Free-radical bromination of propane in the presence of chlorine

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In summary, the presence of chlorine in the free-radical bromination of propane leads to a decrease in the bromination selectivity, as chlorine changes the observed ratio of 1-bromopropane to 2-bromopropane by forming bromine monochloride which is less selective than bromine.
  • #1
scatterbrain8
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When the free-radical bromination of propane is carried out in the presence of chlorine, the bromination selectivity (ratio of 1-bromopropane to 2-bromopropane) is much lower than it is when chlorine is omitted from the reaction (In other words, chlorine changes the observed ratio of 1-bromopropane to 2-bromopropane). In fact, the bromination selectivity is similar to the chlorination selectivity. Explain.


Attempt at solution:

I know that 2-bromopropane is the major product in bromination (without the addition of chlorine), because it has the more subsituted carbocation intermediate. I also know that chlorination won't occur because the enthalpy for that reaction is around -10 kcal/mol. However, I'm not entirely sure as to how the chlorine lowers the ratio. I thought it was something along these lines: without the chlorine, 1-bromopropane is the least favorable product; in the presence of chlorine, a chlorinated product becomes the least favorable product, therefore moving the 1-bromopropane up to be the second least favorable product and thereby lowering the ratio.

But I don't think this accounts for a difference in the ratios. Can anyone explain, or give me a hint as to why this occurs?
 
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  • #2
Ooh I like this question. It really gets at whether or not you're thinking about the mechanism.

I'm not sure where you're going with that answer attempt. Both reactions are exothermic in the end, but you don't need to think about the stability of the products to answer the question.

Here's a hint: which part of free radical bromination makes it selective? How might the presence of chlorine mess this up? The same thing that makes bromination selective also makes chlorination non selective. Remember that all the reactants are in a big soup, and an intermediate from one reaction can bump into a reactant from a "seperate" reaction.

Write out the whole mechanism and think about it. If you haven't already, read up on the hammond postulate and its implications for free radical halogenation.
 
  • #3
An old thread, but I found it in a Google search so it deserves an answer for the sake of anyone else who might find it.

Bromine and chlorine will combine to form bromine monochloride. This compound has properties somewhere between Cl2 and Br2, and will brominate, rather than chlorinate hydrocarbons. (the chlorine preferentially attaches to the hydrogen to form HCl.) Since chlorine is less selective than bromine, bromine monochloride is thus also less selective than bromine.
 

1. What is the purpose of free-radical bromination of propane in the presence of chlorine?

The purpose of this reaction is to convert propane (C3H8) into brominated propane (C3H7Br) by adding a bromine atom (Br) to one of the carbon atoms in the propane molecule. This can be done in the presence of chlorine (Cl2), which acts as a catalyst to initiate the reaction.

2. How does the presence of chlorine affect the free-radical bromination reaction?

Chlorine acts as a catalyst in this reaction by forming free radicals that react with the propane molecule. These free radicals are highly reactive and can easily abstract a hydrogen atom from the propane molecule, initiating the bromination process.

3. What is the mechanism of the free-radical bromination reaction?

The reaction begins with the homolytic cleavage of the Cl-Cl bond in chlorine, forming two chlorine free radicals. One of these free radicals then attacks a hydrogen atom in propane, forming a protonated chlorine radical and a propane radical. The chlorine radical then abstracts a hydrogen atom from the propane radical, forming the desired product, brominated propane, and regenerating the original chlorine radical. This process continues until all three hydrogen atoms in propane have been replaced by bromine atoms.

4. What are the conditions necessary for a successful free-radical bromination reaction?

The reaction requires chlorine gas (Cl2) as a catalyst, as well as a source of bromine atoms, such as elemental bromine (Br2) or a bromine-containing compound. It also requires high temperatures and UV light to initiate the reaction and maintain the free radicals. The reaction must also be carried out in an inert solvent, such as carbon tetrachloride (CCl4), to prevent unwanted side reactions.

5. What are the main applications of free-radical bromination of propane?

This reaction is commonly used in organic synthesis to introduce bromine atoms into molecules, which can then be further modified to create new compounds. Brominated propane can also be used as an intermediate in the production of other chemicals, such as pharmaceuticals, fragrances, and pesticides. Additionally, brominated propane has higher reactivity compared to propane, making it useful in some industrial processes, such as polymerization reactions.

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