Question

explain E1cb elimination mechanism

Answer 1

Mechanism[edit]α and β assignments in a molecule with leaving group, LG.

There are two main requirements to have a reaction proceed down an E1cB mechanistic pathway. The compound must have an acidic hydrogen on its β-carbon and a relatively poor leaving group on the α- carbon. The first step of an E1cB mechanism is the deprotonation of the β-carbon, resulting in the formation of an anionic transition state, such as a carbanion. The greater the stability of this transition state, the more the mechanism will favor an E1cB mechanism. This transition state can be stabilized through induction or delocalization of the electron lone pair through resonance. An example of an E1cB mechanism that has a stable transition state can be seen in the degradation ofethiofencarb - a carbamate insecticide that has a relatively short half-life in earth's atmosphere. Upon deprotonation of the amine, the resulting amide is relatively stable because it is conjugated with the neighboring carbonyl. In addition to containing an acidic hydrogen on the β-carbon, a relatively poor leaving group is also necessary. A bad leaving group is necessary because a good leaving group will leave before the ionization of the molecule. As a result, the compound will likely proceed through an E2 pathway. Some examples of compounds that contain poor leaving groups and can undergo the E1cB mechanism are alcohols and fluoroalkanes. It has also been suggested that the E1cB mechanism is more common among alkenes eliminating to alkynes than from an alkane to alkene.[2] One possible explanation for this is that the sp2 hybridization creates slightly more acidic protons. Although it should be noted that this mechanism is not limited to carbon-based eliminations. It has been observed with other heteroatoms, such as nitrogen in the elimination of a phenol derivative from ethiofencarb.[3]

Degradation of ethiofencarb illustrating the presence of a stable anion due to resonance between the amide functional group and the carbonyl group.Distinguishing E1cB-elimination reactions from E1 and E2-elimination reactions[edit]See also: Elimination reaction

All elimination reactions involve the removal of two substituents from a pair of adjacent atoms in a compound. Alkene, alkynes, or similar heteroatom variations (such as carbonyl and cyano) will form. The E1cB mechanism is just one of three types of elimination reaction. The other two elimination reactions are E1 and E2 reactions. Although the mechanisms are similar, they vary in the timing of the deprotonation of the α-carbon and the loss of the leaving group. E1 stands for unimolecular elimination, and E2 stands for bimolecular elimination. In an E1 mechanism, the molecule contains a good leaving group that departs before deprotonation of the α-carbon. This results in the formation of a carbocation intermediate. The carbocation is then deprotonated resulting in the formation of a new pi bond. The molecule involved must also have a very good leaving group such as bromine or chlorine, and it should have a relatively less acidic α-carbon.

Example of the preferential elimination of fluorine in an E1cB-elimination reaction.

In an E2-elimination reaction, both the deprotonation of the α-carbon and the loss of the leaving group occur simultaneously in one concerted step. Molecules that undergo E2-elimination mechanisms have more acidic α-carbons that undergo E1 mechanisms, but their α-carbons are not as acidic as those of molecules that undergo E1cB mechanisms. The key difference between the E2 vs E1cb pathways is a distinctcarbanion intermediate as opposed to one concerted mechanism. Studies have been shown that the pathways differ by using different halogen leaving groups. One example uses chlorine as a better stabilizing halogen for the anion than fluorine,[4] which makes fluorine the leaving group even though chlorine is a much better leaving group.[5] This provides evidence that the carbanion is formed because the products are not possible through the most stable concerted E2 mechanism. The following table summarizes the key differences between the three elimination reactions; however, the best way to identify which mechanism is playing a key role in a particular reaction involves the application of chemical kinetics.

E1E2E1cBStepwise reactionConcerted reactionStepwise reactionCarbocation intermediateSimultaneous removal of proton, formation of double bond, and loss of leaving groupCarbanion intermediateno kind of conclusionNo preferenceno kind of conclusionGood leaving groupsLeaving groupPoor leaving groupsLess acidic α-carbonAcidic α-carbonMore acidic α-carbon