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The answer is coming out as Z but given is E. I’m not sure where I’m wrong at this point as I moved along one chain at a time and wanted to follow heaviest atom as earliest as possible ! enter image description hereenter image description here

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    $\begingroup$ It all should have ended at the very step where we break down the double bond and add the dummy C atoms. $\endgroup$ – Ivan Neretin Feb 20 '19 at 15:34
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Welcome to Chem SE, A Singh: You have done a nice job laying out the digraphs for the cyclopropene (A) and cyclohexene (B) rings. The double bond in A is accessed early in both directions but only in one direction in B does access to the double bond occur early. Which means that A>B and the double bond is of the (E)-configuration. But I offer you the more rigorous treatment in the diagram below.

As you move outward from C1 and C4, sphere by sphere, compare sphere 2, C1 has [2,3,H] while C4 has [9,5,H]. This is a tie. Now move to sphere 3 (marked in red). The C1 path reveals [3,(3),H;2,(2),H] while the C4 path displays [6,(6),H;8,H,H]. Comparing atom to atom, C1 has one more carbon than C4 (See bold (2) and H above). Therefore, the cyclopropene ring (A) has priority over the cyclohexene ring B. The double bond is of the (E)-configuration given that Cl>C. Recall that all "real" carbons are equal to duplicate carbons (in parentheses) but all duplicate atoms are attached to three atoms of atomic number zero. In some cases, one has to go this far to make a decision.

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