comlete elimination reactions
TRANSCRIPT
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ELIMINATION REACTIONSELIMINATION REACTIONS
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An elimination reaction is one where startingmaterial loses the elements of a small moleculesuch as HCl or H2O or Cl2 during the course of the reaction to form the product.
ELIMINATION REACTIONSELIMINATION REACTIONS
- HClC CH Cl
C C
TWO EXAMPLES FOLLOW
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CH3CH2CH2CH2 Cl CH3CH2 CH CH2NaOH
CH3CH2 CH CH3Cl
CH3 CH CH CH3NaOH
ELIMINATION REACTIONSELIMINATION REACTIONS
CH3CH2CH2CH2 OH CH3CH2 CH CH2H2SO4
Alkyl halide + strong base and heat
Alcohol + strong acid and heat
LOSS OF HCl
LOSS OF H2O
TWO EXAMPLESE2
E1
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+ HClC H3C H2 C H C H3
C lC H3 C H C H C H3
- HCl
ELIMINATION IS THE REVERSE OF ADDITIONELIMINATION IS THE REVERSE OF ADDITION
basic conditions + heat
acidic conditionsconc. HCl
NaOH + heat
interconversions of alkyl halides and alkenes
C CH Cl
C C
ALKYL HALIDE ALKENE
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CH3 CH CH CH3CH3CH2 CH CH3
OH + H2O
- H2O
ELIMINATION IS THE REVERSE OF ADDITIONELIMINATION IS THE REVERSE OF ADDITION
strong acid conditions + heat
dilute aqueous acid conditions2-6 M H2SO4
conc. H2SO4
interconversions of alcohols and alkenes
C CH OH
C C
ALCOHOL ALKENE
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STRONG BASESSTRONG BASES
FIRST WE MUST LEARN WHAT IS A STRONG BASE
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WHAT ARE STRONG BASES ?WHAT ARE STRONG BASES ?
Na + 2 H2O2 2 NaOH + H2 (g)
NaOH sodium hydroxideused with H2O solvent, but most RCl insoluble in H2O
KOH potassium hydroxidesoluble in H2O, methanol or ethanol, most RCl soluble in methanol and ethanol
2 K + 2 H2O 2 KOH + H2 (g)
H-O....:
H-O....:
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NaOR sodium alkoxides
Na + 2 ROH2 2 NaOR + H2 (g)
examples:NaOCH3
NaOCH2CH3
(NaOMe)
(NaOEt)
sodium methoxide
sodium ethoxide
always used with the parent alcohol as solvent, most alkyl halides are soluble
STRONG BASES (continued)STRONG BASES (continued)
sodium t-butoxide
R-O....:
NaOC(CH3)3 (NaOtBu)Stronger basesthan hydroxides….. why?
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STRONG BASES (continued)STRONG BASES (continued)
NaNH2 sodium amidealways used with liquid ammonia solvent
2 Na + 2 NH3 2 NaNH2 + H2FeCl3NH3 (liq)
NH2
..:
ammoniabp -33.4 oC mp -77.7 oC
-33 oC
A gas at room temp
liquifiessolidifies
A stronger base thanthe hydroxides or thealkoxides …. why?
NH3
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SUMMARY OF STRONG BASESSUMMARY OF STRONG BASES
NaOH water
Base Solvents Allowed
KOH water, MeOH, EtOH
NaOR ROH (same R group)
NaNH2 NH3 (liq) -33o C
USED FOR ELIMINATION REACTIONS
Halides (RX) are not soluble in water, but aresoluble in most alcohols, therefore, KOH orsodium alkoxides in alcohol are most often used.
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ALKYL HALIDE + STRONG BASE
E2
FOR A DEHYDROHALOGENATION
defined later...
+ HEAT
LEARN THIS ! IT IS THE FORMULA
REACTION
shorthand designationfor this type of reaction
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THE REACTION IS A THE REACTION IS A -ELIMINATION-ELIMINATION
C CCl
H
The functional group is attached
to the -carbon.
-carbon
-carbon
The-hydrogen is attached to the
-carbon.
Since the -hydrogen is lost this reaction is
called a -elimination.
Reagent = a strong base
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C CC l
H
B:
THE BASE TAKES THE THE BASE TAKES THE -HYDROGEN-HYDROGEN
C C
C l
HB
: :..
: :..
..
MECHANISMMECHANISM
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REGIOSELECTIVITYREGIOSELECTIVITY
ALKYL HALIDE + STRONG BASE (E2)
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WHAT HAPPENS IF THERE IS MORE WHAT HAPPENS IF THERE IS MORE THAN ONE THAN ONE -HYDROGEN ?-HYDROGEN ?
C CCl
CHH
’
Which one do we lose ?
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CH3CH2 CH CH2
CH3CH2 CH CH3Br
CH3 CH CH CH3
ELIMINATION IS REGIOSELECTIVEELIMINATION IS REGIOSELECTIVE
major product81 %
minor product19 %
’
-H
’-HThe major product isthe one which has thelowest energy.See the next slide.
2-bromobutane
2-butene
1-butene
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Usually the pathwayleading to the lowestenergy product is the lowest energypathway (lower TS).
…. but not always
CH3-CH=CH-CH3
CH3CH2-CH=CH2
you get more trans than cis
(exceptions later)
HAMMOND POSTULATEHAMMOND POSTULATE
1-butene
2-butene
lowestenergypathway
lowestenergyproduct
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-30.3 -28.6 -27.6H
CH3CH2CH2CH3
+H2 +H2 +H2
kcal / mole
BUTENE ISOMERS - HEATS OF HYDROGENATIONBUTENE ISOMERS - HEATS OF HYDROGENATION
All are hydrogenated to the same product therefore theirenergies may be compared.
butane
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CH3
ClCH3 CH2
+NaOCH3
CH3OH /
major product minor product
’’ ’
CH3
’’=
’
MORE REGIOSELECTIVITYMORE REGIOSELECTIVITY
identical to-product
’’
1-chloro-1-methylcyclohexane1-methylcyclohexene
methylenecyclohexane
1-methylcyclohexene
three possibilitiesto lose -hydrogens
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-27.8 -25.4H+H2 +H2kcal
mole
METHYLCYCLOHEXENE ISOMERSMETHYLCYCLOHEXENE ISOMERS
Both are hydrogenated to the same product therefore their energies may be compared.
