Halogenoalkanes

1. Overview

A halogenoalkane is an alkane in which a halogen atom, X (Cl, Br or I), has replaced a hydrogen atom. The polar C-X bond leaves the carbon atom attached to the halogen open to attack by nucleophiles, so halogenoalkanes are a key starting point for making alcohols, nitriles, amines and alkenes. This section covers how they are made and classified, their substitution and elimination reactions, the SN1 and SN2 mechanisms, and why iodoalkanes react faster than chloroalkanes.

Key Definitions

• Halogenoalkane: a compound in which one or more hydrogen atoms of an alkane have been replaced by halogen atoms; with one halogen atom, X, the general formula is CnH2n+1X.
• Primary, secondary and tertiary halogenoalkane: a classification by the number of alkyl groups bonded to the carbon atom that carries the halogen: one for primary, two for secondary and three for tertiary.
• Nucleophile: a species that donates a lone pair of electrons to an electron-deficient atom (in organic reactions, usually a δ+ carbon atom), forming a new covalent bond; examples are OH⁻, CN⁻, NH₃ and H₂O.
• Nucleophilic substitution: a reaction in which a nucleophile replaces an atom or group in a molecule; in a halogenoalkane, the nucleophile replaces the halogen atom, which leaves as a halide ion.
• Hydrolysis: a reaction in which a covalent bond is broken by reaction with water (or with hydroxide ions in alkaline hydrolysis); hydrolysis of a halogenoalkane gives an alcohol.
• Elimination reaction: a reaction in which a small molecule is removed from a larger one; a halogenoalkane loses HX to form an alkene with a C=C double bond.
• Carbocation: an ion in which a carbon atom carries a positive charge and forms only three covalent bonds.
• Inductive effect of alkyl groups: the donation of electron density by alkyl groups along their sigma bonds towards a neighbouring atom; it spreads out the charge on a positive carbon atom and so stabilises a carbocation.
• SN2 mechanism: nucleophilic substitution in one step, in which the nucleophile bonds to the carbon atom as the C-X bond breaks, through a single transition state; two species, the halogenoalkane and the nucleophile, take part in this step.
• SN1 mechanism: nucleophilic substitution in two steps: the C-X bond first breaks heterolytically to give a carbocation intermediate (the slow step), and the nucleophile then bonds to it; only one species, the halogenoalkane, takes part in the slow step.

Content

Making halogenoalkanes

Organic equations are normally written without state symbols, as here; R stands for an alkyl group.

1. From an alkane, by free-radical substitution. Ethane reacts with chlorine or bromine in ultraviolet (UV) light:

C₂H₆ + Cl₂ → C₂H₅Cl + HCl

The UV light splits the halogen molecules into free radicals (mechanism in 14.1). Further substitution also gives products such as C₂H₄Cl₂, so this is a poor way to make one pure halogenoalkane.

2. From an alkene, by electrophilic addition. At room temperature, an alkene adds a halogen, X₂, or a hydrogen halide, HX(g). The mechanism, and which product is major for an unsymmetrical alkene, are in 14.2.

CH₂=CH₂ + Br₂ → CH₂BrCH₂Br (1,2-dibromoethane)

CH₂=CH₂ + HBr → CH₃CH₂Br (bromoethane)

3. From an alcohol, by substituting the OH group. A halogen atom replaces the OH group:

• HX(g): ROH + HX → RX + H₂O.
• KCl with concentrated H₂SO₄ or concentrated H₃PO₄: the acid makes HCl in the flask (KCl(s) + H₂SO₄(l) → KHSO₄(s) + HCl(g)), which then reacts as above. For an iodoalkane, use KI with concentrated H₃PO₄: concentrated H₂SO₄ would oxidise the HI to iodine (see 11.3).
• PCl₃ and heat: 3ROH + PCl₃ → 3RCl + H₃PO₃
• PCl₅ at room temperature: ROH + PCl₅ → RCl + POCl₃ + HCl. Steamy fumes of HCl are given off.
• SOCl₂: ROH + SOCl₂ → RCl + SO₂ + HCl. Both by-products are gases and escape, so the chloroalkane is easy to separate.

Primary, secondary and tertiary

Count the alkyl groups (carbon atoms) attached to the carbon atom that carries the halogen.

