Halogenoalkanes and Nucleophilic Substitution

Halogenoalkanes (haloalkanes) contain a halogen atom (F, Cl, Br, or I) bonded to a saturated carbon atom. The polar C–X bond makes them susceptible to attack by nucleophiles. This lesson covers classification, nucleophilic substitution mechanisms (SN1 and SN2), elimination reactions, and the factors affecting reaction pathways.

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Classification of Halogenoalkanes

Halogenoalkanes are classified as primary (1°), secondary (2°), or tertiary (3°) depending on how many carbon atoms are bonded to the carbon bearing the halogen.

Classification	Structure	Example
Primary (1°)	Halogen on a C bonded to 1 other C (or 0 for CH₃X)	CH₃CH₂Br (bromoethane)
Secondary (2°)	Halogen on a C bonded to 2 other C atoms	(CH₃)₂CHBr (2-bromopropane)
Tertiary (3°)	Halogen on a C bonded to 3 other C atoms	(CH₃)₃CBr (2-bromo-2-methylpropane)

Bond Polarity

The C–X bond is polar because halogens are more electronegative than carbon. The carbon is δ+ and the halogen is δ−. This makes the carbon susceptible to attack by nucleophiles (electron-pair donors).

Bond Enthalpy and Reactivity

Bond	Bond enthalpy / kJ mol⁻¹	Reactivity
C–F	484	Least reactive (strongest bond)
C–Cl	338	Moderate
C–Br	276	Reactive
C–I	238	Most reactive (weakest bond)

Reactivity increases from C–F to C–I because the bond gets weaker (easier to break) as the halogen atom gets larger. Although C–F is the most polar bond, it is the least reactive because the very high bond enthalpy makes it harder to break.

Key Definition: A nucleophile is a species that donates a lone pair of electrons to an electron-deficient atom. Nucleophiles are either negatively charged ions or neutral molecules with a lone pair. Examples: OH⁻, CN⁻, NH₃, H₂O.

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Nucleophilic Substitution Reactions

In nucleophilic substitution, a nucleophile replaces the halogen atom. The halogen leaves as a halide ion (X⁻), which is the leaving group.

Reaction 1: Hydrolysis with OH⁻ (to form an alcohol)

CH₃CH₂Br + NaOH → CH₃CH₂OH + NaBr

Reagent: Aqueous sodium hydroxide (NaOH(aq)) Conditions: Reflux (heat under reflux with aqueous NaOH) Nucleophile: OH⁻ (hydroxide ion)

Reaction 2: With CN⁻ (to form a nitrile — chain extension)

CH₃CH₂Br + KCN → CH₃CH₂CN + KBr

Reagent: Potassium cyanide in ethanol (KCN in ethanol) Conditions: Reflux Nucleophile: CN⁻ (cyanide ion)

This reaction is important because it extends the carbon chain by one carbon. The nitrile product (propanenitrile in this case) can then be hydrolysed to a carboxylic acid or reduced to an amine.

Reaction 3: With NH₃ (to form an amine)

CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br

Reagent: Excess concentrated ammonia in ethanol Conditions: Heat in a sealed tube (to prevent NH₃ escaping) Nucleophile: NH₃ (lone pair on nitrogen)

The excess NH₃ is needed to minimise further substitution. If insufficient NH₃ is used, the primary amine product (which is also a nucleophile) can react with more halogenoalkane to form secondary amines, tertiary amines, and eventually quaternary ammonium salts.

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SN1 vs SN2 Mechanisms

SN2 — Bimolecular Nucleophilic Substitution

Occurs with primary halogenoalkanes (and some secondary).

"SN2" means: Substitution, Nucleophilic, 2 molecules in the rate-determining step.

Mechanism (one step):

1. The nucleophile (e.g. OH⁻) attacks the δ+ carbon from the opposite side to the leaving group (backside attack).
2. A curly arrow from the lone pair on OH⁻ points to the δ+ carbon.
3. Simultaneously, a curly arrow from the C–Br bond points to the Br atom.
4. The C–Br bond breaks (heterolytic fission) as the new C–OH bond forms.
5. This occurs in a single concerted step — bond making and bond breaking happen simultaneously.

Key features of SN2:

• One-step mechanism (no intermediate)
• Rate = k[halogenoalkane][nucleophile] — rate depends on both concentrations
• Inversion of configuration occurs (the nucleophile attacks from the back, "flipping" the molecule like an umbrella turning inside out)
• Favoured by primary halogenoalkanes (less steric hindrance at the δ+ carbon)