I deduced the name as cyclodeca-1,6-dien-3-yne. Is it correct?
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ Related: Why is it -en-yne and not yn-ene? What is the explanation for this IUPAC specification? $\endgroup$– user7951Jan 5, 2016 at 12:22
-
$\begingroup$ The sketch has been replaced by a diagram. However, the accepted answer still refers to the sketch. $\endgroup$– NobodySep 20, 2018 at 18:06
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
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. (1Z,6Z)-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 (1Z,6E)-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.
$\begingroup$
$\endgroup$
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.
