Question: Draw a graph, similar to Figure 3-9, of the torsional strain of $\mathbf{2}\mathbf{-}\mathbf{\text{methylpropane}}$ as it rotates about the bond between and . Show the dihedral angle and draw a Newman projection for each staggered and eclipsed conformation.

Newman projections are often used in drawing conformations. In Newman projection, one is viewing the $\text{C}-\text{C}$ bond end on along the axis of connection. The front carbon is represented by the intersection of bonds from it, while the rear carbon appears as a circle.

Dihedral angle$\text{(}\theta \text{)}$ gives the identity of any conformation. It may be defined as the angle between $\text{C}-\text{H}$bonds on the front carbon atom and C-H the bonds on the back carbon atom in the Newman projection. Eclipsed conformation has a dihedral angle,$\theta = 0^0$ and the staggered conformation has a dihedral angle, $\theta = {60}^0$

The potential energy of a molecule increases when it rotates toward an eclipsed conformation, and also there is a resistance to the rotation. The resistance to twisting (torsion) is known as torsional strain.

eclipsed staggered eclipsed