organic chemistry - Heats of combustion and stability of rings

Which isomer would have the largest heat of combustion?

A) Propylcyclopropane
B) Ethylcyclobutane
C) Methylcyclopentane
D) Cyclohexane

E) Since they are all isomers, all would have the same heat of combustion.

Can anyone explain why?

If you think about it, without having the data of the heat of combustion of each type of bond, you can't really answer this question. I know that cyclopropane has a higher heat of combustion per bond than cyclobutane. However, that, in my sense of viewing the problem, does not prevent the fact ethylcyclobutane could have a higher heat of combustion as it is a cycloalkane with 4 bonds rather than 3.

It could happen that the three bonds in cyclopropane and the three bonds in the alkyl branch are lower in energy that the four bonds in cyclobutane and the two bonds in the alkyl branch.

For example:

Let's say a cyclopropane C−C bond is 10 J.
A cyclobutane C−C bond is 9.5 J.
A regular alkane bond is 8 J.

For propylcyclopropane:

Total energy is 54 J
For ethylcyclobutane:
Total energy is 54 J

The same! I know this sounds really simplistic and dumb, but I just don't understand how you can assume the answer is A.

How do you analyze this situation? Please help.

Answer

Compounds A-D all have the same molecular formula, $\ce{C6H12}$. We can burn each compound and measure the heat given off (heat of combustion). Since they are isomers, they will each burn according to the same equation

$$\ce{C6H12 + 9O2 -> 6CO2 + 6H2O + heat}$$

Any differences in the heat given off can be used to say that a compound is more stable (it had a lower energy to begin with, so less heat is given off) or less stable (it had a higher energy to begin with, so more heat is given off).

This link provides the heats of combustion for some useful model compounds. Look at the last column ("Total Strain") in Table I, it shows that cyclopropane is slightly more strained than cyclobutane, while cyclopentane and cyclohexane are both much less strained. The strain energy (SE) in cyclopropane will not change appreciably when we add a propyl group to the ring, nor will the SE in cyclobutane change appreciably when we add an ethyl group to the ring.

Therefore, since cyclopropane has the most ring strain and since propyl, ethyl and methyl groups don't contain any SE, the correct answer to the question is "A".