IB DP Chemistry: HL复习笔记20.1.1 Nucleophilic Substitution Reactions

• In nucleophilic substitution reactions involving halogenoalkanes, the halogen atom is replaced by a nucleophile
• The strength of any nucleophile depends on its ability to make its lone pair of electrons available for reaction
• 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
• A water molecule only has partial charges, δ+ and δ-
  • This means that its lone pair of electrons is less available than the hydroxide ions
  • The lone pairs of electrons in a water molecule are still available to react

Lewis structures of the hydroxide ion and water molecule - illustrating the lone pairs of electrons and charges within their structures

• Nucleophilic substitution reactions can occur in two different ways (known as S_N2 and S_N1 reactions) depending on the structure of 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 (which is determined by the slowest step of the reaction) depends on the concentration of only one reagent, the halogenoalkane

• The S_N1 mechanism is a two-step reaction
• In the first step, the C-X bond breaks heterolytically and the halogen leaves the halogenoalkane as an X^- ion (this is the slow and rate-determining step)
  • As the rate-determining step only depends on the concentration of the halogenoalkane, the rate equation for an S_N1 reaction is rate = k[halogenoalkane]
  • In terms of molecularity, an S_N1 reaction is unimolecular
  • This forms a tertiary carbocation (which is a tertiary carbon atom with a positive charge)
  • In the second step, the tertiary carbocation is attacked by the nucleophile

• For example, the nucleophilic substitution of 2-bromo-2-methylpropane by hydroxide ions to form 2-methyl-2-propanol

The mechanism of nucleophilic substitution in 2-bromo-2-methylpropane which is a tertiary halogenoalkane

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 (which is determined by the slowest step of the reaction) depends on the concentration of both the halogenoalkane and the nucleophile ions

• The S_N2 mechanism is a one-step reaction
  • The nucleophile donates a pair of electrons to the δ+ carbon atom of the halogenoalkane to form a new bond
    • As this is a one-step reaction, the rate-determining step depends on the concentrations of the halogenoalkane and nucleophile, the rate equation for an S_N2 reaction is rate = k[halogenoalkane][nucleophile]
    • In terms of molecularity, an S_N2 reaction is bimolecular
  • At the same time, the C-X bond is breaking and the halogen (X) takes both electrons in the bond (heterolytic fission)
  • The halogen leaves the halogenoalkane as an X^- ion
• For example, the nucleophilic substitution of bromoethane by hydroxide ions to form ethanol

The S_N2 mechanism of bromoethane with hydroxide causing an inversion of configuration

• The bromine atom of the bromoethane molecule causes steric hindrance
• This means that the hydroxide ion nucleophile can only attack from the opposite side of the C-Br bond
  • Attack from the same side as the bromine atom is sometimes called frontal attack
  • While attack from the opposite side is sometimes called backside or rear-side attack
• As the C-OH bond forms, the C-Br bond breaks causing the bromine atom to leave as a bromide ion
  • As a result of this, the molecule has undergone an inversion of configuration
  • The common comparison for this is an umbrella turning inside out in the wind

Inversion of configuration - umbrella analogy

Factors affecting nucleophilic substitution

• Various factors affect the rate of nucleophilic substitution, regardless of S_N1 or S_N2, involving a halogenoalkane:
  1. The nature of the nucleophile
  2. The halogen involved (leaving group)
  3. The structure (class) of the halogenoalkane
  4. Protic & aprotic solvents

1. The nature of the nucleophile

• The most effective nucleophiles are neutral or negatively charged species that have a lone pair of electrons available to donate to the δ+ carbon in the halogenoalkane
• The greater the electron density on the nucleophile ion or molecule; the stronger the nucleophile
  • Consequently, negative anions tend to be more reactive than their corresponding neutral species, e.g. hydroxide ions and water molecules (as previously discussed)
• When nucleophiles have the same charge, the electronegativity of the atom carrying the lone pair becomes the deciding factor
  • The less electronegative the atom carrying the lone pair; the stronger the nucleophile
  • For example:
    • Ammonia is a stronger electrophile than water because the nitrogen atom in ammonia is less electronegative than the oxygen atom in water
  • This is because a less electronegative atom has a weaker grip on its lone pair of electrons, which means that they are more available for reaction
• The effectiveness of nucleophiles is as follows:

