Halogen Compounds

Organic Chemistry Weightage: AS · Papers 1 and 2 (assumed in Paper 4) Topic 15

“Welcome! A halogenoalkane has a polar C–X bond, so the carbon is δ+ and attracts nucleophiles. The same compound can substitute or eliminate depending on the solvent, and substitution can follow two mechanisms, SN2 or SN1, depending on whether the halogenoalkane is primary or tertiary. Learn the reagents with their conditions, the two mechanisms with their rate equations, and the bond-energy reason for the rate order.”

— SCORECHEM ACADEMIC TEAM

0. What Examiners Want

How marks are earned in Halogen Compounds

Reagents and conditions: NaOH(aq) or NaOH in ethanol, heat under reflux; KCN in ethanol; excess NH₃ in ethanol under pressure; AgNO₃(aq) in ethanol for the rate test.
Classification: decide primary, secondary or tertiary by the carbon attached to the halogen, then choose the mechanism.
Mechanisms: SN1 and SN2 with curly arrows, partial charges, a carbocation intermediate or a transition state, and the rate equation.
Explanations: why the rate order is I > Br > Cl (bond energy, not polarity), why a tertiary halogenoalkane uses SN1, why NaOH in ethanol eliminates.
Environment: CFCs, the chlorine-radical chain that destroys ozone, and why HFCs are safer.

Where this topic appears. Recent AS papers have asked for: the type of reaction and the product when 1-chloropropane is heated with NaOH in ethanol; the relative speed of hydrolysis and the SN1 or SN2 mechanism for three given halogenoalkanes; which halogenoalkane reacts fastest with aqueous silver nitrate by SN1; the reagents that convert a halogenoalkane into a nitrile; a comparison of the rate of hydrolysis of 1-chlorobutane and 1-bromobutane; the identification of the nucleophile and leaving group; and the number of primary isomers of C₅H₁₁Br. PCl₅ also appears in the context of phosphorus chemistry. Topic 14 (HBr or Cl₂ with propene) and Topic 16 (alcohols) are closely linked.

1. Making Halogenoalkanes

A halogenoalkane contains a halogen atom bonded to an sp³ carbon: R–X, where X is F, Cl, Br or I. There are three ways to make one.

Three ways to make a halogenoalkane. Substituting an alcohol is the cleanest method. The alcohol −OH is replaced by −X; PCl₅ gives fumes of HCl at room temperature, and with SOCl₂ the by-products are both gases, which makes purification easy.

From an alkane: free-radical substitution with Cl₂ or Br₂ in UV light (Topic 14). It gives a mixture, so it is not a useful route to a pure compound.
From an alkene: electrophilic addition of HX or X₂ at room temperature. For example CH₂=CH₂ + HBr → CH₃CH₂Br. With an unsymmetrical alkene the major product follows Markovnikov's rule.
From an alcohol: the −OH group is replaced by −X (a substitution). The reagents are HX, or KBr (or KCl) with concentrated H₂SO₄ or H₃PO₄ (which makes HX in the flask), or a phosphorus or sulfur halide:
- C₂H₅OH + PCl₅ → C₂H₅Cl + HCl + POCl₃ (room temperature; misty fumes of HCl)
- 3C₂H₅OH + PCl₃ → 3C₂H₅Cl + H₃PO₃ (heat)
- C₂H₅OH + SOCl₂ → C₂H₅Cl + HCl + SO₂ (both by-products are gases, so the product is easy to separate)

2. Primary, Secondary and Tertiary

Halogenoalkanes are classified by the number of carbon atoms attached to the carbon bonded to the halogen.

Classify by the number of carbon atoms attached to the carbon that carries the halogen (not by counting all the carbons). Classification decides the mechanism: primary reacts by SN2, tertiary by SN1, secondary by a mixture.

For example, 1-bromobutane is primary, 2-bromobutane is secondary and 2-bromo-2-methylpropane is tertiary. The number of isomers of C₅H₁₁Br that are primary is four: 1-bromopentane, 1-bromo-2-methylbutane, 1-bromo-3-methylbutane and 1-bromo-2,2-dimethylpropane. The classification also controls which mechanism is followed (section 5).

⚠️ Examiner Trap: Count from the carbon that has the halogen. Do not classify by the total number of carbons in the molecule. Look at the C–X carbon and count how many other carbons are directly bonded to it: one (primary), two (secondary), three (tertiary). Methyl halides, such as CH₃Br, are treated as primary.

