Select the expected major product(s)from the treatment of 1-pentyne with 1 equivalent of HBr in the presence of peroxides.

Question

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When 1-pentyne is treated with 1 equivalent of HBr in the presence of peroxides, the reaction undergoes anti-Markovnikov addition of HBr to form a mixture of both (E)-1-bromo-1-pentene and (Z)-1-bromo-1-pentene. This is because the peroxides generate free radicals, which attack the alkyne in a syn-addition manner, leading to the formation of a radical intermediate. The hydrogen atom from HBr adds to the less substituted carbon, and the bromine atom adds to the more substituted carbon of the intermediate. The resulting alkyl radical can then add to another molecule of HBr to form the final product. Because the addition of HBr is syn, the two bromine atoms will be trans to each other, resulting in a mixture of (E)- and (Z)-isomers.

Select the Expected Major Product(s)from the Treatment of 1-Pentyne with 1