chapter 5 structure and preparation of alkenes: elimination reactions

Chapter 5Chapter 5Structure and Preparation of Alkenes:Structure and Preparation of Alkenes:

Elimination ReactionsElimination Reactions

5.15.1Alkene NomenclatureAlkene Nomenclature

Alkenes

Alkenes are hydrocarbons that contain a carbon-carbon double bond

also called "olefins"

characterized by molecular formula CnH2n

said to be "unsaturated"

H2C CH2 H2C CHCH3

Ethene Propene

Alkene Nomenclature

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

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

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

Alkenyl Groups

methylene

vinyl

allyl

isopropenyl

CHH2C

H2C CHCH2

H2C CCH3

H2C

Cycloalkene Nomenclature

1) Replace the -ane ending of the cycloalkane having the same number of carbons with -ene.

Cyclohexene

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

5.25.2Structure and Bonding in AlkenesStructure and Bonding in Alkenes

Structure of Ethylene

bond angles: H-C-H = 117°

H-C-C = 121°

bond distances: C—H = 110 pm

C=C = 134 pm

planar

Bonding in Ethylene

Framework of bonds

Each carbon is sp2 hybridized

Bonding in Ethylene

Each carbon has a half-filled p orbital.

Bonding in Ethylene

Side-by-side overlap of half-filled p orbitals gives a bond.

5.35.3Isomerism in AlkenesIsomerism in Alkenes

Isomers are different compounds thathave the same molecular formula.

Isomers

Isomers

Constitutional isomersConstitutional isomers StereoisomersStereoisomers

different connectivity same connectivity;different arrangementof atoms in space

Isomers

Constitutional isomersConstitutional isomers StereoisomersStereoisomers

consider the isomeric alkenes of molecular formula C4H8

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


cis-2-Butene

C C

H

H H

CH2CH3

H

CH3

C C

H3C

H

C C

H

HH3C

H3C

Constitutional isomers


trans-2-Butene

C C

H

H H

CH2CH3

H3C

C C

CH3

H

H

C C

H

HH3C

H3C

Constitutional isomers

cis-2-Butene trans-2-Butene

H3C

C C

CH3

H

HH

CH3

C C

H3C

H

Stereoisomers

Stereochemical Notation

trans (identical or analogous substituents on opposite sides)

cis (identical or analogous substituents on same side)

transcis

Interconversion of stereoisomericalkenes does not normally occur.

Requires that component of doublebond be broken.

Rotation Barrier in Alkenes

Rotation Barrier in Alkenes

5.45.4Naming Steroisomeric AlkenesNaming Steroisomeric Alkenesby the E-Z Notational Systemby the E-Z Notational System

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

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

C C

The E-Z Notational System

E : higher ranked substituents on opposite sides

Z : higher ranked substituents on same side

higher

lower

C C


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

C C


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?

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.

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

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

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)

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)

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)

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)

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.

5.55.5

Physical Properties of AlkenesPhysical Properties of Alkenes

= 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?

= 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.

= 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

= 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.

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

5.65.6Relative Stabilities of AlkenesRelative Stabilities of Alkenes

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

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

Substituent Effects on Alkene Stability

Electronic

Disubstituted alkenes are more stable than monosubstituted alkenes.

Steric

Trans alkenes are more stable than cis alkenes.

Figure 5.3 Heats of Combustion of C4H8

Isomers


Alkyl groups stabilize double bonds more than H.

More highly substituted double bonds are morestable than less highly substituted ones.

Electronic

Problem 5.12

C C

H3C

H3C CH3

CH3


Trans alkenes are more stable than cis alkenes.

Cis alkenes are destabilized by van der Waalsstrain.

Steric

cis-2-butene trans-2-butene

van der Waals straindue to crowding ofcis-methyl groups

Figure 5.4 cis and trans-2-Butene

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

5.75.7CycloalkenesCycloalkenes

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.

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


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

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.


Cis and trans-cyclododeceneare approximately equal in stability.

trans-Cyclododecenecis-Cyclododecene

5.8

Preparation of Alkenes:

Elimination Reactions

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

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

5.9Dehydration of Alcohols

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

R

R'

R"

OHC

R

R'

H

OHC

R

H

H

OHC

Relative Reactivity

tertiary:most reactive

primary:least reactive

5.10Regioselectivity in Alcohol Dehydration:

The Zaitsev Rule

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.

