Module 5C

Question 1

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).

Question 2

In a concerted elimination reaction?

Answer 2

1. Base takes proton from B-carbon (a-carbon is the one with the LG)
2. 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

Question 3

How does a stepwise elimination reaction occur?

Answer 3

1. Bond is polarized enough that the LG leaves taking the two e- from the X-a-C bond
2. This leaves a carbocation
3. 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

Question 4

In some cases, a good nuc can also be a base, so a substitution and elimination competition may arise

Question 5

Why is an elimination reaction called E2?

Answer 5

E-elimination
2-biomolecular (LG and Base-H)

Question 6

How is the rate of reactio calculated for an Er reaction?

Answer 6

Rate = k[alkyl halide][base]

Question 7

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

Answer 7

E2 preferred when molecule sterically hindered

Question 8

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

Answer 8

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-)

Question 9

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

Answer 9

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