Chapter 8 Alkenes and Alkynes II: Addition Reactions Alkenes are ...
Chapter 5 Structure and Preparation of Alkenes: Elimination Reactions
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Transcript of Chapter 5 Structure and Preparation of Alkenes: Elimination Reactions
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Chapter 5Chapter 5Structure and Preparation of Alkenes:Structure and Preparation of Alkenes:
Elimination ReactionsElimination Reactions
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5.15.1Alkene NomenclatureAlkene Nomenclature
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Alkenes
Alkenes are hydrocarbons that contain a carbon-carbon double bond
also called "olefins"
characterized by molecular formula CnH2n
said to be "unsaturated"
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H2C CH2 H2C CHCH3
Ethene Propene
Alkene Nomenclature
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1) Find the longest continuous chain that includes the double bond.
2) Replace the -ane ending of the unbranched alkane having the same number of carbons with -ene.
3) Number the chain in the direction that gives the lowest number to the doubly bonded carbon.
H2C CHCH2CH3 1-Butene orBut-1-ene
Alkene Nomenclature
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4) If a substituent is present, identify its position by number. The double bond takes precedence over alkyl groups and halogens when the chain is numbered.
The compound shown above is4-bromo-3-methyl-1-butene.
(or 4-bromo-3-methylbut-1-ene)
H2C CHCHCH2Br
CH3
Alkene Nomenclature
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4) If a substituent is present, identify its position by number. Hydroxyl groups take precedence over the double bond when the chain is numbered.
The compound shown above is2-methyl-3-buten-1-ol.
(or 2-methyl-but-3-en-1-ol)
H2C CHCHCH2OH
CH3
Alkene Nomenclature
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Alkenyl Groups
methylene
vinyl
allyl
isopropenyl
CHH2C
H2C CHCH2
H2C CCH3
H2C
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Cycloalkene Nomenclature
1) Replace the -ane ending of the cycloalkane having the same number of carbons with -ene.
Cyclohexene
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Cycloalkene Nomenclature
1) Replace the -ane ending of the cycloalkane having the same number of carbons with -ene.
2) Number through the double bond in thedirection that gives the lower number to the first-appearing substituent.
6-Ethyl-1-methylcyclohexeneCH3
CH2CH3
1
23
4
5 6
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5.25.2Structure and Bonding in AlkenesStructure and Bonding in Alkenes
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Structure of Ethylene
bond angles: H-C-H = 117°
H-C-C = 121°
bond distances: C—H = 110 pm
C=C = 134 pm
planar
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Bonding in Ethylene
Framework of bonds
Each carbon is sp2 hybridized
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Bonding in Ethylene
Each carbon has a half-filled p orbital.
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Bonding in Ethylene
Side-by-side overlap of half-filled p orbitals gives a bond.
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5.35.3Isomerism in AlkenesIsomerism in Alkenes
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Isomers are different compounds thathave the same molecular formula.
Isomers
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Isomers
Constitutional isomersConstitutional isomers StereoisomersStereoisomers
different connectivity same connectivity;different arrangementof atoms in space
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Isomers
Constitutional isomersConstitutional isomers StereoisomersStereoisomers
consider the isomeric alkenes of molecular formula C4H8
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2-Methylpropene1-Butene
cis-2-Butene trans-2-Butene
C C
H
H H
CH2CH3
H3C
C C
CH3
H
HH
CH3
C C
H3C
H
C C
H
HH3C
H3C
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2-Methylpropene1-Butene
cis-2-Butene
C C
H
H H
CH2CH3
H
CH3
C C
H3C
H
C C
H
HH3C
H3C
Constitutional isomers
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2-Methylpropene1-Butene
trans-2-Butene
C C
H
H H
CH2CH3
H3C
C C
CH3
H
H
C C
H
HH3C
H3C
Constitutional isomers
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cis-2-Butene trans-2-Butene
H3C
C C
CH3
H
HH
CH3
C C
H3C
H
Stereoisomers
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Stereochemical Notation
trans (identical or analogous substituents on opposite sides)
cis (identical or analogous substituents on same side)
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transcis
Interconversion of stereoisomericalkenes does not normally occur.
