A) \[C{{H}_{4}}\]
B) \[C{{H}_{3}}I\]
C) \[C{{H}_{2}}C{{l}_{2}}\]
D) \[CHIBrCl\]
Correct Answer: D
Solution :
Key Idea: optically active compound has asymmetric or chiral carbon atom that is a carbon atom attached to four different atoms. \[H-\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}}\,-H\] \[H-\underset{\begin{smallmatrix} | \\ I \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}}\,-H\] \[\because \]all four atoms \[\because \]3 atoms attached attached to to carbon are carbon are same same \[\therefore \]No chiral carbon \[\therefore \]No chiral carbon present present \[Cl-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}}\,-H\] \[Br-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}}\,-I\] \[\because \]2 atoms are identical \[\because \]all 4 atoms attached to carbon are different \[\therefore \]No chiral carbon \[\therefore \]chiral carbon present present
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