chapter 8: alkene structure and preparation via elimination reactions ...€¦ · chapter 8: alkene...
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Chapter 8: Alkene Structure and Preparation via Elimination Reactions
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
• unlike sigma bonds where rotation takes place readily, rotation about a pi bond wouldrequire breaking the pi bond (loss of overlap of the P-orbitals) and therefore does nottake place under normal condition• this gives rise to two different stereoisomers. Relationship?
Nature of the pi bond
HH
HCC
H
• 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.
[Sections: 8.1-8.13]
H3C
HCH3
HH3C
H3CH
HH3C
H
CH3H
H3C CH3
H2C CH2
bondlength
bondstrength
Alkene Stereochemistry and Eyesight
Alkene Nomenclature With More Than Two Substituents on the Double Bond
Cl CH3
Cl CH3
H3C
• look at one "end" of the double bond• assign priority of one group or atom over the otherbased 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
HO H
ClH2N
OHCl
BrCl
Z or E? Draw the other isomer
Problems: 1,2
Plan of Attack for naming alkenes
Cl CH3
H3C H
same endsame end
same side
same side
Br
complete name?
Relative Stability of Alkene Isomers
cis-2-buteneO2
CO2 + H2O + kJ/mol
trans-2-buteneO2
CO2 + H2O + kJ/mol
E
2,712 kJ/mol 2,710 kJ/mol 2,707 kJ/mol 2,700 kJ/mol
4 CO2 + 4 H2O (heats of combustion)
Alkenes and Pheromones
OH
1-octene-3-ol
mosquitosand fly
9-methyl-α-himalachenemale sex pheromone
oriental fruit moth
O
O
(E)-8-dodecen-1-yl acetate 7%
O
O
(Z)-8-dodecen-1-yl acetate 93%
sex phermonone mixture
H
H
H
H
H
H
R
H
R
H
R
H
H
R
R
H
R
R
H
H
R
R
R
H
R
R
R
R< < < < < <
Alkene Substitution Patterns
unsub monodisubstituted
tri tetra
stability
A B C D
Predict the relative stabilities of the following isomeric alkenes
Bimolecular Elimination Mechanism
H
HH Br
HH
HO
E
• identify starting materials and products• exothermic or endothermic?• multistep or concerted?• RDS = unimolecular or bimolecular?
rate law:• dependent upon concentrations of compounds during (and prior to) the RDS
rate = k
reaction name:
Problems: 3, 11
Predict the products of the following E2 reactions:
KOHEtOH, heat
Br
• E2 elimination results in the formation of all possible alkene products, including stereoisomers• E2 elimination using strong, small bases (HO–, MeO–, EtO–) 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:
OTs
Br
NaOtBu
tBuOH
NaOCH2CH3
ethanol, heat
NaNH2NH3, heat
• 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!
Plan of Attack for E2 reactions
I
Problems: 6, 8
• 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, SN2 is impossible, so E2 is the only possible reaction route• there are ways to influence reactivity towards one direction or another:
The SN2 reaction prefers strong nucleophiles that are weak bases, polar aprotic solvents and minimal, or no, heatThe E2 pathway prefers strong bases, polar protic solvents, and heat
Always Nucleophiles
X–, HS–, RS–, H2SRSH, CN–
Always BasestBuO–, LDANaH, NaNH2
Either/Or
HO–, RO–
Br
OCH3
NaOH
DMSO
NaOH
CH3OH, heat
H
HH Br
HH
HO
Substitution (SN2) or Elimination (E2)?solvent heat
NH3
O
N Li
Unimolecular Elimination Mechanism
Most commonly used E1 type reaction is the dehydration of alcohols
OH
H3PO4or
H2SO4heat
• alcohols may react via either the SN1 or the E1 reaction• in both cases an acid is required to protonate the OH group to convert it to a good leaving group• the intermediate carbocation is then either trapped by a nucleophile (SN1) or loses a β-hydrogen (E1)• elimination is guaranteed to occur if the acid is H3PO4 or H2SO4 since the counterions from these acids are not nucleophilic• heat also favors the elimination process
OH
H3PO4heat
Draw all the products expected. Circle the major product.
Problems: 4,5,7,9,10,12
Br
EtOH
Brtert-butanol
heat
H3PO4or
H2SO4NaBr
The Four Basic Reaction Mechanisms
SN2
Nu R LG Nu R
I- > Br- > Cl- or OTs- or H2O [from protonation of OH]
Leaving groups
NucleophilesX-, HS-, H2S, RSH, CN-, N3
- always act as nucleophilesHO-, RO-, NH3 can act as nucleophiles or bases
BasesHO-, RO-, NH3 can act as bases or nucleophiles
tBuO-, LDA, NaH, NaNH2, always act as bases
• methyl > 1° > 2° (NO 3°)• no intermediates, one step reaction (concerted)• rate = k [Nu][R-X]• inversion of configuration• favored by strong nucleophiles• favored by polar aprotic solvents
SN1
R LG Nu R
• 3° > 2° (NO 1° or methyl)• carbocation intermediate (multistep reaction)• carbocation subject to rearrangement• rate = k [R-X]• usually weak nucleophiles (solvent is often the nucleophile as well = solvolysis reaction)• favored by polar protic solvents
R Nu
E2 • can occur with all types of substrates• no intermediates, one step reaction (concerted)• rate = k [Base][R-X]• favored by strong bases (small bases favor Zaitsev elimination)• favored by polar protic solvents• heating favors elimination over substitution
E1 • 3° > 2° (NO 1° or methyl)• carbocation intermediate (multistep reaction)• carbocation subject to rearrangement• rate = k [R-X]• usually very weak base present (solvent molecule or, for dehydration reactions, H2PO4
- or HSO4-)
• favored by polar protic solvents• heating favors elimination over substitution
BLG
Hβ
RR
RR R
R
R
R
LG
Hβ
RR
RR
R
Hβ
RR
R R
R
R
R
-H+