Module 5C Flashcards
In an elimination reaction, the products are an alkene, a base bonded to an H and the LG.
The H and LG come from the original molecule and the base induces the reaction as it has a lone pair and is negatively charged (attracted to delta + a-carbon).
In a concerted elimination reaction?
- Base takes proton from B-carbon (a-carbon is the one with the LG)
- As the base takes the proton, there are 2 electron left on the beta carbon which forms a pi bond to make the alkene and LG must go for this to occur, taking the 2 e- that were part of the polar covalent bond between X and a-C
How does a stepwise elimination reaction occur?
- Bond is polarized enough that the LG leaves taking the two e- from the X-a-C bond
- This leaves a carbocation
- Base can then bond with H+ and C is left with 2 electrons to form alkene pi bond between carbocation and carbon with e- pair
In some cases, a good nuc can also be a base, so a substitution and elimination competition may arise
Why is an elimination reaction called E2?
E-elimination
2-biomolecular (LG and Base-H)
How is the rate of reactio calculated for an Er reaction?
Rate = k[alkyl halide][base]
In the case of a tertiary alkyl halide, nucleophile cannot get close enough to approach carbon from backside, sn2 cannot occur, so nuc can act as a base and elimination reaction can occur
E2 preferred when molecule sterically hindered
Regioselective means giving a competition between two possible H+ than can be attacked by a base, double bond prefers to form more stable product
Based on alkene properties
Stability: trans > cis
Alkene stability depends on how many other groups are attached to it: more highly substituted = more substituents on carbons (not H) of a pi bond/alkene = more stable
Tetrasubstituted >trisubstituted > disubstituted > monosubstituted
Product distribution of an e2 reaction as a function of base
Small base - more highly substituted alkene is preferred
Large (sterically hindered) base - base has trouble approaching molecule - less substituted is preferred (ex t-BuO-)
If the B-carbon has two B-hydrogens there are 2 different rotamers where a B-hydrogen is anti periplanar to the leaving group and so two stereoisomers will be formed using rotation around sigma bond
The different between stereospeciff and speteroselectiove, stereospetific: substrate is stereoisomeric and results in one stereoisomer as the product
Sterepsepective: substrate can produce two stereoisomers as products where one is the major product
