How do alkenes decolourise bromine water?

Answer 1

By electrophilic addition to give the halohydrin, #RC(OH)H-CH_2Br#.

Electrophiles or potential electrophiles react with olefins, a species rich in electrons:

i.e. #RCH=CH_2 + Br_2 rarr RCHBr-CH_2Br#.

The bromine's brown hue would fade, becoming nearly colorless.

But (of course) there is an added sting in the tail of this question. It asked for the reaction of the olefin with bromine water, i.e. #Br_2(aq)# not #Br_2# per se.

Considering the intermediate steps, then:

#RCH=CH_2 + Br_2(aq) rarr RC^(+)H-CH_2Br + Br^(-)(aq)#.
So in the first addition, the olefin reacts as a nucleophile, as an electron rich species. The intermediate carbocation reacts of course as an electrophile. Because in bromine water #Br_2(aq)#, by far the most concentrated nucleophile is the WATER molecule, this reaction would give #RC(OH)H-CH_2Br#, the halohydrin, as the major product.

Instead of A levels, this is at a first- or second-year level.

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Answer 2

Alkenes decolourise bromine water through an addition reaction, where bromine adds across the carbon-carbon double bond, breaking it and forming a colorless solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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