Haloalkanes and Haloarenes

What this chapter is about

This chapter introduces organic compounds in which one or more hydrogen atoms of a hydrocarbon are replaced by halogen atoms (F, Cl, Br or I). When the halogen is bonded to an sp³-hybridised carbon of an alkyl group, the compound is a haloalkane (alkyl halide). When it is bonded directly to a carbon of a benzene ring, it is a haloarene (aryl halide). These compounds are important intermediates in organic synthesis because the carbon–halogen bond can be broken and replaced by other functional groups.

You will learn how to name these compounds using IUPAC rules, understand the nature of the C–X bond (its polarity and bond strength), and study two broad classes of reactions: nucleophilic substitution (SN1 and SN2) and elimination (E1 and E2). For haloarenes, you will see why direct nucleophilic substitution is difficult and why the halogen is ortho- and para-directing in electrophilic aromatic substitution. By the end, you should be able to predict products, compare reactivities, and write mechanisms with curly arrows.

Key ideas

• The C–X bond is polar (carbon is δ+, halogen is δ−) because halogens are more electronegative than carbon; this polarity makes the carbon susceptible to attack by nucleophiles.
• Bond dissociation enthalpy follows the order C–F > C–Cl > C–Br > C–I, so iodoalkanes react fastest and fluoroalkanes slowest in substitution and elimination.
• In the SN2 mechanism the nucleophile attacks from the side opposite to the leaving group in one concerted step, giving inversion of configuration at a chiral centre (Walden inversion).
• In the SN1 mechanism the C–X bond breaks first to form a carbocation; the nucleophile then attacks this planar intermediate, producing a racemic mixture if the carbon was chiral.
• Elimination reactions (dehydrohalogenation) occur when a strong base abstracts a β-hydrogen; E2 is concerted, E1 proceeds via a carbocation, and Zaitsev's rule predicts the more substituted alkene as the major product.
• Haloarenes are less reactive toward nucleophilic substitution because the C–X bond has partial double-bond character due to resonance between the lone pair on the halogen and the benzene ring.
• In electrophilic aromatic substitution, halogens are deactivating (withdraw electrons by induction) yet ortho- and para-directing (donate electrons by resonance to those positions).

Formulas and facts to remember

1. General formula of haloalkanes: CₙH₂ₙ₊₁X (for mono-substituted, saturated, open-chain).
2. IUPAC naming: the halogen is named as a prefix (fluoro, chloro, bromo, iodo) with position numbers; for example, CH₃–CHBr–CH₃ is 2-bromopropane.
3. Order of reactivity of alkyl halides toward SN2: CH₃X > primary > secondary > tertiary (steric hindrance increases).
4. Order of reactivity toward SN1: tertiary > secondary > primary > CH₃X (carbocation stability increases).
5. Grignard reagent formation: R–X + Mg (dry ether) → R–Mg–X. This reagent reacts with water or acids to give alkanes, and with carbonyl compounds to give alcohols.
6. Wurtz reaction: 2 R–X + 2 Na (dry ether) → R–R + 2 NaX. Used to prepare symmetrical alkanes.
7. Friedel–Crafts alkylation of benzene with R–Cl in presence of anhydrous AlCl₃ gives alkylbenzene.
8. Chlorobenzene requires very harsh conditions (high temperature, pressure, NaOH) for nucleophilic substitution; the presence of strong electron-withdrawing groups (–NO₂) at ortho or para positions activates the ring toward substitution.

Worked examples

Example 1 — IUPAC naming

Problem: Write the IUPAC name of CH₃–CH₂–CHCl–CH₂–CH₃.

Solution:

1. Identify the longest carbon chain containing the halogen: five carbons (pentane).
2. Number from the end that gives the lowest locant to the substituent: chlorine is on carbon 3 whether you number from left or right, so either direction works.
3. Name: 3-chloropentane.

Example 2 — Predicting SN1 vs SN2

Problem: Predict the predominant mechanism when (CH₃)₃C–Br reacts with aqueous ethanol.

Solution:

1. The substrate is a tertiary alkyl bromide; steric hindrance blocks backside attack by a nucleophile.
2. Aqueous ethanol is a weak nucleophile and a polar protic solvent, which stabilises carbocations.
3. Therefore the SN1 pathway dominates: the C–Br bond breaks first to form the (CH₃)₃C⁺ carbocation, which then combines with water or ethanol to give 2-methylpropan-2-ol or its ethyl ether.

Example 3 — Elimination (Zaitsev product)

Problem: What is the major product when 2-bromobutane is heated with alcoholic KOH?

Solution:

1. Alcoholic KOH is a strong base and favours elimination over substitution.
2. Two β-hydrogens are available: on C-1 (giving but-1-ene) and on C-3 (giving but-2-ene).
3. According to Zaitsev's rule, the more substituted alkene is favoured.
4. Major product: but-2-ene (CH₃–CH=CH–CH₃).

Common mistakes

• Confusing inversion with racemisation → Inversion (one flip) occurs in SN2; racemisation (50-50 mixture) occurs in SN1 because the carbocation is planar.
• Thinking chlorobenzene reacts easily with NaOH like chloroethane → Resonance strengthens the C–Cl bond in haloarenes; special conditions are required.
• Numbering the chain to give the halogen the highest number → Always number to give the lowest locant to the first point of difference.
• Forgetting that Grignard reagents react with even traces of water → Dry ether and anhydrous conditions are essential.
• Believing a stronger C–X bond means faster reaction → A weaker bond (C–I) breaks more easily, so iodoalkanes react fastest.

Quick revision

1. Haloalkanes: halogen on sp³ C; haloarenes: halogen on benzene C.
2. SN2: one step, inversion, favoured by primary substrates and strong nucleophiles.
3. SN1: two steps, racemisation, favoured by tertiary substrates and polar protic solvents.
4. Elimination with alcoholic KOH gives the more substituted alkene (Zaitsev rule).
5. Haloarenes resist nucleophilic substitution but are ortho-para directing in electrophilic substitution.
6. Grignard reagent (R–Mg–X) is a versatile nucleophile for forming new C–C bonds.