E2 Elimination Reactions of Alkyl Halides

This Demonstration considers the chemical kinetics of the E2 elimination reaction, a reaction of alkyl halides that proceeds in competition with nucleophilic substitution SN2 [1].
In this molecular model, carbon atoms are represented by black spheres, hydrogen atoms by white spheres, bromine atoms by a purple spheres and the nucleophile ( base) by a blue sphere.
Move the "reaction progress" slider to observe how the C-Br and C-H() bonds in 2-Bromobutane break and form the double bond (in addition to the H-Nu bond), thus producing an alkene. Bromine is dispersed in the reaction vessel.
The reaction is regiospecific, as it produces the most highly substituted alkene in greater quantities: 2-Butene (80%) compared to 1-Butene (20%). In accordance with Zaitsev's rule, it is also stereospecific, mainly forming trans-2-Butene (66.7%), compared to cis-2-butene (13.3%). This is known as Hofmann elimination [2]. This selectivity results from the differing stability of the alkenes due: (i) to the inductive effect (+I); and (ii) to the steric hindrance of the cis and trans isomers. The differing stability is reflected in the three transition states of the reactions (Hammond's postulate).
This is why trans-2-butene is produced in greater quantities (66%): its synthesis requires the lowest activation energy. This is evident from the reaction pathway.
Inductive electron donating effects (2 for 2-Butene, 1 for 1-Butene) can be visualized by checking "inductive effect." Arrows represent the displacement of electrons, which stabilizes both carbocation and double bonding. To avoid a redundant description, the arrows were not superimposed with the bond, but drawn parallel to it.
You can also focus on the bonds and atoms involved in the reaction by selecting "bond breaking and formation." This will allow you to identify the bonds that break (colored in green) and those that form, including the double bond (colored in red). The -hydrogen was colored green in the reaction leading to 1-butene, but for the reaction leading to 2-butene, two colors (orange and yellow) are used to distinguish the -hydrogens in the two conformations. Cis and trans are the two possible results, depending on the rotation around the C(2)-C(3) bond.
Only at the end of the reaction is the double bond in a plane with six atoms. You can identify them by selecting "bonds in same plane," thereby showing the planar structure outlined in red.


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Snapshot 1: bonds breaking (green) and forming (red) in the course of the reaction. Faster compound formation requires a lower activation energy and gives a greater yield. The halogen that comes off must be a good leaving group, such as bromine, and must be in an anti-position with respect to the hydrogen that binds to the nucleophile (), in order to promote the formation of the bond .
Snapshot 2: when the reaction is complete, it is possible to identify the atoms in the same plane; cis and trans structures for the 2-Butene and no geometric isomerism for the 1-Butene compound.
Snapshot 3: as the inductive effect helps to stabilize the final product of the reaction, the relative yield of each depends on the stability
[1] S. Z. Lavagnino. Eliminazioni E1 E2 [Video]. (Sep 15, 2022) www.youtube.com/watch?v=rnPxLZ44uO4&list=PLswwssc6Q2yb9Eo2i42BmIGR_Yo58tL7z&index=6.
[2] W. Reusch. "Elimination Reactions of Alkyl Halides." (Sep 15, 2022) www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alhalrx3.htm.
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