zum Directory-modus

SN/E Competition (overall)

The Role of the Substrate Structure in the SN2 / E2 Competition

The structure of the substrate is one important parameter in SN2/E2 competition. Depending on the number, size, and the degree of branching of alkyl substituents at the heteroatom-substituted α carbon atom, the reaction rates of SN2 reaction and E2 elimination are considerably differently influenced by various factors.

The reaction rate of the E2 elimination increases greatly in the order of primary, secondary, tertiary heteroalkyl compounds, while the reaction rate of SN2 reactions decreases in the same order.

Mouse
Mouse
Mouse
Fig.1

primary

α carbon: green

Fig.2

secondary

α carbon: green

Fig.3

tertiary

α carbon: green

Fig.4
kE2
Fig.5
kSN2

Therefore, E2 eliminations (E2 instead of SN2) with primary substrates can only be obtained by the application of extremely bulky and sterically-demanding bases. E2 eliminations with primary alkyl halides or alkyl sulphonates, for example, can be achieved with DBN or DBU, and in case of primary epoxides with LDA or LHMDS. SN2 reactions at teriary heteroalkyl compounds virtually never occur (in contrast to SN1) even if the β carbon atom is tertiary, instead of the α carbon atom.

What are the reasons of these opposite substituent effects on the reaction rates of SN2 reaction and E2 elimination?

Effects of the α-alkyl substituents on the SN2 reaction

  • The transition state of the SN2 reaction displays a trigonal bipyramidal structure with the nucleophile (Nu) and the nucleofuge (NuF) located at the tops of the pyramids and the three remaining substituents located at the corners of the central trigonal plane. In contrast to this, the substituents of the starting product's (substrate) α carbon in the SN2 reaction are arranged tetrahedrally. That is, all bond angles between the four substituents measure about 109 degrees (ideally 109° 28'). In the trigonal bipyramide of the transition state, the bond angles between the substituents at the top of the pyramids and the substituents in the central trigonal plane amount to only 90°. Thus, the spatial distance of the substituents at the α carbon in the transition state is noticeably smaller than that in the starting product. Since steric interactions between the substituents in the transition state are therefore stronger, the energy of the transition state, and thus the activation energy of the reaction, increases according to the size and the steric demand of the substituents. This effect is not compensated by the larger bond angles (120°) between the substituents in the central trigonal plane.
Fig.6
Mechanism of the SN2 reaction.
Mouse
Mouse
Mouse
Fig.7

starting materials

Fig.8

transition state

Fig.9

products

Three-dimensional interactive molecular models (Chime) of the SN2 mechanism. Nu: green; NuF: brown.

  • Due to this, the SN2 reaction with tertiary heteroalkyl compounds virtually never occurs, as the activation energy is way too high. Secondary, and particularly primary heteroalkyl compounds, possess less bulky substituents. As a result, the activation energy is considerably lower than in the case of tertiary heteroalkyl compounds. Therefore, with secondary and primary heteroalkyl compounds, the SN2 reaction can proceed quickly enough in order to exceed the E2 elimination.

Effects of the α-alkyl substituents on the E2 elimination

  • In an E2 elimination, the base abstracts a proton from the periphery and not from inside the substrate molecule. Therefore, in an E2 elimination, the base (or nucleophile) and the substrate molecule do not need to approach each other as closely as they would have to in an SN2 reaction. Steric interactions do thus not influence the reaction rate of E2 elimination very much. In contrast to SN2 reactions, E2 eliminations are even more rapid with an increasing number of alkyl substituents at the α carbon. What are the reasons of these contrasting results? In the transition state of the E2 elimination, the bonds of the β hydrogen atom and the α heteroatom have already been partially cleaved. In addition, a new π bond between the α carbon and β carbon atom has been partially formed. By analogy with the well-known stabilization of double bonds of alkenes, the partially formed double bond in the transition state of the E2 elimination is also stabilized by alkyl substituents. Therefore, the activation energy of the E2 elimination with tertiary heteroalkyl compounds is much lower than it is with secondary or primary heteroalkyl compounds. As a result, E2 eliminations proceed most rapidly when tertiary heteroalkyl compounds are applied, while they run slowest with primary heteroalkyl compounds.
Fig.10
E2 reaction mechanism.
Mouse
Mouse
Mouse
Fig.11

starting materials

Fig.12

transition state

Fig.13

products

Three-dimensional interactive model (Chime) of the E2 mechanism. B: red; X: green.

Hint
Alkyl substituents exhibit opposite effects on the reaction rate of SN2 reactions and E2 eliminations!

Achieving chemoselective E2 eliminations with primary heteroalkyl compounds is extremely difficult, because of the considerably higher rate with which the competing SN2 reaction proceeds. This particular effect of the substrate structure on the SN2/E2 competition can only be overcompensated for by the application of extremely bulky and sterically-demanding bases, such as DBN, DBU, LDA, and LHMDS, in high concentrations.

Due to the SN2 reaction's characteristically low reaction rate with tertiary heteroalkyl compounds, these reactions hardly occur even if there would not be any competition among them and E2 eliminations.

Page 4 of 9