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For the following compound

3-methoxy-2-(3-methylcyclopentylidene)-3-oxopropanoic acid

Right hand side of double bond: $\ce{CO2CH3}$ is highest priority due to the extra carbon (as we work along both chains)

But for the left hand side oft he double bond I am stuck.

I thought maybe we take the blue path, and as there is 3 carbons as opposed to 2 on the orange path, the blue path is the highest priority :

3-methoxy-2-(3-methylcyclopentylidene)-3-oxopropanoic acid

But my book gives the opposite answer : (2Z)-3-methoxy-2-(3-methylcyclopentylidene)-3-oxopropanoic acid

There is a similar question here How to assign E/Z configuration according to the Cahn-Ingold-Prelog rules when subsituents differ only by stereochemistry but in the case of my question the carbons on the cyclo which are directly after the double bond do not have groups attached.

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The reasoning you have used is clearly faulty as there is no meaning in assigning more number of carbon to one end in a cyclic system.

Just look at the both the branches you have marked using orange and blue. The logic followed is in accordance to Cahn-Ingold-Prelog(CIP) rules.

  1. The first atom in branches, marked as blue and as orange , have two hydrogen and one carbon each.So, there is nothing to clearly distinguish between the two branches.
  2. So, we look at the only distinguishing feature between the two branches , which is the substituents on the carbon atom(marked as 2) attached to the carbon atom(marked as 1) on both sides.
  3. We can see that the carbon atom marked as 2 on orange marked side has two carbon atoms and one hydrogen atom attached to it. This is of higher priority than the carbon on the blue marked side which has one carbon atom and two hydrogen atom attached to it.

Hence the Z-nomenclature as answered in the book.

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  • $\begingroup$ That is fantastic. I understand now. Before you explained it I didn't realise the orange C2 carbon connects to the blue C2 carbon again (and for some reason thought we don't go back around the cycle when assigning priorities). Thanks! $\endgroup$ – K-Feldspar Jul 28 '16 at 19:14
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The simple way to solve it is : look at the carbon number 1 from each side they are the same , then look at the carbon number 2 above there are three bonds but carbon number 2 down there is only 2 bonds So above take the highest priority .. The highest priority doesn't depend on the number of carbon it's depend on the same carbon in each side I hope my answer is clear for you I tried my best to explain it . Good luck

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