The hydrocarbon with molecular formula C5H10 has a degree of unsaturation of 1. Since its isomers do not decolourise KMnO4 solution, they do not contain any carbon-carbon double bonds. Thus, they must be cycloalkanes.
The possible structural isomers of cyclic C5H10 and the number of their monochloro structural isomers are as follows:
- Cyclopentane:
All carbon atoms are equivalent. Chlorination gives 1 monochloro product (chlorocyclopentane).
- Methylcyclobutane:
There are 4 different types of hydrogen atoms (on the methyl group, on C1, on C2/C4, and on C3). Chlorination gives 4 monochloro products.
- Ethylcyclopropane:
There are 4 different types of hydrogen atoms (on the CH3 of ethyl, on the CH2 of ethyl, on C1 of the ring, and on C2/C3 of the ring). Chlorination gives 4 monochloro products.
- 1,1-Dimethylcyclopropane:
There are 2 different types of hydrogen atoms (on the two equivalent methyl groups, and on the two equivalent CH2 groups of the ring). Chlorination gives 2 monochloro products.
- 1,2-Dimethylcyclopropane:
There are 3 different types of hydrogen atoms (on the two equivalent methyl groups, on the two equivalent CH groups of the ring, and on the CH2 group of the ring). Chlorination gives 3 monochloro products.
Since each of these cycloalkanes has a different carbon skeleton, their monochloro derivatives will all be distinct structural isomers.
Total number of monochloro compounds (structural isomers only) = 1+4+4+2+3=14.
Answer: 14