Requires that component of doublebond be broken.
Rotation Barrier in Alkenes
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Rotation Barrier in Alkenes
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5.45.4Naming Steroisomeric AlkenesNaming Steroisomeric Alkenesby the E-Z Notational Systemby the E-Z Notational System
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Stereochemical Notation
Cis and trans are useful when substituents are identical or analogous (oleic acid has a cis double bond).
Cis and trans are ambiguous when analogies are not obvious.
C C
CH3(CH2)6CH2 CH2(CH2)6CO2H
H H
Oleic acid
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Example
What is needed:
1) Systematic body of rules for ranking substituents
2) New set of stereochemical symbols otherthan cis and trans
C C
H F
Cl Br
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C C
The E-Z Notational System
E : higher ranked substituents on opposite sides
Z : higher ranked substituents on same side
higher
lower
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C C
The E-Z Notational System
E : higher ranked substituents on opposite sides
Z : higher ranked substituents on same side
higher
lower
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C C
The E-Z Notational System
E : higher ranked substituents on opposite sides
Z : higher ranked substituents on same side
Entgegen
higher
higherlower
lower
C C
Zusammen
lower
higher
lower
higher
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C C
The E-Z Notational System
Entgegen
higher
higherlower
lower
C C
Zusammen
lower
higher
lower
higher
Answer: They are ranked in order of decreasing atomic number.
Question: How are substituents ranked?
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The Cahn-Ingold-Prelog (CIP) System
The system that we use was devised byR. S. CahnSir Christopher IngoldVladimir Prelog
Their rules for ranking groups were devised in connection with a different kind of stereochemistry—one that we will discuss in Chapter 7—but have been adapted to alkene stereochemistry.
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Table 5.1 CIP Rules
(1) Higher atomic number outranks lower atomic number
Br > F Cl > H
higher
lower
Br
F
Cl
H
higher
lower
C C
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Table 5.1 CIP Rules
(1) Higher atomic number outranks lower atomic number
Br > F Cl > H
higher
lower
Br
F
Cl
H
higher
lower
C C
(Z )-1-Bromo-2-chloro-1-fluoroethene
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Table 5.1 CIP Rules
(2) When two atoms are identical, compare the atoms attached to them on the basis of their atomic numbers. Precedence is established at the first point of difference.
—CH2CH3 outranks —CH3
—C(C,H,H) —C(H,H,H)
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Table 5.1 CIP Rules
(3) Work outward from the point of attachment, comparing all the atoms attached to a particular atom before proceeding furtheralong the chain.
—CH(CH3)2 outranks —CH2CH2OH
—C(C,C,H) —C(C,H,H)
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Table 5.1 CIP Rules
(4) Evaluate substituents one by one. Don't add atomic numbers within groups.
—CH2OH outranks —C(CH3)3
—C(O,H,H) —C(C,C,C)
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Table 5.1 CIP Rules
(5) An atom that is multiply-bonded to another atom is considered to be replicated as a
substituent on that atom.
—CH=O outranks —CH2OH
—C(O,O,H) —C(O,H,H)
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Table 5.1 CIP Rules
A table of commonly encountered substituents ranked according to precedence is given on the inside back cover of the text.
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5.55.5
Physical Properties of AlkenesPhysical Properties of Alkenes
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= 0 D
C C
H H
HH
= 0.3 D
H
H H
C C
H3C
Dipole Moments
What is direction of dipole moment?
Does a methyl group donate electrons to the double bond, or does it withdraw them?
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= 0 D
C C
H H
HH
= 0.3 D
H
H H
C C
H3C
Dipole Moments
= 1.4 D
C C
H H
ClHChlorine is electronegative and attracts electrons.
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= 0.3 D
H
H H
C C
H3C
Dipole Moments
= 1.4 D
C C
H H
ClH
Dipole moment of 1-chloropropene is equal to the sum of the dipole moments of vinyl chloride and propene. = 1.7 D
H
H Cl
C C
H3C
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= 0.3 D
H
H H
C C
H3C
Dipole Moments
= 1.4 D
C C
H H
ClH
= 1.7 D
H
H Cl
C C
H3CTherefore, a methyl group donates electrons to the double bond.
