[limonade]

How would you assign the E Z configuration for the following alkene?

What would be the fastest way for the left portion of the molecule. I know the right side would have OTs as priority?

[antimatter101]

E means entgegen. Z means zusammen. Z means "the same side". E means "not the same side".

Remember to assign the priorities first.

GOod luck.

[limonade]

i understand that. is there a faster way though for doing this problem on the left side?

[Big-Daddy]

My instinctive feeling is that the upper portion is "heavier" (it just looks so, it has 1 more C atom and 2 more H atoms) so I would tempted just to write Z (as obviously if the top is heaver and OT is heavier then the two higher priorities are both on the top so we are looking at the same side, cis, or Z). But I am actually not sure.

[discodermolide]

According to ChemDraw it is
(E)-2-(9,9-dimethyldecahydro-1H-cyclopentanaphthalen-4(2H)-ylidene)-3-methoxypropyl 4-methylbenzenesulfonate.

[limonade]

Ok I get it now. Starting from the double bond, going to the left by the pentagon, we hit the carbon bonded to the two methyl groups first so carbon bound to two carbons has higher priority than the carbon bound to two Hs.

[Big-Daddy]

Ok I get it now. Starting from the double bond, going to the left by the pentagon, we hit the carbon bonded to the two methyl groups first so carbon bound to two carbons has higher priority than the carbon bound to two Hs.

Do we always have to take the long way round (i.e. going around the polygons, rather than down the quickest line to the two methyl groups)?

[discodermolide]

Going the quickest way does not bring any differentiation. The rings are obviously different, 5 membered on one side 6 on the other. That is the difference which helps us here.

[Big-Daddy]

Going the quickest way does not bring any differentiation. The rings are obviously different, 5 membered on one side 6 on the other. That is the difference which helps us here.

Yes but it is counterintuitive to see the smaller group as being the higher priority. This only becomes the case because, counting around the molecule in both directions (around the two cycles - taking the quickest route reaches the 2 methyls at the same time), on the lower group we reach the C with 4 substituted Cs 1 step more quickly than we reach it counting from the other side, so the lower group on the page ends up with a higher priority and in this case that translates to (E) nomenclature.