alkyne with cyclic substituent

My attempt: 1-(1-methylcyclopentyl)-3-methylpent-1-yne. Looks so messy though.

  • 3
    $\begingroup$ It gets even a bit messier than that. Also, note that −≡− should be a straight line: whoever drew this structure doesn't know what they are doing. $\endgroup$
    – andselisk
    Jan 10, 2021 at 12:08

1 Answer 1


As far as I am concerned, your attempt was incorrect.

You must use a parent chain with the highest number of carbon molecules. If I am not mistaken, if a cyclic carbon chain has the same number of carbon molecules as a straight carbon chain, the cyclic carbon chain should be the parent chain.

In addition to what I said above, according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), rings/cyclic carbon chains are always selected as the parent chain.

P- Systems composed of rings and chains (exclusive of linear phanes)

Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).

(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

Thus, the name should be: 1-methyl-1-(3-methylpent-1-yn-1-yl)cyclopentane.

Note that an alkyl substituent that is also an alkyne (i.e. 3-methylpent-1-yn-1-yl in your question above) is called an alkynyl group. It is also worth noting that rings/cyclic carbon chains are always selected as the parent chain for the preferred IUPAC name as I mentioned previously.


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.