question_answer

What will be the position number of isomers in \[{{C}_{2}}{{H}_{4}}C{{l}_{2}}\]?

A)  \[3\]                                    

B)                         \[2\]    

C)                         \[1\]                                    

D)                         zero

Correct Answer: B

Solution :

The position isomer is 2 of \[{{C}_{2}}{{H}_{4}}C{{l}_{2}}\] \[\underset{\begin{smallmatrix}  1-2\,\,dichloro\,ethane \\  ethylene\,chloride \end{smallmatrix}}{\mathop{\underset{C{{H}_{2}}-Cl}{\overset{C{{H}_{2}}-Cl}{\mathop{|}}}\,}}\,\]