate (and fastest with dilute NH₃) → 1-chlorobutane (AgCl)
• Cream precipitate → 1-bromobutane (AgBr)
• Yellow precipitate (forms fastest overall) → 1-iodobutane (AgI)

Confirm by adding dilute then concentrated ammonia: AgCl dissolves in dilute NH₃; AgBr requires conc. NH₃; AgI does not dissolve.

4.6

Uses of Halogenoalkanes and CFCs

Compound	Use	Notes
Chloromethane (CH₃Cl)	Manufacture of silicones; methylating agent	Industrial chemical synthesis
Dichloromethane (CH₂Cl₂)	Solvent (paint stripper, extraction solvent, decaffeination of coffee)	Low boiling point (40°C); good solvent for organics
Trichloromethane (CHCl₃, chloroform)	Formerly used as anaesthetic; solvent; intermediate in synthesis	Now replaced due to toxicity
Tetrachloromethane (CCl₄)	Formerly used as dry-cleaning solvent and fire extinguisher	Phased out due to ozone depletion and toxicity
Chloroethane (C₂H₅Cl)	Local anaesthetic (skin cooling spray); ethylating agent	Rapid evaporation cools the skin
1,2-dichloroethane	Manufacture of vinyl chloride (PVC monomer); solvent	Important petrochemical intermediate
Chloroethene (vinyl chloride)	Monomer for PVC	Pipes, cables, flooring
Halothane (CF₃CHBrCl)	General anaesthetic	Replaced by newer agents
Teflon (PTFE, –[CF₂CF₂]_n–)	Non-stick coatings; electrical insulation	Extremely chemically inert

CFCs and the Ozone Layer

CFCs (Chlorofluorocarbons) CFCs are halogenoalkanes containing both chlorine and fluorine atoms (and no H). Examples: CCl₃F (CFC-11, trichlorofluoromethane), CCl₂F₂ (CFC-12, dichlorodifluoromethane, Freon-12). They were widely used as refrigerants, aerosol propellants, and foam-blowing agents.

Why CFCs Were Considered Safe (Initially)

CFCs are non-toxic, non-flammable, chemically inert in the lower atmosphere, and cheap to produce. These properties made them seem ideal for industrial use.

How CFCs Destroy the Ozone Layer

In the stratosphere (15–50 km altitude), UV radiation is intense enough to break C–Cl bonds homolytically in CFCs, releasing chlorine radicals (Cl•) that catalytically destroy ozone:

CFC photolysis: CCl2F2 --(UV)--> CClF2* + Cl* (Cl radical generated) Ozone destruction (catalytic cycle): Cl* + O3 --> ClO* + O2 (step 1) ClO* + O --> Cl* + O2 (step 2) ------------------------------------------- Net: O3 + O --> 2 O2 (Cl* is regenerated -- acts as catalyst) One Cl* can destroy ~100,000 ozone molecules before being removed.

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Consequences of Ozone Depletion The ozone layer absorbs harmful UV-B radiation (280–315 nm). Depletion leads to: increased UV-B reaching Earth's surface → higher rates of skin cancer, cataracts, and immune suppression; damage to marine phytoplankton and food chains.

The Montreal Protocol (1987) and Replacements

The Montreal Protocol phased out the production and use of CFCs internationally. CFCs have been replaced by HCFCs (hydrochlorofluorocarbons, e.g. CHClF₂) and then HFCs (hydrofluorocarbons, no Cl atoms, e.g. CH₂FCF₃). HFCs do not deplete ozone but are still potent greenhouse gases and are being phased out under the Kigali Amendment (2016).

Example 5

Explaining CFC Ozone Destruction

Explain why a small amount of CFC can destroy a large amount of ozone.

UV radiation in the stratosphere breaks the C–Cl bond in CFCs: CCl₂F₂ →(UV) •CClF₂ + Cl•

Cl• reacts with O₃: Cl• + O₃ → ClO• + O₂

ClO• reacts with an oxygen atom: ClO• + O → Cl• + O₂

Cl• is regenerated — it acts as a catalyst. It is not consumed in the overall reaction. One Cl• can destroy ~100,000 O₃ molecules before being removed by another reaction.

✓

This is why even tiny amounts of CFC cause massive ozone loss.

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Exercises

Classify each as primary, secondary, or tertiary, and give the IUPAC name:
(a) (CH₃)₃CCl (b) CH₃CH₂CHBrCH₃ (c) CH₃CH₂CH₂I

(a) Tertiary; 2-chloro-2-methylpropane (C bearing Cl attached to 3 other C atoms)
(b) Secondary; 2-bromobutane (C bearing Br attached to 2 other C atoms)
(c) Primary; 1-iodopropane (C bearing I attached to 1 other C atom)
Write equations for the reaction of 1-bromopropane with: (a) aqueous KOH; (b) alcoholic KOH; (c) KCN in ethanol. Name all organic products.

