Electrophilic Addition (AQA AS Chemistry): Revision Note

Mechanism: Electrophilic Addition

Electrophilic addition

• Alkenes undergo electrophilic addition reactions

• In an electrophilic addition reaction, two reactants combine to form a single product, giving a 100% atom economy.

• Alkenes are particularly reactive because of the C=C double bond, which is an electron-rich region of the molecule

• This double bond is readily attacked by positively charged electrophiles.

• Alkenes undergo electrophilic addition reactions with hydrogen halides, halogens, and concentrated sulfuric acid (followed by reaction with steam)

 Electrophilic addition of hydrogen bromide

• A molecule of hydrogen bromide (HBr) is polar as the hydrogen and bromine atoms have different electronegativities

• The bromine atom has a stronger pull on the electrons in the H-Br bond

• As a result of this, the Br atom has a partial negative charge and the H atom a partial positive charge

• The H atom acts as an electrophile and accepts a pair of electrons from the C-C bond in the alkene

• The H-Br bond breaks heterolytically, forming a Br^- ion

• This results in the formation of a highly reactive carbocation intermediate, which reacts with the bromide ion, Br^-

• The reaction product is bromoethane:

Electrophilic addition of bromine

• Bromine (Br₂) is a nonpolar molecule because the two bromine atoms have the same electronegativity and share the bonding electrons equally

• However, when a bromine molecule approaches the C=C double bond of an alkene, the high electron density in the double bond repels the electron pair in the Br–Br bond away from the nearer bromine atom

  • This effect is known as induced polarity

• As a result, the bromine atom closest to the double bond becomes slightly positively charged, while the other bromine atom becomes slightly negatively charged

  

• During the addition reaction, the bromine atom closest to the alkene acts as an electrophile and accepts a pair of electrons from the C=C bond

• The Br–Br bond then breaks heterolytically, forming a Br⁻ ion

• This leads to the formation of a highly reactive carbocation intermediate, which is quickly attacked by the Br⁻ ion acting as a nucleophile

• The final product is a dihaloalkane, in this case, dibromoethane

Electrophilic addition reaction with sulfuric acid

• Water is a weak electrophile, so it does not readily undergo addition reactions with alkenes unless a strong acid is present to act as a catalyst

• A suitable strong acid for this reaction is sulfuric acid, H₂SO₄

• In acidic conditions, H₃O⁺ acts as the electrophile

• The reaction occurs in two steps:

  • Step 1:
    The π electrons in the C=C bond are attracted to H₃O⁺. Heterolytic fission occurs, forming a carbocation intermediate

  • Step 2:
    Water acts as a nucleophile and donates a lone pair of electrons to the positively charged carbon atom, forming a C–O bond. An equilibrium is established between the protonated alcohol and the deprotonated product (the alcohol)

• The H₃O⁺ ion is regenerated, so it acts as a catalyst

• The reaction product is ethanol

Major and minor products from unsymmetrical alkenes

• When the reaction takes place with an asymmetrical alkene, two products are possible, and you can predict which one will be the major product

• It depends on the stability of the carbocation formed as the intermediate

• The nucleophile will bond to the positive carbon atom of the carbocations

  • The more stable carbocation produces the major product

  • The less stable carbocation produces the minor product

• The stability of carbon carbocations is as follows:

  • Tertiary > secondary > primary

First step in the reaction with an unsymmetrical alkene

• The π electrons in the C=C bond attack the electrophilic hydrogen:

• There are two possible intermediates formed:

• The major product formed will be from the intermediate with the more stable carbocation, but some of the minor product from the less stable carbocation intermediate will also form

Formation of major and minor products

• The complete mechanism for the addition of HBr to an unsymmetrical alkene is as follows:

• In the mechanism above, the secondary halogenoalkane is the major product, because a secondary carbocation is more stable than a primary carbocation