A) 1-bromopropane
B) 2-bromopropane
C) 1, 2-dibromopropane
D) 2, 2-dibromopropane
Correct Answer: A
Solution :
In the presence of peroxide, anri-Markownikoffs rule is followed, ie, the negative part of reagent goes to that carbon atom which carry more number of H atoms. \[C{{H}_{3}}CH=C{{H}_{2}}+HBr\xrightarrow{\text{Benzoyl}\,\text{peroxide}}\] \[\underset{1-\text{bromopropane}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br}}\,\]
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