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[(2E)-4-ethylhept-2-en-3-yl]cyclopropane why it is named E Where according to cahn ingold prelog rules At left side Methyl is prior and on the right side the butane should be prior than cyclopropane

Mohammed Mostafa
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  • One side of the double bond is decorated by the methyl group and the hydrogen atom. The other side is decorated by cyclopropyl and the the branched alkyl chain. For either side you apply the CIP rules, and then, by the book, the double bond either is (E)-, or (Z)-configurated. If you are unsure how to trade with double bonds/cyclic substituents, see e.g. the answers here on chemistry.se) and then post your solution as self-answer. – Buttonwood Aug 30 '21 at 15:49

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Welcome to ChemSE, M. Mostafa. While @Buttonwood has given you ample direction to solve the configuration of alkene 1, visual inspection reveals a "butane" chain at C1 (arbitrary numbering) versus a cyclopropane ring. Accordingly, the temptation is to assign a (Z)-configuration to the alkene because butane (C4)>cyclopropane (C3) and CH3>H. But you and others would fall into a trap. According to the CIP Rules the cyclopropane ring is deconstructed into an open chain species as illustrated in the digraphs. Starting from C6 (red dot, non-duplicate atom) in alkene 1, two paths [6$\rightarrow$7$\rightarrow$10$\rightarrow$(6) and 6$\rightarrow$8$\rightarrow$9$\rightarrow$(6)] are traced around the cyclopropane ring terminating in duplicate atoms [red dots, (6)]. Duplicate atoms have the same atomic number as non-duplicate atoms but the former are attached to three atoms of atomic number zero.

Sphere 3, which contains C5 and C12 in one substituent and C9 and C10 in the other, allows for a one-to-one comparison. C5 and C9 are a tie because {C,H,H}={(C),H,H}. But C10>C12 {(C),H,H}>{H,H,H}. Therefore, the cyclopropane ring outranks the "butane", or hexyl chain. Structure 1 is an (E)-alkene.

user55119
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