As mentioned in the introduction, one of the purposes of this chapter is to use alkyne chemistry as a vehicle to begin looking at some of the general strategies used in organic synthesis—the construction of complex molecules in the laboratory. There are many reasons for carrying out the laboratory synthesis of an organic compound. In the pharmaceutical industry, new molecules are designed and synthesized in the hope that some might be useful new drugs. In the chemical industry, syntheses are done to devise more economical routes to known compounds. In academic laboratories, the synthesis of extremely complex molecules is sometimes done just for the intellectual challenge involved in mastering so difficult a subject. The successful synthesis route is a highly creative work that is sometimes described by such subjective terms as elegant or beautiful . In this book, too, we will often devise syntheses of molecules from simpler precursors, but the purpose here is to learn. The ability to plan a successful multistep synthetic sequence requires a working knowledge of the uses and limitations of many different organic reactions. Furthermore, it requires the practical ability to piece together the steps in a sequence such that each reaction does only what is desired without causing changes elsewhere in the molecule. Planning a synthesis makes you approach a chemical problem in a logical way, draw on your knowledge of chemical reactions, and organize that knowledge into a workable plan—it helps you learn organic chemistry. There’s no secret to planning an organic synthesis: all it takes is a knowledge of the different reactions and some practice. The only real trick is to work backward in what is often called a retrosynthetic direction . Don’t look at a potential starting material and ask yourself what reactions it might undergo. Instead, look at the final product and ask, “What was the immediate precursor of that product?” For example, if the final product is an alkyl halide, the immediate precursor might be an alkene, to which you could add HX. If the final product is a cis alkene, the immediate precursor might be an alkyne, which you could hydrogenate using the Lindlar catalyst. Having found an immediate precursor, work backward again, one step at a time, until you get back to the starting material. You have to keep the starting material in mind, of course, so that you can work back to it, but you don’t want that starting material to be your main focus. Let’s work several examples of increasing complexity. Worked Example \(\PageIndex{1}\): Devising a Synthesis Route How would you synthesize cis -2-hexene from 1-pentyne and an alkyl halide? More than one step is needed. Strategy When undertaking any synthesis problem, you should look at the product, identify the functional groups it contains, and then ask yourself how those functional groups can be prepared. Always work retrosynthetically, one step at a time. The product in this case is a cis-disubstituted alkene, so the first question is, “What is an immediate precursor of a cis-disubstituted alkene?” We know that an alkene can be prepared from an alkyne by reduction and that the right choice of experimental conditions will allow us to prepare either a trans-disubstituted alkene (using lithium in liquid ammonia) or a cis-disubstituted alkene (using catalytic hydrogenation over the Lindlar catalyst). Thus, reduction of 2-hexyne by catalytic hydrogenation using the Lindlar catalyst should yield cis -2-hexene. Next ask, “What is an immediate precursor of 2-hexyne?” We’ve seen that an internal alkyne can be prepared by alkylation of a terminal alkyne anion. In the present instance, we’re told to start with 1-pentyne and an alkyl halide. Thus, alkylation of the anion of 1-pentyne with iodomethane should yield 2-hexyne. Solution cis -2-Hexene can be synthesized from the given starting materials in three steps. Worked Example \(\PageIndex{2}\): Devising a Synthesis Route How would you synthesize 2-bromopentane from acetylene and an alkyl halide? More than one step is needed. Strategy Identify the functional group in the product (an alkyl bromide) and work the problem retrosynthetically. What is an immediate precursor of an alkyl bromide? Perhaps an alkene plus HBr. Of the two possibilities, Markovnikov addition of HBr to 1-pentene looks like a better choice than addition to 2-pentene because the latter reaction would give a mixture of isomers. What is an immediate precursor of an alkene? Perhaps an alkyne, which could be reduced. What is an immediate precursor of a terminal alkyne? Perhaps sodium acetylide and an alkyl halide. Solution The desired product can be synthesized in four steps from acetylene and 1-bromopropane. Worked Example \(\PageIndex{3}\): Devising a Synthesis Route How would you synthesize 5-methyl-1-hexanol (5-methyl-1-hydroxyhexane) from acetylene and an alkyl halide? Strategy What is an immediate precursor of a primary alcohol? Perhaps a terminal alkene, which could be hydrated with non-Markovnikov regiochemistry by reaction with borane followed by oxidation with H 2 O 2 . What is an immediate precursor of a terminal alkene? Perhaps a terminal alkyne, which could be reduced. What is an immediate precursor of 5-methyl-1-hexyne? Perhaps acetylene and 1-bromo-3-methylbutane. Solution The synthesis can be completed in four steps from acetylene and 1-bromo-3-methylbutane: Exercise \(\PageIndex{1}\) Beginning with 4-octyne as your only source of carbon, and using any inorganic reagents necessary, how would you synthesize the following compounds? cis -4-Octene Butanal 4-Bromooctane 4-Octanol 4,5-Dichlorooctane Butanoic acid Answer \(\ce{H2}\), Lindlar catalyst 1. \(\ce{H2}\), Lindlar catalyst; 2. \(\ce{O3}\); 3. \(\ce{Zn}\) \(\ce{CH3CO2H}\) 1. \(\ce{H2}\)/Lindlar; 2. \(\ce{HBr}\) 1. \(\ce{H2}\)/Lindlar; 2. \(\ce{BH3}\); 3. \(\ce{NaOH}\), \(\ce{H2O2}\) 1. \(\ce{H2}\)/Lindlar; 2. \(\ce{Cl2}\) \(\ce{KMnO4}\), \(\ce{H3O^{+}}\) Exercise \(\PageIndex{2}\) Beginning with acetylene and any alkyl halide needed, how would you synthesize the following compounds? Decane 2,2-Dimethylhexane Hexanal 2-Heptanone Answer 1. \(\ce{HC≡CH + NaNH2}\) ; 2. \(\ce{CH3(CH2)6CH2Br}\); 3. \(\ce{H2/Pd}\) 1. \(\ce{HC≡CH + NaNH2}\); 2. \(\ce{(CH3)3CCH2CH2I}\); 3. \(\ce{H2/Pd}\) 1. \(\ce{HC≡CH + NaNH2}\); 2. \(\ce{CH3CH2CH2CH2I}\); 3. \(\ce{BH3}\); 4. \(\ce{H2O2}\) 1. \(\ce{HC≡CH + NaNH2}\); 2. \(\ce{CH3CH2CH2CH2CH2I}\); 3. \(\ce{HgSO4}\), \(\ce{H3O^{+}}\)