Find the enantiomers of the structures in Column-$(I)$ from Column-$(II)$.
Column-$(I)$ Column-$(II)$ $(A)$ $C_2H_5-C(CH_3)(OH)-H$ $(p)$ $Cl-C(Br)(I)-H$ $(B)$ $Br-C(Cl)(I)-H$ $(q)$ $OH-C(CH_3)(C_2H_5)-H$ $(r)$ $H_5C_2-C(OH)(CH_3)-H$ $(s)$ $Cl-C(I)(Br)-H$
| Column-$(I)$ | Column-$(II)$ |
|---|---|
| $(A)$ $C_2H_5-C(CH_3)(OH)-H$ | $(p)$ $Cl-C(Br)(I)-H$ |
| $(B)$ $Br-C(Cl)(I)-H$ | $(q)$ $OH-C(CH_3)(C_2H_5)-H$ |
| $(r)$ $H_5C_2-C(OH)(CH_3)-H$ | |
| $(s)$ $Cl-C(I)(Br)-H$ |
(A-Q, B-S) Enantiomers are non-superimposable mirror images of each other.
For structure $(A)$,the chiral center is bonded to $-CH_3$,$-C_2H_5$,$-OH$,and $-H$. Its mirror image is $(q)$.
For structure $(B)$,the chiral center is bonded to $-Br$,$-Cl$,$-I$,and $-H$. Its mirror image is $(s)$.
Therefore,the correct matches are $(A)-(q)$ and $(B)-(s)$.
For structure $(A)$,the chiral center is bonded to $-CH_3$,$-C_2H_5$,$-OH$,and $-H$. Its mirror image is $(q)$.
For structure $(B)$,the chiral center is bonded to $-Br$,$-Cl$,$-I$,and $-H$. Its mirror image is $(s)$.
Therefore,the correct matches are $(A)-(q)$ and $(B)-(s)$.
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