A) \[{{C}_{2}}{{H}_{4}}+HCl\]
B) \[{{C}_{3}}{{H}_{6}}+B{{r}_{2}}\]
C) \[{{C}_{3}}{{H}_{6}}+HBr\]
D) \[{{C}_{3}}{{H}_{8}}+C{{l}_{2}}\]
E) \[{{C}_{2}}{{H}_{4}}+{{I}_{2}}\]
Correct Answer: C
Solution :
According to Markownikoff s rule, the addition of a unsymmetrical reagent (HX) to an unsymmetric alkene takes place in such a way that the negative part of the reagent will be attached to the carbon atom which containing lesser number of H-atom. Hence, it is best applicable to the reaction between\[{{C}_{3}}{{H}_{6}}\]and\[HBr\]. \[C{{H}_{3}}CH=C{{H}_{2}}+HBr\xrightarrow[{}]{{}}\] \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}\]
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