CH2
CH3
CH3
methylcyclohexane
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SAYTZEV RULESAYTZEV RULE
The reaction gives the most highly-substituted (lowest energy) alkene as the major product.
ZaitsevTsayseff
etc.
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H
H
R
H
H
H
R
RR
H
R
HH
R
R
H
R
H
R
R
monosubstitutedtrisubstituted
disubstituted
decreasing energy
tetrasubstituted
increasing substitution
R
R
R
R
THE MORE SUBSTITUTED ISOMER IS MORE STABLE
ALKENE ISOMERSALKENE ISOMERS Different positionsof the double bond.
cis
trans
1,1-
1,2-
1,2-
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CH3
ClCH3 CH2
+NaOCH3
CH3OH /
CH3 CH CH CH3CH3CH2 CH CH3Br
+
CH3CH2 CH CH2major product
81 %minor product
19 %
major product minor product
APPLICATIONS OF THE ZAITZEV RULEAPPLICATIONS OF THE ZAITZEV RULE
DISUBSTITUTED
TRISUBSTITUTED DISUBSTITUTED
MONOSUBSTITUTED
2-bromobutane
1-chloro-1-methylcyclohexane
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STEREOCHEMISTRY STEREOCHEMISTRY OF THE REACTIONOF THE REACTION
ALKYL HALIDE + STRONG BASE (E2)
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STEREOCHEMISTRYSTEREOCHEMISTRYTWO EXTREME POSSIBILITIES FOR THE ELIMINATION PROCESS
C CH Cl
syn elimination
C CH
Cl
anti elimination
not common
observedmost often
HH
Cl
Cl anti-coplanar
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ACYCLIC HALIDESACYCLIC HALIDES
STEREOCHEMISTRY
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HCl
H
Cl
ACYCLIC MOLECULES MAY HAVE TO ROTATE ACYCLIC MOLECULES MAY HAVE TO ROTATE IN ORDER TO REACTIN ORDER TO REACT
anti-coplanar
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CH CHCH3
CH3Br
C CHCH3
CH3 CH CHCH3
CH2
NaOEtEtOH
+
Major Product Zaitsev
Minor Product
2-Bromo-3-phenylbutane2-Bromo-3-phenylbutane
cis or trans ?
* *
*
two stereocenters
Is stereochemistry important?
RS SSSR RR
diastereomers
enantiomers
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C CH Br
HPhCH3 CH3 C C
Ph H
CH3CH3
C CBr
HCH3
H
CH3Ph C C
CH3 H
CH3Ph
R S
rotate
NaOMe
MeOH
anti-coplanar
not observed
observedmajor product
2S,3R-DIASTEREOMER2S,3R-DIASTEREOMER
trans
(plus some 1-butene due to -H on the CH3)
trans-2-phenyl-2-butene
(Z)
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C CH B r
HPhC H3 C H3
C CB r
HC H3
H
C H3Ph
C CC H3 H
C H3Ph
R S
C CH B r
HC H3
Ph C H3S S
makediastereomer
R S
?
Will the 2S,3S-diastereomergive the same product asits 2S,3R diastereomer?
trans (Z)
cis ortrans?
MAKING THE 2S,3S-DIASTEREOMERMAKING THE 2S,3S-DIASTEREOMER
( change one stereocenter )
WHAT DO YOU THINK?
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C CH Br
HCH3
Ph CH3 C CCH3 H
CH3Ph
C CBr
HCH3
H
PhCH3 C C
Ph H
CH3CH3
S S
rotate
NaOMe
MeOH
anti-coplanar
not observed
observedmajor product
2S,3S-DIASTEREOMER2S,3S-DIASTEREOMER
cis
(plus some 1-butene)
(E)
cis-2-phenyl-2-buteneA DIFFERENT PRODUCT ISFORMED THAN WITH 2S,3R !
NONO
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CH3CH3
Ph HPh H
CH3 CH3H
Cl
H
Cl
(top)
(bottom)
The methyl groups (blue) are in back in both structures.The phenyl and the hydrogen (black) are in front in both.
CONVERTING THE ALKYL HALIDE TO AN ALKENECONVERTING THE ALKYL HALIDE TO AN ALKENE
alkyl halide alkene
VISUALIZING THE PRODUCT THAT FORMS
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ANOTHER VISUALIZATION OF THE REACTION
H
H
Cl
Cl
H
Ph
CH3
CH3
CH3Ph
CH3H
2S,3R
2S,3S
CC
Ph CH3
HCH3
sameside
sameside
backcarbon
frontcarbon
CC
CH3 Ph
HCH3
sameside
sameside
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CYCLIC HALIDESCYCLIC HALIDES
STEREOCHEMISTRY REARS ITS UGLY HEAD AGAIN !
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CH3
Br
CH3 CH3+
1-Bromo-2-methylcyclohexane1-Bromo-2-methylcyclohexane
major product Zaitsev
The expected result (naïve) :
minor
The result actually depends on the stereochemistry of the starting material ( cis or trans ).
drawn flat withoutstereochemistry
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H
CH3
CH3
H
CH3
HBr
H
H
CH3
H
HBr
H
cis
The elimination needs to have H and Br anti-coplanar
plus a smallamount of
major product (Zaitsev)
Br
CH3
trans shown on next slide
THE THE CISCIS STEREOISOMER STEREOISOMER
CH3H
The other chairwon’t work. Why?
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H
CH3
CH3
H
CH3
HBr
H
H
CH3
H
HBr
H
transonly product
Br
CH3
THE THE TRANSTRANS STEREOISOMER STEREOISOMER
The other chairwon’t work. Why?
