Homolysis of Bu3SnH is promoted by the initiator AIBN
المؤلف:
Jonathan Clayden , Nick Greeves , Stuart Warren
المصدر:
ORGANIC CHEMISTRY
الجزء والصفحة:
ص991-992
2025-07-29
485
Homolysis of Bu3SnH is promoted by the initiator AIBN
As you would imagine, the weakest C–Hal bonds are the easiest to cleave, so alkyl bromides are reduced more rapidly than alkyl chlorides, and alkyl fluorides are unreactive. With alkyl iodides and bromides, daylight can be sufficient to initiate the reaction, but with alkyl chlorides, and often with alkyl bromides as well, it is generally necessary to produce a higher concentration of Bu3Sn• radicals by adding an initiator to the reaction. The best choice is usually AIBN, which you met earlier in the chapter. This compound undergoes thermal homolysis above 60 °C to give nitrile-stabilized radicals that abstract the hydrogen atom from Bu3SnH.
Why use AIBN as an initiator; why not a peroxide? Since we want to cleave only a weak Sn–H bond, we can get away with using a relatively unreactive nitrile-stabilized radical. Peroxides, on the other hand, generate RO• radicals. These are highly reactive and will abstract hydrogen from almost any organic molecule, not just the weakly bonded hydrogen atom of Bu3SnH, and this would lead to side reactions and lack of selectivity. AIBN is needed only in sufficient quantities to be an initiator of the reaction; it is the Bu3SnH that provides the hydro gen atoms that end up in the product, so usually you need only 0.02 to 0.05 equivalents of AIBN and a slight excess (1.2 equivalents) of Bu3SnH.
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