Bromine adds across ethene's C=C double bond and makes 1,2-dibromoethane, with one bromine on each carbon. Curly arrows show how, in two steps.
The same first step is behind the bromine water test in lesson 8, where orange bromine water turns colourless. Ethene's double bond is electron-rich. A bromine molecule, Br–Br, shares its pair evenly. As it comes close to the C=C, the crowd of electrons in the double bond pushes that pair away, toward the far end. The nearer bromine becomes δ+ and the far one δ−.
Step 1
Arrow 1 starts on the C=C double bond and points to the δ+ bromine: the second pair of the double bond becomes a new C–Br bond. A bromine atom's outer shell is full with 8 electrons, so as the new bond forms, the Br–Br bond breaks. Arrow 2 carries its pair to the far bromine, which leaves as a bromide ion, Br⁻.
The pair that went to bromine was shared by both carbons. The left carbon has lost its share, so it has only three bonds and a + charge. A carbon with a + charge like this is a carbocation, and it is very electron-poor.
Step 2
The bromide ion is a nucleophile, with lone pairs and a − charge, and the carbocation is an electrophile. One arrow carries a lone pair from the bromide ion to the + carbon and makes the second C–Br bond. The + and − charges cancel, and the product is 1,2-dibromoethane.
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Quick check
That's the end of lesson 21
In this lesson you found out:
- What a curly arrow shows, from its tail to its head
- Why electrons flow from electron-rich places to electron-poor ones
- How a base takes H⁺ from an acid, in two arrows
- How bromine adds to ethene, step by step