Class	Alkyl groups on the C-X carbon	Example
primary	one	CH₃CH₂CH₂Cl, 1-chloropropane
secondary	two	CH₃CHClCH₃, 2-chloropropane
tertiary	three	(CH₃)₃CCl, 2-chloro-2-methylpropane

Branches elsewhere in the chain do not count. Halomethanes such as CH₃Cl have no alkyl group on that carbon, and they react like primary halogenoalkanes.

The polar C-X bond and nucleophiles

A halogen is more electronegative than carbon (Data Booklet values: C 2.5, Cl 3.0, Br 2.6), so the C-X bond is polar: Cδ+ and Xδ−. The δ+ carbon attracts nucleophiles: species with a lone pair of electrons they can donate to form a new bond, such as OH⁻, CN⁻, NH₃ and H₂O.

In every substitution, the C-X bond breaks heterolytically: the halogen takes both bonding electrons and leaves as a halide ion, X⁻ (the terms are in 13.2).

Nucleophilic substitution reactions

Bromoethane, CH₃CH₂Br, reacts with four nucleophiles. "Heat" means heat under reflux: boil with a condenser fitted so that volatile liquids return to the flask.

Reagent and conditions	Nucleophile	Organic product
NaOH(aq), heat	OH⁻	CH₃CH₂OH, ethanol
KCN in ethanol, heat	CN⁻	CH₃CH₂CN, propanenitrile
NH₃ in ethanol, heat in a sealed tube	NH₃	CH₃CH₂NH₂, ethylamine
AgNO₃(aq) in ethanol	H₂O	CH₃CH₂OH, ethanol

• Hydroxide ions (alkaline hydrolysis): CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻
• Cyanide ions: CH₃CH₂Br + CN⁻ → CH₃CH₂CN + Br⁻. The carbon atom of CN⁻ forms the new bond, so the carbon chain grows by one atom. Count the CN carbon in the name: two carbons plus one gives propanenitrile. The solvent is ethanol, not water, so that the alcohol does not form instead.
• Ammonia: CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br. The first NH₃ forms the C-N bond; a second NH₃ removes H⁺ from the nitrogen and ends up in NH₄Br. Ammonia is a gas, so the mixture is heated in a sealed tube, under pressure, to keep it in. An excess of ammonia makes the primary amine the main product.
• Water (the silver nitrate test, below): CH₃CH₂Br + H₂O → CH₃CH₂OH + H⁺ + Br⁻

Elimination with NaOH in ethanol

Heated with NaOH dissolved in ethanol, not water, a halogenoalkane gives an alkene. The OH⁻ ion acts as a base: it removes H⁺ from a carbon atom next to the C-X carbon. The halogen leaves as X⁻, and a C=C double bond forms between those two carbons. Overall, HX is eliminated. Bromoethane gives ethene:

CH₃CH₂Br + NaOH → CH₂=CH₂ + NaBr + H₂O (NaOH in ethanol, heat)

So the solvent decides the answer:

• NaOH(aq), heat: substitution, giving the alcohol (bromoethane gives ethanol).
• NaOH in ethanol, heat: elimination, giving the alkene (bromoethane gives ethene).

When the halogen is in the middle of a chain, H⁺ can be removed from a carbon on either side, so more than one alkene can form, and some have cis/trans isomers (13.4). You do not need the elimination mechanism.

Beyond the course: in a real flask the two reactions compete, so some substitution product forms alongside the alkene. In your answers, NaOH in ethanol with heat gives the alkene.

Worked example 1: from an alcohol to alkenes

Question: In step 1, hexan-3-ol, CH₃CH₂CH(OH)CH₂CH₂CH₃, is converted into 3-chlorohexane. In step 2, the 3-chlorohexane is heated with NaOH in ethanol.
(a) Give two different reagents for step 1 and an equation for one of them.
(b) Name the type of reaction in step 2 and identify every alkene that can form, including stereoisomers.

(a) Answer: any two of SOCl₂; PCl₅; PCl₃ with heat; HCl(g); KCl with concentrated H₂SO₄. With SOCl₂:

CH₃CH₂CH(OH)CH₂CH₂CH₃ + SOCl₂ → CH₃CH₂CHClCH₂CH₂CH₃ + SO₂ + HCl

(b) Elimination of HCl. Number the chain: Cl is on C3, and both neighbours, C2 and C4, carry H atoms.

