Bromine Water Equilibrium / Saturation Testing

[Njn]

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I'm confused as to what happens what happens to the equilibrium after bromine water reacts with an unsaturated hydrocarbon...

Like, I know that bromine water is an equilibrium of the following:

Br₂ + H₂O <--> H⁺ + Br^- + HOBr

But if it is the reaction of HOBr with the hydrocarbon that causes the colour change, what happens to the rest of the equilibrium? Or do I have this all wrong?

[Pwnage101]

i think you have it all wrong, bromine does not react with the water (at least at HSC level, i believe)

i have very good knowledge of this expt, so if u have any specific q's, ask them

[LordPc]

Bromine breaks the double bonds in the unsaturated hydrocarbon and then attatches itself onto the chain. so that you have a hydrocarbon with some bonds filled by Br instead of the usual H atom.

colour change comes from the bromine moving out of the bromine water and attaching to the hydrocarbon chain

I dont remember any equilibrium in there

[Trebla]

Originally Posted by Njn

I'm confused as to what happens what happens to the equilibrium after bromine water reacts with an unsaturated hydrocarbon...

Like, I know that bromine water is an equilibrium of the following:

Br₂ + H₂O <--> H⁺ + Br^- + HOBr

But if it is the reaction of HOBr with the hydrocarbon that causes the colour change, what happens to the rest of the equilibrium? Or do I have this all wrong?

The reaction at equilibrium is Br₂ getting pulled apart by H⁺ and OH^- in water.

Two things happen here:

1) The Br₂ can split into radicals and consume the double bond of the alkene in halogen addition.
R - CH = CH - R' + Br₂ --> R - CHBr - CHBr - R'

2) On the other hand the HOBr splits into ions Br⁺ and OH^- and both attack the double bond of the alkene, giving a halohydrin addition.
R - CH = CH - R' + HOBr --> R - CHOH - CHBr - R'

HOWEVER, the reaction 1) occurs more readily because the equilibrium is favoured to the left due to the low solubility of Br₂, so the major product would be the dibromo compound. The halohydrin DOES form, but it is the minor product (forms in very small concentrations) because in equilibrium the HOBr occurs at very low concentration.
Most textbooks simply state the dibromo compound because it is the major product.

[Pwnage101]

Originally Posted by Trebla

The reaction at equilibrium is Br₂ getting pulled apart by H⁺ and OH^- in water.

Two things happen here:

1) The Br₂ can split into radicals and consume the double bond of the alkene in halogen addition.
R - CH = CH - R' + Br₂ --> R - CHBr - CHBr - R'

2) On the other hand the HOBr splits into ions Br⁺ and OH^- and both attack the double bond of the alkene, giving a halohydrin addition.
R - CH = CH - R' + HOBr --> R - CHOH - CHBr - R'

HOWEVER, the reaction 1) occurs more readily because the equilibrium is favoured to the left due to the low solubility of Br₂, so the major product would be the dibromo compound. The halohydrin DOES form, but it is the minor product (forms in very small concentrations) because in equilibrium the HOBr occurs at very low concentration.
Most textbooks simply state the dibromo compound because it is the major product.

Thanks for the info Trebla!!

[wendy_]

For a prac report, we were asked to research the preferential solubility of (Br-Br) for the non polar solvent (rather than the polar). I don't really understand what I'm supposed to look for. I've browsed the net, checked up heaps of old textbooks but I can't seem to find anything. Can anyone help me out?

[brenton1987]

Originally Posted by wendy_

For a prac report, we were asked to research the preferential solubility of (Br-Br) for the non polar solvent (rather than the polar). I don't really understand what I'm supposed to look for. I've browsed the net, checked up heaps of old textbooks but I can't seem to find anything. Can anyone help me out?

Atomic bromine has a large electronegativity so it attracts electrons from neighbouring molecules to form a dipole. However molecular bromine is linear and diatomic so the two dipoles in opposite directions cancel eachother resulting in no net dipole. The molecule is non polar. Non polar substances dissolve better in non polar solvents rather than polar solvents.

[MrMMMan]

Examples of non-polar solvents are cyclohexane and cyclohexene that are relevant to this experiment