# Steric Strain

Organic Chemistry | Stereoisomerism Energy.

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• Steric strain is the increase in potential energy of a molecule that is caused by van der Waals repulsions and cannot be reduced by rotation around a single bond.

We often find steric strain in alkenes and ring systems.

Alkenes

van der Waals repulsions between the eclipsed chlorine atoms cannot be reduced by rotation around the rigid carbon-carbon bond.

This is why cis alkenes are less stable than trans alkenes.

Ring systems

Groups in the axial positions of cyclohexane experience van der Waals repulsions from the other axial groups on the same side of the ring.

No such problems exist for equatorial substituents

This is why equatorial positions are more stable than axial ones.

• Steric strain is most evident in cyclic molecules because the ring structure prevents the groups from getting away from each other.

Steric strain is the increase in potential energy of a molecule due to repulsion between groups on non-neighbouring carbons.

Consider a non-cyclic molecule such as 1,3-dichloropropane. We could draw it as

Here, the electron clouds of the Cl atoms are close together, and they repel each other. But there is free rotation about the C-C σ bonds. The CH₂Cl groups can rotate to get the Cl atoms further apart.

This rotation is severely hindered when the C atoms are in a ring.

For example, a cyclohexane ring is most stable in a chair conformation.

If you had a cyclohexane with two groups in axial positions, you would get steric strain because these groups would be too close and they would repel each other.

The molecule below has steric strain from many 1,3-diaxial interactions.

Most of these can be relieved by flipping the cyclohexane chair. But then the Br would be axial and have its own 1,3-diaxial steric repulsions.

• 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 $\text{C"-"C}$ 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.

NEWMAN PROJECTIONS

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 $\text{C"-"C}$ bond freely rotates.

The structures shown are ${60}^{\circ}$ 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.

DETERMINING STABILITIES

As general hierarchical rules:

1. 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.
2. 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.

GENERAL PROCESS/RATIONALE

For the staggered conformations: ${E}_{\text{I" < E_"III" < E_"V}}$

• 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: ${E}_{\text{II" < E_"IV" < E_"VI}}$

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:

$\textcolor{b l u e}{{E}_{\text{I" < E_"III" < E_"V" < E_"II" < E_"IV" < E_"VI}}}$

## Questions

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