Nucleophilic Substitution Answers
Qu 1:
2-methyl-2-propanol or t-butanol (3o), most reactive
2-butanol (2o)
2-methyl-1-propanol (1o)
1-butanol (1o), least reactive

Since the reaction of an alcohol with HBr favors an SN1 type reaction, the systems generating the more stable carbocation will react fastest.

Qu 2:
(a) Since the -OH group is such a poor leaving group, it can't be displaced directly.  In the reaction with HBr the -OH is protonated first and the leaving group is a neutral water molecule which is much better. In the reaction with NaBr, there is no acid catalyst in order to activate the leaving group and therefore no reaction is observed.

(b) In the presence of sulfuric acid, H2SO4, the -OH becomes protonated and can now leave to be replaced by the bromide ion providing the alkyl bromide.

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 if 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:
 		 Hexane has 5 possible isomers, but of these only 2 have have three types of H and would give
three monochlorides: hexane and 2,2-dimethylbutane
hexane isomers
For hexane:
% 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
% 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 %
% 4-chloro-2,2-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 realized 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

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