Question

The number of different substitution products possible when bromine and ethane are allowed to react is?

Answer 1

Since halogenation of alkanes is mostly uncontrolled, you are likely to get a mixture of mono and polysubstituted products. These will be-

Monosubstituted: 1

Disubstituted: 2

Trisubstituted: 2

Tetrasubstituted: 2

Pentasubstituted: 1

Hexasubstituted: 1

Thus, you get a total of 9 structural isomers.



C₂H₆(g) +Br₂(l) --> C₂H₅Br(l) +HBr(g).

C₂H₆(g) +Br₂(l) --> C₂H4Br(l) +2HBr(g).

C₂H₆(g) +2Br₂(l) --> C₂H2(l) +4HBr(g).

C₂H₆(g) +3Br₂(l) --> C₂H3Br3(l) +3HBr(g).

C₂H₆(g) +Br₂(l) --> CHBr(l) +HBr(g)+ CH4

C₂H₆(g) +4Br₂(l) --> C₂H₅Br(l) + HBr(g) + 3Br2

C₂H₆(g) +2Br₂(l) + H2 --> C₂H₅Br(l) +3HBr(g).

2C₂H₆(g) +Br₂(l) --> C₂H₅Br(l) + HBr(g) + C2H6

C₂H₆(g) +3Br₂(l) --> C₂HBr(l) +5HBr(g).