A) 11
B) 12
C) 9
D) 10
Correct Answer: B
Solution :
\[{{C}_{4}}{{H}_{7}}Cl\] is a monochloro derivative of \[{{C}_{4}}{{H}_{8}}\] which itself exists in three isomeric forms. (i) \[C{{H}_{3}}-C{{H}_{2}}-CH=C{{H}_{2}}\] : Its possible mono-chloro derivatives are : \[C{{H}_{3}}-C{{H}_{2}}-CH=CH-Cl\] 2 isomers : cis and trans forms \[C{{H}_{3}}-\underset{Cl\ }{\mathop{\underset{|}{\mathop{\overset{\bullet }{\mathop{C}}\,}}\,H}}\,-CH=C{{H}_{2}}\] optically active (exists in two forms) \[ClC{{H}_{2}}-C{{H}_{2}}-CH=C{{H}_{2}}\] (one form) \[{{H}_{3}}C-C{{H}_{2}}-\overset{Cl\ }{\mathop{\overset{|}{\mathop{C}}\,=}}\,C{{H}_{2}}\] (one form) (ii) \[C{{H}_{3}}-CH=CH-C{{H}_{3}}\] : Its possible monochloro derivatives are : \[C{{H}_{3}}-CH=\underset{Cl\ }{\mathop{\underset{|}{\mathop{C}}\,-}}\,C{{H}_{3}}\] Exists in two geometrical forms \[C{{H}_{3}}-CH=CH-C{{H}_{2}}Cl\] Exists in two geometrical forms (iii) \[C{{H}_{3}}-\underset{C{{H}_{3}}\ \ \ \ \ \,}{\mathop{\underset{|}{\mathop{C}}\,=C{{H}_{2}}}}\,\] : Its possible monochloro derivatives are \[C{{H}_{3}}-\underset{C{{H}_{3}}\ \ \ \ }{\mathop{\underset{|}{\mathop{C}}\,=CH}}\,-Cl\] Only one form \[ClC{{H}_{2}}-\underset{C{{H}_{3}}\ \ \ \ \ \,}{\mathop{\underset{|}{\mathop{C}}\,=C{{H}_{2}}}}\,\] Only one form Thus, the total acylic isomers forms of \[{{C}_{4}}{{H}_{7}}Cl\] are 12.
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