Nature of the pi bond Chapter 8: Alkene Structure and Preparation via Elimination eactions [Sections: 8.1-8.13] C C 3 C C 3 bond length bond strength 2 C C 2 a C=C double bond is stronger than a C C single bond the pi bond component, however, is generally weaker than the sigma bond component. Nomenclature of Alkenes the longest continuous carbon chain containing both carbons of the C=C bond is the parent chain assign the C=C bond the lowest locant value in cyclic molecules the C=C bond is always given the 1,2-designation Stereochemistry of Alkenes 3 C C 3 3 C C 3 3 C 3 C unlike sigma bonds where rotation takes place readily, rotation about a pi bond would require breaking the pi bond (loss of overlap of the P-orbitals) and therefore does not take place under normal condition this gives rise to two different stereoisomers. elationship?
Alkene Stereochemistry and Eyesight same side same end Cl C 3 same end 3 C same side Alkene Nomenclature With More Than Two Substituents on the Double Bond Cl C 3 Plan of Attack for naming alkenes look at one "end" of the double bond assign priority of one group or atom over the other based on atomic number repeat the process for the other end of the double bond if the two high-priority groups are on: the same side of the double bond = Z isomer the opposite sides of the double bond = E isomer cis isomers can also be called Z isomers, trans = E Cl C 3 3 C Z or E? Draw the other isomer 2 N Cl Cl complete name? Cl Problems: 1,2
Alkenes and Pheromones mosquito 1-octene-3-ol sand fly 9-methyl-α-himalachene male sex pheromone sex phermonone mixture (E)-8-dodecen-1-yl acetate 7% oriental fruit moth (Z)-8-dodecen-1-yl acetate 93% elative Stability of Alkene Isomers cis-2-butene 2 C 2 + 2 + kj/mol trans-2-butene 2 C 2 + 2 + kj/mol E 2,712 kj/mol 2,710 kj/mol 2,707 kj/mol 2,700 kj/mol 4 C 2 + 4 2 (heats of combustion)
Alkene Substitution Patterns unsub mono disubstituted tri tetra < < < < < < stability Predict the relative stabilities of the following isomeric alkenes A B C D Problems: 3, 11 Bimolecular Elimination Mechanism E rate law: dependent upon concentrations of compounds during (and prior to) the DS rate = k identify starting materials and products exothermic or endothermic? multistep or concerted? DS = unimolecular or bimolecular? reaction name:
Predict the products of the following E2 reactions: K Et, heat E2 elimination results in the formation of all possible alkene products, including stereoisomers E2 elimination using strong, small bases (, Me, Et ) results in formation of the most stable alkene product (Zaitsev's rule) the most stable alkene is the most substituted Predict all of the products of the following E2 reactions. Circle the major product: Ts NatBu tbu Plan of Attack for E2 reactions identify the base identify the leaving group locate all β-hydrogens draw products from removal of a β-hydrogen and the leaving group draw stereoisomers where relevant check to make sure no products are duplicated! NaN 2 N 3, heat I NaC 2 C 3 ethanol, heat Problems: 6, 8
Substitution (S N 2) or Elimination (E2)? solvent heat 1 and 2 substrates could go via either pathway, and often the two reaction mechanisms compete with one another such that a little substitution occurs even though elimination is the preferred pathway, and a little elimination occurs even though substitution is the preferred pathway for 3 substrates, S N 2 is impossible, so E2 is the only possible reaction route there are ways to influence reactivity towards one direction or another: The S N 2 reaction prefers strong nucleophiles that are weak bases, polar aprotic solvents and minimal, or no, heat The E2 pathway prefers strong bases, polar protic solvents, and heat Always Nucleophiles X, S, S, 2 S S, CN Either/r, N 3 Always Bases tbu, LDA Na, NaN 2 N Li Na DMS Na C 3, heat C 3
Unimolecular Elimination Mechanism Et tert-butanol heat Most commonly used E1 type reaction is the dehydration of alcohols 3 P 4 or 2 S 4 Na 3 P 4 or 2 S 4 heat alcohols may react via either the S N 1 or the E1 reaction in both cases an acid is required to protonate the group to convert it to a good leaving group the intermediate carbocation is then either trapped by a nucleophile (S N 1) or loses a β-hydrogen (E1) elimination is guaranteed to occur if the acid is 3 P 4 or 2 S 4 since the counterions from these acids are not nucleophilic heat also favors the elimination process Draw all the products expected. Circle the major product. 3 P 4 heat Problems: 4,5,7,9,10,12
The Four Basic eaction Mechanisms Leaving groups Nucleophiles I - > - > Cl - or Ts - or 2 [from protonation of ] X -, S -, 2 S, S, CN-, N 3 - -, -, N 3 Bases -, -, N 3 tbu -, LDA, Na, NaN 2, always act as nucleophiles can act as nucleophiles or bases can act as bases or nucleophiles always act as bases S N 2 Nu LG Nu methyl > 1 > 2 (N 3 ) no intermediates, one step reaction (concerted) rate = k [Nu][-X] inversion of configuration favored by strong nucleophiles favored by polar aprotic solvents S N 1 LG Nu Nu 3 > 2 (N 1 or methyl) carbocation intermediate (multistep reaction) carbocation subject to rearrangement rate = k [-X] usually weak nucleophiles (solvent is often the nucleophile as well = solvolysis reaction) favored by polar protic solvents E2 B β LG can occur with all types of substrates no intermediates, one step reaction (concerted) rate = k [Base][-X] favored by strong bases (small bases favor Zaitsev elimination) favored by polar protic solvents heating favors elimination over substitution E1 β LG β - + 3 > 2 (N 1 or methyl) carbocation intermediate (multistep reaction) carbocation subject to rearrangement rate = k [-X] usually very weak base present (solvent molecule or, for dehydration reactions, 2 P - 4 or S - 4 ) favored by polar protic solvents heating favors elimination over substitution