CH3
no methylcyclohexeneis formed
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THE RING MAY HAVE TO INVERT FOR REACTIONTHE RING MAY HAVE TO INVERT FOR REACTION
CH3
H
ClH H
CH3
Cl
HH
H
HCH3
chlorine is not anti-coplanar to any hydrogen
anti-coplanar
KOH / EtOH
Cl
CH3
this ringcannot react
invert
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Br
H
CCH3
CH3
CH3
HH
Br
CCH3
CH3
CH3
HCCH3
CH3
CH3
NaOEt
EtOH
EtOH
NaOEt
no reaction
trans e,e
cis e,a
REACTION DOES NOT OCCUR EASILY IF REACTION DOES NOT OCCUR EASILY IF ANTI-COPLANAR GEOMETRY CANNOT BE ACHIEVEDANTI-COPLANAR GEOMETRY CANNOT BE ACHIEVED
THESE RINGS WILL NOT INVERT( WHYNOT ? )
Br is not axial,no anti-coplanar H
hydrogensequivalent
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HAMMOND POSTULATEHAMMOND POSTULATE
The activation energy leading to the product of lower energy will be lower than the activation energy leading to the product of higher energy
….. unless stereochemistry or some other important factor interferes.
A CASE THAT FOLLOWS THE HAMMOND POSTULATE1
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H
BrCH3
HH H
Br
CH3H
CH3CH3 CH3
CH3
FOLLOWS HAMMOND POSTULATE *
ZAITSEV RULE
NaOEt / EtOHE2
major product
REGIOSELECTIVE
* The activation energy leading to the product of lower energy will be lower than the activation energy leading to the product of higher energy.
AND
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HAMMOND POSTULATEHAMMOND POSTULATE
A CASE THAT CAN NOT FOLLOW THE HAMMOND POSTULATE2
OR….STEREOCHEMISTRY REARS ITS UGLY HEAD !
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E2
H
BrCH3
HH H
Br
CH3H
CH3CH3CH3
CH3
H
DOES NOT FOLLOW HAMMOND POSTULATE
ORZAITSEV RULE *
NaOEt / EtOH
only product
REGIOSPECIFIC
* Due to stereochemical complications.
incorrectstereochemistryraises Ea
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ALKYL HALIDES + STRONG BASE + HEATALKYL HALIDES + STRONG BASE + HEAT....... continued
E2
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SUMMARYSUMMARY
STRONG BASE Required
REGIOSELECTIVE
Follows Zaitsev Rule (Unless Stereochemistry Prevents) - favors most substituted alkene
STEREOSPECIFIC
-H and X must be ANTI-COPLANAR
ELIMINATION REACTIONS OF ALKYL HALIDES
- acyclics may have to rotate- rings may have to invert
HEAT Usuallyrequired
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KINETICSKINETICS
Examination of the rate expression often helpsto understand the mechanism.The rate expression is determined by experiment.
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RATE EXPRESSIONSRATE EXPRESSIONS
RATE = K [A] [B]n m
REACTION ORDER = SUM OF EXPONENTS = n + m
rate “constant” concentrations of reactants A and B in moles / liter
exponents
Not all reactants will necessarily show up in the rate expression.
Fractional rate orders are possible :
Actually will change with temperature and solvent, the specific molecule, etc.
RATE = K [A] [B]3/2 2
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CH3CH2CH2 CH CH3
Br
+ NaOCH3
CH3CH2 CH CHCH3 + CH3OH
Rate [RBr] [OCH3]
x1 x1 x1x2 x2 x1x2 x1 x2x4 x2 x2x8 x4 x2x9 x3 x3
KINETICSKINETICS
Rate = + d ( CH3CH2 CH CHCH3)dt
second order rate
REACTION RATE DATA
35oC
Rate = K [RBr] [OCH3]
This equation explains therate behavior :
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MECHANISMMECHANISM
CH CHBr
H
B:
CONCERTED = only one stepAll bonds are broken and formed withoutthe formation of any intermediates.
EliminationBimolecular
E 2strong base
alkylhalide
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Reaction Order
Molecularity
Rate-determining Step
Transition State
Activated Complex
Sum of the exponents of the concentration termsin the rate expression.
Number of species that come together in therate-determining step.
The slowest step in the reaction sequence.
An energy highpoint in the energy profile of a reaction.
The species that exists at the transition state.
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CONCERTED REACTIONCONCERTED REACTION
E2
One Step - No Intermediates
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CH CHBr
H
B: BH
CH CH
Brmechanism activated complex
E2 ELIMINATIONE2 ELIMINATION
Concerted : everythinghappens at once with-out any intermediates.
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Concerted (one step) reactionConcerted (one step) reaction
product
startingmaterial
transition state TS
activationenergy Ea
heat ofreactionH
ENERGY
This is what E2 looks like.
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The anti-coplanar arrangement of the -H andthe halide leaving group X places the orbitalsthat undergo change in a perfect alignment.
The coplanar arrangement allows a continuous movement of electrons from one end of the systemto the other, much like a stack of dominoes each pushing the next one over.
The two orbitals that will form the pi bond arealready parallel (anti-coplanar) so that the doublecan form easily.
ORBITAL ALIGNMENTORBITAL ALIGNMENTIN MOST CONCERTED REACTIONS THEORBITALS BECOME PREALIGNED FOR A SMOOTH PROGRESSION OF EVENTS
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C C
H
BrR
RH H
OCH3 :....
..
..
..: :The attack of the base on the-hydrogen starts the reaction.
sp3
sp3
When these electrons enter theback lobe of the adjacent orbitalthey “push” the bonding pair outthe other end (along with Br).
The critical event isthe removal of the -H.
Notice the parallel aligmentof the two sp3 orbitals.
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C C
H
Br
RRH
H
OCH3 ..
..
..: :
..
. .
The formation of the double bondand the loss of bromide finish it.
2p 2p
Note the parallel orbitalsin the pi bond.
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FRONTIER MO THEORY
The LUMO is present on the -H only in the anti -coplanar arrangement.
