Organic Chemistry: Examine different types of strain present & identify more stable (lower energy) compound?

Question

For each pair of compounds examine the different types of strain present, then identify the more stable (lower energy) isomer or compound and explain your reasoning.

2.a) cyclopentane b) methylcyclobutane

Answer 1

a typical carbon-carbon bond would like to be 109 degrees. Cyclopentane is 108 degrees which is near ideal. You might say that it has a tiny bit of angle strain but in reality cyclopentane bends itself to relieve the strain. Thus lower in energy.

Cyclobutane on the other hand is only 90 degrees. It is very strained thus high in potential energy.

Answer 2

Look at the strain presented in the ring - draw them out and look at the angles present in the species - cyclobutane has four carbons in the ring, giving a 'square' angles of 90o. This is strained - compared to the cyclopentane, which has angles closer to the desired tetrahedral angle of 109.5.
This is in a textbook if you looked hard enough!

Answer 3

there are lots of types of strain and you need to draw the molecules to see them H and all. remember ring strain, and steric hindrance are not the answers to everything!