# Why is halogenation anti addition?

Jun 30, 2017

Why? Because as far as anyone knows, when...........

#### Explanation:

........when a halogen adds to a cyclic olefin, it forms a so-called bromonium ion, a three-membered ring......

The picture depicts the supposed intermediate after the bromine adds (as an electrophile) to the olefin. We might now invoke a carbocation intermediate, which of course will undergo subsequent reaction with the newly delivered bromide nucleophile....

${R}_{2} C = C {H}_{2} + B {r}_{2} \rightarrow {R}_{2} B r C - \stackrel{+}{C} {R}_{2} + B {r}^{-}$

The formal carbocation MAY be stabilized somewhat by the formation of a bromonium ion, in which the added bromine atom forms a three-membered ring with the potential or conceptual carbocation, thereby somewhat stabilizing it. The newly delivered bromide ion is DIRECTED AWAY from that side of the ring, and thus we get a trans-disubstituted alicycle.

Note that this stereochemical peculiarity is also observed when $B {r}_{2}$ is added to $\text{cis-2-butene}$ versus $\text{trans-2-butene}$.

Because of the anti-addition, $\text{cis-2-butene}$ gives the $R , R$ and $S , S$ enantiomers, whereas addition to $\text{trans-2-butene}$ gives the optically inactive $R , S$, and $S , R$ $\text{meso}$ compound. This simple experiment is good evidence for the intermediacy of a three-membered ring, which blocks off ONE side of the molecule, and results in the given stereochemistry.

Anyway, look in your text, because this is an excellent example of a so-called $\text{stereospecific reaction}$ in which the mechanism of the reaction determines the stereochemical outcome of the reaction.