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B:
LUMO
-
ANTI CONFORMATIONANTI CONFORMATION
DENSITY-ELPOT
LUMOHOMOB:-
LUMO hasdensity on H Frontier theory
requires a baseor nucleophileto add to theLUMO.
RECALL :
negative endof molecule
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SYN CONFORMATIONSYN CONFORMATION
B:-
LUMO
DENSITY-ELPOT
LUMO has no density on any H
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KINETIC ISOTOPE EFFECT
This “kinetic isotope effect” shows that breaking
the C-H bond is a part of the rate-determiningstep.
The reaction slows if the -H is replaced by D.
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ISOTOPES OF HYDROGENISOTOPES OF HYDROGEN
PROTIUM
DEUTERIUM
TRITIUM
HH
DD
TT
1
2
3
NAME SYMBOL MASS COMPOSITION
1 proton + 1 electron
1 proton + 1 neutron + 1 electron
1 proton + 2 neutrons + 1 electronradioactive
99.985%
0.015%
( - ) 12.26 yrs
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BD
CH CH
Br
BH
CH CH
Br
ISOTOPE EFFECTISOTOPE EFFECT
C-D bond is stronger than C-H
Slows the reaction if breaking this bond is part of the rate-determining step.
kHkD
approx. 5-8
C-H C-D
for an isotope effect
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ORIGIN OF THE ISOTOPE EFFECT
The effect is due to differences in C-H and C-D bond strengths.
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+ +
+ +
+ +
+ +
ENERGY
r (bond distance)
r
oo
CH
vibrationalenergylevels
average bondlength
zero-pointenergy CD
(H)1sC(sp3)
+ = nucleus
(D) 1sBONDING CURVEBONDING CURVE
Since D is heavier than H theC-D bond vibrates slower overa shorter distance.
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C-D BONDS ARE STRONGER THAN C-H BONDS
ENERGY
r (bond distance)oo
bond dissociationenergies (CD > CH)CH
CD
D is heavier than H and theCD bond vibrates more slowlyover a shorter distance than CH.
bond breakshere
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OTHER ELIMINATION MECHANISMSOTHER ELIMINATION MECHANISMS
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Three types of elimination reactions are conceivableThree types of elimination reactions are conceivable
C CH
X
C CH
X
C CH
X
B:
C C
C CH
+
C CX
C C
C C
B:
B:
E2
E1
E1cb
concerted
halogenfirst
protonsecond
protonfirst
halogensecond
carbocation
carbanion
just studied
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ELIMINATIONELIMINATIONOTHER POSSIBLE MECHANISMSOTHER POSSIBLE MECHANISMS
Do some elimination reactions occur in a different fashion?
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Three types of elimination reactions are conceivable
C CH
X
C CH
X
C CH
X
B:
C C
C CH
+
C CX
C C
C C
B:
B:
E2
E1
E1cb
concerted
halogenfirst
protonsecond
protonfirst
halogensecond
carbocation
carbanion
juststudied
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ALKYL HALIDES + WEAK BASE ALKYL HALIDES + WEAK BASE (SOLVOLYSIS)(SOLVOLYSIS)
E1
The removal of a -hydrogen becomes difficult withouta strong base and a different mechanism (ionization) begins to take place
….. if the substrate is capable.
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The E1 Elimination Reaction (The E1 Elimination Reaction (two stepstwo steps))
+ :Xslow
fast
B:
C C
+C CH
C CH
X
rate = k [RX]
carbocation
3o > 2o > 1o
Works best in apolar solvent.
IONSFORMED
unimolecular also favored if a resonancestabilized carbocationis formed
step one
steptwo
weakbase
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startingmaterial
product
Ea1
Ea2
H
intermediate
TS2
TS1
ENERGY PROFILEENERGY PROFILEtwo step reaction
ENERGY
step 1 step 2
carbocation E1
slow
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STEREOSPECIFICITYSTEREOSPECIFICITY
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CX
CH
C
XC
H
CCH
CC
H
+
+
Carbocation issp2 hybridized ( planar )and can reactfrom either side.
These twocarbocationsare equivalentby rotation and by symmetry.
anti
syn
THE E1 REACTION IS NOT STEREOSPECIFICTHE E1 REACTION IS NOT STEREOSPECIFIC
rotation
rate of C-C rotation= 1010 to 1012 / sec
( THE OPEN CARBOCATION IS PLANAR AND CAN ROTATE)
Elimination can be either syn or anti.
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REGIOSELECTIVITYREGIOSELECTIVITY
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E1 REACTION IS REGIOSELECTIVEE1 REACTION IS REGIOSELECTIVE
THE ZAITSEV RULE IS FOLLOWED
CH3
BrCH3 CH2
+
major minor
0.001 M
KOH / EtOH
tertiary trisubstituted
(stereochemistry is not a problem as in E2)
disubstituted
very dilute base
Zaitsev
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DIFFERENCES BETWEEN DIFFERENCES BETWEEN E1 AND E2E1 AND E2
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[RX] constant, [B] increasing
Raterate = k1 [RX]
E1
rate = k2 [RX] [B]E2
BEHAVIOR OF THE RATE BEHAVIOR OF THE RATE WITH INCREASING BASE CONCENTRATIONWITH INCREASING BASE CONCENTRATION
second order
first order
E1 dominatesat low baseconcentration E2 dominates
at higher baseconcentration
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[RX] constant, [Base] increasing
Rate
EFFECT OF BASE CONCENTRATION ON E1/E2 REACTIONSEFFECT OF BASE CONCENTRATION ON E1/E2 REACTIONS
secondary RX, k’
tertiary RX, k’’
primary RX, k
k1 [RX]
E1
For E1 elimination : k’’ (tertiary) > k’ (secondary) > k (primary).
k2 [RX] [B]E2At high base concentration
E1 never has a chance.
