A) 12
B) 11
C) 10
D) 9
Correct Answer: A
Solution :
Twelve in all Six geometrical isomers (i) (ii) (iii) (iv) (v) (vi) Two optical isomers (vii) \[C{{H}_{3}}-\underset{H}{\overset{Cl}{\mathop{\underset{|}{\overset{*|}{\mathop{C}}}\,}}}\,-CH=C{{H}_{2}}\] (viii) \[C{{H}_{3}}-\underset{Cl}{\overset{H}{\mathop{\underset{|}{\overset{*|}{\mathop{C}}}\,}}}\,-CH=C{{H}_{2}}\] Four structural isomers (ix) \[C{{H}_{2}}C{{H}_{2}}-\overset{Cl}{\mathop{\overset{|}{\mathop{C}}\,}}\,=C{{H}_{2}}\] (x) \[C{{H}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{C}}\,}}\,=CHCl\] (xi) \[C{{H}_{3}}-\overset{C{{H}_{2}}Cl}{\mathop{\overset{|}{\mathop{C}}\,}}\,=C{{H}_{2}}\] (xii) \[ClC{{H}_{2}}-C{{H}_{2}}-CH=C{{H}_{2}}\]You need to login to perform this action.
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