[beheada]

I was given a picture which depicted: 3-ethyl-2-fluoro-2,4-hexadiene

My question comes in when determining whether it is an E or Z for the 2 double bond. I am currently under the impression that on the H-F side, the F is obviously high priority. On the ethyl-C side, I originally thought the ethyl would naturally be higher than a carbon... BUT... the carbon is DOUBLE bonded to another carbon. In accordance with the CIP rule, this means that the carbon side should be higher priority, making the 2 double bond "ze zame zide".

The 4 double bond is easily identified as an E.

So my final answer came up as: (2Z,4E)-3-ethyl-2-fluoro-2,4-hexadiene

Correct?

[Albert]

Yes, I think so.

[zcm5000]

It depends on how the molecule was depicted.  We can't figure out if each double bond is Z or E just by saying it was a picture of 3-ethyl-2-fluoro-2,4-hexadiene, because it could be shown a number of ways.

[beheada]

For further clarification, just so I know I'm doing this correctly... the molecule was depicted as such:

C₂ on the right side of the bond has a methyl in the top right position and a F in the bottom right. On the left side, attached to C₃ is an ethyl group and C₄. C₄ is double bonded to C₅. The lower right bond to C₄ is a sole hydrogen. The top bond to C₅ is a hydrogen and the bottom is a methyl group.

Therefore, I ask is this (2Z,4E)-3-ethyl-2-fluoro-2,4-hexadiene?

Hope that was elaborate enough to solve.

[Albert]

Is it something like this? 

[beheada]

EDIT: Yes, that's kind of it.

Except, in my example, the F and ethyl group are on opposite sides than yours. Starting from top right to bottom left, you have the 6 carbons. You then have a F on the bottom-right of C2 and an ethyl on the top left of C3. That's it.