92STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONSproblem specifies that the pheromone has the cis configuration, the first 8 carbons and the last 13must be on the same side of the C-9–C-10 double bond.5.6(b)One of the carbons of the double bond bears a methyl group and a hydrogen; methyl is ofhigher rank than hydrogen. The other doubly bonded carbon bears the groups @CH2CH2Fand @CH2CH2CH2CH3. At the first point of difference between these two, fluorine is ofhigher atomic number than carbon, and so @CH2CH2F is of higher precedence.Higher ranked substituents are on the same side of the double bond; the alkene has the Zcon-figuration.(c)One of the carbons of the double bond bears a methyl group and a hydrogen; as we have seen,methyl is of higher rank. The other doubly bonded carbon bears @CH2CH2OH and @C(CH3)3.Let’s analyze these two groups to determine their order of precedence.We examine the atoms one by one at the point of attachment before proceeding down thechain. Therefore, @C(CH3)3outranks @CH2CH2OH.Higher ranked groups are on opposite sides; the con
This is the end of the preview.
access the rest of the document.