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What is the major product on chlorination of 2-methylbutane?
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Malcolm Rahuaha
What is the major product on chlorination of 2-methylbutane?
Because of the great reactivity of the chlorine radical, abstractions of primary, secondary, and tertiary hydrogen atoms are all exothermic. Therefore, the stability of the product radical has less influence on the activation energy of the reaction. Thus, according to the Hammond postulate, the transition state is more reactant-like.
According to R. Brückner Advanced Organic Chemistry, the chlorination of isopentane has four monochlorination products:
Because of the great reactivity of the chlorine radical, abstractions of primary, secondary, and tertiary hydrogen atoms are all exothermic. Therefore, the stability of the product radical has less influence on the activation energy of the reaction. Thus, according to the Hammond postulate, the transition state is more reactant-like.
According to R. Brückner Advanced Organic Chemistry, the chlorination of isopentane has four monochlorination products:
Yes but selectivity factor + probability factor together decide the major and the minor product. The question you have asked shouldnt have been a high school organic chemistry question unless they mention relative reactivity of the hydrogen atoms.More
Thanks a lot for the help. In my book (high school standard though) it is written that the chlorination favours in the following order Teritiary hydrogen>secondary hydrogen>primary hydrogen.More
Because of the great reactivity of the chlorine radical, abstractions of primary, secondary, and tertiary hydrogen atoms are all exothermic. Therefore, the stability of the product radical has less influence on the activation energy of the reaction. Thus, according to the Hammond postulate, the transition state is more reactant-like.
According to R. Brückner Advanced Organic Chemistry, the chlorination of isopentane has four monochlorination products:
22 % 2-chloro-2-methylbutane
33 % 2-chloro-3-methylbutane
30 % 1-chloro-2-methylbutane
15 % 1-chloro-3-methylbutane
The statistical factor is responsible for this.
If each hydrogen atom of isopentane could be substituted at the same rate, the ratio would be
1 hydrogen atom (8 %) → 2-chloro-2-methylbutane
2 hydrogen atoms (17 %) → 2-chloro-3-methylbutane
6 hydrogen atoms (50 %) → 1-chloro-2-methylbutane
3 hydrogen atoms (25 %) → 1-chloro-3-methylbutane
In terms per hydrogen atom, the experimental yield is
22 % / 1 = 22 % → 2-chloro-2-methylbutane
33 % / 2 = 16.5 % → 2-chloro-3-methylbutane
30 % / 6 = 5 % → 1-chloro-2-methylbutane
15 % / 3 = 5 % → 1-chloro-3-methylbutane
Thus, 2-chloro-2-methylbutane is indeed slightly preferred – as you have expected.
Because of the great reactivity of the chlorine radical, abstractions of primary, secondary, and tertiary hydrogen atoms are all exothermic. Therefore, the stability of the product radical has less influence on the activation energy of the reaction. Thus, according to the Hammond postulate, the transition state is more reactant-like.
According to R. Brückner Advanced Organic Chemistry, the chlorination of isopentane has four monochlorination products:
22 % 2-chloro-2-methylbutane
33 % 2-chloro-3-methylbutane
30 % 1-chloro-2-methylbutane
15 % 1-chloro-3-methylbutane
The statistical factor is responsible for this.
If each hydrogen atom of isopentane could be substituted at the same rate, the ratio would be
1 hydrogen atom (8 %) → 2-chloro-2-methylbutane
2 hydrogen atoms (17 %) → 2-chloro-3-methylbutane
6 hydrogen atoms (50 %) → 1-chloro-2-methylbutane
3 hydrogen atoms (25 %) → 1-chloro-3-methylbutane
In terms per hydrogen atom, the experimental yield is
22 % / 1 = 22 % → 2-chloro-2-methylbutane
33 % / 2 = 16.5 % → 2-chloro-3-methylbutane
30 % / 6 = 5 % → 1-chloro-2-methylbutane
15 % / 3 = 5 % → 1-chloro-3-methylbutane
Thus, 2-chloro-2-methylbutane is indeed slightly preferred – as you have expected.
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