OK, assuming xiao1985 is correct and this is supposed to indicate the reactions of an alkane with bromine in the presence of UV light, then NEITHER equation is correct.
The substitution reaction should be:
hexane + bromine ---UV Light---> an isomer of bromohexane + hydrobromic acid
Note that by an isomer of bromohexane, I mean one of 1-bromohexane, 2-bromohexane or 3-bromohexane. In any of these cases, the reaction will have a molecular equation of:
C<sub>6</sub>H<sub>14</sub> + Br<sub>2</sub> ---UV Light---> C<sub>6</sub>H<sub>13</sub>Br + HBr
This reaction is NOT subject to the debate about the nature of bromine water that I have discussed in the other thread on this topic for the reaction of bromine water with alkenes. The reaction of a halogen with an alkane follows a radical chain reaction mechanism which requires a halogen (X<sub>2</sub>) as a reacting species. Neither a hydrohalic acid (HX) nor any oxyhalic acid (HOX<sub>n</sub>, where n = 1, 3, 5 or 7) is an acceptable substitute in this mechanism.
Assuming that you MUST use one of these equations, then I would say that the first is objectionable, and the second is ridiculous / preposterous / absurd / chemically evil / ... . The reason is that there is NO WAY that a system like the one described would produce hydrogen gas as a product, and this could easily be demonstrated by noting that no bubbles form in the reaction mixture.
A further note: Assuming that the reaction is in the absence of UV light, then there is no way ANY reaction will take place. Alkanes do not undergo substitution reactions under anything but fairly extreme conditions. If anyone can find a way to make alkanes into a product like a haloalkane or a haloalkanol in a controlled way under mild conditions - such as is suggested by these equations in the absence of UV - then that person is well on their way to a Nobel Prize.
CKO: I will chack back on this thread later tonight if you want to post any other questions about this before your exam tomorrow, and in any case, good luck!
