Question 27.IE: The female of a species of worm produces the sex attractant ......
The female of a species of worm produces the sex attractant spodoptol, which has the following structure.
\begin{matrix} {HOCH_2CH_2CH_2CH_2CH_2CH_2CH_2CH_2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {CH_2CH_2CH_2CH_3}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \overset{\diagdown }{\ \ \ \ C} =\overset{ \ \ \ \ \ \diagup }{C}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \overset{ \ \ \ \ \diagup }{\underset{\text{Spodoptol}}{H} } \ \ \ \ \ \ \ \ \ \ \ \ \overset{\diagdown }{ \ \ \ H} \end{matrix}Spodoptol is a pheromone that attracts the male worms of the species. Synthetic spodoptol in traps can be used to control the population of worms. Devise a synthesis of spodoptol starting from the alcohol shown below.
HOCH_{2}CH_{2}CH_{2}CH_{2}CH_{2}CH_{2}CH_{2}CH_{2}C≡CH
Analyze
In designing the synthesis, we break down the spodoptol molecule into the starting material and the necessary alkyl group that has to be attached to a double bond, namely a n-butyl group. In order to produce the alkene skeleton structure, we will have to convert the triple bond in the starting material to a double bond with cis stereochemistry.
Solve
Working backward, this cis isomer is obtained by using Lindlar’s catalyst (recall page 1300).
HOCH_{2}(CH_{2})_{6}CH_{2}C≡C(CH_{2})_{3}CH_{3} \xrightarrow[\begin{matrix} Lindlar’s \\ catalyst \end{matrix} ]{H_{2}}
\begin{matrix} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {HO(CH_2 )_7CH_2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ {(CH_2 )_3CH_3}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \overset{\diagdown }{\ \ \ \ C} =\overset{ \ \ \ \ \ \diagup }{C}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \overset{ \ \ \ \ \diagup }{{H} } \ \ \ \ \ \ \ \ \ \ \ \ \overset{\diagdown }{ \ \ \ H} \end{matrix}
Now we need to make the alkyne needed in the above hydrogenation. The starting material provided lacks the butyl group at the triple bond. To attach the alkyl group, we can perform a nucleophilic substitution at
1-bromobutane with the acetylide ion generated from the starting material, by using sodium amide:
Because the alcohol proton of the acetylide is also acidic, we need to add an excess of sodium amide to give the combined alkoxide and acetylide. (Ammonia is also produced in the reaction.)
HO(CH_{2})_{8}C≡CH \xrightarrow[NaNH_{2} ]{\text{excess}} ^{−}O(CH_{2})_{8}C≡C:We can now combine these steps into a complete synthesis where we have converted the alkoxide to the alcohol in the next-to-last step by adding ethanol. The ethanol gives up a proton to the alkoxide we have synthesized (the ethoxide ion is the weaker base).
HO(CH_{2})_{8}C≡CH \xrightarrow[NaNH_{2} ]{\text{excess}} ^{−}O(CH_{2})_{8}C≡C:^{−} \xrightarrow[]{CH_{3}(CH_{2})_{3}Br}
^{−}O(CH_{2})_{8}C≡C(CH_{2})_{3}CH_{3} \xrightarrow[]{C_{2}H_{5}OH} \underset{H_{2}\downarrow \begin{matrix} \text{Lindlar’s} \\ \text{catalyst} \end{matrix} }{HO(CH_{2})_{8}}C≡C(CH_{2})_{3}CH_{3}
\begin{matrix} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {HO(CH_2 )_8} \ \ \ \ \ \ \ \ \ \ \ \ \ \ {(CH_2 )_3CH_3}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \overset{\diagdown }{\ \ \ \ C} =\overset{ \ \ \ \ \ \diagup }{C}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \overset{ \ \ \ \ \diagup }{{H} } \ \ \ \ \ \ \ \ \ \ \ \ \overset{\diagdown }{ \ \ \ H} \end{matrix}
Assess
By using the backward synthetic approach—retrosynthesis—we have established a possible synthetic pathway for the synthesis of spodoptol from the starting material. This is a common approach in the development of a synthetic route to a target compound. A proposed synthesis should always be carefully examined to assess whether side reactions are possible because they can decrease the yield of the desired product. In the present case, a side reaction is possible in the second step above. The —O(CH_{2})_{8}CC— ion could also react with CH_{3}(CH_{2})_{3}Br to give an ether.