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Radical Substitution: Allylic Bromination
In organic synthesis, the formation of products can be altered by changing the reaction conditions. For example, a dibromo addition product is formed…
Allylic halogenation of an alkene requires a low concentration of halogen and a high temperature.
To carry out allylic bromination at room temperature, N-bromosuccinimide, or NBS, is used as the reagent. It maintains a low bromine concentration and acts as a source of bromine radicals.
For example, propene reacts with NBS in the presence of light or peroxide to form 3-bromopropene.
The mechanism of allylic bromination involves initiation, propagation, and termination steps.
During the initiation step, NBS undergoes light-induced homolytic cleavage of the N–Br bond to form a bromine radical.
In the first propagation step, the bromine radical reacts with propene and abstracts allylic hydrogen, forming an allylic radical and HBr.
The generated HBr reacts with NBS to produce the bromine molecule required for further propagation of the reaction.
In the second propagation step, the bromine molecule reacts with the allylic radical to give allyl bromide and a bromine atom.
Finally, in the termination step, the different radicals couple with each other to give nonradical products, stopping further reaction.
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Q1: Why is N-bromosuccinimide used instead of molecular bromine for allylic bromination?
NBS maintains a low bromine concentration throughout the reaction, preventing unwanted addition reactions. Molecular bromine at room temperature causes dibromo addition to alkenes. NBS acts as a bromine radical source, enabling selective allylic substitution in non-polar solvents while keeping both HBr and Br2 concentrations minimal, allowing radical bromination to outcompete ionic addition.
Q2: What are the three main steps in the allylic bromination mechanism?
Allylic bromination involves initiation, propagation, and termination. During initiation, light or peroxide causes NBS to undergo homolytic cleavage, forming a bromine radical. Propagation includes two steps: the bromine radical abstracts allylic hydrogen to form an allylic radical and HBr, then Br2 reacts with the allylic radical to yield allyl bromide. Termination occurs when radicals couple to form nonradical products.
Q3: How does HBr participate in the propagation cycle of allylic bromination?
HBr generated in the first propagation step reacts with NBS in an ionic reaction to produce Br2, which is essential for the second propagation step. This regeneration of bromine maintains the radical chain reaction. By consuming HBr, NBS ensures that bromine concentration stays low while continuously supplying Br2 needed for the reaction to proceed efficiently.
Q4: Why does allylic bromination of substituted alkenes produce a mixture of products?
The allylic radical intermediate formed during bromination is resonance stabilized, allowing the unpaired electron to delocalize across multiple carbon atoms. This resonance stabilization enables the radical to abstract bromine from either resonance site, generating two different allyl bromide products. The distribution of products depends on the stability and reactivity of each resonance form.
Q5: What reaction conditions favor allylic substitution over addition when treating propene with bromine?
Allylic substitution occurs in non-polar solvents at high temperatures with very low bromine concentration, producing 3-bromopropene. In contrast, room temperature treatment with bromine in polar solvents favors ionic addition, forming a dibromo product. Using NBS as the reagent achieves selective allylic substitution at room temperature by maintaining minimal bromine concentration throughout.
Q6: What is the role of light or peroxide in allylic bromination with NBS?
Light or peroxide initiates the radical chain reaction by providing energy to cleave the weak N–Br bond in NBS homolytically, generating the initial bromine radical. This radical then abstracts allylic hydrogen from the alkene, starting the propagation cycle. Without photochemical activation or peroxide, NBS cannot efficiently generate the bromine radicals needed to begin the reaction.
Q7: How does keeping bromine concentration low prevent the ionic addition reaction?
At low bromine concentration in non-polar solvents, the ionic addition pathway cannot effectively compete with radical bromination. Radical reactions proceed through chain propagation with minimal bromine present, while ionic addition requires higher Br2 concentrations to form stable bromonium intermediates. By using NBS to maintain low Br2 levels, radical substitution becomes the dominant pathway, selectively forming allyl bromide.