Kimmeh: As has already been pointed out, there is no 'benzene ring' in cyclohexane, or any other cyclic compound dealt with in the HSC. The reactions of cyclohexane and hexane with bromine have already been provided, and here are the reactions for hexane and 1-hexene:
C<sub>6</sub>H<sub>14</sub> + Br<sub>2</sub> ---UV Light---> C<sub>6</sub>H<sub>13</sub>Br + HBr
The product here is one of the isomers of bromohexane - ie. one of 1-bromohexane, 2-bromohexane or 3-bromohexane.
C<sub>4</sub>H<sub>9</sub>-CH=CH<sub>2</sub> + Br<sub>2</sub> ---> C<sub>4</sub>H<sub>9</sub>CHBr-CH<sub>2</sub>Br
The product here is 1,2-dibromohexane.
OK, now to the bromine water issue.
Let's start with the easy bit. Bromine on its own, or bromine in a non-polar solvent (organic chemists like to use carbon tetrachloride, CCl<sub>4</sub>) is unequivocally Br<sub>2</sub>. It reacts in these circumstances by adding across a double bond in the usual way, ie:
R<sup>1</sup>R<sup>2</sup>C=CR<sup>3</sup>R<sup>4</sup> + Br<sub>2</sub> ---> R<sup>1</sup>R<sup>2</sup>CBr-CBrR<sup>3</sup>R<sup>4</sup>
where R<sup>1</sup>, R<sup>2</sup>, R<sup>3</sup> and R<sup>4</sup> are all either hydrogen or carbon chanins.
In water, the situation becomes more complex. As has been implied above, we have a problem as bromine exists in equilibrium in water with hypobromous acid and hydrobromic acid:
Br<sub>2 (aq)</sub> + H<sub>2</sub>O<sub> (l)</sub> <---> HOBr<sub> (aq)</sub> + HBr<sub> (aq)</sub>
That is, there is present the strong acid HBr, the weak acid HOBr, and in addition, unreacted Br<sub>2</sub>. Conquering Chemistry maintains that the reactive species in this mixture is the HOBr, and thus that bromine water should be formulated as HOBr. This completely ignores the fact that HBr (or, to be more precise, its ionised form) is a capable of adding across a double bond, and that this equilibrium lies to the left.
Furthermore, it ignores the purpose of the bromine water test. The use of bromine water is as a simply, qualitative, diagnostic test for the presence of unsaturations in the carbon chain - that is, carbon-to-carbon double and triple bonds - and this is done by the colour change that occurs due to the reaction. Since the coloured species in this equilibrium mixture is the Br<sub>2</sub>, and not either of the others, in my opinion the critical reaction in this case is still the addition of Br<sub>2</sub>, and not the reaction of minor species produced due to the hydrolysis of bromine. Note that the fact that bromine water retains significant orange / brown colour is proof that bromine remains present at equilibrium in substantial quantities.
I might add that I have discussed this matter in the past with a member of the USyd academic staff who is an organic chemist, and he concurs that the important reaction is the one that leads to the colour change, and thus is the reaction of bromine itself.
Now, what about exam questions? Well, if you feel it is necessary to offer an explanation in an exam, then I would note that the equilibrium does exist, but that the colour change is the basis for the use of the test, and thus it is the reaction of the coloured species that should be considered, and not any other reactions.
In conclusion, whilst I would not go as far as to say that the mechanism discussed by Smith is wrong, I would say that it does not focus on the critical issue, and thus I would still write reactions of bromine water using Br<sub>2</sub>.
Please post more questions if you are still having troubles with this.
