Part 3 · Chapter 26

Haloalkanes and Haloarenes

A halogen on carbon — the polar bond that makes alkyl halides react readily by substitution and elimination, and the resonance that makes aryl halides stubbornly inert

Fundamentals of Chemistry Prof. Mithun Mondal Reading time ≈ 55 min
i What you'll learn
  • How haloalkanes and haloarenes are classified, and why the C–X bond is polar.
  • Preparation from alcohols (\(\ce{PCl5},\ \ce{SOCl2}\)), hydrocarbons, and by halogen exchange.
  • The SN1 and SN2 mechanisms — rate laws, stereochemistry, and what favours each.
  • The reactivity order of halides and the substitution-vs-elimination contest.
  • E1 and E2 elimination and Saytzeff's rule.
  • Why haloarenes resist nucleophilic substitution, and the uses of polyhalogen compounds.
Section 26-1

Classification & the C–X Bond

Replacing a hydrogen of a hydrocarbon with a halogen gives a haloalkane (halogen on an \(sp^3\) carbon) or a haloarene (halogen on an aromatic \(sp^2\) carbon). Because the halogen is more electronegative than carbon, the C–X bond is polar — the carbon bears a \(\delta^+\) charge that invites attack by nucleophiles. This single feature drives nearly all the chemistry of alkyl halides.

ClassHalogen onExample
Haloalkane (1°, 2°, 3°)\(sp^3\) carbon\(\ce{CH3CH2Cl}\)
AllylicC next to \(\ce{C=C}\)\(\ce{CH2=CH-CH2Cl}\)
Vinylic\(sp^2\) C of \(\ce{C=C}\)\(\ce{CH2=CHCl}\)
Haloarene (aryl)aromatic \(sp^2\) C\(\ce{C6H5Cl}\)
Section 26-2

Preparation

Alkyl halides are most often made from alcohols, while halogen-exchange reactions convert one halide into another.

RouteReactionNote
From alcohol (\(\ce{HX}\))\(\ce{R-OH + HX -> R-X + H2O}\)order 3° > 2° > 1°
From alcohol (\(\ce{SOCl2}\))\(\ce{R-OH + SOCl2 -> R-Cl + SO2 + HCl}\)best — by-products are gases
From alkene (\(\ce{HX}\))\(\ce{C=C + HX -> }\) haloalkaneMarkovnikov
Finkelstein\(\ce{R-Cl + NaI ->[acetone] R-I + NaCl}\)makes iodides
Swarts\(\ce{R-Cl + AgF -> R-F + AgCl}\)makes fluorides
Section 26-3

Nucleophilic Substitution: SN2

In an SN2 (substitution, nucleophilic, bimolecular) reaction, the nucleophile attacks the carbon from the side opposite the leaving group in a single concerted step. Bond-making and bond-breaking happen together, so the rate depends on both reactants.

Nu⁻ → C → X⁻ backside attack → umbrella inversion (Walden)
SN2 — concerted backside attack inverts the configuration
🔄
SN2 fingerprints
rate \(= k[\ce{RX}][\ce{Nu}]\) · one step · inversion of configuration

Because the nucleophile attacks from behind, the molecule turns inside-out like an umbrella in a gale — Walden inversion. SN2 is favoured by unhindered \(1^\circ\) halides, strong nucleophiles and polar aprotic solvents. Steric crowding slows it: \(\ce{CH3} > 1^\circ > 2^\circ > 3^\circ\).

Section 26-4

Nucleophilic Substitution: SN1

An SN1 (substitution, nucleophilic, unimolecular) reaction goes in two steps. The leaving group departs first, forming a planar carbocation; the nucleophile then attacks. The slow, rate-determining step is the first one, so the rate depends only on the halide.

R–X slow R⁺ (planar) fast, Nu⁻ R–Nu Nu⁻ attacks the flat cation from either face → racemic product
SN1 — via a planar carbocation, giving racemisation
⚖️
SN1 fingerprints
rate \(= k[\ce{RX}]\) · two steps · racemisation

The flat carbocation can be attacked from either face, so a single enantiomer gives a near-racemic mixture. SN1 is favoured by \(3^\circ\) halides (stable cation), weak nucleophiles and polar protic solvents. Reactivity tracks carbocation stability: \(3^\circ > 2^\circ > 1^\circ\).

