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I deduced the name as cyclodeca-1,6-dien-3-yne. Is it correct?


2 Answers 2


You have the right name unless you need to specify the geometric isomer.

If I may stretch you a little, consider whether to specify the relative positions of the atoms at the double-bonds, e.g. (1​Z,6​Z)-cyclodeca-1,6-dien-3-yne. It's not possible to tell which geometric isomer it is from your drawing, because you would need a clear depiction of the bond angles around the double-bonds. If this drawing is all you have to go on, you can't specify the isomer and your answer is spot on.

The below image is a model of your molecule showing the (1​Z,6​E)-cyclodeca-1,6-dien-3-yne form. You can see the difference between E ("opposite" or in this case "trans"), on the left, and Z ("together" or in this case "cis"), on the right, by looking at the shape of the chain of the four carbons nearest each double-bond. Of course, there is more to geometric isomerism than this but I hope this is a useful introduction if you've not encountered it before.



Your answer is correct. In case of cyclic unsaturated hydrocarbon organic compound with no side group attached, then numbering of multiple bonds should be done in such a way that there will be minimum number to multiple bonds.


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