Choosing Between SN1, SN2, E1, E2

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The sad fact is that in reality, many reactions can do both elimination and substitution. For example, if we treated the electrophile below with methoxide (–OCH₃), that methoxide could potentially come directly into the carbon that’s bonded to the Br, kick off the Br leaving group, and give us the product shown below, with inverted stereochemistry, via an S_N2 mechanism:

In contrast, the electrophile could instead interact with the methoxide so that the methoxide grabbed an H next-door, dumped the electrons down, and kicked off the Br leaving group, E2-style, to give our major trans (E) Zaitsev alkene, as well as its cis (Z) isomer:

What about S_N1 versus E1? Well, analogously, you could imagine subjecting this same electrophile to S_N1 or E1 conditions, as shown below. In this scenario, the Br leaving group departs first. Then a weak nucleophile/base—in this case, methanol (CH₃OH or HOCH₃) instead of methoxide uses its lone pairs to form a bond with that positively charged carbon center. After subsequent deprotonation by another molecule of CH₃OH, we end up with the S_N1 substitution product with (of course) a <mark 50="" class="rollover-highlighted-text" data-block-id="1aa73288-71cd-434c-9d9d-151c76c42f5a" data-body="A 50/50 mixture of two enantiomers. Racemic mixtures are optically inactive. The phrase " racemized="" stereochemistry"="" here="" refers="" to="" the="" products="" being="" a="" racemic="" mixture."="" data-for="1e609119-b0ab-43ea-a08d-71a845cff7be" data-rollover-id="1e609119-b0ab-43ea-a08d-71a845cff7be" data-tip="true" data-title="Racemic Mixture" currentitem="false">racemized stereochemistry</mark>:

Alternatively, if the methanol instead reached up, grabbed the adjacent H, and dumped the electrons down, then this reaction would traverse an E1 pathway, giving us the same alkene products that we saw before:

As much as I wish that substitutions and eliminations were always cut-and-dried and that every set of reaction conditions resulted in one—and only one—of our four potential pathways (S_N1, S_N2, E1, or E2), this usually is not the case. In reality, almost any time we do a substitution or elimination reaction, there is—going on in the background—at least a little bit of one or more of the other pathways.

        The point is: in reality, there is not always a cut-and-dried, singular pathway down which any given set of substitution-elimination reaction conditions will universally go. Sometimes we see competitive substitution-elimination happening. Sometimes we get a mixture of products, which can complicate matters. Nevertheless, for me—as both a teacher and a learner—I like hard-and-fast, cut-and-dried rules that I can use, wherever possible.

        Hence, this is the way I like to teach this subject. I would much rather teach you some hard-and-fast, cut-and-dried rules that will take you to the correct answer than teach you all kinds of ambiguities that make this concept so difficult that you can’t understand what’s going on. 

        To be honest, there are such ambiguities in the world of substitution-elimination. To address this, we’ll discuss them in greater detail later on. Nevertheless, for right now, I’m just going to teach you some hard-and-fast rules that will take you to the correct answer in the majority of cases. Again, we’ll deal with exceptions and nuances in a later section of this chapter. 

If you are given a multiple-choice question that requires you to pick the correct substitution or elimination pathway (S_N1, S_N2, E1, or E2), here is a list of questions you can ask yourself (in this order) to choose the correct one:

• Question #1: Does your multiple-choice question specifically tell you if the reaction is substitution or elimination or (even more conveniently) if your reaction is S_N1, S_N2, E1, or E2? In other words, if you see—embedded within a multiple-choice question—language that steers you toward the pathway the question “wants,” then you should definitely just use whichever reaction pathway (S_N1, S_N2, E1, or E2) is indicated by the question.

For example, you might see a question that looks something like this:

In this example, the question tells us that we’re looking for an “elimination” pathway because the word “elimination” is embedded in the question. Now, this question does not necessarily mean that the elimination pathway is the major/most-favored pathway in this set of conditions. It might be. It might not be. That isn’t what the question is asking, anyway. It’s just asking us to identify the major product of the “elimination” pathway.

        If, for instance, the elimination pathway happened to be a minor pathway for this reaction, it would not invalidate the question. The question is not asking us to identify the major product of this reaction. It’s specifically asking what is the major elimination product of this reaction. This subtlety is why Question #1 (above) is so important. After we answer Question #1, we of course move on to Question #2:

• Question #2: Is the carbon bonded to your leaving group (LG) a methyl, 1°, 2°, 3°, or resonance-stabilized carbon? If it’s methyl, then the reaction must be S_N2 (see Figure 7.18 below). In contrast, if it’s 1°, then the reaction is either S_N2 or E2. (Primaries cannot undergo S_N1's or E1's.) Even more confusingly, if your carbon is 3°, then the reaction can potentially be S_N1, E1, or E2. (Tertiaries do not do S_N2’s.) And finally, if your carbon is 2° or a potentially resonance-stabilized carbocation (like an allyl or benzyl carbocation), then your reaction could be any of the above (S_N1, S_N2, E1, or E2).

