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Chemistry 2131:
Organic Chemistry for the Life Sciences (3)
Beta-Elimination
1. Summary of Nucleophilic Substitution Reactions:
- to summarize nucleophilic substitution, we saw that it can proceed by two possible mechanisms, SN1 and SN2.
- the major difference between these mechanisms is the timing of bond breakage and bond formation. In SN1, the C-X bond breaks completely, generating a caarbocation intermediate, before the next bond is formed by attack of the nucleophile. In SN2, the bond formation and bond breakage occurs in a concerted fashion.
- factors that affect these reactions include the strength of the nucleophile, the nature of the leaving group, the structure of the halide and the solvent that the reaction takes place in.
- we can summarize these factors in the following table:
Type of Alkyl Halide SN2 SN1 methyl favoured does not occur because methyl cations are so unstable they are never observed in solution primary favoured rarely occurs because primary carbocations are so unstable secondary favoured in aprotic solvents with good nucleophiles favoured in protic solvents with poor nucleophiles tertiary does not occur because of steric hindrance favoured because of the ease of formation of tertiary carbocations - let's use this to work through a few problems: 2-chlorobutane in methanol/water gives 2-butanol and butyl methyl ether. This is a secondary alkyl halide with a poor nucleophile in a protic solvent, this means that it must be SN1
- another example is 1-bromo-2-methylpropane with sodium iodide in DMSO to give 1-iodo-2-methylpropane. We have a primary halide, and 2 beta branches, a good nucleophile in an aprotic solvent. On balance this must be SN2.
2. Beta-Elimination:
- so far we have concentrated on nucleophilic substitution, now we turn to the other characteristic reaction of alkyl halides, beta-elimination. In this reaction the halide leaves, and a proton is removed from the beta-carbon (the adjacent one) to give a double bond. It is the opposite of the addition of HX across a double bond.
- the origin of the pi bond electrons is the C-H bond on the beta carbon.
- like nucleophilic substitution, there are two possible mechanisms, E1 and E2
3. Zaitsev's Rule:
- before getting into the mechanisms, let's look at Russian Rule #2, Zaitsev's Rule. This is a regioselective rule like Markovnikov's Rule.
- for a simple primary alkyl halide, there is only one choice of where to put the double bond, but when the halide is secondary or tertiary, there can be more possibilties.
- Zaitsev's rule states that the double bond will form between the more substituted carbon and the halide carbon
- for example, if we look at 2-chlorobutane, there are two possible products 1-butene and 2-butene. The major product is 2-butene.
4. The E1 and E2 Mechanisms:
- E1 stands for elimination unimolecular. This means that only one molecule is involved in progression ot the transition state.
- in the first step of this reaction the C-X bond breaks to give a carbocation intermediate. If you have a couple of neurons firing you should realize that this is exactly the same first step as for SN1.
- in the second step a strong base removes the beta-proton and the electrons from the C-H bond go to form a pi bond between the alpha and beta carbons.
- what about E2? It stands for elimination bimolecular, meaning that two molecules must interact to progress to the transition state. Like SN2, bond breakage and new bond formation is a concerted process.
- so we have concerted removal of the beta-proton, the electrons go to form the pi bond and the halide leaves all in one step.
- both E1 and E2 follow Zaitsev's Rule
5. Substitution versus Elimination:
- we can see that it different reactions can occur to the same reactants, sometimes at the same time. How do we determine which reaction will take place when reagents are mixed?
- we must weigh the factors that influence these reactions. Let's start by constrating SN1 and E1.
- both of these reactions start with the same step, loss of the halide to form a carbocation intermediate.
- we know that more highly substituted halides react more easily by these mechanisms because the carbocation is formed more easily.
- so, secondary and tertiary alkyl halides reacts in protic solvents to give a mixture of elimination and substitution products.
- for instance when tert-butylchloride (2-chloro-2-methylpropane) is mixed with 80% aqueous ethanol, three products are observed 2-methylpropene, 2-methly-propanol and ethyl tert-butyl ether.
- it is tricky to predict the major and minor products and in fact all are formed.
- turning to SN2 and E2, the situation is a little easier.
- primary alkyl halides react with reagents that are strong bases and good nucleophiles to give the substitution products mainly
- secondary halides can go either way depending on the nucleophile/base, the solvent and even the temperature.
- with good nucleophiles that are weak bases, such as I-, CH3S-, SN2 generally predominates
- with strong bases such as HO- and RO-, elimination predominates
- tertiary halides react with strong bases regardless of nucleophilicity to give E2 because of steric hindrance of the SN2 reaction.
- these observations can be summarized in a table:
| alkyl halide | Reaction | Comments |
|---|---|---|
| methyl | SN2 | methyl carbocations are too unstable to form in solution elimination can't occur |
| primary | SN2 | the main reaction with good nucleophiles/weak bases (I- and CH3COO-) |
| E2 | the main reaction with strong sterically hindered bases such as potassium tert-butoxide
SN1 and E1 rarely occur because of carbocation instability | |
| secondary | SN2 | main reaction with strong nucleophiles/weak bases |
| E2 | main reaction with strong bases such as HO- and RO- | |
| SN1/E1 | common in reactions with weak nucleophiles in polar protic solvents such as water, methanol, and ethanol | |
| tertiary | E2 | main reactions with strong bases such as HO- and RO- |
| SN1/E1 | main reaction with poor nucleophiles SN2 rarely occurs because of steric hindrance |