• H removed from C2: the double bond is C2=C3, giving CH₃CH=CHCH₂CH₂CH₃, hex-2-ene.
• H removed from C4: the double bond is C3=C4, giving CH₃CH₂CH=CHCH₂CH₃, hex-3-ene.

In each alkene, both carbons of the C=C carry two different groups (an H atom and an alkyl group), so each has cis and trans isomers.

Answer: four alkenes: cis-hex-2-ene, trans-hex-2-ene, cis-hex-3-ene and trans-hex-3-ene. With NaOH(aq) instead, substitution would give back hexan-3-ol.

The SN2 mechanism

SN2 stands for substitution, nucleophilic, bimolecular: two species, the halogenoalkane and the nucleophile, take part in the only step. Bromoethane with OH⁻:

1. Dipole: show Cδ+ on the carbon bonded to Br and Brδ− on Br.
2. Arrow 1: from a lone pair on the O atom of OH⁻ to the δ+ carbon. The OH⁻ approaches from the side opposite the Br atom.
3. Arrow 2: from the middle of the C-Br bond to the Br atom. The C-O bond forms as the C-Br bond breaks.
4. Transition state: drawn in square brackets with one negative charge outside. The carbon has partial (dotted) bonds to both O and Br, which lie in a straight line on opposite sides of it. Its other three groups (two H atoms and CH₃) lie flat in a plane around it.
5. Products: CH₃CH₂OH and Br⁻.

A curly arrow shows a pair of electrons moving. It starts at a lone pair or the middle of a bond (never at an atom symbol or a charge sign) and ends at the atom that receives the pair.

The transition state exists only at the top of the barrier and cannot be collected. SN2 has no intermediate; SN1, below, has one.

The SN1 mechanism

SN1 stands for substitution, nucleophilic, unimolecular: only the halogenoalkane takes part in the slow step. The tertiary halogenoalkane 2-bromo-2-methylpropane, (CH₃)₃CBr, with OH⁻:

Step 1 (slow). Show Cδ+ and Brδ− on the C-Br bond. One curly arrow goes from the middle of the C-Br bond to the Br atom. The bond breaks heterolytically:

(CH₃)₃CBr → (CH₃)₃C⁺ + Br⁻

Step 2 (fast). A curly arrow goes from a lone pair on the O atom of OH⁻ to the positive carbon atom of the carbocation:

(CH₃)₃C⁺ + OH⁻ → (CH₃)₃COH (2-methylpropan-2-ol)

The carbocation is an intermediate: a real species that forms in step 1 and is used up in step 2. Its positive carbon has three bonds in a flat (trigonal planar) arrangement at 120°, so the nucleophile can attack from either side.

Which mechanism: primary, secondary or tertiary

Alkyl groups donate electron density along their bonds towards a positive carbon atom. This positive inductive effect spreads out the positive charge and stabilises a carbocation. More alkyl groups give more stabilisation, so the order of carbocation stability is: tertiary (three alkyl groups) > secondary (two) > primary (one).

• Tertiary: SN1. The tertiary carbocation is stable enough to form readily, so step 1 has a relatively low $E_{\text{A}}$. Also, three alkyl groups crowd the back of the C-X carbon and block a nucleophile's approach, so SN2 is difficult.
• Primary: SN2. A primary carbocation, with only one alkyl group, is too unstable to form readily. The C-X carbon carries two small H atoms, so a nucleophile can reach it from the back.
• Secondary: a mixture of both. The balance depends on the structure of the molecule.

Worked example 2: the SN2 mechanism with cyanide ions

Question: 1-Chloropentane, CH₃CH₂CH₂CH₂CH₂Cl, is heated with KCN in ethanol.
(a) Name the organic product and write an equation.
(b) Describe the mechanism, stating where each curly arrow starts and ends.
(c) Explain why this reaction goes by SN2, not SN1.

(a) CN⁻ replaces Cl, and its carbon bonds to the chain, so the product has 5 + 1 = 6 carbon atoms.