Now, after going through Question #2, if you’re in one of the cases from Figure 7.18 that has two or more options, how do you narrow them down? By going on to Question #3:

• Question #3: Is your nucleophile/base strong or weak? At this point, we do not yet have to distinguish if our nucleophile/base is either a nucleophile or a base (we'll tackle this in Question #4). We just have to determine if it’s strong or weak. Now, strong nucleophiles/bases have negative charges. (Exceptions: negative charges that are resonance-stabilized are weak.) Hence:
  • strong nucleophile/base = a “2” reaction (either S_N2 or E2)
  • weak nucleophile/base = a “1” reaction (either S_N1 or E1). For this, I like to remember that the words “weak” and “one” both start with a W sound.

Figure 7.19 below (compare with Figure 7.10 from earlier) shows some generic examples of both strong and weak nucleophiles/bases. The top half of Figure 7.19 shows some generic strong (S_N2/E2) nucleophiles/bases. These are molecules that have carbons, oxygens, nitrogens, sulfurs, etc., being bonded directly to metals, where the letter M stands for metals like Li, Na, K, or Mg (in the case of Grignard reagents, discussed later). For all of these metals, you can replace the metal atom with a negative charge because in substitution-elimination reactions, for all intents and purposes, this is how such metals behave. 

By contrast, the bottom half of Figure 7.19 shows some generic weak (S_N1/E1) nucleophiles/bases. These have either resonance-delocalized negative charges (in the case of carboxylates–CO₂^–) or are molecules that do not have metals bonded to the nucleophilic atom. For the latter group, these molecules’ nucleophilic atoms (typically carbon, oxygen, sulfur, or nitrogen) have no formal negative charges (click here to review formal charge) but instead have just neutral lone pairs.

        If by this point you still haven’t determined which of these four pathways (S_N1, S_N2, E1, or E2) your reaction will do, you then have to move on to Question #4:

• Question #4: Is your nucleophile/base a nucleophile or a base? 

To explain, in substitution and elimination reactions, a nucleophile is something that does substitution, while a base is something that does elimination. 

        Now, larger nucleophiles/bases tend to behave more as bases (they prefer elimination) because they cannot fit as easily into the carbon bonded to the leaving group to do a substitution. They therefore prefer to remove a next-door hydrogen to do an elimination reaction.

        It follows, then, that smaller nucleophiles/bases tend to behave more as nucleophiles (preferring substitution) because they can fit more easily into the carbon bonded to the leaving group to do a substitution.

As you might imagine, if we treat the alkyl bromide below with a small nucleophile, such as the ^–O—small alkoxide nucleophile shown, then that nucleophile can fit much more easily into the carbon, kicking off the Br leaving group S_N2-style to form our product:

In contrast, if we react that same starting material with an alkoxide nucleophile that’s bonded to a large carbon group (^–O—large in the figure below), it is much more difficult for the negative charge to get into the carbon attached to the Br. Hence, size prohibits some groups from being able to do nucleophilic attack (nucleophilic substitution). In such cases, then, the negative charge will much more easily reach over, form a bond with an adjacent H (which is a little further away from the <mark class="rollover-highlighted-text" data-block-id="619e5d92-b258-4fb7-8a08-d05d01b49654" data-body="Synonymous with " sterically="" hindered."="" this="" phrase="" refers="" to="" an="" atom="" or="" group="" being="" surrounded="" by="" other="" groups="" that="" are="" bulky="" enough="" "get="" in="" the="" way"="" inhibit="" incoming="" nucleophile="" from="" coming="" in.="" more="" "sterically="" encroached"="" is,="" it="" has="" around="" it."="" data-for="84747501-fbc3-4965-8b9a-e66af5aa081f" data-rollover-id="84747501-fbc3-4965-8b9a-e66af5aa081f" data-tip="true" data-title="Sterically Encumbered" currentitem="false">sterically encumbered</mark> leaving group), slap the electrons down, and push the Br off, E2-style, to give our alkene products:

By analogy, the same kind of thing applies to S_N1 scenarios. If we have the same starting material, and the Br takes off first to form a carbocation (see the figure below), and we then react it with a less reactive (or weak) nucleophile/base (in this case, one that is small) like this, . . . 

. . . then lone pairs on the oxygen come in, form a bond with that positively charged carbon to give us a charged intermediate, which then gets deprotonated by a subsequent molecule of HO—small and pushes the electrons in to form our final S_N1 product (above figure, right side).

