Because of the extra electron density in a C=C double bond, alkenes are attacked by species which ‘like’ electrons.
The electrophile, having some positive character, is attracted to the alkene.
The electrons in the pi bond come out to form a bond to the positive end.
Because hydrogen can only have two electrons in its orbital, its other bond breaks heterolytically. The H attaches to one of the carbon atoms.
A carbocation is formed. The species that left now has a lone pair.
It acts as nucleophile and attacks the carbocation using its lone pair to form a covalent bond. Overall, there is ADDITION
Reagent:
Condition: Room temperature. NO Catalyst
Equation:
C2H4(g) + HBr(g) ———> C2H5Br(l) bromoethane
Mechanism
Reagent:
Condition:
Equation:
C2H4(g) + Br2(l) ——> CH2BrCH2Br(l) 1,2 - dibromoethane
Mechanism
The addition of bromine dissolved in tetrachloromethane (CCl4) or water (known as bromine water) is used as a test for unsaturation. If the reddish-brown colour is removed from the bromine solution, the substance possesses a C=C bond.
Method for Testing
2. Add the Hydrocarbon to be tested and gently shake
3. If the brown colour disappears then the Hydrocarbon is an Alkene
1 2 3
Because the bromine adds to the alkene, it no longer exists as molecular bromine and the typical red-brown colour disappears
Chemically, the alkynes are similar to the alkenes. Since the C≡C functional group has two π bonds, alkynes typically react even more readily, and react with twice as much reagent in addition reactions. The reaction of acetylene with bromine is a typical example:
Draw the Displayed formula and name the products of the following reactions:
• addition of HBr to propene gives two isomeric brominated compounds
• HBr is unsymmetrical and can add in two ways
• products are not formed to the same extent
• the problem doesn't arise in ethene because it is symmetrical.
Possibility 1
Possibility 2
In the electrophilic addition to alkenes the major product is
formed via the more stable carbocation (carbonium ion)
In the addition to propene, path A involves a 2° carbocation, path B a 1° carbocation.
As the 2° ion is more stable, the major product (i.e. 2-bromopropane) is formed this way.
Reagent:
Conditions:
Catalyst:
Product:
Equation:
C2H4(g) + H2O(g) --> C2H5OH(g) ethanol
Use:
Comments:
Reagent:
Conditions:
Product:
Equation:
C2H4(g) + H2(g) ———> C2H6(g) ethane
Use:
margarine manufacture