Consider the isomers of hydrocarbon with molecular formula C5H10. These isomers do not decolourise KMnO4 solution. These isomers are subjected to chlorination with chlorine in presence of light to give monochloro compounds. The total number of monochloro compounds (structural isomers only) formed is .
Answer
14
Solution
C5H10 (one degree of unsaturation) that does not decolourise KMnO4 ⇒ no C=C ⇒ the isomers are cycloalkanes.
Number of monochloro products = number of different kinds of hydrogen (structural isomers only):
[figure]
Cyclopentane: all 10 H equivalent ⇒ 1
Methylcyclobutane: CH3, ring C-1, C-2 (= C-4), C-3 ⇒ 4
Ethylcyclopropane: CH3, CH2, ring C-1, C-2 (= C-3) ⇒ 4
1,1-Dimethylcyclopropane: CH3 (both same), ring CH2 ⇒ 2
1,2-Dimethylcyclopropane: CH3, ring C-1 (= C-2), ring C-3 ⇒ 3
∴ Total = 1 + 4 + 4 + 2 + 3 = 14
Answer: 14