At low base concentrationE2 is nonexistent
1
1
1
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secondary RX, k’
tertiary RX, k’’
primary RX, k
[RX] constant, [B]
Rate
k1 [RX]
E1
EFFECT OF BASE CONCENTRATION ON E1/E2 REACTIONSEFFECT OF BASE CONCENTRATION ON E1/E2 REACTIONS
k2 [RX] [B]E2
For E2 elimination : line slopes k2 differ for 1o,2o,3o .Different substrates react at different rates,
primary
secondary
tertiary
1
1
2k2k’
k’’2
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Obviously for E1 which forms a carbocation intermediaterate : tertiary > secondary > primary > methyl
But this same order holds for E2 also.
STRUCTURE OF SUBSTRATESTRUCTURE OF SUBSTRATE
R-C-X
R
RR-C-X
R
HR-C-X
H
H
primary secondary tertiary
tertiary has more -hydrogens
C CC
C
HH
H
H H
HH H
Br
HEtO-
more opportunitesfor reaction
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E1 occurs only
1) at zero or low base concentration
2) with solvolysis (the solvent is the base)3) with tertiary and resonance capable substrates (alkyl halides)
If a strong base is present in moderate to high concentration, or the substrate is a primary halide, the E2 reaction dominates.
WHEN THE E1 MECHANISM OCCURSWHEN THE E1 MECHANISM OCCURS
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E2 mechanism E1 mechanism
strong base high base conc.
weak base low base conc.
ALKYL HALIDE + BASEALKYL HALIDE + BASE
solvolysis
must be able to make“good” carbocation
or
anti-coplanarrequirementstereospecific not stereospecific
(solvent is base)
regioselective regioselective
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EXAMPLES
NaOE tE tOH
E tOH
6M
K OH0.01 M
C HB r
C H3
C HB r
C H3
C H C H2
C H C H3+
E2
E1
rate = k [RBr]
rate = k [RBr] [OEt]..
:OE t..
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SOLVOLYSISSOLVOLYSIS
The solvent is the thing !
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CH3
ClEtOH
CH3
+CH3
OEt
CH3
HH
+ O EtH
MANY E1 REACTIONS ARE SOLVOLYSIS REACTIONSSOLVOLYSISSOLVOLYSIS
SOLVOLYSIS = THE SOLVENT IS THE REAGENT (BASE)
E1competing product
EtOH adds to thecarbocation
O EtH
+CH3
HHEtOH solvent acts as base -no other baseis present
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SOMETIMES E1 AND E2 RESULTS DIFFERSOMETIMES E1 AND E2 RESULTS DIFFER
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A COMPARISON OF E1 AND E2A COMPARISON OF E1 AND E2
CH3
CH3CH3
Br H
H
HH
H
Zaitsev
Anti-ZaitsevNaOEt
EtOH /
E2
stereospecific anti
not stereospecific
E1EtOH / anti
syn
major product
E1 doesn’t requireanti-coplanarity
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ALCOHOLS WITH SULFURIC ACIDALCOHOLS WITH SULFURIC ACID
“Acid-assisted” E1
E1
Carbocation Rearrangements
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DEHYDRATION OF AN ALCOHOLDEHYDRATION OF AN ALCOHOLAn alcohol can be “dehydrated” by treatment with concentrated sulfuric acid.
CH CH3CH3C
H3C+ H-O-H
H2SO4heat
CH CH3
OC
H
H3C
H3C H
-hydrogen
This is a beta-elimination reaction, similar to loss of HCl, but requiring acid conditions, rather than a strong base.
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H O SO
OO H
CH CH3
OCH
H
H3C
H3C
H
CH CH3CH3C
H3C HCH CH3C
H3C
H3C
H O SO
OO
+
+
CH CH3
OCH
H
H3C
H3C
ACID-ASSISTED E1 ELIMINATIONACID-ASSISTED E1 ELIMINATION
FOLLOWS ZAITSEV RULE
slow
O HH
fast
fast
-(or alcohol or water)
LIKE E1FROM HERE
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ROLE OF THE ACIDROLE OF THE ACID
Alcohols do not ionize because OH- is a strong
base (that is, OH- has a high energy).
R-O-H R+ +O-H-
However, if you protonate the OH group, water leaves.
R-O-H R+ + O-HH H
+
Water is a stable, low energy, molecule.
..
......:
:....
+ H+
NO
YES
“ACID ASSISTED”R+
R+
OH-
H2O
ROH
ROH2+
ionization
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EXAMPLESEXAMPLESCH3OH
CH3
CH CH CH3OH
CH3CH C CH3
CH3
H3PO4heat
H2SO4heat
CH CCH3CH3
CH3 CH3OHC C
CH3CH3
CH3 CH3
H3PO4heat
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CARBOCATION CARBOCATION REARRANGEMENTSREARRANGEMENTS
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CH3CCH3
CH3
CH CH3
OHC C
CH3
CH3
CH3
CH3+ H2O
REARRANGEMENT OF A CARBOCATIONREARRANGEMENT OF A CARBOCATION
H2SO4
SURPRISE!
differentskeleton !
CH3CCH3
CH3
CH CH3
OH
WHY NOT ?
no -H
-H here
CCH3
CH3
CH3
CH CH2
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CH3CCH3
CH3
CH CH3
OHC C
CH3
CH3
CH3
CH3+ H2O
REARRANGEMENT OF A CARBOCATIONREARRANGEMENT OF A CARBOCATION
CH3CCH3
CH3
CH CH3+ C CCH3
CH3
CH3
CH3 H+
protonation andloss of water
OSO3H
loss of H+
C C
CH3
+:REARRANGEMENTmethyl shifts withits pair of electrons
H2SO4
differentskeleton !