3. Nucleophilic Substitution Reactions

The halogen is more electronegative than carbon, so the C–X bond is polar and the carbon is δ+. A nucleophile (a species that donates a lone pair) attacks this carbon and the halogen leaves as the halide ion (the leaving group). The halogen is substituted by the nucleophile.

A halogenoalkane has four main fates. Aqueous NaOH substitutes; ethanolic NaOH eliminates. KCN lengthens the carbon chain by one, which makes it useful in synthesis. NH₃ must be in excess or the amine product reacts again to give a mixture.

Nucleophile	Reagent and conditions	Product (from bromoethane)	Equation
OH⁻	NaOH(aq), heat under reflux	ethanol (an alcohol)	CH₃CH₂Br + NaOH → CH₃CH₂OH + NaBr
CN⁻	KCN in ethanol, heat under reflux	propanenitrile (a nitrile)	CH₃CH₂Br + KCN → CH₃CH₂CN + KBr
NH₃	excess NH₃ in ethanol, heat under pressure	ethylamine (a primary amine)	CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br

Hydroxide. The hydrolysis is carried out with aqueous NaOH (hydroxide is a better nucleophile than water). Heating under reflux stops the volatile halogenoalkane from escaping.
Cyanide. The product has one more carbon than the starting material, so this is a way to lengthen a carbon chain. The nitrile can be hydrolysed to a carboxylic acid or reduced to an amine (later topics).
Ammonia. The NH₃ must be in excess. If it is not, the amine formed is itself a nucleophile and reacts with more halogenoalkane, giving a mixture of secondary and tertiary amines.

⚠️ Examiner Trap: Name the nucleophile by what it donates. A nucleophile has a lone pair and attacks a δ+ carbon. Write OH⁻, CN⁻ or NH₃ (with the lone pair shown on the curly arrow), and name the leaving group as the halide ion (Br⁻). The nucleophile is not the sodium or potassium ion.

4. Rate of Hydrolysis and Bond Energy

Water (or hydroxide) replaces the halogen, and the halide ion is released. To compare the rates, warm each halogenoalkane with aqueous silver nitrate in ethanol (ethanol is the common solvent that lets the two layers mix) in a water bath at the same temperature, and time the appearance of the precipitate.

Halogenoalkane	Precipitate	Speed	Average bond energy (kJ mol⁻¹)
chloroalkane	white AgCl	slowest	C–Cl 340
bromoalkane	cream AgBr	medium	C–Br 280
iodoalkane	pale yellow AgI	fastest	C–I 240

Rate of hydrolysis of RX: R–I > R–Br > R–Cl >> R–F. Water replaces X and the X⁻ released is precipitated by Ag⁺: the time for the precipitate to appear measures the rate.

The reaction is fastest for the iodoalkane because the C–I bond is the weakest and breaks most easily. This is true even though C–Cl is the most polar bond: bond energy, not polarity, controls the rate. Fluoroalkanes have the strongest C–X bond (no value is given in the Data Booklet) and do not hydrolyse at a measurable rate.

⚠️ Examiner Trap: Do not use polarity to explain the order. Chlorine is the most electronegative of Cl, Br and I, so C–Cl is the most polar bond, but chloroalkanes react slowest. The correct explanation is the bond energy: C–I < C–Br < C–Cl, so the weakest bond breaks fastest. State that bond breaking is part of the rate-determining step.

5. SN2 and SN1 Mechanisms

Nucleophilic substitution can follow two different mechanisms. The name gives the type: S for substitution, N for nucleophilic, and the number is the molecularity of the rate-determining step (how many species are involved in it).

SN2 (primary halogenoalkanes). One step. The nucleophile attacks the δ+ carbon from the side opposite the halogen while the C–X bond breaks. A transition state forms in which the carbon is partly bonded to both. There is no intermediate. Both the halogenoalkane and the nucleophile are in the rate-determining step, so rate = k[RX][OH⁻].

SN2 (nucleophilic substitution, bimolecular). OH⁻ donates a lone pair to the δ+ carbon while the C–Br bond breaks, in a single step through a transition state in which both bonds are partly formed. Primary halogenoalkanes react this way because the δ+ carbon is not blocked by bulky groups.

SN1 (tertiary halogenoalkanes). Two steps. First the C–X bond breaks slowly to give a carbocation and X⁻ (the rate-determining step). The nucleophile then attacks C⁺ quickly. Only the halogenoalkane is in the slow step, so rate = k[RX] and the concentration of the nucleophile does not affect the rate.

SN1 (nucleophilic substitution, unimolecular). The C–Br bond breaks first (slowly) to give a tertiary carbocation and Br⁻; the nucleophile then bonds to C⁺ quickly. Because the slow step involves only the halogenoalkane, changing [OH⁻] does not change the rate.

