I'm confused about halogenation pathways. From what I understand, there are multiple halogenation pathways. I'll try summarizing them below; let me know if I do anything incorrectly.
- Halogenation with the diatomic halogen - i.e. $\ce{Br2}$. This reaction proceeds through a halonium intermediate.
- Halogenation with the diatomic halogen in the presence of light or heat. This reaction proceeds via radical intermediates. From what I understand, the light or heat cleaves the weak halogen-halogen bond and we get two radicals, which can initiate the radical reactions. Also the cool thing about this reaction is that it doesn't require an nucleophilic pi bond, so if we were starting from an alkane, we don't even need to first form an alkene or alkyne; we can free-radical halogenate that alkane and then eliminate a beta proton as to make the alkene. All the other reactions mentioned here require some sort of nucleophile to start with.
- Halogenation in the presence of water - i.e. $\ce{Br2}$ with $\ce{H2O}$. Again, we go through a bromonium intermediate. The $\ce{Br-Br}$ bond is instantaneously polarized and at that point we can form the bridged halonium ion and that gives water an entrance - the ring is highly strained and has partial charges all over the place - perfect for water to do something.
- Are these all the halogenation pathways I need to know for an elementary understanding of organic chemistry? Did I miss any? I think I finally clarified all of them to myself through typing up this post ...
- Oops yes I missed a few. NBS for example is used for allylic (or benzylic) bromination, and this directly implies the need for a starting alkene ... so this isn't as synthetically useful as free-radical bromination in the presence of light.
- Addition of $\ce{HX}$ - the pi bond acts as a simple nucleophile and attacks the proton. Then the anionic halogen adds to the resultant carbocation.
- Addition of $\ce{HX}$ in the presence of peroxides - the pi bond again acts as a simple nucleophile but this reaction proceeds via a radical intermediate and as a result the reaction is overall anti-Markovnikov.