Draw the major product formed when HBr reacts with the following epoxide.
The concept used to solve this question is cleavage of epoxide and nucleophilic substitution in the presence of acid catalyst.
Under acidic conditions, the nucleophile preferentially attacks the more substituted ring carbon.
Under basic or neutral conditions, the nucleophile preferentially attacks at the less substituted (or less sterically hindered ring carbon).
Protonation is shown as follows:
The formation of product is as follows:
Ans:The major product is as follows:
The given reaction is a ring opening reaction of epoxides in the presence of HBr and an alkyl bromide as the final product in obtained.
The epoxide is a three-membered ring in which oxygen atom is present. The three-membered ring is highly reactive because three membered rings is highly strained. Thus, in presence of nucleophile its undergo ring opening reaction. When epoxide reacts with hydrogen halide (HBr), the reaction occurs with two steps. In the first step, protonation of an oxygen atom takes place. In the second step, the protonated epoxide reacts with halide ion and final product, that is, halohydrin (organic molecule contains both halogen and OH group) is obtained.
The general reaction mechanism of epoxide with HBr is as follows:
In the second step of mechanism, for unsymmetrical epoxide two possibilities are there for halide ion to attack. the halide ion attacks more substituted carbon atom rather than less substituted carbon atom because protonated epoxide generated partially positive charge on the carbon atom and more substituted carbon stabilized the partial positive charge more easily. Therefore, halide ion attacks more substituted carbon atom in protonated epoxide.
Protonation of oxygen atom present in epoxide.
The reaction is as follows:
Protonated epoxide attacked by the bromide ion from more substituted side.
Ans:Draw the major product formed when HBr reacts with the following epoxide. Draw the major product...
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