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In the presence of the electron rich (pi bond) of ethene, the bromine molecule becomes a temporary induced dipole, one end being electron deficient and acting as the electrophile. (The pi bond in) ethene attacks the electrophile.

At the same time, one of the lone pair of electrons in bromine attacks one of the C atom. Why does this happen?

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Normally, I'm familiar with the same process, but with the carbocation being formed without the lone pair of electrons of bromine attacking the other C atom:

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Does this have something to do with resonance as shown by the diagram below?

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    $\begingroup$ The alkene's filled π orbital is higher in energy than the empty σ* orbital of the $\ce{X-X}$ bond. Also, the $\ce{X-}$ can attack the halonium ion but the result is just regeneration of halogen and the alkene: the first step of the reaction is reversible. $\endgroup$ – Apoorv Potnis Feb 11 '18 at 7:28
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There are too many bond lines to explain the reaction; the question suggests that we have to explain the bond lines!

Let's start from scratch, with a different viewpoint. Ethylene isn't the attacker; it doesn't burst into flame in oxygen or do anything aggressive. It may be reactive, but in a "normal" range. Bromine, on the other hand, is a potent oxidizer and does go around interacting with relatively inert materials - it's a powerful bleach.

So, when bromine comes near the ethylene pi bond, it attacks the electron pair and grabs it, but the second bromine steals it away and departs as Br-. After all, bromine is more electronegative than carbon; the first bromine is just a link in the chain of events that allows the second bromine to escape with an electron. The first bromine winds up stuck to the C-C double bond; the two carbons and the bromine share the positive charge. The bromine certainly gets the lion's share of electron density, leaving the carbons with most of the positive charge. That's why another bromine can attack the carbons from the rear to give a dibromide. (If a Br- attacked the bromine attached to the double bond, it would just regenerate the reactants.) I bet that the second bromine to bond to the ethylene is not from the same bromine molecule as the first bromine. (But since all bromine molecules look the same, I can't prove it.) Doing the bromination in a solution with an overwhelming abundance of another nucleophile would probably give a mixed product.

A C-C double bond is about 1.34 Angstroms long; a bromine atom (in Br2) is about 2.29 Angstroms in diameter. Rather than envisioning the cation as a triangle of three grapefruit, I see it as a cantaloupe sitting on two large oranges. Just consider the empty p orbital of bromine and the pi lobe of the carbons; no need or benefit from making a three-membered ring.

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