Answer
Starting from a point connected to the back carbon (behind the circle), the clockwise or counterclockwise order for the most stable should be:
(Me)-(Me)-(H)-(Me)-(H)-(H).
And for the least stable:
(Me)-(H)-(H)-(Me)-(H)-(Me)
Work Step by Step
1. For both most and less stable conformations, draw the atoms/groups connected to the back carbon. (Represented by the circle.)
In this case:*
Up: (Me)
Right:(H)
Left:(H)
2. Identify the largest group connected to the front carbon, and for the back carbons:
In this case:
Front carbon: Methyl
Back carbon: Methyl
3. Now, for the most stable one, we will rotate only the groups/atoms on the front carbon (The point on the center.), trying to put the largest groups we identified as distant as possible.
In this case, the front carbon will have:*
Down: (Me)
Right: (Me)
Left: (H)
4. Now, for the most stable one, we will rotate only the groups/atoms on the front carbon (The point on the center.), trying to put the largest groups we identified as close as possible.
In this case, the front carbon will have:*
Up: (Me)
Right: (H)
Left: (Me)
* This is just an example, but if your drawing isn't equal, it should be at least a rotation of this drawing.