Answer: hexanenitrile. CH₃CH₂CH₂CH₂CH₂Cl + CN⁻ → CH₃CH₂CH₂CH₂CH₂CN + Cl⁻

(b) Answer:

• Dipole: Cδ+ on C1 (the carbon bonded to Cl) and Clδ− on Cl.
• Arrow 1: from the lone pair on the carbon atom of CN⁻ to C1, approaching from the side opposite Cl.
• Arrow 2: from the middle of the C-Cl bond to the Cl atom.
• Transition state, in square brackets with a negative charge outside: C1 has partial bonds to the CN carbon and to Cl, in a straight line, with its two H atoms and the C₄H₉ chain in a plane around it.
• Products: CH₃CH₂CH₂CH₂CH₂CN and Cl⁻.

(c) C1 carries one alkyl group, so 1-chloropentane is primary.

Answer: a primary carbocation would have only one alkyl group to stabilise its charge by the inductive effect, so it is too unstable to form readily. C1 carries two small H atoms, so CN⁻ can attack it from the back. One step with one transition state is the easier route: SN2.

Reactivity: C-X bond strength and the silver nitrate test

In both mechanisms, the C-X bond breaks in the slow step. A weaker C-X bond breaks more easily, with a lower $E_{\text{A}}$, so the halogenoalkane reacts faster. Data Booklet values (average bond energies):

Bond	Bond energy / kJ mol⁻¹	Rate of hydrolysis
C-Cl	340	slowest
C-Br	280	between
C-I	240	fastest

So, for the same carbon chain and conditions, iodoalkanes react fastest. The C-F bond is stronger still (it is not in this Data Booklet table), so fluoroalkanes hardly react.

Polarity points the other way. With Data Booklet values, the electronegativity difference is 0.5 for C-Cl but only 0.1 for C-Br, so the C-Cl carbon is more δ+. If polarity controlled the rate, chloroalkanes would react fastest. They react slowest, so bond strength is the deciding factor.

The silver nitrate test. Into each of several test tubes, put the same volumes of ethanol and AgNO₃(aq), and stand them in the same warm water bath. Add the same number of drops of a different halogenoalkane to each, and time how long a precipitate takes to appear.

• Ethanol dissolves both the halogenoalkane and the water, so they can mix and react.
• Water is the nucleophile, and this hydrolysis is slow: RX + H₂O → ROH + H⁺ + X⁻.
• Each X⁻ ion precipitates at once: Ag⁺(aq) + X⁻(aq) → AgX(s). So the time taken measures the rate of hydrolysis.

Halogen	Precipitate	Order of appearance
Cl	white, AgCl	last
Br	cream, AgBr	second
I	pale yellow, AgI	first

The colour identifies the halogen: bromoethane, for example, gives a cream precipitate of AgBr, so it contains bromine. Aqueous ammonia confirms the halide (11.3). To compare halogens fairly, use the same carbon chain and the same class, because a tertiary halogenoalkane is hydrolysed faster than a primary one with the same halogen.

Worked example 3: predicting the order of precipitation

Question: Three isomers of C₆H₁₃Cl are:

• P: (CH₃)₂CHCH₂CH₂CH₂Cl, 1-chloro-4-methylpentane
• Q: CH₃CH₂CH₂CH₂CHClCH₃, 2-chlorohexane
• R: (CH₃)₂CClCH₂CH₂CH₃, 2-chloro-2-methylpentane

(a) Classify each and state its main mechanism of hydrolysis.
(b) Each is added to aqueous silver nitrate in ethanol at the same temperature. Describe what you see, predict the order in which precipitates appear, and explain.
(c) Would 1-bromo-4-methylpentane give a precipitate faster or more slowly than P? Explain.

(a) Count the carbon atoms bonded to the C-Cl carbon.

• P: one (the branch at C4 is far from Cl), so primary: SN2.
• Q: two, so secondary: a mixture of SN1 and SN2.
• R: three, so tertiary: SN1.

(b) All three are chloroalkanes, so each gives a white precipitate of AgCl, and only the class differs. R forms a tertiary carbocation, stabilised by the inductive effect of three alkyl groups, so its C-Cl bond breaks fastest. Q, secondary, forms a less stable carbocation, so it reacts more slowly. P cannot form a stable carbocation, so it reacts by SN2 with water, a weaker nucleophile than OH⁻, and is slowest.

Answer: R first, then Q, then P last; a white precipitate in each.

