What happens when $H - X$ $H-X$ adds to the double bond of an alkene?

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What happens when $H - X$ $H-X$ adds to the double bond of an alkene?

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Answer 1 · 2015-01-10T01:11:32

The end result will be that one of the two bonds opens up. $H$ $H$ adds to one side, and $X$ $X$ to the other. This is called mono-halogenation.

( $X$ $X$ stands for $F$ $F$ , $C l$ $Cl$ , $B r$ $Br$ or $I$ $I$ )

This is an example of an addition reaction, because the whole $H X$ $HX$ molecule is added tot the alkene. There are no other reaction products.

The mechanism is as follows:
$H - X$ $H-X$ is always very polar, the $X$ $X$ -side drawing the electrons more to its side of the molecule. The extra electron pair in the double bond matches up with the (positive) $H$ $H$ and then the $X$ $X$ has 'no other choice' but matching up with the other side of the now open bond.

Example :
Ethene + hydrogen bromide

$(H_{2} C = C H_{2}) + (H - B r) \to (H_{3} C - C H_{2} B r)$ $(H_2C=CH_2)+ (H-Br)->(H_3C-CH_2Br)$

Extra :
Contrary to mono-halogenation of alkAnes, this reaction occurs under rather normal conditions.

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What happens when $H - X$ $H-X$ adds to the double bond of an alkene?

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