Strongest     CN^- > OH^- > NH₃ > H₂O     Weakest

2. The halogen involved (leaving group)

• The halogenoalkanes have different rates of substitution reactions
• Since substitution reactions involve breaking the carbon-halogen bond, the bond energies can be used to explain their different reactivities

Approximate Halogenoalkane Bond Energy Table

• The table above shows that the C-I bond requires the least energy to break, and is therefore the weakest carbon-halogen bond
  • During substitution reactions, the C-I bond will breaks heterolytically as follows:

R₃C-I + OH^-     →    R₃C-OH + I^-

• The C-F bond, on the other hand, requires the most energy to break and is, therefore, the strongest carbon-halogen bond
  • Fluoroalkanes will therefore be less likely to undergo substitution reactions

• This idea can be confirmed by reacting the product formed by nucleophilic substitution of the halogenoalkane with aqueous silver nitrate solution
• As a halide ion is released, this results in the formation of a precipitate
• The rate of formation of these precipitates can also be used to determine the reactivity of the halogenoalkanes

Halogenoalkane Precipitates Table

• The formation of the pale yellow silver iodide is the fastest (fastest nucleophilic substitution reaction) whereas the formation of the silver fluoride is the slowest (slowest nucleophilic substitution reaction)
• This confirms that fluoroalkanes are the least reactive and iodoalkanes are the most reactive halogenoalkanes

The trend in reactivity of halogenoalkanes

3. The structure (class) of the halogenoalkane

• Tertiary halogenoalkanes undergo S_N1 reactions, forming stable tertiary carbocations
• Secondary halogenoalkanes undergo a mixture of both S_N1 and S_N2 reactions depending on their structure
• Primary halogenoalkanes undergo S_N2 reactions, forming the less stable primary carbocations
• This has to do with the positive inductive effect of the alkyl groups attached to the carbon which is bonded to the halogen atom
  • The alkyl groups push electron density towards the positively charged carbon, reducing the charge density
  • In tertiary carbocations, there are three alkyl groups stabilising the carbocation
  • In primary carbocations, there is only one alkyl group
    • This is why tertiary carbocations are much more stable than primary ones

The diagram shows the trend in stability of primary, secondary and tertiary carbocations

• Overall, the structure (class) has a direct effect on the formation of the carbocation and, therefore, the rate-determining step
• Consequently, this affects the overall rate of the nucleophilic substitution reaction

4. Protic & Aprotic Solvents

    Hydrogen bonding

• Protic, polar solvents contain a hydrogen atom bonded to a very electronegative nitrogen or oxygen atom
  • This means that they are capable of hydrogen bonding
  • Examples of protic solvents include ammonia, carboxylic acids, ethanol and water
• Aprotic, polar solvents contain hydrogen atoms but they are not bonded to an electronegative atom
  • This means that they cannot participate in hydrogen bonding
  • Examples of aprotic solvents include ethanenitrile, ethyl ethanoate and propanone

    Solvation

• Solvation is where solvent molecules surround a dissolved ion
  • In S_N1 reactions, the rate-determining step is not the attack of the nucleophile
  • The rate-determining step is the formation of the carbocation intermediates and halide ion
  • Both ions could be stabilised by the use of a protic solvent, as shown in the following example:

Protic polar solvent stabilising carbocation intermediates and halide ions

• • In S_N2 reactions, the rate-determining step is the attack of the nucleophile
  • The use of aprotic solvents does not solvate the nucleophile
  • This means that the nucleophile is more able to react and form the transition state
• S_N1 reactions are best conducted using protic, polar solvents
• S_N2 reactions are best conducted using aprotic, non-polar solvents