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Grade:
        
Calculate percentage of mono chlorinated product obtained by 2-methyl butane
7 months ago

Answers : (1)

Arun
20915 Points
							
The distribution of monochloro products is determined by the number of hydrogens of each type: 

Product 1: The three hydrogens on C-4 are equivalent, so substitution of one gives one of the products. Because there are three of these hydrogens, this product. 1-chloro-3-methylbutane, counts (statistically) as "3" 

.........CH3 
.........| 
H3C-CH-CH2-CH2Cl 
....4..3....2......1 

Product 2: The two hydrogens on C-3 are equivalent, so substitution of one gives one of the products. Because there are two of these hydrogens, this product, 2-chloro-3-methylbutane, counts (statistically) as "2" 

.........CH3 
.........| 
H3C-CH-CHCl-CH3 
....4..3....2.......1 

Product 3: The one hydrogen on C-2 gives one substitution product. Because there is one of these hydrogens, this product, 2-chloro-2-methylbutane, counts (statistically) as "1" 

.........CH3 
.........| 
H3C-CHCl-CH2-CH3 
....1..2.......3......4 

Product 4: The three hydrogens on C-1 and the three hydrogens on the 2-methyl group are equivalent ((CH3)2CH- is an isopropyl group), so substitution of one gives one of the products. Because there are six of these hydrogens, this product, 1-chloro-2-methylbutane, counts (statistically) as "6" 

...........CH3 
............| 
ClCH2-CH-CH2-CH3 
....1.....2....3......4 


So here is the realtive distributions of products: 

Product 1.....1-chloro-3-methylbutane....3 
Product 2.....2-chloro-3-methylbutane....2 
Product 3.....2-chloro-2-methylbutane....1 
Product 4.....1-chloro-2-methylbutane....6 

These can be expressed as percentages (the sum of all = 12) 

Product 1.....1-chloro-3-methylbutane....3/12 = 25% 
Product 2.....2-chloro-3-methylbutane....2/12 = 16.7% 
Product 3.....2-chloro-2-methylbutane....1/12 = 8.3% 
Product 4.....1-chloro-2-methylbutane....6/12 = 50%
7 months ago
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