[monkeyfeet228]

So my Ochem prof gave me (3R,4E)-3-ethyl-4-iodo-2-methyl-4-heptene and I think he's screwing with me because wouldn't a double bond at the 4th carbon be impossible given that it's on a 7 carbon chain? Wouldn't it just become 3? I tried drawing this and this is the closest I can get.

And yes I understand that the double bond placement throws off the entire numbering but it does work if you number from the other (incorrect) side...I think.

[monkeyfeet228]

Sorry, that last picture doesn't have the iodine but it still doesn't really change the nature of the question :p

[orgopete]

I believe you are correct. An alkene has the highest priority, therefore it should have the lowest locant, 3. The locants for iodo and methyl must change as well. You need to review the orientation for E and Z-double bonds.

[monkeyfeet228]

I asked a friend and they said it can work because the double bond isn't prioritized in the numbering...is that true?

[Arctic-Nation]

Don't take my word for it, but I don't think that the double bond is always prioritized, with the reason being that you should try to get the lowest possible combination of numbers for a given configuration. Or something like that, it's been a long, long time since I last indulged myself in this matter.

[Smrt guy]

The highest priority functional group always gets the lowest number possible.

[monkeyfeet228]

Wouldn't the alkene be that in this case or does it still work like an alkane where I number the subs the lowest I can?

[Smrt guy]

The alkene is the highest priority.  Teachers make mistakes too.

[pfnm]

Then is the name of the drawn structure (3Z,5R)-5-ethyl-4-iodo-6-methylhept-3-ene?

[orgopete]

Then is the name of the drawn structure (3Z,5R)-5-ethyl-4-iodo-6-methylhept-3-ene?

Yes