A) 1-bromo-2-methylbutane
B) 1-bromo-3-methylbutane
C) 2-bromo-3-methylbutane
D) 2-bromo-2-methylbutane
Correct Answer: D
Solution :
The order of substitution in different alkanes is 3° > 2° > 1° Thus the bromination of 2-methyl butane mainly gives 2-Bromo - 2 - methyl butane \[C{{H}_{3}}-\underset{2-methyl\,bu\tan e}{\mathop{C{{H}_{2}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{CH}}\,-}}\,C{{H}_{3}}\xrightarrow[{}]{B{{r}_{2}}}\] \[C{{H}_{3}}-\underset{\begin{smallmatrix} 1-Bromo-2methyl\,bu\tan e \\ (\min or) \end{smallmatrix}}{\mathop{C{{H}_{2}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{CH}}\,-}}\,C{{H}_{2}}Br+\underset{\begin{smallmatrix} 2-Bromo-2-methyl\,bu\tan e \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(major) \end{smallmatrix}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{3}}}}\,\]
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