Identify the Hydrocarbon C5H10

A hydrocarbon C5H10 does not react with Chlorine in dark but it gives a single monochlorocompound C5H9Cl in bright sunlight. Identify the hydrocarbon.

Prerequisites- Double bond unsaturation (DOU), Types of Hydrogen, Addition reactions of alkene, free radical halogenation of alkanes
 
The first step would be to find the unsaturation in the molecule from the given molecular formula, C5H10. The degree of unsaturation can be found using the formula-
(2C + 2 + N - H - X)/2 = (2 x 5 + 2 - 10)/2 = (12-10)/2 = 1 
 
where, 
C is number of Carbon
N is number of Nitrogen
H is number of Hydrogen
X is number of Halogen
 
As the degree of unsaturation is one, it is not a saturated alkane but carries unsaturation either in the form of a double bond or as a cycloalkane ring.
 
But as the compound does not react with Cl2 in dark, it signifies that it is not a double bonded alkene. An alkene would undergo Cl2 addition reaction to give vicinal dihalides even in dark.
 
addition of halogen to alkene
 
The next assumption is it is a cycloalkane ring, a five carbon cyclopentane ring.
type of hydrogen cyclopentane
 
As it has only one type of hydrogen, it would give a single monochloro product on free radical halogenation in the presence of sunlight (UV light). 
halogenation of cyclopentane in UV light
 
 
The hydrocarbon therefore is cyclopentane and the product is 1-chlorocyclopentane. 
 
 

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