Chapter 9 - Really great lecture notes PDF

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Chapter 9: Sn1, Sn2, E1, E2 Part 2 This chapter addresses the confusion we have introduced in how to decide whether an Sn1, Sn2, E1, or E2 reaction will predominate in any given circumstances. These can always be in competition Already know a little bit about this: • 1˚ substrate favors Sn2 over Sn1 • 3˚ subtstrate means Sn2 won’t happen • Strong, bulky base favors E2 • Weak-base-weak-Nucleophile favors E1/Sn1 over Sn2/E2 Complications: • all bases are nucleophiles and the converse. (Solution is often to find a Nu that isn’t very basic or a base that isn’t very nuclophilic) • Carbocation is the same intermediate in both E1 and Sn1 reaction Will examine some of these things one by one now.

Consider a particular tricky example from text. Using acetate as base/nucleophile and a secondary/benzyl substrate [write on board, not on screen. Too big.] Sn1

E1

E2

Sn2:

When there are legitimately competing reactions, usually we can decide how the results are determined: either “kinetic control” or “thermodynamic control” Kinetic control: assumes that the reactions are effectively irreversible, so that whichever one happens fastest gets the most. This happens when reactions are strongly favored and when there’s not an easy mechanism by which the products can interconvert. (Strongly favored matters because it means the back reaction is sufficiently slow that it doesn’t matter) Under kinetic control, it’s simple: the fastest reaction predominates. This is sometimes the most stable product, but not always.

Thermodynamic control: Sometimes ∆G for a reaction is small (so the forward rate and backwards rate are similar) or there is a mechanism by which possible products can interconvert. In this case, since the products can “go back and forth” all they want, the products are also determined simply: the major product is always the most stable product.

In competitions between E1, E2, Sn1, Sn2, we generally have kinetic control. Fastest = mostest. So let’s see what makes reactions fastest. Rate determining steps are the single step the ionization step for the “2” and “1” rxns, respectively.

This brings us to consider The Hammond Postulate: consecultive structures of similar energy in the same reaction process have similar geometries. Often restated: For an exothermic reaction, the transition state tends to resemble the starting materials, both in structure and energy • Often called an “early” transition state • Corrolary that generally holds: for closely related reactions, the barrier is lower for reactions that are more exothermic. Converse also holds: For an endergonic/endothermic reaction, the transition state tends to resemble the products, both in structure and energy • Late TS, More endothermic generally leads to slower.

Nucleophiles: Strong nucleophiles favor Sn2 reactions. Nothing else requires strong Nu.

All of these are Lone Pair nucleophiles. (All pi bond nucleophiles are weaker!) Notice inherent trends: negative > neutral, left on periodic table > right on periodic table (solvents can affect the absolute nucleophilicity through differential solvation of one ion over another, and often S, P, and halogen nuleophiles are stronger than higher congeners) Key predictor, then: If you want to do Sn2, you use a strong nucleophile. If you see a strong nucleophile in a problem, it’s probably Sn2 reaction. If you see a very weak nucleophile like ROH or H2O, it’s probably not intended to be an Sn2. Making Carbon nucleophiles: In general, wthe most important reactions you can do are the ones that make a molecule bigger over ones that change one functional group into another. That means carbon nucleophiles

Cyanide, –CN RCCH + NaH à RCC– (Not shown in text, but…) R-X (X = halogen) + M (Li/Na/Mg/K) -à RM + MX (different kind of reaction…not a deprotonation, which is why it’s not here)

Bases: Naturally, the rate of E2 reactions goes up with stronger base:

Sometimes we do things to make these bases even less nucleophilic by messing with their structures, e.g., 2,6-dimethylpyridine and ethyl diisopropylamine and LDA. t-BuO-

Steric hindrance affects nucleophile vs base because of hindrance;

Consider Leaving Group: Notice that LG is involved in all four rate determining steps, so you can’t tell how this affects the choice of reaction mechanism, per se. But effect is greater when a full ion is formed. Recall that strongest correlation with LG ability is being a weak base.

Useful group of leaving groups, as we will see, are sulfonate derivatives: This brings us to the topic of turning a bad LG into a good LG. Classic case is ROH. Hydroxide is a pretty bad LG, but water is a good LG.

Suflonate LG is used in the same way in a 2-step reaction that is also an Sn2 reaction (not shown in text yet):

Substrate: 1. 2. 3. 4.

Sn1 reaction dominated by stability of the cation: 3˚ > 2˚ >> 1˚ (benzyl good) Sn2 reaction dominated by sterics: CH3 > 1˚ > 2˚ >> 3˚ (benzyl also good) E1 is more or less same as Sn1, since RDS is the same! E2 just needs accessible adjacent H, but stereochemistry can be important:

Example = cyclohexane derivatives H should be anti to LG

Solvent effects In the Sn2 reaction, you usually have an ionic nucleophile and non/semi-ionic transition state. In Sn1, you have non-ionic starting materials, but nearly ionic TS (because it resembles the fully ionic intermediate. As a rule of thumb, then, solvation by hydroxylic solvents hinders the Sn2 reaction because it stabilizes the starting materials more than it stabilizes the transition state.

Generally, we use “polar aprotic” solvents for Sn2 reactions, e.g., THF, acetonitrile, acetone, DMSO, DMF

Heat Haven’t spoken much about E1 vs Sn1 and they typically compete To favor E1, get as non-nucleophilic as possible. Also, heat reaction up. Why heat?

∆G = ∆H – T∆S

Thus at high T, ∆S becomes more important… and elimination favors entropy because it gives 2 molecules instead of 1.

Two other factors to work through for this chapter: Intramolecular reactions Sometimes the same molecule will have “both ends” of a reaction. Sn2 is a good example of that.

In the case of a molecule like this, you have to decide whether the “intramolecular” or “intermolecular reaction will be faster.

Doing the intramolecular reaction makes a ring. • Intermolecular reaction is 2nd order: faster at higher concentration, slower at lower concentration • Intramolecular reaction independent of concentration. • Ring strain slows intramolecular reaction (obviously) • Entropy favors intramolecular over intermolecular in that the head and tail are “right there”, but the favoring falls off as the length of the chain gets longer. Result: Almost any cyclization that can make a 3, 5, or 6 membered ring is fast by intramolecular reactions and can be favored by doing the reaction under relatively dilute conditions. Most other ring sizes are much more difficult . Example above is one in which intramolecular cyclization would almost certainly be favord under all reasonable conditions.

Final consideration we haven’t really touched: regiochemistry of E1/E2 when not driven by stereochemical considerations. Zaitsev’s rule and regiochemistry Simply stated: major product is generally the most stable product in and E2 reaction. This is driven by kinetics of the second step. They are not rate determining, but follow Hammond’s postulate. Regiochemistry = which atom is attacked, if two or more are possible.

E2 example:

E1 elimination example (same general principle)

Exception that proves the rule. Use a super-big (sterically hindered) leaving group or supersterically hindered base...