7.9     What Can We Do with These Reactions? How to Do Organic Synthesis

Your ability to make molecules has just increased enormously. From rather easily accessible compounds, such as the alkyl halides and inorganic materials, myriad other structural types are suddenly available. Figure 7.89 summarizes the synthetic potential of the substitution reaction. Not all reactions will work for all R―LG molecules, and as we will see in Chapter 8, other reactions often compete with substitution.

 

Figure 7.89 adds several reactions to those we have discussed specifically. For example, it sneaks in a brand-new synthesis of substituted alkynes by using the acetylide ion as a nucleophile (usually effective only with primary R―LG compounds). We mentioned the acidity of acetylenes in Chapter 3 (p. 133) and even warned that more was coming about this reaction. Monosubstituted (terminal) alkynes are decent acids (pK_a ~ 25), which means that treatment of a terminal alkyne with a strong base can result in removal of a proton and formation of the acetylide. A favorite base for doing this is the amide ion, ⁻NH₂ (Fig. 7.90). There are some other new reactions in Figure 7.89. Go over each of them carefully and be sure you see how each works.

 

It is relatively easy (note that we didn’t say it was easy!) to keep track of the available reactions and therefore not too difficult to think up one-step transformations. Once synthetic chemistry goes beyond one-step reactions, however, it gets much harder. Now, the intellectual challenge of synthetic chemistry increases sharply, and synthesis problems, whether in the real world or in this book, become not only more challenging but also much more fun. You can begin to see why the art of making molecules fascinates so many people. Let’s look at a few examples.

 

7.9a Sulfur as Nucleophile Suppose we are set the task of making a thiol from an alkyl halide. We will use the nucleophilicity of sulfur to accomplish this transformation. The mercaptide ion itself (⁻SH) and alkyl thiolates (⁻SR) are powerfully nucleophilic and can be used to great effect in the S_N2 reaction. This nucleophilicity leads to the most common synthesis of thiols and sulfides. The only serious restriction is that the alkyl halide must be active in the S_N2 reaction (Fig. 7.91).

 

Sulfides have two lone pairs of electrons, and these relatives of ethers are quite nucleophilic. S_N2 reactions of sulfides give sulfonium ions (Fig. 7.92).

 

Sulfonium ions and the related, less stable oxonium ions are used as alkylating agents. Dimethyl sulfide and dimethyl ether are excellent leaving groups and can be displaced by nucleophiles in S_N2 reactions (Fig. 7.93). As we have already seen (p. 304), nature uses a variant of this reaction, displacement on the amino acid derivative S-adenosylmethionine, to do many methylation reactions.

 

So, if you are set the task of synthesizing a specific thiol, say, propane-2-thiol, you have an answer right at hand: Just treat isopropyl bromide with thiolate, HS⁻. You might get a little elimination product as well (Chapter 8), but the reaction should work quite well. That’s an easy problem, at least if you know the reaction in Figure 7.94.

Now, however, suppose we had to make propane-2-thiol not from an alkyl halide, but from propene. We have no direct way to do this transformation, and Figure 7.89 is no help. But we do have a way to convert propene into 2-bromopropane (Chapter 3, p. 135). So, a two-step synthesis is possible (Fig. 7.95), but it is much harder to conjure up than the simple one-step method. By far, the best way to find it is to work backward from the desired product, asking yourself the question, From what immediate precursor molecule or molecules can I get the ultimate product? In the reaction we are looking at here, the answer to that question should lead you to 2-bromopropane, among others. Now ask again, From what immediate molecule or molecules can I get 2-bromopropane? The answer to that question should include propene plus HBr. Working backward in this way is the only successful route to solving problems in synthesis, and all successful students and professional chemists do it this way.

Now let’s make the sequence even harder and add stereochemistry to the mix. How would you make (S)-2-butanethiol from (R)-2-butanol? Working backward tells us that we want to add thiolate (HS⁻) to a butyl group with a leaving group on carbon 2 and, because we want the (S) configuration in the product, that leaving group position must be the (R) configuration. You might think that we can simply protonate the (R)-2-butanol with hydrochloric acid to make it a leaving group that then will react with thiolate. But this approach will have S_N1 results in addition to S_N2. Another problem with this approach is that the thiolate will simply become protonated in the acidic conditions, which will make it a much less efficient nucleophile. And don’t forget, the chloride is able to compete as a nucleophile. Instead, the best way to make the alcohol into a leaving group without altering the stereochemistry is to convert it into a tosylate (p. 299). The nucleophilic thiolate can then cleanly do S_N2 chemistry as shown in Figure 7.96. Working backward carefully will always lead you to a correct solution.

7.9b Oxygen as Nucleophile: The Williamson Ether Synthesis Here’s another synthetic challenge that uses propyl bromide: Make methyl propyl ether. Given our discussion of Figure 7.94, you have almost certainly come up with the idea of displacing bromide ion from 1-bromopropane with methoxide in a straightforward S_N2 sequence (Fig. 7.97). Good idea! In fact, you have just invented a good way to make ethers—well, not exactly, because A. W. Williamson (1824–1904) discovered this process, now called the Williamson ether synthesis, more than 100 years ago (Fig. 7.98). This synthesis has both limitations and opportunities, and they rather nicely illustrate some of the kinds of thinking that synthetic chemists have to go through. So let’s explore the Williamson synthesis a bit.

