On free radical monochlorination of 2-methylbutane, the total number…

The correct answer is Option (2) → 4

• In free radical monochlorination, Cl replaces one H. Each kind of equivalent hydrogen gives one product.

• A product is chiral when it has a carbon joined to four different groups. A chiral product exists as two enantiomers (non-superimposable mirror images), and both are formed.

2-Methylbutane (main chain numbered C1 to C4 from left to right):

    \( \displaystyle \ce{CH3-CH(CH3)-CH2-CH3} \)

C1 and the methyl group on C2 are the same, so there are 4 kinds of hydrogen: on C1, C2, C3 and C4. Putting Cl on each in turn:

• Cl on C1: 1-chloro-2-methylbutane, \( \ce{ClCH2-CH(CH3)-CH2CH3} \) — C2 has H, \( \ce{CH3} \), \( \ce{CH2Cl} \), \( \ce{C2H5} \). Chiral: 2 enantiomers.

• Cl on C2: 2-chloro-2-methylbutane, \( \ce{(CH3)2C(Cl)-CH2CH3} \) — C2 has two \( \ce{CH3} \) groups. Not chiral: 0.

• Cl on C3: 2-chloro-3-methylbutane, \( \ce{(CH3)2CH-CH(Cl)-CH3} \) — this carbon has H, Cl, \( \ce{CH3} \), isopropyl. Chiral: 2 enantiomers.

• Cl on C4: 1-chloro-3-methylbutane, \( \ce{(CH3)2CH-CH2-CH2Cl} \) — no carbon has four different groups. Not chiral: 0.

Total: 2 chiral structures × 2 enantiomers = 4 chiral products.

NCERT (2026-27 edition):
6.4.2 From Hydrocarbons – “Free radical chlorination or bromination of alkanes gives a complex mixture of isomeric mono- and polyhaloalkanes, which is difficult to separate as pure compounds. Consequently, the yield of any single compound is low.”

(ii) Molecular asymmetry, chirality and enantiomers – “… Dutch scientist, J. Van’t Hoff and French scientist, C. Le Bel in the same year (1874), independently argued that the spatial arrangement of four groups (valencies) around a central carbon is tetrahedral and if all the substituents attached to that carbon are different, the mirror image of the molecule is not superimposed (overlapped) on the molecule; such a carbon is called asymmetric carbon or stereocentre.”