Chapter 4 : Alcohols and Alkyl Halides Ch 4 contents
Radical Halogenation Answers
Qu 1: In butane we have only two types of H, the 6 x 1o in 2 equivalent -CH3 and the 4 x 2o in 2 equivalent -CH2- .
Use the diagrams below to highlight this is you are unsure. 
1o CH 
2o CH 
Reset
butane
1-chlorobutane
2-chlorobutane
For chlorination: 
% 1-chlorobutane = 100 x (6 x 1) / (6 x 1 + 4 x 3.9) = 100 x 6 / 21.6 = 27.8 %
% 2-chlorobutane = 100 x (4 x 3.9) / (6 x 1 + 4 x 3.9) = 100 x 15.6 / 21.6 = 72.2 %
The only thing that changes for bromination are the selectivity factors: 
% 1-bromobutane = 100 x (6 x 1) / (6 x 1 + 4 x 82) = 100 x 6 / 334 = 1.8 % 
% 2-bromoobutane = 100 x (4 x 82) / (6 x 1 + 4 x 82) = 100 x 164 / 334 = 98.2 %

Qu 2:  As shown below, hexane has 5 possible isomers. Of these 5 isomers, only 2 have have three types of H and hence would give 
three monochlorides : hexane and 2,2-dimethylbutane
hexane isomers
For hexane the relative yields of the 3 possible chlorination products are:
% 1-chlorohexane = 100 x (6 x 1) / (6 x 1 + 8 x 3.9) = 100 x 6 / 37.2 = 16.1 %
% 2-chlorohexane = 100 x (4 x 3.9) / (6 x 1 + 8 x 3.9) = 100 x 15.6 / 21.6 = 41.9 %
% 3-chlorohexane = 100 x (4 x 3.9) / (6 x 1 + 8 x 3.9) = 100 x 15.6 / 21.6 = 41.9 %
For 2,2-dimethylbutane the relative yields of the 3 possible chlorination products are:
% 1-chloro-2,2-dimethylbutane = 100 x (9 x 1) / (12 x 1 + 2 x 3.9) = 100 x 9 / 19.8 = 45.5 % 
% 3-chloro-2,2-dimethylbutane = 100 x (2 x 3.9) / (12 x 1 + 2 x 3.9) = 100 x 7.8 / 19.8 =  39.4 %
% 1-chloro-3,3-dimethylbutane = 100 x (3 x 1) / (12 x 1 + 2 x 3.9) = 100 x 3 / 19.8 = 15.1 %
Hence the isomer must be 2,2-dimethylbutane. 
Note we could also have concluded this without doing the calculation if we realised that the yields of 2-chloro- and 3-chlorohexane would have to be the same since each arises from the substitution of 1 of 4 2o H atoms.
One of the important issues to emerge is that 2 H atoms can be of different types even if they are both 1o / 2o / 3o