Question:
An alkyl halide C\(_4\)H\(_8\)Br(X) undergoes hydrolysis preferably in polar protic solvents. X can be prepared from which of the following reactants?
An alkyl halide C\(_4\)H\(_8\)Br(X) undergoes hydrolysis preferably in polar protic solvents. X can be prepared from which of the following reactants?
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In the presence of HBr alone, alkenes undergo Markovnikov addition, while in the presence of peroxides, they undergo anti-Markovnikov addition.
Updated On: Mar 25, 2025
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The Correct Option is A
Solution and Explanation
To determine the correct preparation method for C4H8Br, let's analyze the reactants:
1. First option: The reaction of an alkene with HBr follows Markovnikov’s rule, leading to the formation of the most stable alkyl halide. This is the correct method.
2. Second option: The presence of peroxide (C6H5CO)2O2 leads to anti-Markovnikov addition, forming a different product.
3. Third option: Bromination with Br2/heat leads to the formation of a different type of brominated product (not an alkyl halide via hydrohalogenation).
4. Fourth option: Again, HBr in the presence of peroxides gives anti-Markovnikov addition, which is not the desired product.
Thus, the correct reaction is HBr addition following Markovnikov’s rule, making option (1) the correct choice.
1. First option: The reaction of an alkene with HBr follows Markovnikov’s rule, leading to the formation of the most stable alkyl halide. This is the correct method.
2. Second option: The presence of peroxide (C6H5CO)2O2 leads to anti-Markovnikov addition, forming a different product.
3. Third option: Bromination with Br2/heat leads to the formation of a different type of brominated product (not an alkyl halide via hydrohalogenation).
4. Fourth option: Again, HBr in the presence of peroxides gives anti-Markovnikov addition, which is not the desired product.
Thus, the correct reaction is HBr addition following Markovnikov’s rule, making option (1) the correct choice.
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