A) \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}CH(Br)C{{H}_{3}}\]
B) \[{{C}_{6}}{{H}_{5}}CH(Br)C{{H}_{2}}C{{H}_{3}}\]
C) \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br\]
D) \[{{C}_{6}}{{H}_{5}}CH(Br)CH=C{{H}_{2}}\]
Correct Answer: B
Solution :
According to Markownikoff's rule, the negative part of the unsymmetrical reagent adds to less hydrogenated (more substituted) carbon atom of the double bond. C6H5CH = CH - CH3 + HBr ® \[\overset{\overset{Br\,\,\,\,\,\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\,}{\mathop{{{C}_{6}}{{H}_{5}}CHC{{H}_{_{2}}}C{{H}_{3}}}}\,\]
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