Ch4: Radical answers

Chapter 4 : Alcohols and Alkyl Halides

Radical Halogenation Answers

Qu 1:

In butane we have only two types of H, the 6 x 1^o in 2 equivalent -CH₃ and the 4 x 2^o in 2 equivalent -CH₂- .

Use the diagrams below to highlight this is you are unsure.


	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-bromobutane = 100 x (4 x 82) / (6 x 1 + 4 x 82) = 100 x 328 / 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

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 2^o H atoms.
One of the important issues to emerge is that 2 H atoms can be of different types even if they are both 1^o / 2^o/ 3^o