With alkanes, a free-radical halogenation (substitution) is observed.

I was told that there will be no reaction in a similar situation with ethene and ethyne — and I am having trouble understanding why.

Is it because the halogen radical will not be able to abstract a hydrogen atom from carbon because it is influenced by the greater electron density in the carbon-carbon double/triple bonds? Or because the $\ce{C-H}$ bonds are shorter in these molecules compared to alkanes? (resonance?)

I've assumed that the problem arises in the propagation step of the most common reaction mechanism seen with alkanes.

Or is it due to sterics? ($\unicode[Times]{x3c0}$-bonds get in the way of the attacking bromine radical or restricted rotation in double/triple bonds get in way of ideal orientation for reaction to proceed to the transition state)

The necessary heat/$h \nu$ is provided for the reaction. Additionally, no other reactants/catalysts are present.

Please keep your answers on a novice college student level.


2 Answers 2


Both substance classes have an inherently different reactivity. Alkanes are very unreactive — in fact, save radical reactions, there is not much that will adequately attack them. Alkenes and alkynes have $\unicode[Times]{x3c0}$-bonds which are much less energetically stable than the alkanes’ $\unicode[Times]{x3c3}$-bonds.

$\unicode[Times]{x3c0}$-bonds can be attacked by halogens directly, the mechanism is called electrophilic addition and the usual product is an anti-dihalide with the halogen atoms attached to the carbons previously involved in the double bond. An intermediate species of this reaction is the halonium ion, a three-membered ring consisting of the ex-double-bonded carbons and one positively charged halogen. The other halogen can then attack according to $\mathrm{S_N2}$ from behind.

However, if you try hard enough you will also manage to create halogen radicals in the presence of alkenes. The radical will now not attack a $\ce{C-H}$ bond, but rather the energetically less stable $\ce{C=C}$ bond forming a $\ce{C-Hal}$ bond and a new carbon-centred radical. This carbon radical will then most likely polymerise with further double bonds by the same mechanism, creating a carbon polymer rather than a haloalkene.

I’m not entirely sure how alkynes will react. Their triple bond is more stable than a double bond, so selective halogenation of double bonds is possible in the presence of triple bonds. However, in general there is no reason why they couldn’t react according to the same pathways.

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    $\begingroup$ @SufyanNaeem Thanks, I keep reconsidering becoming a teacher ;D $\endgroup$
    – Jan
    Dec 16, 2016 at 22:44

It's because the addition of a halogen (at least chlorine, bromine and iodine) with an alkene or alkyne is not a free-radical halogenation, it's (quite rapid) electrophilic addition. For instance, see this tutorial: http://www.chemguide.co.uk/mechanisms/eladd/symbr2.html

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    $\begingroup$ It would be helpful to provide a summary of the link in the answer, in case the link becomes invalid in the future. $\endgroup$
    – bon
    Oct 10, 2015 at 10:27
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    $\begingroup$ If you generated a halogen radical, wouldn't you end up with the more competitive polymerisation occurring as compared to the benzoyl radical (from benzoyl peroxide) $\endgroup$
    – Beerhunter
    Oct 10, 2015 at 11:29

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