Download presentation
Presentation is loading. Please wait.
1
Elimination reactions
Chapter 6
2
Elimination reactions
A problem that often occurs in substitution reactions is elimination
3
E1-elimination n = k [substrate]
The mechanism is similar to that of the SN1 reaction: the first step is formation of the carbocation via heterolytic cleavage
4
Substitution vs elimination
Elimination and substitution reactions are competing processes: in this case, the tertiary carbocation gives primarily elimination in the presence of a strong base (EtO–) mainly substitution if there is only a good nucleophile present This can be considered as a ‘rule of thumb’
5
pKa’s revisited *The pK values are those of the conjugated acids
HI H3O+ HF CH3CO2H H2S NH4+ MeOH HCCH H2C=CH2 pKa –10 –1.7 3.2 4.2 7.0 9.4 15.2 25 45 Base CH3CO2¯ CN ¯ NH3 (CH3CH2)3N CH3O¯ HO¯ CH3CH2O¯ (CH3)3CO¯ NH2¯ pK* 4.2 9.1 9.4 10.7 15.2 15.7 15.9 19 35 *The pK values are those of the conjugated acids If the pK of the conjugated base is higher than the pKa of the acid deprotonation is possible (pK > 12-15: strong base)
6
The Saytzeff elimination
Saytzeff elimination: in the elimination process, formation of the most substituted double bond is favored
7
Explanation In the transition state for elimination, the increased stability of the most substituted double bond is already felt so that there is a lower energy barrier for elimination of Ha
8
Problems 6.20: Predict the products of the E1 and SN1 reactions of the following molecules in water:
9
n = k [substrate] [base]
The E2 elimination n = k [substrate] [base] As in the SN2 reaction, the rate is dependent on the concentration of both reaction partners
10
E2 vs SN2 There is a strong similarity between the E2 and SN2 reactions: strong nucleophiles favor substitution, while strong bases favor elimination
11
Steric bulk favors elimination
We already saw that SN2 substitution on a tertiary carbon is not possible, therefore E2 elimination will prevail (beside E1 elimination)
12
Other examples Note that at a primary carbon atom, only SN2 and no SN1 substitution is possible
13
Stereochemistry of the E2 rxn
Possible orientations of the proton that eliminates and the leaving group
14
E2 and the anti-conformation
The E2 elimination takes place in a single process in the anti-conformation: proton abstraction, double bond formation and cleavage of the C–L bond occur simultaneously
15
Resemblance to SN2 The mechanism is ‘more or less’ analogous to the SN2 substitution The process is also called anti-elimination or antiperiplanar elimination
16
How to visualize this aspect ?
What are the products of E2 elimination of both diastereomers ?
17
The answer…..
18
Regiochemistry Generally, formation of the most substituted double bond is favored Note the occurrence of E- and Z-isomers
19
Saytzeff vs Hofmann elimination
20
Influence of the base
21
Effect of the leaving group
Especially quarternary ammonium leaving groups favor the Hofmann product
22
The E1cB elimination The E1cB reaction resembles the SN2 reaction, with the difference that there is an anion formed prior to the loss of the leaving group
23
The three eliminations
E1cB: the proton is removed first, an anion is formed E1: the leaving group departs first, a cation is formed E2: all processes occur at the same time
24
Example of an E1cB reaction
This reaction is possible if there is a group present that can stabilize the negative charge
25
Stability of anions The more substituted the carbanion, the less stable it is; this is a result of the inductively electron donating alkyl groups
26
The synthetic outlook
27
Formation of CC-bonds Formation of CC-bonds is a synthetically important reaction
28
Summary
29
Summary (2)
30
Problems Make problems: 6.30, 6.31, 6.32, 6.38, 6.53, 6.54, 6.56, 6.59
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.