A) hexane
B) bromopropane
C) propane
D) ethane.
Correct Answer: B
Solution :
\[HBr\] adds on the double bond of propene according to Markownikoffs rule. \[\underset{2-Bromopropane}{\mathop{\underset{\Pr opene}{\mathop{\text{C}{{\text{H}}_{\text{3}}}\text{CH=C}{{\text{H}}_{\text{2}}}\text{HBr}}}\,\to \text{C}{{\text{H}}_{\text{3}}}\text{-}\underset{Br}{\mathop{\underset{|}{\mathop{\text{CH}}}\,}}\,\text{-}\underset{H}{\mathop{\underset{|}{\mathop{\text{C}{{\text{H}}_{\text{2}}}}}\,}}\,\equiv \text{C}{{\text{H}}_{\text{3}}}\underset{Br}{\mathop{\underset{|}{\mathop{\text{CH}}}\,}}\,\text{.C}{{\text{H}}_{\text{3}}}}}\,\]
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