# Bromination Pathways with alkane, alkene, and alkyne substrates

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.
• Radical bromination is not very selective and hardly controllable. – Martin - マーチン Aug 28 '14 at 3:28
• That's true but it also gives you at least something if you have a plain alkane. – Dissenter Aug 28 '14 at 3:29
• Basically everything you've said here is correct. You will probably cover more halogenation reactions as you continue to study organic chemistry, but if you're limiting yourself to starting from alkanes/alkenes, then you've got it. – jerepierre Aug 28 '14 at 3:47

1. You mentioned halogenation of alkenes in the presence of $\ce{H2O}$. This reaction is known as halohydrin formation. It's worth mentioning that this reaction can also be performed in the presence of other nucleophiles, perhaps most commonly alcohols and salts of other halogens to give mixed dihalogenated products. (Other types of nucleophiles may also successfully react, but I can't recall seeing any specific examples.) The reaction isn't always efficacious. Frequently, addition of both halogen atoms (i.e., ordinary halogen addition) will occur; the proximity of the halide leaving group to the halonium bridge will often result in a rapid reaction before any other nucleophile can intercept.
2. Radical halogenation can be very difficult to control, even when dealing with substrates that should exhibit high regioselectivity. NBS (and its chlorinated counterpart, NCS) is actually an extremely useful and versatile reagent which mitigates some of the practical pitfalls of radical halogenation reactions. Under photocatalytic conditions, or with sufficient heating and a radical initiator, it can react by free radical pathways. It's often preferable to using the diatomic halogen directly (unless exhaustive halogenation is really the goal) because it generates lower concentrations of $\ce{Br2}$ in situ, so the reactions are somewhat more controllable and side-products from direct addition to $\pi$-bonds are minimized. It can also react analogously to $\ce{Br2}$ with alkenes to give a halonium ion, which makes it preferable for halohydrin formation (since there is no $\ce{Br-}$ leaving group present to immediately attack the bromonium bridge).