Section 26-5

SN1 vs SN2 & Reactivity

The two mechanisms pull in opposite directions on substrate structure — and on stereochemistry. Knowing which dominates lets you predict both the rate and the product's configuration.

FeatureSN1SN2
Molecularityunimolecular (2 steps)bimolecular (1 step)
Rate law\(k[\ce{RX}]\)\(k[\ce{RX}][\ce{Nu}]\)
Substrate\(3^\circ > 2^\circ > 1^\circ\)\(\ce{CH3} > 1^\circ > 2^\circ > 3^\circ\)
Stereochemistryracemisationinversion
Favoured byweak Nu, protic solventstrong Nu, aprotic solvent
Halide reactivity. For a given alkyl group, reactivity toward substitution is \(\ce{R-I} > \ce{R-Br} > \ce{R-Cl}\). The C–I bond is the weakest and iodide the best leaving group, so it breaks most easily — bond strength, not electronegativity, sets the order.
Section 26-6

Elimination Reactions

A base can remove a \(\beta\)-hydrogen and the halogen together, forming an alkene — dehydrohalogenation. Like substitution, elimination comes in bimolecular (E2) and unimolecular (E1) flavours, and when more than one alkene is possible, Saytzeff's rule predicts the more substituted one.

Saytzeff elimination
\(\ce{CH3-CHBr-CH2-CH3 ->[alc. KOH] CH3-CH=CH-CH3}\) (major)

The more substituted (more stable) alkene — but-2-ene here — dominates over the terminal alkene. E2 is concerted and favoured by strong bases; E1 goes via a carbocation like SN1. Strong, bulky bases and high temperature push toward elimination over substitution.

Section 26-7

Why Haloarenes Are Inert

Aryl halides barely react with nucleophiles under ordinary conditions — a stark contrast to alkyl halides. Several reinforcing factors explain this stubbornness.

Cl partial C=Cl lone-pair delocalisation shortens & strengthens C–Cl
Resonance gives the C–X bond of a haloarene partial double-bond character
FactorEffect
ResonanceC–X gains partial double-bond character → shorter, stronger
\(sp^2\) carbonmore electronegative, holds the C–X electrons tighter
Unstable phenyl cationSN1 route is blocked
Ring electron densityrepels the incoming nucleophile
Section 26-8

Reactions of Haloarenes

Haloarenes do react — but only under forcing conditions, or when an electron-withdrawing group activates the ring toward nucleophilic attack.

🔥
Forcing the substitution
\(\ce{C6H5Cl + NaOH ->[623\,K][300\,atm] C6H5OH}\) (Dow process)

Only high temperature and pressure displace the halogen of plain chlorobenzene. But an ortho/para nitro group makes substitution far easier (addition–elimination): \(p\)-nitrochlorobenzene reacts under much milder conditions because the \(\ce{-NO2}\) stabilises the intermediate. Haloarenes also undergo normal electrophilic substitution, with halogen as an \(o/p\)-director.

Section 26-9

Polyhalogen Compounds

Compounds with several halogens have wide industrial and historical use — though many are now restricted for safety or environmental reasons.

CompoundFormulaUse / note
Dichloromethane\(\ce{CH2Cl2}\)solvent, paint remover
Chloroform\(\ce{CHCl3}\)solvent; oxidises in air to toxic phosgene \(\ce{COCl2}\)
Iodoform\(\ce{CHI3}\)former antiseptic
Carbon tetrachloride\(\ce{CCl4}\)solvent, once a fire extinguisher
FreonsCFCsrefrigerants; deplete ozone
DDTinsecticide; persistent, largely banned
Why chloroform is stored carefully. In light and air, chloroform is slowly oxidised to phosgene (\(\ce{COCl2}\)), a poisonous gas. It is therefore kept in dark bottles filled to the brim and stabilised with a little ethanol, which converts any phosgene to harmless products.
Worked Examples

Putting It to Work

1 Predict the mechanism

Problem. Will \(\ce{(CH3)3C-Br}\) hydrolyse mainly by SN1 or SN2? Explain.

