A) 5
B) 7
C) 4
D) 6
Correct Answer: B
Solution :
\[{{C}_{5}}{{H}_{11}}Br\]has seven structural isomers. (I) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}-Br\] (II) \[C{{H}_{3}}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,C{{H}_{2}}C{{H}_{2}}Br\] (III) \[C{{H}_{3}}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{\overset{\begin{smallmatrix} Br \\ | \end{smallmatrix}}{\mathop{CH}}\,}}\,C{{H}_{2}}C{{H}_{3}}\] (IV) \[C{{H}_{3}}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}\] (V) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}C{{H}_{3}}\] (VI) \[C{{H}_{3}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}C{{H}_{3}}\] (VII) \[C{{H}_{3}}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}\,}}\,-C{{H}_{2}}Br\]
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