A) 2-bromo-2-methylbutane
B) 1-bromo-2-methylbutane
C) 1-bromo-3-methylbutane
D) 2-bromo-3-methylbutane
Correct Answer: A
Solution :
\[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}-C{{H}_{3}}\xrightarrow[B{{r}_{2}}]{h\upsilon }\] \[\underset{\begin{smallmatrix} 2-bromo-2-methyl\,bu\tan e \\ (major\,product) \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}-C{{H}_{3}}}}\,\] The reactivity of different types of hydrogen follows the order : benzylic \[\approx \]allylic > tertiary > secondary > primary > vinylic \[\approx \] aryl.
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