(c) Both compounds are primary with the same chain, so only the halogen differs. The C-Br bond (280 kJ mol⁻¹) is weaker than the C-Cl bond (340 kJ mol⁻¹) by 60 kJ mol⁻¹, so it breaks more easily.

Answer: faster, giving a cream precipitate of AgBr.

Key Equations

R is an alkyl group and X is Cl, Br or I.

Making halogenoalkanes from alcohols:

• ROH + HX → RX + H₂O
• 3ROH + PCl₃ → 3RCl + H₃PO₃ (heat)
• ROH + PCl₅ → RCl + POCl₃ + HCl
• ROH + SOCl₂ → RCl + SO₂ + HCl

Nucleophilic substitution (the nucleophile replaces X):

• RX + OH⁻ → ROH + X⁻ (NaOH(aq), heat)
• RX + CN⁻ → RCN + X⁻ (KCN in ethanol, heat; one extra carbon)
• RX + 2NH₃ → RNH₂ + NH₄X (NH₃ in ethanol, heat in a sealed tube)
• RX + H₂O → ROH + H⁺ + X⁻, then Ag⁺(aq) + X⁻(aq) → AgX(s)

Elimination (NaOH in ethanol, heat):

• CH₃CH₂Br + NaOH → CH₂=CH₂ + NaBr + H₂O

Bond energies (Data Booklet, kJ mol⁻¹): C-Cl 340, C-Br 280, C-I 240. Weaker bond, faster hydrolysis: RI > RBr > RCl.

Common Mistakes to Avoid

• Mistake: leaving out the solvent, or mixing up the two NaOH reactions. Fix: always write the solvent. NaOH(aq) and heat turns 3-chlorohexane into hexan-3-ol (substitution); NaOH in ethanol and heat turns it into hexenes (elimination).
• Mistake: giving HCN, or KCN with acid, to make a nitrile from a halogenoalkane. Fix: HCN with a little KCN adds to aldehydes and ketones (17.1). For a halogenoalkane, write "KCN in ethanol, heat".
• Mistake: joining the nitrile through nitrogen, or miscounting its carbons. Fix: the carbon atom of CN⁻ bonds to the chain, and the CN carbon counts in the name: 1-chloropentane gives hexanenitrile, CH₃(CH₂)₄CN, not pentanenitrile.
• Mistake: starting a curly arrow at an atom symbol, at the H of OH⁻ or at the "−" sign. Fix: in the SN2 reaction of bromoethane with OH⁻, arrow 1 starts at a lone pair on oxygen and arrow 2 at the middle of the C-Br bond.
• Mistake: saying a tertiary halogenoalkane reacts by SN1 because the halogenoalkane is more stable. Fix: it is the carbocation that is more stable: (CH₃)₃C⁺ is stabilised by the inductive effect of three alkyl groups.
• Mistake: classifying by branching anywhere in the molecule. Fix: count only the alkyl groups on the C-X carbon. 1-Chloro-4-methylpentane is branched, but it is primary.
• Mistake: explaining the reactivity order by bond polarity. Fix: use bond strength: C-I (240 kJ mol⁻¹) breaks more easily than C-Cl (340 kJ mol⁻¹), even though C-Cl is the more polar bond.
• Mistake: writing "white solution" or calling AgI cream. Fix: describe a precipitate with its colour: AgCl white, AgBr cream, AgI pale yellow.

Exam Tips

• State the reagents and conditions in full, with the solvent: "KCN in ethanol, heat" or "NH₃ in ethanol, heat in a sealed tube". "KCN" alone is incomplete.
• Name the type of reaction with both words: "nucleophilic substitution", not just "substitution". Use "elimination" for NaOH in ethanol with heat, and "hydrolysis" when water or OH⁻ gives an alcohol.
• Draw a mechanism with the C-X dipole, a lone pair on the nucleophile, curly arrows from a lone pair or a bond, the halide ion product, and either the bracketed transition state with its charge (SN2) or the carbocation with its + charge (SN1).
• Compare rates by quoting the C-X bond energies from the Data Booklet and linking them: weaker bond, broken more easily, faster hydrolysis, precipitate appears sooner.
• Sketch a reaction pathway: SN2 has one maximum (the transition state) and no intermediate; SN1 has two maxima with the carbocation in the dip between them, and the first maximum is higher.