Organic Chemistry
Calculate percentage of mono chlorinated product obtained by 2-methyl butane
Calculate percentage of mono chlorinated product obtained by 2-methyl butane
1 Answer
Arun
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%
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%
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