(a) CH₃CH₂CH₂Br + KOH(aq) → CH₃CH₂CH₂OH + KBr
Product: propan-1-ol

(b) CH₃CH₂CH₂Br + KOH(alc) → CH₃CH=CH₂ + KBr + H₂O
Product: propene

(c) CH₃CH₂CH₂Br + KCN → CH₃CH₂CH₂CN + KBr
Product: butanenitrile (chain extended from 3 C to 4 C)
Explain why iodoalkanes react faster with nucleophiles than chloroalkanes, even though C–I is less polar than C–Cl.

Although C–I is less polar than C–Cl, the C–I bond is much weaker (238 kJ/mol vs 338 kJ/mol for C–Cl). In nucleophilic substitution, the bond breaking step (C–X cleavage) is rate-determining. The weaker C–I bond requires less activation energy to break, so iodoalkanes react faster despite being less polar.
Describe the test you would use to distinguish between 1-chlorobutane and 1-iodobutane. State clearly what you would observe with each compound.

Add each compound to ethanolic AgNO₃ solution and warm in a water bath.

1-chlorobutane: white precipitate of AgCl (dissolves in dilute ammonia).
1-iodobutane: yellow precipitate of AgI (insoluble in both dilute and concentrated ammonia). Also forms faster than the chloro compound.
Write the two steps of the catalytic cycle by which chlorine radicals destroy ozone. Explain the term "catalytic" in this context.

Step 1: Cl• + O₃ → ClO• + O₂
Step 2: ClO• + O → Cl• + O₂
Net: O₃ + O → 2 O₂

Catalytic: The Cl• radical is regenerated in step 2 and is not consumed overall. It speeds up the destruction of ozone without being permanently used up, so one Cl• can destroy thousands of ozone molecules.
A student wants to prepare chloroethane from ethanol. Give two possible reagents they could use, and write the equation for one of them.

Reagent 1: PCl₅
CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl

Reagent 2: SOCl₂ (thionyl chloride)
CH₃CH₂OH + SOCl₂ → CH₃CH₂Cl + SO₂ + HCl

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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]

Classify each compound as primary, secondary, or tertiary, and give its IUPAC name:
(a) CH₃CH₂CH₂Cl (b) (CH₃)₂CHCH₂Br (c) CH₃C(CH₃)(Cl)CH₂CH₃ (d) CH₃CHClCH₂CH₃

(a) Primary; 1-chloropropane
(b) Primary; 1-bromo-2-methylpropane (Br on C1, methyl branch on C2)
(c) Tertiary; 2-chloro-2-methylbutane
(d) Secondary; 2-chlorobutane

Q2 [4 marks]

Explain the trend in boiling points of CH₃Cl, CH₃Br, and CH₃I. Why does CH₃F have an anomalously low boiling point compared to the other three?

Trend CH₃Cl → CH₃Br → CH₃I: boiling points increase (−24°C → +4°C → +43°C) because atomic mass (and electron count) of the halogen increases from Cl to I → greater polarisability → stronger London dispersion forces → more energy needed to separate molecules.

CH₃F anomaly: F has a much smaller atomic radius and fewer electrons than Cl, giving very weak dispersion forces. Despite C–F being the most polar bond, fluoromethane has a much lower boiling point (−78°C) because dispersion forces dominate over dipole-dipole effects in these small molecules.

Q3 [6 marks]

Write equations, with reagents and conditions, for the reactions of 2-bromobutane with:
(a) aqueous NaOH (b) alcoholic KOH (c) KCN in ethanol (d) excess NH₃ in ethanol
Name all organic products.

(a) CH₃CHBrCH₂CH₃ + NaOH(aq) →^heat CH₃CH(OH)CH₂CH₃ + NaBr
Product: butan-2-ol

(b) CH₃CHBrCH₂CH₃ + KOH(alc) →^heat CH₃CH=CHCH₃ + KBr + H₂O (major: but-2-ene, Zaitsev)
Product: but-2-ene (major) / but-1-ene (minor)

(c) CH₃CHBrCH₂CH₃ + KCN(alc) →^heat CH₃CH(CN)CH₂CH₃ + KBr
Product: 2-methylbutanenitrile (5 C total)

(d) CH₃CHBrCH₂CH₃ + NH₃(excess, alc) →^heat CH₃CH(NH₂)CH₂CH₃ + HBr
Product: butan-2-amine (primary amine, major with excess NH₃)

Q4 [5 marks]

Describe fully the test to identify the halogen in an unknown halogenoalkane. Include reagents, observations, and how you would confirm the result.

Reagent: Ethanolic silver nitrate (AgNO₃) solution.
Procedure: Add a few drops of the halogenoalkane to ethanolic AgNO₃ in a test tube. Warm in a water bath (~60°C). Observe the precipitate.

Observations:
• White precipitate → chlorine present (AgCl)
• Cream precipitate → bromine present (AgBr)
• Yellow precipitate → iodine present (AgI)

Confirmation: Add dilute nitric acid (to remove interfering ions), then add dilute ammonia solution:
• AgCl dissolves in dilute NH₃ → confirms Cl
• AgBr dissolves only in concentrated NH₃ → confirms Br
• AgI does not dissolve in dilute or concentrated NH₃ → confirms I

Q5 [5 marks]

Give three different reagents that can be used to convert ethanol into chloroethane. Write a balanced equation for each reaction.

