Unit 21: Halogenoalkanes — Short Questions

The C–X bond in halogenoalkanes is polar because of the electronegativity difference between carbon and the halogen, giving the carbon atom a partial positive charge. This makes carbon an electrophilic centre that is readily attacked by nucleophiles (giving substitution) or allows a base to remove a β-hydrogen (giving elimination), making halogenoalkanes very reactive.

The C–F bond (467 kJ/mol) is the strongest of all the C–X bonds, much stronger than C–Cl, C–Br or C–I. Because this bond is so strong and difficult to break, fluoride is a very poor leaving group, so fluoroalkanes essentially do not react under ordinary nucleophilic substitution conditions.

In a primary halogenoalkane, the carbon bearing the halogen is attached only to hydrogen atoms or one other carbon, so there is little steric hindrance. This allows the nucleophile to approach and attack the electrophilic carbon easily from the side opposite the leaving group, so bond formation and bond breaking occur simultaneously in a single step, i.e. Sɴ2.

No, it cannot undergo elimination. 1-bromo-2,2-dimethylpropane is (CH₃)₃C–CH₂–Br; elimination requires removal of a hydrogen atom from the β-carbon along with the halogen from the α-carbon. Here the β-carbon is the quaternary carbon bonded to three methyl groups and the CH₂Br carbon, so it has no hydrogen atom attached to it. Since no β-hydrogen is available, elimination (E1 or E2) cannot occur; this molecule can only undergo substitution.

Tertiary halogenoalkanes ionize readily because the resulting tertiary carbocation is the most stable (due to the +I effect of three alkyl groups), so the slow ionization step of E1 is favoured. Also, the three bulky alkyl groups around the electrophilic carbon sterically hinder backside attack needed for a bimolecular mechanism, so instead of substitution/elimination occurring in one step, the base simply removes a β-hydrogen from the stable carbocation intermediate in a separate fast step, i.e. E1.

The OH⁻ ion carries a negative charge and lone pairs of electrons on oxygen. As a nucleophile, it uses this lone pair to attack the electrophilic carbon of a halogenoalkane, displacing the halide and forming an alcohol (substitution). As a base, it instead removes a β-hydrogen atom, causing the halide to leave simultaneously and forming an alkene (elimination). Because it can do both, bromoethane with aqueous NaOH gives both ethanol (substitution) and ethene (elimination).

FeCl₃ acts as a catalyst in the electrophilic aromatic substitution of benzene by chlorine. It polarizes the Cl–Cl bond of the chlorine molecule, generating a strongly electrophilic chlorine species that can attack the benzene ring, giving chlorobenzene and HCl: C₆H₆ + Cl₂ →(FeCl₃) C₆H₅Cl + HCl.

Haloalkanes are more reactive towards nucleophiles. In haloarenes, the lone pair on the halogen atom is delocalised into the aromatic ring, giving the C–X bond partial double-bond character; this strengthens and shortens the bond, making it hard to break, so haloarenes are far less reactive to nucleophilic substitution than haloalkanes.

CH₃CH₂CH₂Cl + H₂O → CH₃CH₂CH₂OH + H⁺ + Cl⁻ (water, as nucleophile, in ethanol-water mixture); the released Cl⁻ then reacts with AgNO₃: Ag⁺ + Cl⁻ → AgCl↓ (white precipitate). Overall: CH₃CH₂CH₂Cl + AgNO₃ + H₂O → CH₃CH₂CH₂OH + AgCl↓ + HNO₃.

A primary chloroalkane always follows the Sɴ2 mechanism because of low steric hindrance around the electrophilic carbon. A tertiary chloroalkane always follows the Sɴ1 mechanism because it forms a stable tertiary carbocation and bulky groups block backside attack. A secondary chloroalkane can follow either Sɴ1 or Sɴ2, depending on the nature of the solvent used.

Many reagents, such as hydroxide ion, can act both as a nucleophile and as a base. When such a reagent attacks the electrophilic carbon it gives substitution, but when it instead removes a β-hydrogen it gives elimination. Since both pathways are available to the same reagent and substrate under the same reaction conditions, the two reactions compete and proceed side by side — for example, bromoethane with aqueous NaOH gives both ethanol (substitution) and ethene (elimination) simultaneously.

In chlorobenzene, a lone pair on the chlorine atom is delocalised into the aromatic ring through resonance. This gives the C–Cl bond partial double-bond character, making it shorter and stronger, so it resists attack by nucleophiles and nucleophilic substitution is difficult. At the same time, this delocalisation increases the electron density on the benzene ring, which makes the ring more attractive to electrophiles, so chlorobenzene readily undergoes electrophilic substitution reactions (e.g. further halogenation) in preference to nucleophilic substitution.

The C–I bond (228 kJ/mol) is much weaker than the C–Cl bond (346 kJ/mol), so it breaks far more easily, releasing I⁻ ions quickly. These I⁻ ions immediately react with Ag⁺ from AgNO₃ to give a precipitate of AgI. Because the C–Cl bond is much stronger, chloroethane releases Cl⁻ only slowly, so AgCl precipitates much later.

Sulphuric acid first reacts with potassium bromide to generate hydrogen bromide (HBr) in situ. This HBr then reacts with the alcohol, substituting the –OH group with –Br to give a bromoalkane: CH₃CH₂OH + KBr + H₂SO₄ → CH₃CH₂Br + H₂O + KHSO₄. This method is commonly used to prepare bromoalkanes and (using the weaker oxidizing H₃PO₄ instead, to avoid oxidizing I⁻ to I₂) iodoalkanes.

The alcohol reacts with thionyl chloride (SOCl₂), using pyridine as a catalyst, to replace the –OH group with –Cl, forming the alkyl chloride: CH₃CH₂OH + SOCl₂ →(pyridine) CH₃CH₂Cl + SO₂ + HCl. This method is especially useful because the by-products, SO₂ and HCl, are gases that escape, leaving behind the pure alkyl chloride.

Phosphorus trichloride reacts with alcohols, replacing the –OH group with a chloro group to give alkyl chlorides along with phosphorous acid: 3CH₃CH₂OH + PCl₃ → 3CH₃CH₂Cl + H₃PO₃.

Phosphorus pentachloride reacts with the alcohol, replacing the –OH group with –Cl to give the alkyl chloride, along with phosphorus oxychloride and hydrogen chloride gas: CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl.

Hydroxide ion acts as a nucleophile and attacks the electrophilic carbon of bromoethane, displacing the bromide ion and forming ethanol: CH₃CH₂Br + NaOH(aq) → CH₃CH₂OH + NaBr. The reaction is carried out under reflux at higher temperature to avoid loss of the volatile reactants/products. (Under the same conditions OH⁻ can also act as a base, giving ethene as an elimination product alongside the substitution product.)

Cyanide ion is a strong nucleophile, so it substitutes the bromine atom of bromoethane to give a nitrile. The reaction is carried out by heating bromoethane with ethanolic potassium cyanide under reflux: CH₃CH₂Br + KCN(ethanolic) → CH₃CH₂CN + KBr, giving propanenitrile. Since CN⁻ is a weak base, elimination is not significant and substitution predominates.