 

First of all, we need to be able to make the necessary alkoxide. As we saw in Chapter 6 (p. 247), when an alcohol is treated with a much stronger base, it can be converted entirely into the alkoxide. Note that the base used must be much stronger than the alkoxide. If that is not the case, an equilibrium mixture of the base and the alkoxide will result. A favorite reagent for formation of alkoxides is sodium hydride (Fig. 7.99). Sodium hydride has the advantage of producing hydrogen gas, which can easily be removed from the reaction mixture.

 

Just as water reacts with metallic sodium or potassium to give the corresponding metal hydroxide, alkoxides can be made through a reaction in which sodium or potassium metal is dissolved directly in the alcohol. Hydrogen is liberated in what can be a most vigorous reaction indeed. Potassium is particularly active in all cases, and the reaction of sodium metal with smaller alcohols, such as methyl and ethyl alcohol, is also rapid and exothermic (Fig. 7.100).

Now let’s use alkoxides to make ethers—the Williamson ether synthesis. Several modifications of the original procedure have made this venerable method quite useful, although there are some restrictions. The reaction works only for alkyl halides that are active in the S_N2 reaction. Therefore, tertiary halides cannot be used. They are too hindered to undergo the crucial S_N2 reaction. Sometimes there is an easy way around this problem, but sometimes there isn’t. For example, tert-butyl methyl ether cannot be made from tert-butyl iodide and sodium methoxide (these conditions lead to elimination as we will see in Chapter 8), but it can be made from tert-butoxide and methyl iodide (Fig. 7.101). However, there is no way to use the Williamson ether synthesis to make di-tert-butyl ether.

 

Even secondary halides can be problematic in the Williamson reaction. Alkoxides are strong bases (high affinity for a hydrogen 1s orbital), and elimination rather than substitution is often the major reaction when a secondary halide is used (Fig. 7.102). We will learn about this “E2” elimination reaction in Chapter 8.

 

Be alert for intramolecular, ring-forming versions of the Williamson ether synthesis. Cyclic ethers can be formed by intramolecular S_N2 reactions. There is no mystery in this reaction; it is simply the acyclic process applied in an intramolecular way (Fig. 7.103).

 

An important use of the intramolecular Williamson ether synthesis is in the making of epoxides (p. 237). A 2-halo-1-hydroxy compound (a halohydrin) is treated with base to form the haloalkoxide, which then undergoes an intramolecular backside S_N2 displacement to give the three-membered ring (Fig. 7.104).

7.9c Nitrogen as Nucleophile: Alkylation of Amines We have just examined a series of synthetically useful reactions that uses oxygen as the nucleophile. Nitrogen has a lone pair of electrons and therefore is also nucleophilic. Let’s look here at some synthetic reactions in which nitrogen is the nucleophile. Conceptually, the following reactions are rather closely related to the Williamson ether synthesis.

Amines are good nucleophiles and can displace leaving groups in the S_N2 reaction. Of course, the limitations of the S_N2 reaction also apply—there can be no displacements at tertiary carbons. Successive displacement reactions yield primary, secondary, and tertiary amines. The initial products are alkylammonium ions, which are subsequently deprotonated by the amine present. These S_N2 reactions produce amines that are also nucleophilic. Further displacement reactions are possible, which can lead to problems. Indeed, the product amine is often a better participant in the S_N2 reaction than the starting material. Therefore, it can be difficult to stop the reaction at the desired point, even if an excess of the starting amine is used to minimize overalkylation (Fig. 7.105).

 

Tertiary amines are also strong nucleophiles, and a final displacement reaction can occur to give a stable quaternary ammonium ion (Fig. 7.106). Thus, the alkylation process can be pushed to its logical conclusion in an effective synthesis of quaternary ammonium ions.

 

These alkylation reactions of amines are so efficient that it is difficult to stop at monoalkylation. However, Theodore Cohen (b. 1929) and his co-workers at the University of Pittsburgh worked out a method that uses ammonia in excess under pressure to generate primary amines efficiently. The vast excess of ammonia means that it is difficult for the initial alkylated product, methylamine, to compete with ammonia for a molecule of methyl iodide (Fig. 7.107).

7.9d “Real-World” Difficulties In the real world, the challenges of synthesis are great—you not only need to find a reaction or, more likely, a sequence of reactions that will make the desired molecule, but also you must do so specifically and, in industry, economically. Specificity becomes increasingly important these days as we recognize how critical chirality is in biological action. Also, the more stereogenic centers (p. 156) a molecule has, the more difficult the synthesis. Imagine making a molecule with only a modest number of stereogenic atoms, say four. There are 16 possible isomers, and you almost certainly can use only one. Every reaction you use must lead to the correct stereochemistry and no others.

Although it is easy to look down your nose at questions of cost, you would be very wrong to do so. Chemists in industry must find sophisticated ways to make those stereochemically rich molecules without using superexpensive reagents that might be fine in very-small-scale work but utterly useless in work designed to produce a molecule to be used by millions of people. Doing clever syntheses with cheap reagents is much more difficult, both intellectually and practically, than making the molecule regardless of cost.

Figure 7.89 fails to show the potential complications we have talked about. There can be competition between the substitution reactions, S_N1 and S_N2, and the elimination reactions we will discuss in Chapter 8 are often a problem. Under some circumstances, the major products of the reactions summarized in Figure 7.89 will be alkenes. The identity of the leaving group, so conveniently avoided by using the symbol LG in the figure, makes a difference, and the structure of R, also deliberately vague in the figure, is vital. To design a successful synthesis, you have to think about all the components of the reaction.