Which of the following will give a single monochloride on photochemical chlorination ?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2,2-dimethyl propane
In photochemical chlorination, a chlorine free radical removes a hydrogen atom and a chlorine atom takes its place. Any hydrogen atom can be replaced, so the number of possible monochlorides equals the number of different kinds of hydrogen atoms in the alkane.
2,2-dimethylpropane (neopentane), \( \ce{C(CH3)4} \), has four identical methyl groups attached to one central carbon. All 12 hydrogen atoms are equivalent, so only one monochloride is formed:
\( \displaystyle \ce{C(CH3)4 + Cl2} \xrightarrow{\text{light}} \ce{(CH3)3C-CH2Cl + HCl} \)
So 2,2-dimethylpropane gives a single monochloride, 1-chloro-2,2-dimethylpropane. Alkanes with more than one kind of hydrogen give a mixture of isomeric monochlorides.
NCERT (2026-27 edition):
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.”