A) \[C{{H}_{3}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\mathop{C}}\,HC{{H}_{3}}\]
B) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Cl\]
C) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CHC{{l}_{2}}\]
D) None of the above
Correct Answer: A
Solution :
Key Idea: Enantiomers are pair of optical isomers which are related as non-superimposable mirror images of each other. The compounds which have at least one asymmetric or chiral carbon atoms form enantiomers (or show optical activity.) Draw structures of all the given choices to find which of them has asymmetric carbon atom and hence, forms enantiomers. [a] \[C{{H}_{3}}C{{H}_{2}}-\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\mathop{C}}\,}}\,*-C{{H}_{3}}\] \[\because \] It has one asymmetric carbon (marked with *) atom. \[\therefore \] It will have enantiomers. [b] \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Cl\] [c] \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CHC{{l}_{2}}\] Choice [b] and [c] does not have asymmetric carbonation. So, they do not have enantiomers.
You need to login to perform this action.
You will be redirected in
3 sec
