As you guys know, the product of bromination should be drawn like this:
Instead of something like:
The reason for this lies in the way the addition reaction takes place
To be precise, this mechanism is known as “electrophilic addition” and applies to addition of electrophiles such as halogens (e.g. bromine) and water to alkenes. This mechanism does not apply to catalytic hydrogenation which will be covered in another post.
What are “electrophiles”?
Electrophiles simply refer to molecules/ions that “like electrons”. On the other hand, an electrophobe is a molecule/ion that “dislikes electrons”.
The C = C portion of the alkene is said to have “high electron density” / “electron-rich” as there are 4 electrons that are clustered around this region. Therefore when a bromine molecule is brought near to the C = C portion of the alkene, the electrons on the bromine atom that is closer to C =C bond will be repelled to the other bromine atom because like charges repel.
The bromine atom that is nearer to the C = C bond acquires a partial positive charge (δ +) while the other bromine atom becomes partially negative (δ – ). This is illustrated in the diagram below.
The partially positive bromine atom is attracted to the “electron-rich” C = C double bond. A pair of electrons is transferred from the C = C bond to the partially positive bromine atom forming a C – Br bond with one of the carbon atoms. The C = C bond becomes a C – C bond. This results in the formation of an intermediate known as a carbocation (think carbon – cation) and a negative bromine ion. See diagram below.
Finally, the negative bromine ion is attracted to the carbocation forming the final product, 1,2-dibromoethane!
Tell me why! (A pair of electrons are transferred from the C = C bond to the partially positive bromine atom..)
As you guys know, most chemical reactions are driven by the quest for greater stability. The C = C bond [BE= 610kJ/mol] while stronger than the C – C bond [BE = 350kJ/mol] is less than twice as strong. You can artificially decompose the C = C bond into two C – C bonds as shown below.
Why is one bond weaker than the other? Since like charges repel each other, you can think of the reason for this as due to higher levels of repulsion as there are more electrons that are “clustered” in the same space. [JC students should think in terms of sigma and pi bonds. Just saying.]
Hence when the partially positive bromine atom that is deficient in electron approaches the C = C bond, the pair of electrons from the C = C bond which is less stable [BE = 260kJ/mol] will “defect” to it, forming the C – Br bond [BE = 280kJ/mol], which is stronger, hence achieving a higher degree of stability!