1. PCl₅:
CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl

2. PCl₃:
3 CH₃CH₂OH + PCl₃ → 3 CH₃CH₂Cl + H₃PO₃

3. SOCl₂:
CH₃CH₂OH + SOCl₂ → CH₃CH₂Cl + SO₂ + HCl

Q6 [6 marks]

Compare the S_N1 and S_N2 mechanisms for nucleophilic substitution in halogenoalkanes. For each mechanism, state: (i) the number of steps; (ii) which class of halogenoalkane it favours; (iii) the rate expression; and (iv) write a brief mechanism for the reaction of the appropriate 2-bromopropane isomer with OH⁻.

S_N2:
(i) One step (concerted)
(ii) Favours primary halogenoalkanes (low steric hindrance)
(iii) Rate = k[R–X][OH⁻] (bimolecular)
(iv) CH₃Br + OH⁻ → [HO···CH₃···Br]⁻ → CH₃OH + Br⁻
(backside attack; Walden inversion of configuration)

S_N1:
(i) Two steps
(ii) Favours tertiary halogenoalkanes (stable carbocation intermediate)
(iii) Rate = k[R–X] (unimolecular; rate depends only on R–X)
(iv) (CH₃)₃CBr → (CH₃)₃C⁺ + Br⁻ (slow, rate-determining)
(CH₃)₃C⁺ + OH⁻ → (CH₃)₃COH (fast)
(Attack from both faces possible → racemic mixture if chiral centre)

Section B — Extended Response

20 marks

Q7 [10 marks]

(a) Describe the role of CFCs in the destruction of the ozone layer. Include: how Cl radicals are formed, the two-step catalytic cycle with equations, and why small amounts of CFC cause large-scale damage. [6 marks]

(b) Explain why HCFCs and HFCs were introduced as replacements for CFCs and evaluate their advantages and disadvantages. [4 marks]

(a) CFCs are stable in the lower atmosphere but rise into the stratosphere. UV radiation (short wavelength, high energy) cleaves the C–Cl bond homolytically:
CCl₂F₂ →(UV) •CClF₂ + Cl•

Catalytic cycle:
Cl• + O₃ → ClO• + O₂
ClO• + O → Cl• + O₂
Net: O₃ + O → 2O₂

Cl• is regenerated in step 2 — it is a catalyst. One Cl atom can destroy ~100,000 O₃ molecules before being removed (e.g. by reacting with CH₄ or NO₂). This amplification effect means even small quantities cause massive O₃ depletion.

(b) HCFCs (e.g. CHClF₂) contain C–H bonds, making them reactive in the troposphere (degraded before reaching stratosphere). They deplete ozone ~5% as much as CFCs. Advantage: less ozone-depleting. Disadvantage: still contains Cl, so not completely safe; also a greenhouse gas.

HFCs (e.g. CH₂FCF₃) contain no Cl atoms → zero ozone depletion potential. Advantage: safe for ozone layer. Disadvantage: very potent greenhouse gases (global warming potential thousands of times that of CO₂); being phased out under the Kigali Amendment (2016).

Q8 [10 marks]

(a) Starting from 2-methylpropan-2-ol, describe with equations how you could prepare: (i) 2-chloro-2-methylpropane; (ii) 2-methylpropene; (iii) 2-methyl-2-methylpropanenitrile (a nitrile). State reagents and conditions for each. [6 marks]

(b) Arrange the following in order of increasing reactivity towards nucleophilic substitution, and explain the trend: CH₃F, CH₃Cl, CH₃Br, CH₃I. [4 marks]

(a)(i) 2-chloro-2-methylpropane:
(CH₃)₃COH + PCl₅ → (CH₃)₃CCl + POCl₃ + HCl
(or SOCl₂: (CH₃)₃COH + SOCl₂ → (CH₃)₃CCl + SO₂ + HCl)

(a)(ii) 2-methylpropene:
(CH₃)₃COH →^{conc. H₂SO₄, 170°C} CH₂=C(CH₃)₂ + H₂O
(dehydration/elimination; or via the chloride + alcoholic KOH)

(a)(iii) Nitrile:
(CH₃)₃CCl + KCN(alc) →^heat/reflux (CH₃)₃CCN + KCl
Product: 2,2-dimethylpropanenitrile

(b) Increasing reactivity: CH₃F < CH₃Cl < CH₃Br < CH₃I

Explanation: Despite F being most electronegative (making C most δ+), the C–F bond has the highest bond dissociation energy (484 kJ/mol). Nucleophilic substitution requires breaking the C–X bond. The weaker the bond, the lower the activation energy, and the faster the reaction. Bond strengths: C–F (484) > C–Cl (338) > C–Br (276) > C–I (238) kJ/mol. Therefore reactivity: C–I > C–Br > C–Cl > C–F.

← Unit 3: Alkenes & Alkynes Unit 5: Alcohols & Ethers →