Identify the alkane which on photochemical chlorination yields a single mono chloro product .
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2,2-Dimethylpropane
In photochemical chlorination, a chlorine free radical can remove any hydrogen atom of the alkane. So the number of monochloro products equals the number of different kinds (non-equivalent sets) of hydrogen atoms in the molecule. A single product forms only if every hydrogen atom is equivalent.
2,2-Dimethylpropane, \( \ce{C(CH3)4} \), has four identical \( \ce{CH3} \) groups on one central carbon. All 12 hydrogen atoms are equivalent, so only one product can form:
\( \displaystyle \ce{C(CH3)4 + Cl2} \xrightarrow{\text{UV light}} \ce{(CH3)3C-CH2Cl + HCl} \)
The product is 1-chloro-2,2-dimethylpropane.
Why the other options are incorrect
Option (1) Hexane — Incorrect. \( \ce{CH3CH2CH2CH2CH2CH3} \) has three kinds of hydrogen (on C-1, C-2 and C-3), so it gives three monochloro products.
Option (3) 2-Methylpentane — Incorrect. \( \ce{(CH3)2CH-CH2-CH2-CH3} \) has five kinds of hydrogen, so it gives five monochloro products.
Option (4) Pentane — Incorrect. \( \ce{CH3CH2CH2CH2CH3} \) has three kinds of hydrogen, so it gives three monochloro products.