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Alkyl Groups Stabilize sp2 HybridizedCarbon by Releasing Electrons
R—C+ H—C+is more stable than
is more stable thanR—C• H—C •
R—C is more stable than H—C
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5.65.6Relative Stabilities of AlkenesRelative Stabilities of Alkenes
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Double Bonds are Classified According tothe Number of Carbons Attached to Them
H
C C
R
H
H
monosubstituted
R'
C C
R
H
H
disubstituted
H
C C
R
H
R'
disubstituted
H
C C
R H
R'
disubstituted
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Double Bonds are Classified According tothe Number of Carbons Attached to Them
R'
C C
R
H
R"
trisubstituted
R
R'
C C
R"'
R"
tetrasubstituted
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Substituent Effects on Alkene Stability
Electronic
Disubstituted alkenes are more stable than monosubstituted alkenes.
Steric
Trans alkenes are more stable than cis alkenes.
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Figure 5.3 Heats of Combustion of C4H8
Isomers
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Substituent Effects on Alkene Stability
Alkyl groups stabilize double bonds more than H.
More highly substituted double bonds are morestable than less highly substituted ones.
Electronic
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Problem 5.12
C C
H3C
H3C CH3
CH3
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Substituent Effects on Alkene Stability
Trans alkenes are more stable than cis alkenes.
Cis alkenes are destabilized by van der Waalsstrain.
Steric
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cis-2-butene trans-2-butene
van der Waals straindue to crowding ofcis-methyl groups
Figure 5.4 cis and trans-2-Butene
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cis-2-butene trans-2-butene
van der Waals straindue to crowding ofcis-methyl groups
Figure 5.4 cis and trans-2-Butene
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van der Waals Strain
Steric effect causes a large difference in stabilitybetween cis and trans-(CH3)3CCH=CHC(CH3)3.
Cis is 44 kJ/mol less stable than trans.
C C
H H
CC CH3
CH3H3C
H3C
H3C CH3
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5.75.7CycloalkenesCycloalkenes
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Cycloalkenes
Cyclopropene and cyclobutene have angle strain.
Larger cycloalkenes, such as cyclopenteneand cyclohexene, can incorporate a double bond into the ring with little or no angle strain.
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Stereoisomeric Cycloalkenes
Cis-cyclooctene and trans-cycloocteneare stereoisomers.
Cis-cyclooctene is 39 kJ/mol more stablethan trans-cyclooctene.
cis-Cyclooctene trans-Cyclooctene
H
H HH
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Stereoisomeric Cycloalkenes
Trans-cyclooctene is smallest trans-cycloalkene that is stable at room temperature.
Cis stereoisomer is more stable than trans through C11 cycloalkenes.
trans-Cyclooctene
HH
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When there are more than 12 carbons in thering, trans-cycloalkenes are more stable than cis.The ring is large enough so the cycloalkene behaves much like a noncyclic one.
Stereoisomeric Cycloalkenes
Cis and trans-cyclododeceneare approximately equal in stability.
trans-Cyclododecenecis-Cyclododecene
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5.8
Preparation of Alkenes:
Elimination Reactions
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X Y
dehydrogenation of alkanes:X = Y = H
dehydration of alcohols:X = H; Y = OH
dehydrohalogenation of alkyl halides:X = H; Y = Br, etc.
C CC C + X Y
-Elimination Reactions Overview
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limited to industrial syntheses of ethylene, propene, 1,3-butadiene, and styrene
important economically, but rarely used in laboratory-scale syntheses
750°CCH3CH3
750°CCH3CH2CH3
H2C CH2 + H2
H2C CHCH3 + H2
Dehydrogenation
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5.9Dehydration of Alcohols
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H2SO4
160°CCH3CH2OH H2C CH2 + H2O
(82%)
C OH
CH3
CH3
H3CH2SO4
heatCH2
H3C
C
H3C
+ H2O
Dehydration of Alcohols
(79-87%)
H2SO4
140°C
+ H2O
OH
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R
R'
R"
OHC
R
R'
H
OHC
R
H
H
OHC
Relative Reactivity
tertiary:most reactive
primary:least reactive
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5.10Regioselectivity in Alcohol Dehydration:
The Zaitsev Rule
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10 % 90 %
HO
H2SO4
80°C+
Regioselectivity
A reaction that can proceed in more than one direction, but in which one direction predominates, is said to be regioselective.
