According to me, it should be 11 and this is how I have worked out as shown in the attached diagram.

However, I am not sure whether all these 11 isomers can be considered as those belonging to cyclic ether.

I had erroneously uploaded incorrect diagram earlier. Now I have uploaded the correct diagram. My specific query is whether the 11th Isomer shown in this diagram, wherein Oxygen is not part of the ring, can be considered as Cyclic Ether.My sketch of 11 Isomers


1 Answer 1


I would not include the last structure, even if it was clever of you to think about it.

I think that the term "cyclic ether" restricts the scope to "cyclic compounds which have an ether moiety inside the ring". Your last compound would be considered as a linear ether with an alkyl and a cycloalkyl substituents.

Oh and your 6th structure does not have 3 isomers but just 2 because the (R,R) an the (S,S) isomers are actually equivalent by rotation of the molecule. This compound is said to be "meso".

  • $\begingroup$ Thanks for your views. Though most of us are of the same view as yours i.e. "cyclic compounds which have an ether moiety inside the ring", should only be termed as cyclic ether, however, I have not been able to locate any standard reference book which would explicitly define it so. I shall highly appreciate, if you can guide me to any reference book wherein this has been explicitly deliberated upon. $\endgroup$
    – Hemant
    Commented Jun 2, 2019 at 4:35
  • $\begingroup$ Unfortunately, I don't think there is a general rule for that. It merely comes from usage, i.e. the chemical function must be part of the cyclic structure, EXCEPT if it is a monovalent moiety (e.g. alcohol, halide, carboxylic acid...) which can never be inside a ring. For example in the Gold Book (goldbook.iupac.org ), lactams are defined as "cyclic amides" $\endgroup$
    – SteffX
    Commented Jun 3, 2019 at 14:18

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.