In the conformer I, both the substituents are in axial positions.
Methyl group has two 1, 3 diaxial (H----CH_3) interactions and isopropyl group
has two 1, 3 diaxial (H----pr) interactions. Therefore, the energy of these
interactions is given as (2x0.9) + (2x1.1) = 4.0kCal/mol.
In conformer II both the substituents are equatorial positions.
So, axial interactions are absent and one gauche interaction between isopropyl
and methyl groups exists therefore, the energy of this interaction is given as
1.1kCal/mol. The difference in the energy of two conformations is calculated as
E = (1.0 – 1.1) kCal/mol = 2.9
kCal/mol
The answer to the question is 2.9 kCal/mol
