Heat of combustion increases as the number of carbons increases, but on comparing ethyl cyclopentane, ethyl cyclohexane, andethyl cycloheptane, shouldn't you consider ring stability too?

In that context, it should be 6-membered ring < 7-membered ring < 5-membered ring.

Why isn't this so?

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    $\begingroup$ Because an additional $\ce{CH2}$ group gives an extra mole of $\ce{CO2}$ on combustion, the energy released in this process trumps the effect of stability of the ring. $\endgroup$ Commented Mar 26, 2020 at 5:20

1 Answer 1


Have a look at this table from Organic Chemistry by Wade and Simek [1, p. 134]:

TABLE 3-5. Heats of Combustion (per Mole) for Some Simple Cycloalkanes

Every time a $\ce{-CH2}$ group is added, the heat of combustion is increased by almost 650 kJ/mol; whereas the maximum relief in ring strain attained by adding that $\ce{-CH2}$ group is not more than 25 kJ/mol.

So there is no competition between increase in heat of combustion due to more number of carbon atoms and decrease in it caused by ring stability. Every time you see two hydrocarbons with different numbers of carbon atoms, be assured that the one with more carbon will release more heat on combustion.

P.S. In case you are unfamiliar with how ring strain is quantitatively measured [1, p. 133]:

The molar heat of combustion of cyclohexane is nearly twice that of cyclopropane, simply because cyclohexane contains twice as many methylene ($\ce{-CH2}$) groups per mole. To compare the relative stabilities of cycloalkanes, we divide the heat of combustion by the number of methylene ($\ce{-CH2}$) groups. The result is the energy per $\ce{-CH2}$ group. These normalized energies allow us to compare the relative amounts of ring strain (per methylene group) in the cycloalkanes. The reference value of 658.6 kJ (157.4 kcal) per mole of $\ce{-CH2}$ groups comes from an unstrained long-chain alkane.


  1. Wade, L. G.; Simek, J. W. Organic Chemistry, 9th ed.; Pearson: Glenview, IL, 2017. ISBN 978-0-321-97137-1.

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