Multiple larger substituents that are close to each other on a single molecule increase the amount of steric strain. This is easiest to see in a Newman projection.
For simplicity, let's say we drew Newman projections of an arbitrary two-carbon alkane with various substituents.
The highlighted #"C"-"C"# bond will be hidden upon rotation as we rotate the structure to view it in such a way that the left carbon in the main chain is in the front.
We get the following conformations:
STAGGERED vs. ECLIPSED
What we have on the left (I, III, V) are the staggered conformations, and on the right (II, IV, VI) are the eclipsed conformations. That simply illustrates the way the highlighted/hidden #"C"-"C"# bond freely rotates.
The structures shown are #60^@# rotations of each other. To determine the steric strain in each of these, we would look at the size of the substituent relative to the sizes of the other substituents nearby.
As general hierarchical rules:
- Staggered positions are usually more stable than eclipsed positions due to less spatial crowding (less torsional strain) from nearby substituents as well as their nonbonding orbitals.
- It is more stable to have smaller substituents near smaller substituents, and less stable for larger ones to be near other large ones, again based on spatial crowding.
A plot of energy vs. degrees of rotation is sinusoidal, and highest energy = least stable.
For the staggered conformations: #E_"I" < E_"III" < E_"V"#
The one with the lowest energy is I, which has the staggered position, with the rear isopropyl group between a front methyl and front hydride, yet the rear methyl group is between a front hydride and another front methyl.
Reason: It is better to be between a methyl and a hydride than two methyls to reduce the spatial crowding.
In contrasting with I, the second lowest in energy is III.
Reason: The rear isopropyl is between a front methyl and front hydride, but the rear methyl is instead between two front methyls.
In contrasting with I and III, the third lowest in energy is V.
Reason: The rear isopropyl is instead between two front methyls, while the rear methyl is between a front methyl and a front hydride.
For the eclipsed conformations: #E_"II" < E_"IV" < E_"VI"#
We would approach the eclipsed conformations using very similar reasoning as before:
- The isopropyl/methyl eclipses are higher in energy than isopropyl/hydride eclipses.
- Having two methyl/hydride eclipses is slightly higher in energy than having one hydride/hydride and one methyl/methyl eclipse.
Overall, the energy ordering becomes roughly:
#color(blue)(E_"I" < E_"III" < E_"V" < E_"II" < E_"IV" < E_"VI")#