Answer to Question #94053 in Organic Chemistry for Victor Idem

Question

Consider compound A= 2 Chloro 2 Methyl butane, B=2 Chloro butane, C=Chloro propane
1) Which of the compound will readily undergo dehydrochlorination
2) what inform your answer

Answer 1

The mechanism of dehydrochlorination proceeds through the stage of formation of the intermediate product with a partly formed double bond and partly broken C-Cl and H-C bonds (figure below).

The more alkyl groups located around future double bond, the lower energy required to form transition state and the formation of such alkene becomes more favorable. In case A there are 3 methyl groups located around future double bond. In case B - there are 2, and in case C only one methyl group stabilizes the transition state. So 2-chloro-2-methylbutane will be most active in the dehydrochlorination reaction as it has the most stable transition state and most thermodynamically stable final structure.