SN1/E1 reactions

Sn1/E1 reactions are favoured over SN2/E2 reactions by using polar protic solvents (which can solvate the carbocation intermediates). 

Sn2/E2 reactions are favoured over SN1/E1 reactions by using strong nucleophiles or bases. 

Sn2/E2 reactions are favoured over SN1/E1 reactions by using high concentrations of the nucleophile/base (as the rate of these bimo-lecular reactions depends on the concentration of the nucleophile or base). 

First we must decide whether the reaction conditions favor Sn2/E2 or Sn1/E1 reactions. (Recall that the conditions that favor an Sn2 reaction also favor an E2 reaction and the conditions that favor an SnI reaction also favor an El reaction.) The decision is easy if the reactant is a primary alkyl halide—it undergoes only Sn2/E2 reactions. Primary carbocations are too unstable to be formed, so primary alkyl halides caimot undergo Sn1/E1 reactions. 

If the reactant is a secondary or a tertiary alkyl halide, it may undergo either Sn2/E2 or Sn1/E1 reactions, depending on the reaction conditions. Sn2/E2 reactions are favored by a high concentration of a good nucleophile/strong base, whereas Sn1/E1 reactions are favored by a poor nucleophile/weak base (Sections 10.9 and... 

Having decided whether the conditions favor Sn2/E2 reactions or Sn1/E1 reactions, we must next decide how much of the product will be the substitution product and how much will be the elimination product. The relative amounts of substitution and elimination products will depend on whether the alkyl halide is primary, secondary, or tertiary, and on the nature of the nucleophile/base. This is discussed in the next section and is summarized later in Table 11.6. 

Now let s look at what happens when conditions favor Sn1/E1 reactions (a poor nucleophile/weak base). In Sn1/E1 reactions, the alkyl halide dissociates to form a carbocation, which can then either combine with the nucleophile to form the substitution product or lose a proton to form the elimination product. 

When the same compound reacts with methoxide ion in a solvent that favors Sn1/E1 reactions, twelve products are formed. Identify the products that are formed under the two sets of conditions. 

Scheme 7.29. A representation of two potential pathways for proton loss from a carbocation intermediate on the Sn1-E1 reaction surface. Both pathways are followed. It is supposed that proton loss via the pathway labeled (a), which results in the most highly substituted alkene (the Saytzeff product), occurs preferentially because that is the product formed in highest yield. The alkene resulting from pathway (b), the Hofmann product, is also formed. A. W. Hofmann (1818-1895) was a German chemist who was professor of chemistry at the Royal College of Chemistry in London (1845-1864) and then accepted a post as professor at the University of Berlin. Most of Hofmann s work dealt with amines (Chapter 10). Hofmann found, in contrast to Saytzeff, that the least highly substituted alkene is formed when the elimination is carried out on amine quaternary salts (so-called onium salts). This is, in part, presumably due to the close association between the base and the positively charged onium salt as well as to the removal of the proton in the rate-determining step (cf. the E2 reaction). (Note the 82 18 ratio of products shown here should be considered identical, within experimental error, to the 79 21 ratio of Table 7.9.)...

An Sn1/E1 reaction of an alkyl halide is favored by a poor nucleophile/ weak base. 

It is fortunate that the Sn1/E1 reactions of tertiary alkyl halides favor the substitution product, because under Sn2/E2 conditions only the elimination product is formed. 

Primary alkyl halide primarily substitution, unless there is steric hindrance in the alkyl halide or nucleophile, in which case elimination is favored cannot undergo Sn1/E1 reactions... 

Under these conditions, unimoiecuiar reactions (Sn1 and E1) are favored. Fiigh temperature favors E1. A tertiary aicohoi wiii undergo an E1 reaction when treated with suifuric acid and heat. 

Some E1 reaction may occur in competition with SN1, product with the double bond in the ring giving mainly the... 

When regioisomers are possible in an El reaction, the product distribution is found to follow Zaitsev s rule. The reaction of 2-bromo-2-methylbutane under Sn1/E1 conditions (in a polar solvent mixture of ethanol and water with no good base or nucleophile present) gives 64% of the substitution products (water acts as the nucleophile to give an alcohol, or ethanol acts as a nucleophile to give an ether), 30% of the more highly substituted alkene, and 6% of the less highly substituted alkene. 

The bromine is bonded to a tertiary carbon and there is not a strong base present, so the reaction will proceed by an SN1 /E1 mechanism. The substitution product should predominate. The El reaction follows Zaitsev s rule, so more 1-methylcyclohexene should be formed than methylenecyclohexane. 

CHAPTER 7 Inversion of Configuration in the Sn2 Reaction 244 Racemization in the Sn1 Reaction 252 Hydride Shift in an Sn1 Reaction 253 Methyl Shift in an Sn1 Reaction 254 Rearrangement in an E1 Reaction 261 Dehydrohalogenation by the E2 Mechanism 304 Stereochemistry of the E2 Reaction 306 E2 Debromination of a Vicinal Dibromide 310... 

Alkyl halides have the same order of reactivity in S l reactions as they do in El reactions because both reactions have the same rate-determining step—dissociation of the alkyl halide (Table 11.5). This means that all alkyl halides that react under Sn1/E1 conditions will give both substitution and elimination products. Remember that primary alkyl halides do not undergo SnI/EI reactions because primary carbocations are too unstable to be formed. 

Remember that reactions in which arenediazonium ions are involved must be carried out at 0 °C because they are unstable at higher temperatures. Alkanediazonium ions are even less stable. They lose molecular N2—even at 0 °C—as they are formed, reacting with whatever nucleophiles are present in the reaction mixture by both Sn1/E1 and Sn2/E2 mechanisms. Because of the mixture of products obtained, alkanediazonium ions are of limited synthetic use. 

Note that the first molecule of methanol was displaced In an Sn1 reaction and the second lost in an E1 reaction. The chemistry of acetals is dominated by the loss of protonated OR or OH groups in the steps with green boxes. Never be tempted to write S 2 mechanisms with acetals. 

Because E1 and Sn1 reactions proceed through the formation of a common intermediate, the two types respond in simiiar ways to factors affecting reactivities. E1 reactions are favored with substrates that can form stabie carbocations (i.e., tertiary haiides) they are also favored by the use of poor nucleophiles (weak bases) and they are generally favored by the use of polar solvents. 

In most unimolecular reactions the Sn1 reaction is favored over the E1 reaction, especially at lower temperatures. In general, however, substitutbn reactbns of tertiary halides do not find wide use as synthetic methods. Such halides undergo eliminations much too easily. 

In this problem, three products are fomied one from an Sn1 reaction and two from E1 reactions. 

When Sn1/E1 conditions are favored, tertiary alkyl halides, allylic halides, and benzylic halides can form both substitution and elimination products primary and secondary alkyl halides do not undergo SnI/EI reactions. 

Fig. 9.1. Simplified reaction mechanisms in the hydrolytic decomposition of organic nitrates. Pathway a Solvolytic reaction (Reaction a) with formation of a carbonium ion, which subsequently undergoes SN1 addition of a nucleophile (e.g., HO ) (Reaction b) or proton E1 elimination to form an olefin (Reaction c). Pathway b HO -catalyzed hydrolysis (,SN2). Pathway c The bimolecular carbonyl-elimination reaction, as catalyzed by a strong base (e.g., HO or RO ), which forms a carbonyl derivative and nitrite.

---