Hey King_sanj:
Jake is temporarily away for a period of time, so I can answer the question here if you wouldnt mind. In Production of Materials alone, we wouldnt need to know anything about Hydrobromous acids (HBr), because when we are doing the Bromine water test we are using only Br2(aq). Because the reaction between bromine water and alkenes is an addition process, it doesnt form hydrobromous acid and hence it is not necessary to know about HBr for the module. However, what you do need to know is the final product between the alkene and the Br2(aq), because oftentimes you would be asked to provide with an equation of the reaction between the alkene and bromine water. For module 2 Acidic Environment you WOULD need to know about HBr and how it is a weak acid compared to other ones such as H2SO4 and HCl.
Best Regards
Happy Physics Land
I know what he's referring to.
In truth, whilst we prepare it using bromine itself, Br
2, what we actually get and use for the practical is what is called bromine water - HOBr.
Observe the following equation:
Br
2(l) + H
2O
(l) -> HOBr
(aq) + HBr
(aq)Now, hydrobromic acid will not affect the experiment in any way whatsoever. It may be worth noting that it's a strong acid, but it just sits there.
We are only interested in bromine water - HOBr, which actually has the same molecular formula as
hyprobromous acid (HBrO) - not hydrobromous acid. Hydrobromous acid isn't actually a thing. Technically, the experiment used to distinguish between the alkane and alkene series actually uses bromine water. This is where the colloquial name "bromine water experiment" is derived from.
For the sake of having an alkene, I will use hex-1-ene, with formula C
6H
12(l).
The HSC (generally) accepts your usage of bromine itself in the equation you use:
Alkene: C
6H
12 + Br
2 -> C
6H
12Br
2But the ACTUAL equation of what goes on, is believe it or not, that with the bromine water
Alkene: C
6H
12 + HOBr -> C
6H
12BrOH
Realistically speaking, the second equation would be more accurate.
However, in terms of what bromine water HOBr actually is? That is beyond the scope of the syllabus. It is merely a tool for us to use in our Production of Materials topic.
Aside:
Naturally, we would use hexane as our contrast for the experiment:
Alkane: C
6H
14 + Br
2 -(UV)-> C
6H
12Br
2 + H
2Alkane: C
6H
14 + HOBr -(UV)-> C
6H
12BrOH + H
2