3.3.2 Reactions of Halogenoalkanes

Halogenoalkanes are versatile molecules that can undergo two main types of reactions: nucleophilic substitution and elimination. The reaction pathway depends on the reaction conditions and the nature of the halogenoalkane. This note explores both types of reactions, focusing on the role of the reagent as both a nucleophile and a base.

Nucleophilic Substitution

In nucleophilic substitution reactions, a nucleophile (an electron-pair donor) attacks the carbon atom in the halogenoalkane, replacing the halogen atom. Common nucleophiles include $OH^-$ , $CN^-$ , and $NH_3$ .

infoNote

Example: Hydrolysis of 2-Bromopropane Reagent: Potassium hydroxide ( $KOH$ ) in water.

Reaction:

CH_3CHBrCH_3 + OH^- \rightarrow CH_3CH(OH)CH_3 + Br^-

The $OH^-$ ion acts as a nucleophile, attacking the slightly positive carbon atom bonded to the bromine.

The $Br^-$ ion (the leaving group) is displaced, and propan-2-ol is formed.

Mechanism

The nucleophile ( $OH^-$ ) donates a pair of electrons to the carbon bonded to the halogen.
The carbon-halogen bond breaks heterolytically, with the halogen ( $Br$ ) taking both electrons to form $Br^-$ .
The halogen is released as a halide ion (e.g., $Br^-$ ), and the product (alcohol) is formed.

Elimination Reaction

In elimination reactions, an atom or group of atoms is removed, forming an alkene. This reaction typically occurs in the presence of a strong base like hydroxide ions in ethanol.

infoNote

Example: Elimination of 2-Bromopropane Reagent: Potassium hydroxide ( $KOH$ ) in ethanol.

Reaction:

CH_3CHBrCH_3 + OH^- \rightarrow CH_3CH=CH_2 + H_2O + Br^-

The $OH^-$ ion acts as a base, removing a proton ( $H^+$ ) from a carbon atom adjacent to the one bonded to the halogen.

A double bond is formed between the carbons, and bromine is released as $Br^-$ .

Mechanism

The hydroxide ion (OH(^-)) removes a hydrogen atom from a carbon adjacent to the carbon bonded to the bromine.
A double bond forms between two carbons, and the bromine atom is eliminated as Br(^-).

Concurrent Substitution and Elimination

Halogenoalkanes can undergo both nucleophilic substitution and elimination reactions simultaneously. The reagent, such as potassium hydroxide, can act as both a nucleophile (substitution) and a base (elimination), depending on the conditions:

Aqueous potassium hydroxide favours nucleophilic substitution, forming alcohols.
Ethanolic potassium hydroxide favours elimination, forming alkenes.

Example: Reaction of 2-Bromopropane with Potassium Hydroxide

In aqueous conditions:
- The hydroxide ion acts as a nucleophile, leading to substitution and forming propan-2-ol.
In ethanolic conditions:
- The hydroxide ion acts as a base, removing a proton and leading to elimination, forming propene.

Reaction Mechanisms

Nucleophilic Substitution: The nucleophile replaces the halogen, forming a new bond with the carbon.
Elimination: The base removes a proton, and a double bond forms, eliminating the halogen. Understanding these mechanisms is key to predicting whether a halogenoalkane will undergo substitution or elimination under different conditions. The choice of solvent (water or ethanol) plays a crucial role in determining the reaction pathway.

Reactions of Halogenoalkanes Simplified Revision Notes for A-Level AQA Chemistry