3.3.3 Halogenoalkanes

Which species could act as a nucleophile?

• BH₃

• NH₄⁺

• PH₃

• SiH₄

Which compound has the fastest rate of reaction with potassium cyanide to form pentanenitrile?

• 1-bromobutane

• 1-chlorobutane

• 1-fluorobutane

• 1-iodobutane

Which of the following mechanisms does not occur in reactions of bromoethane?

• Electrophilic addition

• Elimination

• Nucleophilic substitution

• Radical substitution

Which of the arrows, labelled A, B, C or D in the mechanism in the diagram, is not correct?

• A

• B

• C

• D

This question is about a method that can be used to prepare ethylamine.

CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br

Which of the curly arrows in the mechanism is not correct?

• 1

• 2

• 3

• 4

This question is about a method that can be used to prepare ethylamine.

CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br

Which statement about the reaction is not correct?

• Ethylamine is a primary amine.

• The mechanism is a nucleophilic substitution.

• Using an excess of bromoethane will prevent further reaction to form a mixture of amine products.

• Ammonium bromide is an ionic compound.

Why are fluoroalkanes unreactive?

• Fluorine is highly electronegative.

• The F^– ion is very stable.

• They are polar molecules.

• The C–F bond is very strong.

Pentanenitrile can be made by reaction of 1-bromobutane with potassium cyanide.

Which of these is the correct name for the mechanism of this reaction?

• Electrophilic addition

• Electrophilic substitution

• Nucleophilic addition

• Nucleophilic substitution

Which one of the following statements explains best why fluoroalkanes are the least reactive haloalkanes?

• Fluorine is much more electronegative than carbon.

• The F⁻ ion is the most stable halide ion.

• The C–F bond is the most polar carbon–halogen bond.

• The C–F bond is the strongest carbon–halogen bond.

Which one of the following reactions does not involve donation of an electron pair?

• H⁺ + CH₃NH₂ → CH₃NH₃⁺

• AlCl₃ + Cl⁻ → AlCl₄^-

• CH₃Cl + CN⁻ → CH₃CN + Cl⁻

• 0.5Cl₂ + I⁻ → Cl⁻ + 0.5I₂