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Elimination reactions. Chapter 6. Elimination reactions. A problem that often occurs in substitution reactions is elimination. E1-elimination. n = k [substrate]. The mechanism is similar to that of the S N 1 reaction: the first step is formation of the carbocation via heterolytic cleavage.
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Elimination reactions Chapter 6
Elimination reactions • A problem that often occurs in substitution reactions is elimination
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
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’
pKa’s revisited pKa –10 –1.7 3.2 4.2 7.0 9.4 15.2 25 45 pK* 4.2 9.1 9.4 10.7 15.2 15.7 15.9 19 35 Acid HI H3O+ HF CH3CO2H H2S NH4+ MeOH HCCH H2C=CH2 Base CH3CO2¯ CN ¯ NH3 (CH3CH2)3N CH3O¯ HO¯ CH3CH2O¯ (CH3)3CO¯ NH2¯ *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)
The Saytzeff elimination • Saytzeff elimination: in the elimination process, formation of the most substituted double bond is favored
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
Problems • 6.20: Predict the products of the E1 and SN1 reactions of the following molecules in water:
The E2 elimination • As in the SN2 reaction, the rate is dependent on the concentration of both reaction partners n = k [substrate] [base]
E2 vs SN2 • There is a strong similarity between the E2 and SN2 reactions: strong nucleophiles favor substitution, while strong bases favor elimination
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)
Other examples • Note that at a primary carbon atom, only SN2 and no SN1 substitution is possible
Stereochemistry of the E2 rxn Possible orientations of the proton that eliminates and the leaving group
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
Resemblance to SN2 • The mechanism is ‘more or less’ analogous to the SN2 substitution • The process is also called anti-elimination or antiperiplanar elimination
How to visualize this aspect ? • What are the products of E2 elimination of both diastereomers ?
Regiochemistry • Generally, formation of the most substituted double bond is favored • Note the occurrence of E- and Z-isomers
Effect of the leaving group • Especially quarternary ammonium leaving groups favor the Hofmann product
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
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
Example of an E1cB reaction • This reaction is possible if there is a group present that can stabilize the negative charge
Stability of anions • The more substituted the carbanion, the less stable it is; this is a result of the inductively electron donating alkyl groups
Formation of CC-bonds • Formation of CC-bonds is a synthetically important reaction
Problems • Make problems: 6.30, 6.31, 6.32, 6.38, 6.53, 6.54, 6.56, 6.59
