Question 4.6: What is the configuration (E or Z) of the following alkene? ...

At carbon-2, the methyl group has higher priority, by rule 1a. At carbon-3, rules 1a and 1b allow no decision, because the level-1 atoms (a1 and b1) are identical—both are carbons. Proceeding to step 2, the set of atoms (level-2 set) attached to either carbon a1 or b1 is (C,H,H); again, no decision is possible. According to step 3, we must now consider the atom of highest priority (carbon) in each level-2 set; these atoms are labeled a2 and b2, respectively. The level-3 set of atoms attached to a2 is (C,H,H); the level-3 set attached to b2 is (C,C,H). Notice that carbons a1 and b1 considered in the previous step are not considered as members of these sets, because we always work outward, away from the double bond, by step 2. The difference in the second atoms of each set—C versus H—dictates a decision. Because the level-3 set of atoms at carbon b2 has higher priority, the group containing carbon b2 (the isobutyl group) also has the higher priority. The process used can be summarized as follows:

Notice that, although Br has a higher priority than H, the decision point occurs before we reach the Br (which is in a level-4 set). Therefore, the Br-versus-H comparison is irrelevant in this case.

Because the groups of like priority are on the same side of the double bond, this alkene has the Z configuration; the name is (Z)-6,6-dibromo-3-isobutyl-2-heptene.