Compare stability between the two bicyclic compounds.

I: bicyclo[2.2.1]heptan‐1‐ylium; II: bicyclo[2.2.2]octan‐1‐ylium

Both compounds I and II seem to be unstable because of violation of Bredt’s rule: they are sp2-hybridized. Then how to compare the stability between I and II?


1 Answer 1


The carbocation's energy would be effectively minimised when it adopts an $\ce {sp^2}$ hybridisation scheme, that is, when it adopts a trigonal planar molecular shape. This would ensure that all the precious s orbital character of the carbon atom would be used for bonding with the other three surrounding atoms, leaving a vacant p orbital. However, this bicyclic structure prevents the carbocation from adopting this scheme, destabilising the carbocation formed. This is the basis of Bredt's rule.

On p. 435, Carey & Sundberg (2007) wrote the following:

... bridgehead carbocations become accessible in more flexible structures. The relative solvolysis rates of the bridgehead bromides 1-bromoadamantane, 1-bromobicyclo[2.2.2]octane, and 1-bromobicyclo[2.2.1]heptane illustrate this trend. The relative rates for solvolysis in 80% ethanol at 25 degrees Celsius are shown.

enter image description here

When there is greater flexibility in the structure, the carbocation is better able to adopt a structure that is closer to the preferred planar structure, thus it is better able to be minimise its energy. This results in a lower activation energy and an increased rate of reaction, explaining the data above.


Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.