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Regioselectivity
A reaction that can proceed in more than one direction, but in which one direction predominates, is said to be regioselective.
84 %
H3PO4
heat+
16 %
CH3
OH
CH3 CH3
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When elimination can occur in more than one direction, the principal alkene is the one formed by loss of H from the carbon having thefewest hydrogens.
R OH
CH3
C C
H
R CH2R
three protons on this carbon
The Zaitsev Rule
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When elimination can occur in more than one direction, the principal alkene is the one formed by loss of H from the carbon having thefewest hydrogens.
two protons on this carbon
The Zaitsev Rule
R OH
CH3
C C
H
R CH2R
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When elimination can occur in more than one direction, the principal alkene is the one formed by loss of H from the carbon having thefewest hydrogens.
R
R
CH2R
CH3
C C
only one proton on this carbon
The Zaitsev Rule
R OH
CH3
C C
H
R CH2R
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5.11
Stereoselectivity in Alcohol Dehydration
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A stereoselective reaction is one in which a single starting material can yield two or more stereoisomeric products, but gives one of them in greater amounts than any other.
(25%) (75%)
+
OH
H2SO4
heat
Stereoselectivity
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5.12
The E1 and E2 Mechanisms
of Alcohol Dehydration
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The dehydration of alcohols and the reaction of alcohols with hydrogen halides share thefollowing common features:
1) Both reactions are promoted by acids
2) The relative reactivity decreases in theorder tertiary > secondary > primary
These similarities suggest that carbocations areintermediates in the acid-catalyzed dehydration ofalcohols, just as they are in the reaction of alcoholswith hydrogen halides.
A Connecting Point...
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First two steps of mechanism are identical tothose for the reaction of tert-butyl alcohol withhydrogen halides.
Dehydration of tert-Butyl Alcohol
C OH
CH3
CH3
H3CH2SO4
heatCH2
H3C
C
H3C
+ H2O
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Step 1: Proton transfer to tert-butyl alcohol
(CH3)3C O
H
..: H O+
..
H
H
+
H
O: +(CH3)3C
H+
fast, bimolecular
tert-Butyloxonium ion
O
H
:
H
:
Mechanism
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+
Step 2: Dissociation of tert-butyloxonium ionto carbocation
(CH3)3C O
H
:
H +
slow, unimolecular
(CH3)3C O
H
:
H
:
tert-Butyl cation
+
Mechanism
Because rate-determiningstep is unimolecular, thisis called the E1 mechanism.
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Step 3: Deprotonation of tert-butyl cation
+ O
H
:
H
:
CH2+
H3C
C
H3C
H
fast, bimolecular
CH2
H3C
C
H3C
+ O
H
:
H
H+
Mechanism
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are intermediates in the acid-catalyzed dehydration of tertiary and secondary alcohols
carbocations can:
react with nucleophileslose a -proton to form an alkene
Carbocations
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avoids carbocation because primary carbocations
are too unstable
oxonium ion loses water and a proton in abimolecular step
H2SO4
160°CCH3CH2OH H2C CH2 + H2O
Dehydration of Primary Alcohols
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Step 1: Proton transfer from acid to ethanol
H
..: H O+OCH3CH2 ..
H
H
H
O: +
H+
fast, bimolecular
Ethyloxonium ion
CH3CH2 O
H
:
H
:
Mechanism
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Step 2: Oxonium ion loses both a proton and a water molecule in the same step.
+
H
O:
H+
CH2CH2HO
H
:
H
:
slow, bimolecular
+ O
H
:
H
:O
H
H
: H+
H2C CH2+
Mechanism
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Step 2: Oxonium ion loses both a proton and a water molecule in the same step.
