When looking for exocyclic double bonds in a diene, should a double bond not connected to the conjugated system be considered?

For example, in this compound, should the marked double bond considered as an exocyclic double bond?


2 Answers 2


Woodward's rules are an empirical method for predicting the maximum absorption of a molecule based on the functional groups present (that is, conjugated functional groups which possess a distinctive maximum absorbance such as a diene or an unsaturated carbonyl).

Should the marked double bond considered as an exocyclic double bond? enter image description here

Exocyclic is defined as being outside of the ring system (or in more complex systems, outside of the ring system being considered), the bond highlighted is indeed inside the ring system and is as such endocyclic.

If we start with the alkene at the bottom left, then it is in conjugation with an endocyclic alkene (to the immediate right) and an exocyclic alkene (in the ring to the right hand side). These double bonds are all in conjugation with one another.

enter image description here

The bond you've pointed at isn't in conjugation with the 'triene'

The Woodward rules are specifically related to the effect of groups in conjugation, a simple test of which is to see if you can push curly arrows between the pi bonds. If you try this for your molecule you'll quickly see that the bond you've highlighted is too distant from the conjugated system to have any involvement.

  • 2
    $\begingroup$ Thanks for the answer. But according to the paper by Fieser the pointed double bond is exocyclic to the 5 membered ring or right. What I want to know is whether a double bond must be a part of the conjugated system for it to be considered as an exocyclic double bond in the calculation? $\endgroup$
    – user16096
    Commented Jan 13, 2016 at 7:07

If you find a double bond outside a ring (as in the 4 ring system here), it will be considered as exocyclic. So, the pointed double bond is exocyclic to the adjoining cyclopentane ring.

  • $\begingroup$ Considering the double bond is in a six-membered ring, I have difficulties seeing it as exocyclic … $\endgroup$
    – Jan
    Commented Sep 30, 2017 at 16:50

Your Answer

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