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

Gaurang Tandon
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Tanmay
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2 Answers2

10

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.

(1Z,6E)-cyclodeca-1,6-dien-3-yne

Nobody
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4

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.

solanki...
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