        By comparison, if our S_N1 nucleophilic oxygen is bonded to a very large, sterically cumbersome group, then that group cannot fit into the carbocation center as easily because that center is flanked by two carbons (sometimes more) on either side. Thus, substitution becomes harder for larger nucleophiles/bases. In this case, the oxygen will instead take its lone pairs and (much more easily and with less sterical hindrance) grab the adjacent H, dump the electrons down, and form our elimination products via an E1 pathway:

Again, then, in a substitution-elimination context, the larger the nucleophile/base is, the more of a base (elimination pathway) it is; the smaller the nucleophile/base is, the more of a nucleophile (substitution pathway) it is.

        You might wonder, then, “Where do we draw the line? What is considered small (a nucleophile), and what is considered large (a base)?” Well, for my university classes, my rule is this:

• Any nucleophile/base that looks as big as ethanol (CH₃CH₂OH) or ethoxide (CH₃CH₂O^–) or larger when drawn on paper is a base and will do an “E” reaction (E1 or E2). Anything smaller is a nucleophile and will do an “S” reaction (S_N1 or S_N2).
• Exceptions: 
  • Although acetate (CH₃CO₂^–) looks larger than ethanol when drawn out on paper, acetate is a nucleophile and will do an “S” reaction. The same is true of other carboxylates (RCO₂^–).
  • Also, nucleophilic carbon and sulfur atoms are usually nucleophiles, regardless of size.

Figure 7.20 below (compare with Figures 7.10 and 7.18 from earlier) shows some generic examples of nucleophiles/bases. Figure 7.20’s top half indicates ones that are smaller than ethanol or ethoxide on paper. These can be alcohols, ethers, or alkoxides, where M stands for the familiar metals listed, which we just replace with a – charge. Acetate (CH₃CO₂^–) and other carboxylates (RCO₂^–), despite being larger than ethanol, are still considered to be nucleophiles and will consequently favor substitution over elimination. 

Additional exceptions are also listed in the top half of Figure 7.20. These exceptions (mentioned earlier) are nucleophilic carbon atoms/carbanions (commonly formed by bonding carbon to Mg or Li) or sulfur nucleophiles. Sulfur atoms are interesting because they are polarizable, which means that they have the ability to morph and change shape a little bit (kind of like an amoeba), which allows them to fit into smaller "holes" (electrophilic carbon atoms) even when they’re dragging larger tails behind them. Thus, carbon and sulfur nucleophiles are usually nucleophiles (S_N reactions), not bases (E reactions), in a substitution-elimination context.

        By contrast, the bottom half of Figure 7.20 gives a summary list of generic bases—usually alcohols or alkoxides—where the alkyl “R” group attached to the oxygen is ethyl or larger in size. These prefer elimination reactions.

In reality, there are some situations in which Question #4 falls apart. One significant example is the Williamson ether synthesis, which we will discuss in a later chapter. However, for most situations, Question #4 is sufficient.

        Now, given that Question #4 is not perfect, you might wonder, “Why are you teaching it to me?” The reason is because I would rather teach you a straightforward, but sometimes incorrect, rule and have you understand it than teach you a bunch of cloudy, ambiguous, confusing, and mysterious ideas that will leave you understanding less about substitution and elimination than you knew before even starting this class.

        Thus, as I mentioned in Chapter 1 of this text, regarding this subject, I will teach rules first—even sometimes imperfect rules—and let you get the hang of them by practicing a bunch of examples with them. I’ll then gradually introduce you to exceptions so that you can eventually master both and understand when those rules apply and when they do not.

To determine which pathway a substitution-elimination reaction will go—be it S_N1, S_N2, E1, or E2—we ask the following four questions, in this order:

• Question #1: Does your multiple-choice question tell you which pathway it "wants"?
• Question #2: Is the carbon bonded to your leaving group (LG) a methyl, 1°, 2°, or 3° carbon?
  • If it’s a methyl, then the reaction is S_N2. If it’s 1°, then the reaction is S_N2 or E2. If it’s 3°, then the reaction is S_N1, E1, or E2. If it’s 2° or a resonance-stabilized carbon, then the reaction is potentially any of the above (S_N1, S_N2, E1, or E2). To narrow it down, then, we have to move on to the next question.
• Question #3: Is your nucleophile/base strong or weak? 
  • Strong = 2 (S_N2 or E2)
  • Weak = 1 (S_N1 or E1)
• Question #4: Is your nucleophile/base a nucleophile or a base?
  • This, of course, comes down to size. Ethanol/ethoxide or larger (with the above-mentioned exceptions) is a base too big to get into a carbon “hole,” so it favors elimination. Smaller than ethanol, as well as acetate, other carboxylates, and carbon or sulfur nucleophiles will all favor substitution. Again, then:
    • Nucleophile = S (S_N1 or S_N2)
    • Base = E (E1 or E2)