[DoubleZero]

Are all halogenation reactions radical? I saw some reactions involving monobromination. It seemed like the reaction was started slowly to avoid a radical reaction (Drop by drop addition of a diluted bromine in a solvent into reaction solution in an ice bath). Would you start a reaction slowly to avoid the propagation step and therefore allowing only one site to be brominated? I saw some reactions in an acidic environment that seemed to not have an start to the reaction by uv light or heat like i've seen in radical reactions. Or is what I saw possibly a slow and controlled radical reaction that would favor monobromination.

So, Slow Non-Radical single site halogenation Or Slow radical single site halogenation?

How woud you tell if monobromination or double is occuring in application?

[Schrödinger]

Bromine is less reactive than Chlorine. Hence, it is more selective. Even in free radical reactions, where multiple chlorination is possible, multiple bromination is not very common, atleast as far as I know. I haven't come across such reactions.

Consider allylic bromination. This is carried out using NBS, a very cautious provider of Bromine. It produces Br2 insitu in small quantities. This gives very good monobrominated yield. The reaction is not as 'haywire' as the free radical one. It proceeds in a calm and methodical manner.

Free radical reactions on the other hand, are messy. Atleast with Chlorine. With bromine, yields are very good owing to its low reactivity and high selectivity. Mostly monobromo products, I believe.

Please correct me if I'm wrong.

[DoubleZero]

So performing the reaction slowly and at a low temperature would be to make the radical reaction go slower and more controlled?

What is the most practical way to tell this bromination reaction is complete? When I combine the reactants together, the solution is a dark red color which I assume indicates the presence of free bromine.

I'm not really sure where to go from here. Should I keep the solution in an ice bath and stir for a certain amount of time. Or should I wait until the red color goes away (indicating the free bromine has been used up) Or could I slowly heat the solution to speed the reaction up?

[Schrödinger]

Well, I don't have any laboratory experience, so I can't help you with that. But I guess, if you take something like stoichiometric amounts of bromine and the alkane, an indicator that the reaction is complete would be the fading of the red colour.

[fledarmus]

The best way to tell if your reaction is complete is by TLC, following the disappearance of your starting material. Unless, of course, you have access to some really nice instrumentation like HPLC-MS or GC-MS. This reaction is usually so much slower than addition across an alkene that following the disappearance of the red color isn't as useful as it is in that reaction. Bromine has a high enough vapor pressure that it will evaporate out of the reaction mixture if you leave it there long enough.

Bromine is much more selective than chlorine, and since the transition state is very late in the reaction pathway, the stability of the alkyl radical determines where in the compound it will react. However, if there are two sites on the molecule that have equivalent radical stabilities, you will still get statistical mixtures of mono-, di-, and non-brominated compounds (assuming 1 equivalent of bromine is used).

[Nescafe]

As fledarmus said in his post the stability of the formed radical determines where the reaction takes place. Radicals are electron deficient because they are now missing their octet. If destabilized by their neighboring groups they will try to quickly take back their proton. Inductive and resonance stabilization plays a particularly important role in the stability of the formed radical as well as the branched degree of the carbon where the radical is initially formed.