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While looking for data about ring strain in cycloalkanes, I came across two types of data:

  1. Ring strain per methylene group (let's denote this quantity by X): The difference between heat of combustion of the cycloalkane per methylene group and the reference value of $\pu{658.6 kJ/mol}$.
  2. Total ring strain (let's denote this quantity by Y): The difference between molar heat of combustion of the cycloalkane and the reference value of $\pu{658.6 kJ/mol}$ multiplied by number of carbon atoms in the cycloalkane.

From what I have read till now, it appears that the first quantity is the better representation of ring strain. (I say this on the basis that Y has the same value for cyclopentane as well as cycloheptane but still cycloheptane is considered to be less strained as it has lower value of X).

I want to know why X is a better measure for ring strain than Y. When we talk about a ring, we consider it as a whole; then why do we calculate ring strain per methylene group?

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  • $\begingroup$ Take a look at this explanation: ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/thermo/… $\endgroup$ – user55119 Mar 26 at 20:11
  • $\begingroup$ I read it, and I failed to understand why it talked about cycloheptadecane at the end. What was the conclusion of it? $\endgroup$ – Aumkaar Pranav Mar 27 at 3:53
  • $\begingroup$ The present answer has given a nice reasoning, but it's more about how the value of X confirms with the experimental findings. If anyone could provide a theoretical reason (using thermochemistry maybe) as to why X is a more suitable measure, it would be really helpful. $\endgroup$ – Aumkaar Pranav Mar 27 at 5:42
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Let's look at following table from a Yale website:

$$ \begin{array}{c|ccc} & \ce{(CH2)3} & \ce{(CH2)4} & \ce{(CH2)5} & \ce{(CH2)6} & \ce{(CH2)7} \\ \hline \Delta H_\mathrm{Combusion} \ (\pu{kcal/mol}) & -499.8 & -656 & -793.5 & -944.6 & -1108.3 \\ \Delta H_\mathrm{Combusion} \text{ per } \ce{-CH2}- \ (\pu{kcal/mol}) & -166.6 & -164 & -158.7 & -157.4 & -158.3 \\ \text{Ring strain per } \ce{-CH2}- \ (\pu{kcal/mol}) & 9.2 & 6.6 & 1.3 & 0 & 0.9 \\ \text{Total ring strain} \ (\pu{kcal/mol}) & 27.6 & 26.3 & 6.5 & 0 & 6.3 \\ \hline \end{array} $$

Now, just concentrate on ring strain values for cyclopropane and cyclobutane only:

  1. Ring strain per $\ce{-CH2}-$ (X-values) are $9.2$ and $\pu{6.6 kcal/mol}$, respectively.
  2. Total ring strains (Y-values) are $27.6$ and $\pu{26.3 kcal/mol}$, respectively.

According to Y-values, cyclopropane and cyclobutane are equally unstable (relatively speaking), while according to Y-values, cyclopropane is significantly less stable than cyclobutane. Which valuse should be more reliable? Let's look at experimental findings:

Although cyclopropanes are far less reactive than alkenes, they react with chlorine and bromine under polar conditions to form corresponding addition products, 1,3-dihalopropanes, as the predominant product (Ref.1). To my understanding, cyclobutane and higher cycloalkanes do not give this ring opening reaction.

Cyclopropane also reacts with HCl and HBr to form corresponding addition products, 1-halopropanes (Ref.2; Reactions of bromocyclopropane with hydrobromic acid has given mixture of 1,1-, 1,2- and 1,3-dibromopropanes). Again, to my understanding, cyclobutane and higher cycloalkanes do not give this ring opening reaction.

So which one you think should be more reliable and why?

References:

  1. Joseph B. Lambert, Erik C. Chelius, William J. Schulz, Jr., Nancy E. Carpenter, “Polar bromination and chlorination of cyclopropane,” J. Am. Chem. Soc. 1990, 112(8), 3156–3162. (https://doi.org/10.1021/ja00164a043).
  2. Choi Chuck Lee, Bo-Sup Hahn, Kwok-Ming Wan, D. J. Woodcock, “Reactions of cyclopropane with hydrochloric acid and of bromocyclopropane with hydrobromic acid,” J. Org. Chem. 1969, 34(10), 3210–3211. (https://doi.org/10.1021/jo01262a097).
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  • $\begingroup$ I get this reasoning. Can you also please address this (maybe silly) question: why does a larger cycloalkane show greater total ring strain than another smaller cycloalkane having similar value of X (for example, cyclooctane and cyclopentane)? Does the quantity 'total ring strain' hold any significance? $\endgroup$ – Aumkaar Pranav Mar 27 at 4:01
  • $\begingroup$ The flexibility in the ring release some strain energy. This can be explained by the values of higher ring sizes, some of which gives even negative values for ring strain (See the Yale website given in my answer). $\endgroup$ – Mathew Mahindaratne Mar 27 at 4:40
  • $\begingroup$ Umm... My question was about increased total ring strain in larger cycloalkanes. It seems you have explained why large cycloalkanes have lower ring strain. $\endgroup$ – Aumkaar Pranav Mar 27 at 5:32

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