The number of monobromo structural isomers formed by free radical bromination of isopentane is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4
In free radical halogenation a halogen atom can replace any hydrogen atom of the alkane. Hydrogens in the same environment (equivalent hydrogens) give the same product, so the number of monobromo structural isomers equals the number of different kinds of hydrogen.
Isopentane is 2-methylbutane:
\( \displaystyle \ce{CH3-CH(CH3)-CH2-CH3} \)
It has four kinds of hydrogen: the six H of the two \( \ce{CH3} \) groups on C2 (equivalent), the one H on C2, the two H on C3, and the three H of the C4 methyl group. Replacing each kind by Br gives:
\( \ce{BrCH2-CH(CH3)-CH2-CH3} \): 1-bromo-2-methylbutane
\( \ce{CH3-CBr(CH3)-CH2-CH3} \): 2-bromo-2-methylbutane
\( \ce{CH3-CH(CH3)-CHBr-CH3} \): 2-bromo-3-methylbutane
\( \ce{CH3-CH(CH3)-CH2-CH2Br} \): 1-bromo-3-methylbutane
So 4 monobromo structural isomers are formed. (Bromination strongly favours the tertiary H, so 2-bromo-2-methylbutane is the major product, but all four are formed.)
NCERT (2026-27 edition):
From Alkanes by Free Radical Halogenation – “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.”