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TYPES OF CARBOCATION REARRANGEMENTSTYPES OF CARBOCATION REARRANGEMENTS
C C
CH3
+
+C C
H
C C+
1,2-methyl shift
1,2-hydride shift
1,2-phenyl shift
methyl migration
hydrogen migration
phenyl migration
groups move with their bonded electrons
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methyl shift
+CH3
CH3H+
CH3H3C
CH3
CH3
CH3H3COH
OSO3H_
protonationloss of H2O
loss of H+
REARRANGEMENT - A 1,2-METHYL SHIFTREARRANGEMENT - A 1,2-METHYL SHIFT
H2SO4
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+CH3H3C
+CH3
CH3H
WHY DO THEY REARRANGE ?WHY DO THEY REARRANGE ?
secondary ion
tertiary ion
CARBOCATIONSREARRANGETO ACHIEVE A LOWER ENERGY
Carbocation Energies3o < 2o < 1o < CH3
+
energy decrease
lowest highest
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WHICH GROUP MIGRATES ?WHICH GROUP MIGRATES ?
C CHH
H3C CH3+
C CHH3C CH3
H
C CH CH3
CH3H
C CHH
H3C CH3
+
+
+
tertiarybenzylic
The group that gives thebest carbocation will bethe one that migrates.
yes
secondarybenzylic
secondary
no
no
Competing Options H or Me or Ph ?
BINGO !BINGO !
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RING EXPANSIONRING EXPANSIONCH2 OH CH2
+ +
OMe
CH2+
When a carbocation is formed next toa small strained ring (cyclopropane orcyclobutane) the ring will often expandto the next larger size.
MeOHH2SO4
MeOH
This allows the relief of some of the strain.
CC
CH3+
CC
CH2+
CC
CH2+CC
CH3+
=
JUST LIKE A METHYL MIGRATION !
1o 2o
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DO YOU HAVE A CARBOCATION?DO YOU HAVE A CARBOCATION?
STOPSTOP
E X R E M E
R E G N A D
STOP - LOOK - THINK
ALWAYS
EVALUATE FOR AEVALUATE FOR AREARRANGEMENTREARRANGEMENT
CAN YOU FORM A BETTER CARBOCATION ?
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THE E1cb MECHANISMTHE E1cb MECHANISM
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THE E1cb MECHANISMTHE E1cb MECHANISM
C CH
X
C CH
X
C CH
X
B:
C C
C CH
+
C CX
C C
C C
B:
B:
protonfirst
halogensecond
carbanion
This mechanism is rare since it requires specialcharacteristics for the substrate:1. The proton must be easy to remove (very acidic). This usually requires resonance stabilization in the conjugate base.2. The leaving group must be hesitant to leave. This usually requires it to be a strong base, or to have a strong bond to carbon.
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O
OCH3
HH
acidic because the adjacent carbonyl groupprovides resonance in the conjugate base
An Example of an E1cb SubstrateAn Example of an E1cb Substrate
methoxide, the leaving group, is astrong basestrong bond
to carbon
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O
OCH3
HH
O
OCH3
H
O
OCH3
HO CH3
O
O CH3
NaOCH3
CH3OH
+
slowstep
faststep
conjugate basestabilized by resonance
WHY IT WORKS VIA E1cbWHY IT WORKS VIA E1cb
strong base =poor leaving group
acidic hydrogeneasily removed
UNIMOLECULARSlow step doesnot involve base
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THERE IS A RANGE OF DIFFERENT THERE IS A RANGE OF DIFFERENT MECHANISMS FOR MECHANISMS FOR -ELIMINATION REACTIONS-ELIMINATION REACTIONS
SUMMARY
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E1cb E2 E1 E1 E1 acid assisted
strong strong weak base base base
“solvolysis”
Zaitsev ifstereochemallows
Zaitsev Zaitsev Zaitsev
stereospecificanti-coplanar
alkyl halides alcoholsspecial
special case -not common
acidicneutral
carbocation rearrangements
concerted stepwise - carbocation
COMPARISON OF COMPARISON OF -ELIMINATION MECHANISMS-ELIMINATION MECHANISMS
requires:acidic H andpoor leaving group
stepwise -carbanion
not stereospecific
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K.I.S.S.K.I.S.S.alkyl halide + strong base + heat = E2
alkyl halide + solvent + heat (solvolysis) = E1
alcohol + strong acid + heat = E1 (acid assisted)
CO
CH
CX
typical situation for E1cbH next to C=O (easy to remove)X = strong base (difficult to break bond)
Only E1 reactions have rearrangements (carbocations)
Only E2 reactions require anti-coplanar -hydrogens
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HOFMANN REACTIONHOFMANN REACTION
E2 gone astray
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HOFMANN RULEHOFMANN RULEWhen you have a bulky leaving group like-N(CH3)3
+ the least-substituted alkene will be the major product.
BULKYBULKY = Branched at the first atom attached to the chain
N CH3CH3
H3C + +S CH3H3C
OTHER GROUPS FOLLOW THE ZAITSEV RULE
trimethylammonium
dimethylsulfonium
chain chain
Big is notthe sameas bulky.
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HOFMANN ELIMINATIONHOFMANN ELIMINATIONHofmann found that when the leaving group was -N(CH3)3
+ E2 elimination reactions gave the least-substituted alkene.
CH3 CH2 CHN
CH3CH3
CH3
H3CCH3CH2 CH CH2
CH3 CH CH CH3
+KOH
EtOH +
+ EtOHKOH
CH3 CH CH CH3
CH3CH2 CH CH2CH3 CH2 CHBr
CH3
95%
5%
31%
69%
Hofmann
Zaitsev
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KOHEtOH
CH3CH2CH2CHCH3X
CH3CH2CH2CH CH2 CH3CH2CH CHCH3X
Br
IO
OSO CH3
SCH3
CH3
N CH3CH3
CH3
+
+
31% 69%30% 70%48% 52%
87% 13%
98% 2%
HOFMANN ZAITSEV
EFFECT OF INCREASING SUBSTITUENT BULKEFFECT OF INCREASING SUBSTITUENT BULK
(cis + trans)E2
Big is notthe sameas bulky.