+
H
O:
H+
CH2CH2HO
H
:
H
:
slow, bimolecular
+ O
H
:
H
:O
H
H
: H+
H2C CH2+
Mechanism
Because rate-determiningstep is bimolecular, thisis called the E2 mechanism.
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OH
H3PO4, heat
3%
64% 33%
+ +
Example
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3% CH3
CHCH3CH3
CH3
+C
Carbocation can lose a proton as shown;
or it can undergo a methyl migration.
CH3 group migrates with its pair of electrons to adjacent positively charged carbon.
Rearrangement Involves Alkyl Group Migration
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3%
CH3
CHCH3
CH3
+CH3
97% CH3
CHCH3CH3
CH3
+C C
tertiary carbocation; more stable
Rearrangement Involves Alkyl Group Migration
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3%
CH3
CHCH3
CH3
+CH3
97% CH3
CHCH3CH3
CH3
+C C
Rearrangement Involves Alkyl Group Migration
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CH3CH2CH2CH2OH
H3PO4, heat
12%
+
mixture of cis (32%)and trans-2-butene (56%)
CH2CH3CH2CH CHCH3CH3CH
Another Rearrangement
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Oxonium ion can losewater and a proton(from C-2) to give1-butene;
doesn't give a carbocation directlybecause primarycarbocations are toounstable.
CH3CH2CH2CH2 O
H
H
+:
CH2CH3CH2CH
Rearrangement Involves Hydride Shift
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Hydrogen migrates with its pair of electrons from C-2 to C-1 as water is lost;
carbocation formed by hydride shift is secondary.
CH3CH2CH2CH2 O
H
H
+:
CH2CH3CH2CH
CH3CH2CHCH3+
Rearrangement Involves Hydride Shift
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H
CH3CH2CH2CH2 O
H
+:
CH2CH3CH2CH
CH3CH2CHCH3+
mixture of cisand trans-2-butene
CHCH3CH3CH
Rearrangement Involves Hydride Shift
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H
H
CH3CH2CHCH2 O
H
+:
+CH3CH2CHCH2 +
H
O
H
H
::
Hydride Shift
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• react with nucleophiles
• lose a proton from the -carbon to form an alkene
• rearrange (less stable to more stable)
Carbocations Can...
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5.145.14
Dehydrohalogenation of Dehydrohalogenation of
Alkyl Halides Alkyl Halides
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X Y
dehydrogenation of alkanes:X = Y = H
dehydration of alcohols:X = H; Y = OH
dehydrohalogenation of alkyl halides:X = H; Y = Br, etc.
C CC C + X Y
-Elimination Reactions Overview
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dehydrogenation of alkanes:industrial process; not regioselective
dehydration of alcohols:acid-catalyzed
dehydrohalogenation of alkyl halides:consumes base
-Elimination Reactions Overview
X YC CC C + X Y
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is a useful method for the preparation of alkenes
(100 %)
likewise, NaOCH3 in methanol, or KOH in ethanol
NaOCH2CH3
ethanol, 55°C
Dehydrohalogenation
Cl
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CH3(CH2)15CH2CH2Cl
When the alkyl halide is primary, potassiumtert-butoxide in dimethyl sulfoxide is the base/solvent system that is normally used.
KOC(CH3)3
dimethyl sulfoxide
(86%)
CH2CH3(CH2)15CH
Dehydrohalogenation
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Br
29 % 71 %
+
Regioselectivity
follows Zaitsev's rule:
more highly substituted double bond predominates
KOCH2CH3
ethanol, 70°C
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more stable configurationof double bond predominates
Stereoselectivity
KOCH2CH3
ethanol
Br
+
(23%) (77%)
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more stable configurationof double bond predominates
Stereoselectivity
KOCH2CH3
ethanol
+
(85%) (15%)
Br
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5.155.15 The E2 Mechanism of The E2 Mechanism of
Dehydrohalogenation of Alkyl Dehydrohalogenation of Alkyl HalidesHalides
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Facts
(1) Dehydrohalogenation of alkyl halides exhibits second-order kinetics
first order in alkyl halidefirst order in baserate = k[alkyl halide][base]
implies that rate-determining step involves both base and alkyl halide; i.e., it is bimolecular
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Facts
(2) Rate of elimination depends on halogen
weaker C—X bond; faster raterate: RI > RBr > RCl > RF
implies that carbon-halogen bond breaks in the rate-determining step
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concerted (one-step) bimolecular process
single transition state
C—H bond breaks
component of double bond forms
C—X bond breaks
The E2 Mechanism
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–OR..
