Place a double bond in the carbon skeleton shown so as to represent
(a) (Z)-l-Methylcyclodecene
(b) (E)-1-Methylcyclodecene
(c) (Z)-3-Methylcyclodecene
(d) (E)-3-Methylcyclodecene
(e) (Z)-5-Methylcyclodecene
(f) (E)-5-Methylcyclodecene
Sample Solution (a) and (b) Because the methyl group must be at C-1, there are only two possible places to put the double bond:
In the Z stereoisomer the two lower priority substituents—the methyl group and the hydrogen—are on the same side of the double bond. In the E stereoisomer these substituents are on opposite sides of the double bond. The ring carbons are the higher ranking substituents at each end of the double bond.
We need at least 10 more requests to produce the solution.
0 / 10 have requested this problem solution
The more requests, the faster the answer.