Comlete elimination reactions

ELIMINATION REACTIONSELIMINATION REACTIONS

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.


- HClC CH Cl

C C

TWO EXAMPLES FOLLOW

CH3CH2CH2CH2 Cl CH3CH2 CH CH2NaOH

CH3CH2 CH CH3Cl

CH3 CH CH CH3NaOH


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

+ 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

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

STRONG BASESSTRONG BASES

FIRST WE MUST LEARN WHAT IS A STRONG BASE

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....:

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?

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

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.

ALKYL HALIDE + STRONG BASE

E2

FOR A DEHYDROHALOGENATION

defined later...

+ HEAT

LEARN THIS ! IT IS THE FORMULA

REACTION

shorthand designationfor this type of reaction

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

C CC l

H

B:

THE BASE TAKES THE THE BASE TAKES THE -HYDROGEN-HYDROGEN

C C

C l

HB

: :..

: :..

..

MECHANISMMECHANISM

REGIOSELECTIVITYREGIOSELECTIVITY

ALKYL HALIDE + STRONG BASE (E2)

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 ?

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

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

-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

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

-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

SAYTZEV RULESAYTZEV RULE

The reaction gives the most highly-substituted (lowest energy) alkene as the major product.

ZaitsevTsayseff

etc.

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-

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

STEREOCHEMISTRY STEREOCHEMISTRY OF THE REACTIONOF THE REACTION

ALKYL HALIDE + STRONG BASE (E2)

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

ACYCLIC HALIDESACYCLIC HALIDES

STEREOCHEMISTRY

HCl

H

Cl

ACYCLIC MOLECULES MAY HAVE TO ROTATE ACYCLIC MOLECULES MAY HAVE TO ROTATE IN ORDER TO REACTIN ORDER TO REACT

anti-coplanar

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

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)

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?

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

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

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

CYCLIC HALIDESCYCLIC HALIDES

STEREOCHEMISTRY REARS ITS UGLY HEAD AGAIN !

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

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?

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

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

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


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

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


A CASE THAT CAN NOT FOLLOW THE HAMMOND POSTULATE2

OR….STEREOCHEMISTRY REARS ITS UGLY HEAD !

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

ALKYL HALIDES + STRONG BASE + HEATALKYL HALIDES + STRONG BASE + HEAT....... continued

E2

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

KINETICSKINETICS

Examination of the rate expression often helpsto understand the mechanism.The rate expression is determined by experiment.

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

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 :

MECHANISMMECHANISM

CH CHBr

H

B:

CONCERTED = only one stepAll bonds are broken and formed withoutthe formation of any intermediates.

EliminationBimolecular

E 2strong base

alkylhalide

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.

CONCERTED REACTIONCONCERTED REACTION

E2

One Step - No Intermediates

CH CHBr

H

B: BH

CH CH

Brmechanism activated complex

E2 ELIMINATIONE2 ELIMINATION

Concerted : everythinghappens at once with-out any intermediates.

Concerted (one step) reactionConcerted (one step) reaction

product

startingmaterial

transition state TS

activationenergy Ea

heat ofreactionH

ENERGY

This is what E2 looks like.

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

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.

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.

FRONTIER MO THEORY

The LUMO is present on the -H only in the anti -coplanar arrangement.

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

SYN CONFORMATIONSYN CONFORMATION

B:-

LUMO

DENSITY-ELPOT

LUMO has no density on any H

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.

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

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

ORIGIN OF THE ISOTOPE EFFECT

The effect is due to differences in C-H and C-D bond strengths.

+ +

+ +

+ +

+ +

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.

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

OTHER ELIMINATION MECHANISMSOTHER ELIMINATION MECHANISMS

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

ELIMINATIONELIMINATIONOTHER POSSIBLE MECHANISMSOTHER POSSIBLE MECHANISMS

Do some elimination reactions occur in a different fashion?

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

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.

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

startingmaterial

product

Ea1

Ea2

H

intermediate

TS2

TS1

ENERGY PROFILEENERGY PROFILEtwo step reaction

ENERGY

step 1 step 2

carbocation E1

slow

STEREOSPECIFICITYSTEREOSPECIFICITY

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.

REGIOSELECTIVITYREGIOSELECTIVITY

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

DIFFERENCES BETWEEN DIFFERENCES BETWEEN E1 AND E2E1 AND E2

[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

[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

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

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

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

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

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..

SOLVOLYSISSOLVOLYSIS

The solvent is the thing !

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

SOMETIMES E1 AND E2 RESULTS DIFFERSOMETIMES E1 AND E2 RESULTS DIFFER

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

ALCOHOLS WITH SULFURIC ACIDALCOHOLS WITH SULFURIC ACID

“Acid-assisted” E1

E1

Carbocation Rearrangements

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.

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

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

EXAMPLESEXAMPLESCH3OH

CH3

CH CH CH3OH

CH3CH C CH3

CH3

H3PO4heat

H2SO4heat

CH CCH3CH3

CH3 CH3OHC C

CH3CH3

CH3 CH3

H3PO4heat

CARBOCATION CARBOCATION REARRANGEMENTSREARRANGEMENTS

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

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 !

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

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

+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

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 !

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

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 ?

THE E1cb MECHANISMTHE E1cb MECHANISM

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.

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

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

THERE IS A RANGE OF DIFFERENT THERE IS A RANGE OF DIFFERENT MECHANISMS FOR MECHANISMS FOR -ELIMINATION REACTIONS-ELIMINATION REACTIONS

SUMMARY

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

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

HOFMANN REACTIONHOFMANN REACTION

E2 gone astray

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.

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

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.

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

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

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

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

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

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

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 ...

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.

BREDT’S RULEBREDT’S RULE

A double bond cannot form at a bridgehead.

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 !

SYN ELIMINATIONSYN ELIMINATION

More difficult than anti-coplanar elimination,but does occur in some circumstances.

( Usually requires forced conditions - heat and pressure. )

~~~~

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

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.

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

+

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

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

-ELIMINATION-ELIMINATION

occurs when the substrate has

NO -HYDROGENS C C HX

C

on these carbon atoms

C HCl

ClCl

CCl

ClCl

: -

CCl

Cl : Cl+

KOH

CHCl3

a carbene

-elimination

-ELIMINATION-ELIMINATION

veryreactive

no -hydrogens

::....

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

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

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

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

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

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 !

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

ELIMINATIONS SUMMARY

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

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

MAKING ALKYNESMAKING ALKYNES

“DOUBLE ELIMINATION”

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

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.

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 )

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)

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

Documents

Comlete elimination reactions