Inorganic chemistry,Zaytsev's rule (orZaitsev's rule,Saytzeff's rule,Saytzev's rule) is anempirical rule for predicting the favoredalkene product(s) inelimination reactions. While at theUniversity of Kazan, Russian chemistAlexander Zaytsev studied a variety of different elimination reactions and observed a general trend in the resulting alkenes. Based on this trend, Zaytsev proposed that the alkene formed in greatest amount is that which corresponded to removal of thehydrogen from the alpha-carbon having the fewest hydrogensubstituents. For example, when2-iodobutane is treated with alcoholicpotassium hydroxide (KOH),but-2-ene is the major product andbut-1-ene is the minor product.[1]
More generally, Zaytsev's rule predicts that in an elimination reaction the most substituted product will be the most stable, and therefore the most favored. The rule makes no generalizations about thestereochemistry of the newly formed alkene, but only theregiochemistry of the elimination reaction. While effective at predicting the favored product for many elimination reactions, Zaytsev's rule is subject to many exceptions.Many of them include exceptions under Hofmann product (analogous to Zaytsev product). These include compounds having quaternary nitrogen and leaving groups like NR3+, SO3H, etc. In these eliminations the Hofmann product is preferred. In case the leaving group is halogens, except fluorine; others give the Zaytsev product.[clarification needed]
Alexander Zaytsev first published his observations regarding the products of elimination reactions inJustus Liebigs Annalen der Chemie in 1875.[2][3] Although the paper contained some original research done by Zaytsev's students, it was largely a literature review and drew heavily upon previously published work.[4] In it, Zaytsev proposed a purely empirical rule for predicting the favored regiochemistry in thedehydrohalogenation of alkyl iodides, though it turns out that the rule is applicable to a variety of other elimination reactions as well. While Zaytsev's paper was well referenced throughout the 20th century, it was not until the 1960s that textbooks began using the term "Zaytsev's rule".[3]
Zaytsev was not the first chemist to publish the rule that now bears his name. Aleksandr Nikolaevich Popov published an empirical rule similar to Zaytsev's in 1872,[5] and presented his findings at the University of Kazan in 1873. Zaytsev had cited Popov's 1872 paper in previous work and worked at the University of Kazan, and was thus probably aware of Popov's proposed rule. In spite of this, Zaytsev's 1875Liebigs Annalen paper makes no mention of Popov's work.[3][4]
Any discussion of Zaytsev's rule would be incomplete without mentioningVladimir Vasilyevich Markovnikov. Zaytsev and Markovnikov both studied underAlexander Butlerov, taught at the University of Kazan during the same period, and were bitter rivals. Markovnikov, who published in 1870 what is now known asMarkovnikov's rule, and Zaytsev held conflicting views regarding elimination reactions: the former believed that the least substituted alkene would be favored, whereas the latter felt the most substituted alkene would be the major product. Perhaps one of the main reasons Zaytsev began investigating elimination reactions was to disprove his rival.[3] Zaytsev published his rule for elimination reactions just after Markovnikov published the first article in a three-part series inComptes Rendus detailing his rule for addition reactions.[4]
Thehydrogenation of alkenes toalkanes isexothermic. The amount of energy released during a hydrogenation reaction, known as the heat of hydrogenation, is inversely related to the stability of the starting alkene: the more stable the alkene, the lower its heat of hydrogenation. Examining the heats of hydrogenation for various alkenes reveals that stability increases with the amount of substitution.[6]
| Compound Name | Structure | Molar Heat of Hydrogenation | Degree of Substitution | |
|---|---|---|---|---|
| in kJ/mol | in kcal/mol | |||
| Ethylene | 137 | 32.8 | Unsubstituted | |
| 1-Butene |  | 127 | 30.3 | Monosubstituted |
| trans-2-Butene |  | 116 | 27.6 | Disubstituted |
| 2-Methyl-2-butene |  | 113 | 26.9 | Trisubstituted |
| 2,3-Dimethyl-2-butene |  | 111 | 26.6 | Tetrasubstituted |
The increase in stability associated with additional substitutions is the result of several factors.Alkyl groups are electron donating by inductive effect, and increase the electron density on the sigma bond of the alkene. Also, alkyl groups are sterically large, and are most stable when they are far away from each other. In an alkane, the maximum separation is that of thetetrahedral bond angle, 109.5°. In an alkene, the bond angle increases to near 120°. As a result, the separation between alkyl groups is greatest in the most substituted alkene.[7]
Hyperconjugation, which describes the stabilizing interaction between theHOMO of the alkyl group and theLUMO of the double bond, also helps explain the influence of alkyl substitutions on the stability of alkenes. In regards toorbital hybridization, a bond between ansp2 carbon and ansp3 carbon is stronger than a bond between two sp3-hybridized carbons. Computations reveal a dominant stabilizing hyperconjugation effect of 6 kcal/mol per alkyl group.[8]
InE2 elimination reactions, a base abstracts a proton that is beta to a leaving group, such as a halide. The removal of the proton and the loss of the leaving group occur in a single, concerted step to form a new double bond. When a small, unhindered base – such assodium hydroxide,sodium methoxide, orsodium ethoxide – is used for an E2 elimination, the Zaytsev product is typically favored over the least substituted alkene, known as theHofmann product. For example, treating 2-Bromo-2-methyl butane with sodium ethoxide in ethanol produces the Zaytsev product with moderate selectivity.[9]
Due tosteric interactions, a bulky base – such aspotassiumtert-butoxide,triethylamine, or2,6-lutidine – cannot readily abstract the proton that would lead to the Zaytsev product. In these situations, a less sterically hindered proton is preferentially abstracted instead. As a result, the Hofmann product is typically favored when using bulky bases. When 2-Bromo-2-methyl butane is treated with potassiumtert-butoxide instead of sodium ethoxide, the Hofmann product is favored.[10]
Steric interactions within the substrate also prevent the formation of the Zaytsev product. Theseintramolecular interactions are relevant to the distribution of products in theHofmann elimination reaction, which convertsamines to alkenes. In the Hofmann elimination, treatment of a quaternary ammonium iodide salt withsilver oxide produces hydroxide ions, which act as a base and eliminate the tertiary amine to give an alkene.[11]
In the Hofmann elimination, the least substituted alkene is typically favored due to intramolecular steric interactions. The quaternary ammonium group is large, and interactions with alkyl groups on the rest of the molecule are undesirable. As a result, the conformation necessary for the formation of the Zaytsev product is less energetically favorable than the conformation required for the formation of the Hofmann product. As a result, the Hofmann product is formed preferentially. TheCope elimination is very similar to the Hofmann elimination in principle but occurs under milder conditions. It also favors the formation of the Hofmann product, and for the same reasons.[12]
In some cases, the stereochemistry of the starting material can prevent the formation of the Zaytsev product. For example, when menthyl chloride is treated with sodium ethoxide, the Hofmann product is formed exclusively,[13] but in very low yield:[14]
This result is due to the stereochemistry of the starting material. E2 eliminations requireanti-periplanar geometry, in which the proton and leaving group lie on opposite sides of the C-C bond, but in the same plane. When menthyl chloride is drawn in thechair conformation, it is easy to explain the unusual product distribution.
Formation of the Zaytsev product requires elimination at the 2-position, but theisopropyl group – not the proton – isanti-periplanar to the chloride leaving group; this makes elimination at the 2-position impossible. In order for the Hofmann product to form, elimination must occur at the 6-position. Because the proton at this position has the correct orientation relative to the leaving group, elimination can and does occur. As a result, this particular reaction produces only the Hofmann product.