Question

1. 
   

When 2-chloro-2-methylbutane is treated with a variety of strong bases, the products always seem to contain two isomers (A and B) of formula. When sodium hydroxide is used as base, isomer A predominates. When potassium tert-butoxide is used as the base, isomer B predominates. TheandNMR spectra of A and B are given below.

1. Determine the structures of isomers A and B.
2. Explain why A is the major product when using sodium hydroxide as the base and why B is the major product when using potassium tert-butoxide as the base.

Short answer

a.

b. With sodium hydroxide as the base, the more highly substituted alkene, that is isomer A would be expected to predominate as per Zaitsev rule. With potassium tert-butoxide as the bulky base, the less substituted alkene, that is, isomer B would predominate, that is the Hofmann product.

Step-by-step solution

Explanation of part (a):

When 2-chloro-2-methylbutane is treated with variety of bases such as sodium hydroxide and potassium tert-butoxide, fromation of two products which are isomers of each other occur. As, sodium hydroxide will lead to formation of more substituted alkene which is Zaitsev alkene, 2-methylbut-2-ene (isomer A) and potassium tert-butoxide is bulky base, thus, it will lead to formation of less substituted alkene which is isomer B, that is, 2-methylbut-1-ene.

From the given NMR spectra, we can deduce the following information about splitting pattern and chemical shift values of protons present in isomer A and B which confirms the structures are correct, that is when treated with base, 2-chloro-2-methylbutane gives isomers A and B.

Explanation of part (b):

With sodium hydroxide as the base, the more highly substituted alkene, that is isomer A would be expected to predominate as per Zaitsev rule. With potassium tert-butoxide as the bulky base, the less substituted alkene, that is, isomer B would predominate, that is the Hofmann product.