E2 Flashcards
What is elimination?
Halogenoalkanes react with bases to produce alkenes - a molecules of HX is lost so the reaction is an elimination reaction
What is an E2 reaction?
Elimination taking part in a single concerted step, bimolecular
What is the rate of an E2 reaction?
k x [halogenoalkane] x [base]
Why is the reaction bimolecular?
Both the halogenoalkane and base are involved in the transition state of the rate determining step
What is a description of the mechanism of E2?
The base attacks a hydrogen atom on the carbon atom adjacent to one carrying the halogen atom, even though the hydrogen atom is not acidic loss of H+ can occur because the halogen in a good leaving group, the base forms a new bond to the hydrogen atom at the same time as the C-X bond is broken and the C=C bond is formed.
What conformation does the halogenoalkane adopt?
An anti peri planar conformation in which the H and X groups are int he same plane and on opposite sides of the molecule
what is a description of the orbital overlap during the reaction?
The sigma orbital of the C-H bond is parallel to the sigma star orbital of the C-X bond. When these orbitals are parallel the orbitals can overlap side on as the H and X groups leave and rehybridise to form the overlapping p orbitals of the new C=C pi bond.
What is the stereoselectivity of an E2 reaction?
When the halogenoalkane can spot two anti periplanar conformations, one leading to the E isomer and the other leading to the Z isomer of the alkene, in general the E isomer is formed in higher yield - the stereoselectivity arises because the conformation leading to the E isomer is more stable than the conformation leading to the Z isomer because there is less steric strain
When is E2 stereospecific?
For halogenoalkanes that can only adopt one anti peri planar conformation the E2 reaction is stereospecific, ie different stereoisomers of the halogenoalkane form different stereoisomers of the alkene (eg 1S,2R 1S2S)
When is regioselectivity in E2 important?
The need for an anti periplanar conformation in the E2 mechanism explains why some substituted halocyclohexanes undergo regioselective E2 reactions, ie elimination forms one structural isomer of an alkene in preference to another.
An example is the E2 reaction of trans-1-bromo-2-methylcyclohexane, the product it forms is surprising since we would expect the other product to form due to it being the more stable alkene because the C=C bond is more substituted. However this cannot happen because the H and halogen are both in the equatorial position so there is no anti peri planar conformation, although the conformer where the halogen and H are anti peri planar is less stable and makes up a small fraction of the equilibrium mixture and therefore the rate is slow, it is the only conformer from which E2 can take place.
What is Saytzev elimination?
E2 reactions if aliphatic halogenoalkanes that can rotate freely about C-C bonds normally produce the most substituted alkene. In the transition state of E2 the C=C bond is partially formed so the structure of the transition state resembles the alkene product - the more stable the alkenes the more stable the transition state. Consequently an E2 reaction gives the more substituted alkene with the more stable C=C bond, because the transition state leading to the more substituted alkene will be the more stable ie the lower in energy
What is Hoffman elimination?
Where the less stable alkene is formed selectively, the unexpected results is explained by the size of the base, for example the tert-butoxide ion is much larger than the ethoxide ion and it has difficult in approaching the hydrogen bonding t the Sseoncadary carbon atom. Instead the tert-butoxide ion selectively removes one of the more access hydrogen atoms on the primary carbon atom at the end of the chain