Secondary halogenoalkanes can react by both mechanisms.

	SN2	SN1
Halogenoalkane	primary	tertiary
Number of steps	1	2
Intermediate	none (transition state)	carbocation
Rate equation	k[RX][Nu⁻]	k[RX]
Molecularity of slow step	bimolecular	unimolecular
Reason	δ+ carbon is not shielded	three alkyl groups stabilise C⁺ and block back-side attack

⚠️ Examiner Trap: Match mechanism, rate equation and structure. A tertiary halogenoalkane is SN1 because it forms a stable tertiary carbocation (three electron-donating alkyl groups) and because three alkyl groups hinder attack from the back. Draw curly arrows from a lone pair to the carbon and from the C–X bond to the halogen. An SN2 transition state is in square brackets with partial charges and a dagger, and is not an intermediate.

6. Elimination

Heating a halogenoalkane with NaOH dissolved in ethanol under reflux removes HX and forms an alkene. Here OH⁻ acts as a base (it removes a proton) rather than as a nucleophile.

The same reagents in different solvents give different products. Aqueous NaOH: substitution to an alcohol. Ethanolic NaOH: elimination of HX to an alkene. In elimination, the hydrogen removed is on the carbon adjacent to the C–Br carbon.

Example: CH₃CH₂CH₂CH₂Br + NaOH → CH₃CH₂CH=CH₂ + NaBr + H₂O. Compare 1-chloropropane: with NaOH(aq) it gives propan-1-ol (substitution), but with NaOH in ethanol it gives propene (elimination). The hydrogen that is removed is on a carbon next to the carbon holding the halogen. This reaction is another way to make an alkene.

⚠️ Examiner Trap: Name the type of reaction from the conditions. Aqueous NaOH means substitution (OH⁻ is a nucleophile). Ethanolic NaOH means elimination (OH⁻ is a base). Write both the solvent and ‘heat under reflux’; a reagent given without its solvent will not gain full credit.

7. Uses, CFCs and the Ozone Layer

Halogenoalkanes are used as solvents, refrigerants, aerosol propellants, foam-blowing agents, anaesthetics (halothane) and flame retardants. Chlorofluorocarbons (CFCs) were widely used because they are unreactive (strong C–F and C–Cl bonds), non-flammable and non-toxic. This is also why they persist and reach the stratosphere.

In the stratosphere UV light breaks the weaker C–Cl bond homolytically, producing chlorine radicals. These radicals catalyse the destruction of ozone:

CFCs are unreactive near the ground, so they drift to the stratosphere, where UV light splits the C–Cl bond. The chlorine radical is a catalyst for ozone destruction, because it is regenerated in step 3.

Because Cl• is regenerated, one radical can destroy thousands of ozone molecules. Less ozone means more harmful UV radiation reaches the ground (skin cancer, crop damage). CFCs have been phased out under the Montreal Protocol. Replacements include hydrofluorocarbons (HFCs) such as CH₂FCF₃, which contain no C–Cl bond and so give no Cl•, together with hydrocarbons such as propane and butane and with CO₂.

8. Quick Sheet and Checklist

Topic	Key point
Classification	by the carbons attached to the C–X carbon: 1°, 2°, 3°
From an alcohol	HX; KX + conc. H₂SO₄; PCl₅; PCl₃; SOCl₂
NaOH(aq), reflux	substitution: alcohol
KCN in ethanol, reflux	substitution: nitrile, one carbon longer
excess NH₃, ethanol, pressure	substitution: primary amine
NaOH in ethanol, reflux	elimination: alkene (OH⁻ is a base)
Rate of hydrolysis	R–I > R–Br > R–Cl: weakest bond fastest
Test	AgNO₃(aq) in ethanol: white, cream, pale yellow
SN2	primary; one step; rate = k[RX][OH⁻]
SN1	tertiary; carbocation; rate = k[RX]
CFCs	UV breaks C–Cl; Cl• catalyses ozone destruction

Before you leave the question, check:

Every substitution or elimination has its reagent, solvent and heat under reflux.
The halogenoalkane is classified from the C–X carbon, and the mechanism fits the class.
Curly arrows start at a lone pair or bond; SN2 shows a transition state and SN1 shows a carbocation.
The rate order is explained by bond energy, not polarity.
Equations balance: KCN gives a nitrile with one more carbon; excess NH₃ needs two NH₃ per halogenoalkane (one becomes NH₄Br).
CFC answers say homolytic fission of C–Cl, UV light, and that Cl• is regenerated.