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) \[C{{H}_{3}}-*\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\mathop{C}}\,}}\,-CH=C{{H}_{2}}\] (viii) \[C{{H}_{3}}-*\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\mathop{C}}\,}}\,-CH=C{{H}_{2}}\] Four structural isomers (ix) \[C{{H}_{3}}C{{H}_{2}}-\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{2}}\] (x) \[C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}\,=CHCl\] (xi) \[C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{2}}Cl \\ | \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{2}}\] (xii) \[ClC{{H}_{2}}-C{{H}_{2}}-CH=C{{H}_{2}}\]You need to login to perform this action.
You will be redirected in
3 sec