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C C C C CH3X
H
H
HOFMANN
ZAITSEVcis
trans
( all H equivalent )bulky groupscause crowdingand give Hofmannproducts
ANALYSIS OF 2-SUBSTITUTED PENTANE ELIMINATIONS
less crowding inthis area of the molecule
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cis transZAITSEVPRODUCTS HOFMANN
PRODUCT
N(CH3)3
CH2CH2CH3
H
N(CH3)3
CH3CH2
CH3
H
CH3
CH2CH3
N(CH3)3
H most stericcrowding
no stericcrowding
less stericcrowding
H
H
HH
H
H
H
WOULD MAKE WOULD MAKE
NOT FORMEDFORMED
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CH2H3C C CH3Br
HCHH3C CH CH3 CH3CH2 CH CH2
CH3 O
OCCH3
H3CCH3
-
-
BULKY BASES ALSO INCREASE HOFMANN PRODUCTBULKY BASES ALSO INCREASE HOFMANN PRODUCT
81% 19%
47% 53%
ZAITSEV HOFMANN
bulkybase
t -butoxide
methoxide
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CH2H3C C CH3Br
CH3CHH3C C CH3
CH3CH3CH2 C CH2
CH3
CH C CH2
CH3
CH3
H3CCH3C C CH3
CH3
CH3
CHH3C C CH3Br
CH3
CH3
CH2 C CH2
CH3CCH3
CH3H3CCHC C CH3
CH3
CH3
CH3H3CCH2C C CH3
Br
CH3
CH3
CH3H3C
BULKY BULKY -SUBSTITUENTS-SUBSTITUENTS
80% 20%
79% 21%
14% 86%
ZAITSEV HOFMANN
NaOEt
NaOEt
NaOEt
What constitutes bulky?
NO
NO
YES
t-butyl is bulky !
a methyl groupis not bulky
even two or threeare not bulky
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CH2C C CH3Br
CH3
CH3
CH3H3C
CH3
CH3
C
H
CH3H3C
H3C
CH2
H3C
CH2CCH3
H3CCH3+
H
H
CH3
Br
C
CH3
CH3CH3
CH3
CH
H
CH3
H
Br
CH2
CH3CH3
CH3
crowdedlesscrowded
THE ELIMINATION MOVES TO A LESS CROWDED REGIONTHE ELIMINATION MOVES TO A LESS CROWDED REGION
86%14%
spacer
REACTIVE CONFORMATIONS
crowding
crowding
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Br
CH3 CH3
CH3
N(CH3)3+
I-
CH2
Br
CH3 CH2 CH3
NaOEtEtOH /
KOHEtOH /
NaOtButBuOH / +
~90%
~90%
~60/40%
Zaitsev
Hofmann
NORMAL
BULKYLEAVINGGROUP
BULKYBASE
HOW THE VARIOUS FACTORS AFFECT THE OUTCOMEHOW THE VARIOUS FACTORS AFFECT THE OUTCOME
Bulky base alone not as effective as bulky leaving group
Prototypical “Hofmann” elimination
Bromine is big, not bulky
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Br
CH3
CH3
N(CH3)3+
I-
CH2NaOtButBuOH /
BULKY BASE& LEAVINGGROUP
~100%Hofmann
tBu
CH3NaOEt
EtOH /
BULKY-SUBSTITUENT
HOW THE VARIOUS FACTORS AFFECT THE OUTCOMEHOW THE VARIOUS FACTORS AFFECT THE OUTCOME( CONTINUED )
tBu
either cis or trans to Br - same result
H
CH2
tBu
+
no doublebond here
Double Whammy !
Favors Hofmann products
Bulky base + bulky leaving group
Use a bulky base here and ...
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E2 REACTIONS DEVIATE FROM THE ZAITSEV RULEE2 REACTIONS DEVIATE FROM THE ZAITSEV RULE
1. If the favored -hydrogen can’t achieve anti-coplanar geometry.
2. If the double bond would form at a bridgehead.
3. If there is a bulky leaving group.
4. If there is a bulky base.
5. If there is a bulky -substituent.
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BREDT’S RULEBREDT’S RULE
A double bond cannot form at a bridgehead.
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BREDT’S RULEBREDT’S RULEA double bond cannot form at a bridgehead in abicyclic system with small rings.
Cl
no way !
p orbitals cannotbecome coplanar
y
z
z
y Try a model !
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SYN ELIMINATIONSYN ELIMINATION
More difficult than anti-coplanar elimination,but does occur in some circumstances.
( Usually requires forced conditions - heat and pressure. )
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~~~~
anti- coplanar
syn- coplanar
not coplanar~~
COPLANAR ARRANGEMENTS
( very difficult )
E2
Difficulty Order
(difficult)
(easiest)
180o
0o
otherangles
C C
H X syn-coplanarSECOND-BESTSITUATION
0o
C C
H
Xanti-coplanar
BEST SITUATION
180o
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SYN-COPLANAR ELIMINATIONS SYN-COPLANAR ELIMINATIONS REQUIRE FORCED CONDITIONSREQUIRE FORCED CONDITIONS
C C
H X syn-coplanar (0o)
C C
This often requires heating above the boiling pointof the solvent in a sealed tube (next slide).Temperatures above 100 oC are common.
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SEALED TUBESEALED TUBE
heated oil (bp > 250 oC)
alkyl halide +NaOEt / EtOH (bp 78 oC)inside tube
1.25” D0.25” wall
glasstube
Glass tube sealed at both ends.
Carried out in a hoodbehind a glass shield.
hot plate
Allows reactantsto be heated to ahigh temperature(above bp) without boiling away.
REACTANTS
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+
0%
same productsNO DEUTERIUM
NaOEt
EtOH
NaOEt
EtOH
A CASE OF SYN ELIMINATIONA CASE OF SYN ELIMINATIONSYN ELIMINATION OCCURS BECAUSE
THERE ARE NO ANTI-COPLANAR -H
110 oC
proves the syn hydrogenis the one removed
110 oC
not this one
Bredt’sRule
HH
DBr
H
HH
HBr
HH
H
100% H
H
D
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HCl
Cl
H
ClCl
H
H Cl
H
Cl
H
NaOH 110o
NaOH 110o
very slow
faster syn-elimination
hydrogens are notanti-coplanarto chlorines
SYN ELIMINATION IS SEEN WHEN ANTI-COPLANAR DOES NOT EXIST
difficultreaction
notcoplanar
coplanar
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-ELIMINATION-ELIMINATION
occurs when the substrate has
NO -HYDROGENS C C HX
C
on these carbon atoms
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C HCl
ClCl
CCl
ClCl
: -
CCl
Cl : Cl+
KOH
CHCl3
a carbene
-elimination
-ELIMINATION-ELIMINATION
veryreactive
no -hydrogens
::....
