Nucleophilic Substitution (HL IB Chemistry)

• Haloalkanes will undergo nucleophilic substitution reactions due to the polar C-X bond (where X is a halogen)

Partial positive C atom and partial negative X atom

Hydrolysis of Haloalkanes 

• The nucleophile in this reaction is the hydroxide, OH^– ion
• An aqueous solution of sodium hydroxide (NaOH) or potassium hydroxide (KOH) with ethanol is used
• This reaction is very slow at room temperature, so the reaction mixture is warmed
• This is an example of a hydrolysis reaction and the product is an alcohol

CH₃CH₂Br + OH^– → CH₃CH₂OH + :Br^–

bromoethane      →       ethanol

• :Br^– is the leaving group 
  • Halogens make good leaving groups as they form relatively weak bonds with carbon
  • Their higher electronegativity also means the bonded electrons are drawn towards the halogen atom making the carbon partially positive, δ+, and susceptible to nucleophilic attack
• The rate of this reaction depends on the type of halogen in the haloalkane 
• The stronger the C–X bond, the slower the rate of the reaction
• In terms of bond enthalpy, C–F > C–Cl > C–Br > C–I
• Fluoroalkanes do not react at all, but iodoalkanes have a very fast rate of reaction

The nucleophilic substitution mechanisms for the above reactions are as follows:

Nucleophilic substitution mechanism of bromoethane with a hydroxide ion 

Nucleophilic Substitution with OH^–, the bond that forms and the bond that breaks must both involve the carbon atom that is bonded to the leaving group

Neutral nucleophiles

• When the nucleophile is neutral, e.g. H₂O, the initial product is positive
• The positive product then deprotonates, losing H⁺, and forms a neutral product
  • CH₃CH₂Cl + H₂O → CH₃CH₂OH + :H⁺

Diagram to show water acting as a nucleophile forming a positive product which is then deprotonated

Nucleophilic substitution reactions with neutral nucleophiles involves deprotonation