Solution. A \(3^\circ\) halide forms a stable cation and is too hindered for backside attack:

Working
\[ \textbf{SN1}\ (\text{stable }3^\circ\text{ carbocation}) \]
2 Reactivity order

Problem. Arrange \(\ce{CH3Cl},\ \ce{CH3Br},\ \ce{CH3I}\) by reactivity toward nucleophilic substitution.

Solution. The weaker the C–X bond, the better the leaving group:

Working
\[ \ce{CH3I} > \ce{CH3Br} > \ce{CH3Cl} \]
3 Stereochemistry

Problem. An optically active \(2^\circ\) halide reacts by SN2. What happens to its configuration?

Solution. Backside attack flips the configuration:

Working
\[ \textbf{inversion}\ (\text{Walden inversion}) \]
4 Why chlorobenzene resists

Problem. Give two reasons chlorobenzene does not undergo nucleophilic substitution easily.

Solution. Bond strengthening plus a blocked mechanism:

Working
\[ \text{resonance (partial C=Cl) + unstable phenyl cation} \]
5 Finkelstein reaction

Problem. Write the Finkelstein conversion of bromoethane to iodoethane.

Solution. \(\ce{NaI}\) in acetone exchanges the halogen (NaBr precipitates):

Working
\[ \ce{C2H5Br + NaI ->[acetone] C2H5I + NaBr v} \]
6 Saytzeff product

Problem. Give the major alkene from dehydrohalogenation of 2-bromobutane with alc. KOH.

Solution. Saytzeff favours the more substituted alkene:

Working
\[ \ce{CH3-CH=CH-CH3}\ (\text{but-2-ene, major}) \]
Review

Chapter Summary

The C–X bond

Polar bond, \(\delta^+\) carbon invites nucleophiles; sp³ (alkyl) reactive, sp² (aryl) inert.

Preparation

From alcohols (\(\ce{SOCl2}\) cleanest), alkenes, and halogen exchange (Finkelstein, Swarts).

SN2

One step, \(k[\ce{RX}][\ce{Nu}]\), inversion; favours 1°, strong Nu, aprotic solvent.

SN1

Two steps via carbocation, \(k[\ce{RX}]\), racemisation; favours 3°, protic solvent.

Elimination

E1/E2 dehydrohalogenation; Saytzeff gives the more substituted alkene.

Haloarenes

Inert (resonance, sp² C, unstable phenyl cation); react only under force or with \(-\)M groups.

Practice

Problems

For each item, first decide whether it concerns preparation, a substitution mechanism, elimination, or haloarene reactivity — then apply the relevant rule. Difficulty rises down the list.

  1. Why is the C–X bond polar, and how does this make alkyl halides reactive?
  2. Write the preparation of chloroethane from ethanol using \(\ce{SOCl2}\), and say why this reagent is preferred.
  3. Compare the SN1 and SN2 mechanisms in molecularity, rate law and stereochemistry.
  4. Arrange \(\ce{CH3Cl},\ \ce{CH3Br},\ \ce{CH3I}\) by reactivity toward substitution and explain.
  5. Why does a 3° halide prefer SN1 while a 1° halide prefers SN2?
  6. What is Walden inversion, and which mechanism shows it?
  7. Write the Finkelstein and Swarts reactions.
  8. State Saytzeff's rule and give the major product from 2-bromobutane with alc. KOH.
  9. Give three reasons haloarenes resist nucleophilic substitution.
  10. Under what conditions does chlorobenzene react with \(\ce{NaOH}\), and what activates the ring toward milder substitution?
  11. Why is chloroform stored in dark, filled bottles with a little ethanol?
  12. Explain why halogens are deactivating yet ortho/para-directing on a benzene ring.
Tip: for any alkyl halide, read the substrate first. A \(3^\circ\) centre means a stable carbocation → SN1/E1; an unhindered \(1^\circ\) centre means a clear path for backside attack → SN2. Then the conditions break the tie: strong base/heat tilts toward elimination, a good nucleophile in aprotic solvent toward substitution. Substrate, then conditions — in that order.