A hydrocarbon X, C6H12, does not decolourise a solution of bromine in CCl4 in the dark. In bright sunlight, X reacts with bromine to give a single monobromo compound, C6H11Br. What is X?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Cyclohexane
Two clues decide it.
Bromine in CCl4 adds across a C=C bond and loses its reddish brown colour, so an alkene would react in the dark. X does not, so it has no double bond; with the formula C6H12, two hydrogen atoms fewer than hexane, it must be a ring compound, a cycloalkane.
In sunlight a free radical bromination replaces one H by Br, and a single monobromo product means that all the hydrogen atoms of X are equivalent.
In cyclohexane all six CH2 groups are alike, so replacing any one of its twelve hydrogen atoms gives the same product, bromocyclohexane:
\( \displaystyle \ce{C6H12 + Br2} \xrightarrow{h\nu} \ce{C6H11Br + HBr} \)
Why the other options are incorrect
Option (1) Hex-1-ene — Incorrect. It has a C=C bond, so it adds bromine even in the dark and decolourises the solution, giving the vic-dibromide 1,2-dibromohexane:
\( \displaystyle \ce{CH2=CHCH2CH2CH2CH3 + Br2} \xrightarrow{\ce{CCl4}} \ce{BrCH2CHBrCH2CH2CH2CH3} \)
Option (2) Methylcyclopentane — Incorrect.
It has no double bond, but its hydrogen atoms are not all equivalent.
Those of the CH3 group, the ring CH, the two ring CH2 groups next to it and the two far ring CH2 groups are four different kinds, so it gives four monobromo isomers.
Option (3) 2,3-Dimethylbut-2-ene — Incorrect. Its twelve hydrogen atoms are all alike (four CH3 groups), but it is an alkene, so it adds bromine in the dark, giving 2,3-dibromo-2,3-dimethylbutane:
\( \displaystyle \ce{(CH3)2C=C(CH3)2 + Br2} \xrightarrow{\ce{CCl4}} \ce{(CH3)2CBr-CBr(CH3)2} \)
NCERT (2026-27 edition):
6.4.2 From Hydrocarbons – “(ii) Addition of halogens: In the laboratory, addition of bromine in CCl4 to an alkene resulting in discharge of reddish brown colour of bromine constitutes an important method for the detection of double bond in a molecule.”
Answers to Some Intext Questions – “6.4 All the hydrogen atoms are equivalent and replacement of any hydrogen will give the same product.”