Identify the enantiomer of the molecule given below:

The correct answer is Option (1) → Structure (1)

Enantiomers are stereoisomers that are non-superimposable mirror images of each other. The molecule drawn is 2-bromobutane, \( \ce{CH3-CH(Br)-C2H5} \). Its C-2 carries four different groups (H, Br, \( \ce{CH3} \) and \( \ce{C2H5} \)), so it is a chiral centre and the compound exists as two enantiomers, (R) and (S), which differ only in the direction in which they rotate plane polarised light.

To find the configuration, rank the groups by atomic number: Br > \( \ce{C2H5} \) > \( \ce{CH3} \) > H. In the given drawing H is in the plane (up), \( \ce{C2H5} \) in the plane (left), \( \ce{CH3} \) on the dashed bond (behind) and Br on the wedge (in front). Interchanging H and \( \ce{CH3} \) puts H behind the page; then Br (bottom) → \( \ce{C2H5} \) (left) → \( \ce{CH3} \) (top) runs clockwise, which is R. One interchange inverts the configuration, so the given molecule is (S)-2-bromobutane, and its enantiomer is (R)-2-bromobutane.

A quicker check is the interchange rule: a drawing that differs from the given one by an odd number of interchanges of two groups is the enantiomer; an even number gives the same molecule. All the structures use the same four positions (top, left, dashed, wedge).

• Option (1) Top \( \ce{C2H5} \), left Br, dashed H, wedge \( \ce{CH3} \). With H behind the page, Br (left) → \( \ce{C2H5} \) (top) → \( \ce{CH3} \) (bottom right) runs clockwise: (R). Every group has moved one place round (H top to dashed, \( \ce{CH3} \) dashed to wedge, Br wedge to left, \( \ce{C2H5} \) left to top), which equals three interchanges, an odd number. So this is the enantiomer.

• Option (2) Top \( \ce{CH3} \), left Br, dashed H, wedge \( \ce{C2H5} \). Compared with the given drawing, H and \( \ce{CH3} \) are interchanged, and so are \( \ce{C2H5} \) and Br: two interchanges, so it is the same molecule, (S).

• Option (3) Top \( \ce{C2H5} \), left H, dashed Br, wedge \( \ce{CH3} \). H and \( \ce{C2H5} \) are interchanged, and so are \( \ce{CH3} \) and Br: two interchanges, the same molecule, (S).

• Option (4) This is the given molecule itself, drawn the same way, (S).

So the only enantiomer among the options is the structure in Option (1).

NCERT (2026-27 edition):
Molecular Asymmetry, Chirality and Enantiomers – “… The stereoisomers related to each other as non-superimposable mirror images are called enantiomers … Enantiomers possess identical physical properties namely, melting point, boiling point, refractive index, etc. They only differ with respect to the rotation of plane polarised light. If one of the enantiomer is dextro rotatory, the other will be laevo rotatory.”