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CARBENES ARE ELECTROPHILES !CARBENES ARE ELECTROPHILES !Because of an incomplete octet carbenes areelectrophilic (need electrons to complete theirvalence shell).
Carbenes will react with an alkene (electron pair donor).
CCl
Cl :missing a pairof electrons
CCl
Cl : :electrophile
nucleophile
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CARBENES ADD TO DOUBLE BONDSCARBENES ADD TO DOUBLE BONDS
+ C ClCl:
Cl
Cl
CHCl3
KOH
1
2
Probablyconcerted
syn additionH
H
: steps 1,2 for visualization only
alkene + carbene =cyclopropane ring
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C:H
H..
C ClCl
Cl
ClCl
Cl
: ..-+
stepwise analysis of the concerted process
SP2 hybrid
ANALYSIS OF THE ADDITIONANALYSIS OF THE ADDITION
the intermediate does not exist
2p
syn addition
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H
H
H
H
ClCl
SYNSYN ADDITION ADDITION
KOHCHCl3
EtOH
BOTH NEW RINGBONDS FORM ON THE SAME SIDE OF THE DOUBLE BOND
H
H
C ClCl
Stereospecific
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H
H CH3
CH3 H
H CH3
CH3
ClCl
CH3
H CH3
H CH3
H CH3
H
ClCl
KOHCHCl3
EtOH
KOHCHCl3
EtOH
cis cis
trans trans
SUBSTITUENTSON THE DOUBLEBOND RETAINTHEIR ORIGINAL CIS OR TRANS RELATIONSHIPSIN THE NEW RING
STEREOSPECIFICITY PROVES STEREOSPECIFICITY PROVES THE REACTION IS CONCERTEDTHE REACTION IS CONCERTED
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CH3
CH3
C ClCl
H
H
CH3
CH3
C ClCl
H
H
+
..
An intermediatewould allow rotation.
STEREOSPECIFIC = CONCERTEDSTEREOSPECIFIC = CONCERTED
This does not happen.
Both bonds form on the same side.
Concerted.
Stereospecific!
Would not be stereospecific.
YES !YES !
NO !NO !
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COMPOUNDS WITHOUT COMPOUNDS WITHOUT -HYDROGENS-HYDROGENS
Not a common type of compound !
C H
Ph
Ph
Cl
C H
Cl
Cl
Cl
C H
Cl
H
Cl
C H
Ph
H
Cl
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ELIMINATIONS SUMMARY
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K.I.S.S.K.I.S.S.alkyl halide + strong base + heat = E2
alkyl halide + solvent + heat (solvolysis) = E1
alcohol + strong acid + heat = E1 (acid assisted)
CO
CH
CX
typical situation for E1cbH next to C=O (easy to remove)X = strong base (difficult to break bond)
Only E1 reactions have rearrangements (carbocations)
Only E2 reactions require anti-coplanar -hydrogens
THE MOST BASIC STUFF
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E1cb E2 E1 E1 E1 acid assisted
strong strong weak base base base
“solvolysis”
Zaitsev ifstereochemallows Zaitsev Zaitsev Zaitsev
stereospecificanti-coplanar
alkyl halides alcoholsspecial
special case -not common
acidicneutral
carbocation rearrangements
concerted stepwise - carbocation
requires:acidic H andpoor leaving group
stepwise -carbanion
not stereospecific
-elim. if no -H
Hofmann ifbulky groups
THE BIG PICTURETHE BIG PICTURE
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MAKING ALKYNESMAKING ALKYNES
“DOUBLE ELIMINATION”
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COMPOUNDS WITH TWO HALOGENSCOMPOUNDS WITH TWO HALOGENSIf you have a compound with two halogens it can react twice (two E2 eliminations).
If both halogens are on the same carbon, an alkyneis produced.
The second elimination is more difficult than thefirst one - it requires a stronger base.
C CH
Cl
ClCH3CH3
HC C
ClCH3
H CH3C C CH3CH3
more difficult, requires astronger base like NH2
-most E2 baseswill work
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AMIDE VS. ETHOXIDEAMIDE VS. ETHOXIDE
N:..
HH
NaNH2 / NH3 (liq) NaOEt / EtOH
O CH2CH3:....
more basic (N is less electro- negative then O)
less basic(O accommodates the charge better)
The usual reagent issodium amide in liquidammonia (-33o C):
The usual reagent issodium ethoxide in ethanol :
Both are made by adding sodium metal to the solvent.
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C CClCH3
H CH3
C C CH3CH3
C CH
Cl
ClCH3CH3
HC C
ClCH3
H CH3
KOH
EtOH
C CH
Cl
ClCH3CH3
HC C CH3CH3
NaNH2
NH3 (liq)
NaNH2
NH3 (liq)
EXAMPLESEXAMPLES
trans(the reaction is more difficult for cis )
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THE REACTION CAN BE DONE IN TWO STEPSTHE REACTION CAN BE DONE IN TWO STEPS
C CH
Cl
ClCH3CH3
HC C
ClCH3
H CH3
C C CH3CH3
NaOEtEtOH
NaNH2NH3 (liq)
stops herewith sodiumethoxide
stronger basebrings aboutthe second step
predominantly thetrans isomer
(lower energy product)
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COMPOUNDS WITH TWO HALOGENSCOMPOUNDS WITH TWO HALOGENSIf you have a compound with two halogens it can react twice (two E2 eliminations).
If the halogens are on different carbons, a diene isusually produced.
Cl
Cl
KOHEtOH
BrBrBr2
CCl4
NaOEtEtOH