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

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


two protons on this carbon

The Zaitsev Rule

R OH

CH3

C C

H

R CH2R


R

R

CH2R

CH3

C C

only one proton on this carbon

The Zaitsev Rule

R OH

CH3

C C

H

R CH2R

5.11

Stereoselectivity in Alcohol Dehydration

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

5.12

The E1 and E2 Mechanisms

of Alcohol Dehydration

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...

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

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

+

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.

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

are intermediates in the acid-catalyzed dehydration of tertiary and secondary alcohols

carbocations can:

react with nucleophileslose a -proton to form an alkene

Carbocations

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

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

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

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.

OH

H3PO4, heat

3%

64% 33%

+ +

Example

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

3%

CH3

CHCH3

CH3

+CH3

97% CH3

CHCH3CH3

CH3

+C C

tertiary carbocation; more stable


3%

CH3

CHCH3

CH3

+CH3

97% CH3

CHCH3CH3

CH3

+C C


CH3CH2CH2CH2OH

H3PO4, heat

12%

+

mixture of cis (32%)and trans-2-butene (56%)

CH2CH3CH2CH CHCH3CH3CH

Another Rearrangement

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

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+


H

CH3CH2CH2CH2 O

H

+:

CH2CH3CH2CH

CH3CH2CHCH3+

mixture of cisand trans-2-butene

CHCH3CH3CH


H

H

CH3CH2CHCH2 O

H

+:

+CH3CH2CHCH2 +

H

O

H

H

::

Hydride Shift

• react with nucleophiles

• lose a proton from the -carbon to form an alkene

• rearrange (less stable to more stable)

Carbocations Can...

5.145.14

Dehydrohalogenation of Dehydrohalogenation of

Alkyl Halides Alkyl Halides

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


dehydrogenation of alkanes:industrial process; not regioselective

dehydration of alcohols:acid-catalyzed

dehydrohalogenation of alkyl halides:consumes base


X YC CC C + X Y

is a useful method for the preparation of alkenes

(100 %)

likewise, NaOCH3 in methanol, or KOH in ethanol

NaOCH2CH3

ethanol, 55°C

Dehydrohalogenation

Cl

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

Br

29 % 71 %

+

Regioselectivity

follows Zaitsev's rule:

more highly substituted double bond predominates

KOCH2CH3

ethanol, 70°C

more stable configurationof double bond predominates

Stereoselectivity

KOCH2CH3

ethanol

Br

+

(23%) (77%)

more stable configurationof double bond predominates

Stereoselectivity

KOCH2CH3

ethanol

+

(85%) (15%)

Br

5.155.15 The E2 Mechanism of The E2 Mechanism of

Dehydrohalogenation of Alkyl Dehydrohalogenation of Alkyl HalidesHalides

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

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

concerted (one-step) bimolecular process

single transition state

C—H bond breaks

component of double bond forms

C—X bond breaks

The E2 Mechanism

–OR..

.. :

C C

H

X..::

Reactants

The E2 Mechanism

C C

–

OR..

.. H

X..::–

Transition state

The E2 Mechanism

OR..

.. H

C C

–X..

::..

Products

The E2 Mechanism

5.165.16

Anti Elimination in E2 Reactions: Anti Elimination in E2 Reactions:

Stereoelectronic EffectsStereoelectronic Effects

(CH3)3C

(CH3)3C

Br

KOC(CH3)3

(CH3)3COH

cis-1-Bromo-4-tert- butylcyclohexane

Stereoelectronic Effect

(CH3)3C

(CH3)3CBr KOC(CH3)3

(CH3)3COH

trans-1-Bromo-4-tert- butylcyclohexane


(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.


(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


(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


cis

more reactive

trans

less reactive



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.

5.175.17

Isotopes Effects And The E2 Isotopes Effects And The E2

MechanismMechanism

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

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

5.185.18

The E1 Mechanism ofThe E1 Mechanism of

Dehydrohalogenation of Alkyl Dehydrohalogenation of Alkyl

HalidesHalides

CH3 CH2CH3

Br

CH3

C

Ethanol, heat

+

(25%) (75%)

H3C

CH3

C C

H3C

H

CH2CH3

CH3

CH2C

Example

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

CH3 CH2CH3

Br

CH3

C

..::

CCH2CH3CH3

CH3

+

:..: Br.. –

slow, unimolecular

Step 1

CCH2CH3CH3

CH3

+

CH3CH3

CCH2CH3

CH2

+ CCHCH3

CH3

– H+

Step 2

chapter 5 structure and preparation of alkenes: elimination reactions

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