.. :
C C
H
X..::
Reactants
The E2 Mechanism
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C C
–
OR..
.. H
X..::–
Transition state
The E2 Mechanism
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OR..
.. H
C C
–X..
::..
Products
The E2 Mechanism
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5.165.16
Anti Elimination in E2 Reactions: Anti Elimination in E2 Reactions:
Stereoelectronic EffectsStereoelectronic Effects
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(CH3)3C
(CH3)3C
Br
KOC(CH3)3
(CH3)3COH
cis-1-Bromo-4-tert- butylcyclohexane
Stereoelectronic Effect
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(CH3)3C
(CH3)3CBr KOC(CH3)3
(CH3)3COH
trans-1-Bromo-4-tert- butylcyclohexane
Stereoelectronic Effect
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(CH3)3C
(CH3)3C
Br
(CH3)3C
Br
KOC(CH3)3
(CH3)3COH
KOC(CH3)3
(CH3)3COH
cis
trans
Rate constant for dehydrohalogenation of cis is 500 times greater than that of trans.
Stereoelectronic Effect
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(CH3)3C
(CH3)3C
Br
KOC(CH3)3
(CH3)3COH
cis
H that is removed by base must be anti coplanar to Br.
Two anti coplanar H atoms in cis stereoisomer
HH
Stereoelectronic Effect
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(CH3)3C
KOC(CH3)3
(CH3)3COH
trans
H that is removed by base must be anti coplanar to Br.
No anti coplanar H atoms in trans stereoisomer; all vicinal H atoms are gauche to Br.
HH
(CH3)3CBr
H
H
Stereoelectronic Effect
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cis
more reactive
trans
less reactive
Stereoelectronic Effect
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Stereoelectronic Effect
An effect on reactivity that has its origin in the spatial arrangement of orbitals or bonds is called a stereoelectronic effect.
The preference for an anti coplanar arrangement of H and Br in the transition state for E2 dehydrohalogenation is an example of a stereoelectronic effect.
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5.175.17
Isotopes Effects And The E2 Isotopes Effects And The E2
MechanismMechanism
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A C-D bond is 12 kJ/mol stronger than a C-H bond. The activation energy for breaking a C-D bond is greater than for breaking a C-H bond. The rate constant k for an elementary step where C-D breaks is smaller than for a C-H bond.
The difference in rate is expressed as a ratio kH/kD, and is a kinetic isotope effect. Because it compares 2H to 1H, it is called a deuterium isotope effect.
The Isotope Effect
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In the rate determining step of the E2 mechanism,
a base removes a proton from a carbon.
The mechanism should exhibit a deuterium isotope
effect.
The Isotope Effect
D3CCHCD3
Br
D2C=CHCD3
kH/kD = 6.7
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5.185.18
The E1 Mechanism ofThe E1 Mechanism of
Dehydrohalogenation of Alkyl Dehydrohalogenation of Alkyl
HalidesHalides
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CH3 CH2CH3
Br
CH3
C
Ethanol, heat
+
(25%) (75%)
H3C
CH3
C C
H3C
H
CH2CH3
CH3
CH2C
Example
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1. Alkyl halides can undergo elimination in absence of base.
2. Carbocation is intermediate.
3. Rate-determining step is unimolecular ionization of alkyl halide.
The E1 Mechanism
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CH3 CH2CH3
Br
CH3
C
..::
CCH2CH3CH3
CH3
+
:..: Br.. –
slow, unimolecular
Step 1
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CCH2CH3CH3
CH3
+
CH3CH3
CCH2CH3
CH2
+ CCHCH3
CH3
– H+
Step 2