A) \[9\]
B) \[10\]
C) \[11\]
D) \[12\]
Correct Answer: D
Solution :
\[{{C}_{4}}{{H}_{7}}Cl\]is ammonochloro derivative of \[{{C}_{4}}{{H}_{10}}\] which it self exists in three ispmeric forms. (i) \[C{{H}_{3}}-C{{H}_{2}}-CH=C{{H}_{2}}\]: Its possible monochloro derivatives are: \[\underset{2\,\text{isomers cis and trands (2 from)}}{\mathop{C{{H}_{3}}-C{{H}_{2}}CH=CH-Cl}}\,\] \[\underset{\text{one isomer (one form)}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-\underset{Cl}{\mathop{\underset{|}{\mathop{C}}\,}}\,=C{{H}_{2}}}}\,\] \[\underset{\text{optically active (exists in two forms)}}{\mathop{C{{H}_{3}}-\underset{Cl}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-CH=C{{H}_{2}}}}\,\] (ii)\[C{{H}_{3}}-CH=CH-C{{H}_{3}}\]: Its possible monochloro derivatives are: \[\underset{(\text{Exists in two geometrical forms)}}{\mathop{C{{H}_{3}}-CH=\underset{Cl}{\mathop{\underset{|}{\mathop{C}}\,}}\,-C{{H}_{3}}}}\,\] \[\underset{(\text{Exists in two geometrical forms)}}{\mathop{C{{H}_{3}}-CH=CH-C{{H}_{2}}Cl}}\,\] (iii) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,=C{{H}_{2}}\]: Its possible mono- chloro derivatives are: \[\underset{Only\,one\,form}{\mathop{C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,=CH-Cl;}}\,\] \[\underset{Only\,one\,form}{\mathop{Cl-C{{H}_{2}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,=C{{H}_{2}}}}\,\] Thus, the total acyclic isomeric forms of \[{{C}_{4}}{{H}_{7}}Cl\]are twelve (12).
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