Nucleophilic Substitution in Halogenoalkanes (HL) (DP IB Chemistry): Revision Note

Nucleophilic substitution in halogenoalkanes

• In nucleophilic substitution, the halogen atom in a halogenoalkane is replaced by a nucleophile

  • A nucleophile is more reactive if it can readily donate a lone pair of electrons

• The hydroxide ion, OH^-, is a stronger nucleophile than water because:

  • It has a full negative charge

  • This means that it has a readily available lone pair of electrons

• Water (H₂O) is a neutral molecule with partial charges, δ+ and δ-

  • Its lone pairs are less available for reaction

  • However, it can still act as a nucleophile

• This contrast is shown in the Lewis structures of OH^- and H₂O:

• There are two types of nucleophilic substitution reaction:

  • S_N1 

  • S_N2 

• The type of nucleophilic substitution depends on the halogenoalkane involved

S_N1 reactions

• In tertiary halogenoalkanes, the carbon that is attached to the halogen is also bonded to three alkyl groups

• These halogenoalkanes undergo nucleophilic substitution by an S_N1 mechanism

  • ‘S’ stands for ‘substitution’

  • ‘N’ stands for ‘nucleophilic’

  • ‘1’ means that the rate of the reaction

    • This is determined by the slowest step of the reaction

    • It depends on the concentration of one reagent - the halogenoalkane

S_N1 mechanism

• The S_N1 mechanism is a two-step reaction

• In the first step:

  • The C-X bond breaks heterolytically

  • The halogen leaves the halogenoalkane as an X^- ion

    • This is the slow, rate-determining step

  • A tertiary carbocation intermediate is formed

• In the second step:

  • The nucleophile attacks the positively charged carbon in the tertiary carbocation intermediate

• This two-step process is evident in the energy profile diagram for an S_N1 reaction:

• The rate-determining step depends only on the concentration of the halogenoalkane

  • Therefore, the rate equation for an S_N1 reaction is:

rate = k[halogenoalkane]

• In terms of molecularity, an S_N1 reaction is unimolecular

Example of S_N1 mechanism

• The nucleophilic substitution of 2-bromo-2-methylpropane by hydroxide ions is S_N1

  • The product is 2-methyl-2-propanol

S_N2 reactions

• In primary halogenoalkanes, the carbon that is attached to the halogen is bonded to one alkyl group

• These halogenoalkanes undergo nucleophilic substitution by an S_N2 mechanism

  • ‘S’ stands for ‘substitution’

  • ‘N’ stands for ‘nucleophilic’

  • ‘2’ means that the rate of the reaction

    • This is determined by the slowest step of the reaction

    • It depends on the concentration of two reagents - the halogenoalkane and the nucleophile

S_N2 mechanism

• The S_N2 mechanism is a one-step reaction

  • The nucleophile attacks the δ⁺ carbon atom of the halogenoalkane from the opposite side of the leaving group

    • This forms a new bond between the nucleophile and the carbon atom

  • At the same time, the C–X bond breaks by heterolytic fission

    • The halogen leaves as an X⁻ ion

  • The nucleophile donates a lone pair of electrons to form the new C–Nu bond

  • The halogen takes both bonding electrons as it leaves

  • All of these changes happen simultaneously in a one-step process

    • No intermediate is formed

• The energy profile for an S_N2 reaction shows a single peak representing the transition state:

• As this is a one-step reaction, the rate-determining step depends on the concentrations of the halogenoalkane and nucleophile

  • Therefore, the rate equation for an S_N2 reaction is:

rate = k[halogenoalkane][nucleophile]

• In terms of molecularity, an S_N2 reaction is bimolecular

Example of S_N2 mechanism

• The nucleophilic substitution of bromoethane by hydroxide ions is S_N2

  • The product is ethanol

Inversion of configuration in S_N2

• In S_N2 reactions, the nucleophile always attacks from the opposite side of the leaving group due to steric hindrance

  • An attack from the same side as the bromine atom is a frontal attack

  • An attack from the opposite side to the bromine atom is a backside or rear-side attack

  • In bromoethane, the large bromine atom blocks a frontal attack

• As a result, the hydroxide ion performs a rear-side (backside) attack, rather than a frontal one

• This causes the molecule to undergo inversion of configuration

  • This is when the three other groups around the carbon are flipped

• A common analogy is an umbrella turning inside out in the wind

  • As the new C–OH bond forms, the C–Br bond breaks causing the bromine to leave as a bromide ion, Br⁻