Question

Give the structures of both 1,2 and 1,4 adducts resulting from reaction of 1 equivalent of HBr with the following compound:

Short answer

Halogen halide reacts with alkene hydrogen attached to one ${\mathbf{sp}}^{\mathbf{2}}$carbon of alkene and halide added to another${\mathbf{sp}}^{\mathbf{2}}$alkene carbon; even when the double bond is in the ring compound, the result is the same.

Hydrogen halides, e.g., HCl, HBr, or HI, react with an alkene, e.g., ethylene form alkyl halide as the product

Step-by-step solution

Addition of hydrogen halide to an alkene

Halogen halide reacts with alkene hydrogen attached to one ${\mathbf{sp}}^{\mathbf{2}}$carbon of alkene and halide added to another${\mathbf{sp}}^{\mathbf{2}}$alkene carbon; even when the double bond is in the ring compound, the result is the same.

Hydrogen halides, e.g., HCl, HBr, or HI, react with an alkene, e.g., ethylene form alkyl halide as the product

Formation of 1,2 and 1,4 adducts

2,3-dimethylcyclohexa-1,3-diene having a double bond, reacts with hydrogen halide HBr to give the carbocation, which is resonance stabilized (two resonating structures).

Two carbocation forms get attacked by the nucleophile bromine ion to result in the formation of two different types of products (adducts):

• 1,2 product is 6-bromo-1,6-dimethylcyclohex-1-ene
• 1,4 product is 3-bromo-1,2-dimethylcyclohex-1-ene