Created byProfessor William Tam & Dr. Phillis

Chang Ch. 7 - 1

Chapter 7

Alkenes and Alkynes I:Properties and

Synthesis.Elimination Reactions

of Alkyl Halides

Ch. 7 - 2

About The Authors

These PowerPoint Lecture Slides were created and prepared by Professor William Tam and his wife, Dr. Phillis Chang.

Professor William Tam received his B.Sc. at the University of Hong Kong in 1990 and his Ph.D. at the University of Toronto (Canada) in 1995. He was an NSERC postdoctoral fellow at the Imperial College (UK) and at Harvard University (USA). He joined the Department of Chemistry at the University of Guelph (Ontario, Canada) in 1998 and is currently a Full Professor and Associate Chair in the department. Professor Tam has received several awards in research and teaching, and according to Essential Science Indicators, he is currently ranked as the Top 1% most cited Chemists worldwide. He has published four books and over 80 scientific papers in top international journals such as J. Am. Chem. Soc., Angew. Chem., Org. Lett., and J. Org. Chem.

Dr. Phillis Chang received her B.Sc. at New York University (USA) in 1994, her M.Sc. and Ph.D. in 1997 and 2001 at the University of Guelph (Canada). She lives in Guelph with her husband, William, and their son, Matthew.

Ch. 7 - 3

1. Introduction

Alkenes●Hydrocarbons containing C=C●Old name: olefins

CH2OH

Vitamin A

HO

H3C

H3C

H HCholesterol

Ch. 7 - 4

Alkynes●Hydrocarbons containing C≡C●Common name: acetylenes

O

NH

O

Cl

Efavirenz(antiviral, AIDS therapeutic)

CF3CC

Cl

ClCl

OCC

I

Haloprogin(antifungal, antiseptic)

Ch. 7 - 5

2. The (E) - (Z) System for Designating Alkene Diastereomers

Cis-Trans System●Useful for 1,2-disubstituted

alkenes●Examples:

ClBr

H

H

ClH

Br

H

(1) vs

trans -1-Bromo-2-chloroethene

cis -1-Bromo-2-chloroethene

Ch. 7 - 6

●ExamplesH

H

(2) vs

trans -3-Hexene cis -3-Hexene

HH

Br(3) vs

trans -1,3-Dibromopropene

cis -1,3-Dibromopropene

BrBr

Br

Ch. 7 - 7

ClBr

CH3

H

e.g. cis or trans?

Cl is cis to CH3

and trans to Br

(E) - (Z) System●Difficulties encountered for

trisubstituted and tetrasubstituted alkenes

Ch. 7 - 8

The Cahn-Ingold-Prelog (E) - (Z) Convention

●The system is based on the atomic number of the attached atom

●The higher the atomic number, the higher the priority

Ch. 7 - 9

The Cahn-Ingold-Prelog (E) - (Z) Convention● (E) configuration – the highest

priority groups are on the opposite side of the double bond “E ” stands for “entgegen”; it

means “opposite” in German● (Z) configuration – the highest

priority groups are on the same side of the double bond “Z ” stands for “zusammer”; it

means “together” in German

Ch. 7 - 10

On carbon 2: Priority of Br > COn carbon 1: Priority of Cl > H

highest priority groups are Br (on carbon 2) and Cl (on carbon 1)

ClBr

CH3

H

21

●Examples

Ch. 7 - 11

ClBr

CH3

H (E )-2-Bromo-1-chloropropene

●Examples

ClCH3

Br

H (Z )-2-Bromo-1-chloropropene

Ch. 7 - 12

●Other examples

(1)

(2)

ClCl

H

H

12

(E )-1,2-Dichloroethene[or trans-1,2-Dichloroethene]

ClCl

Br1

2 (Z )-1-Bromo-1,2-dichloroethene

C1: Br > ClC2: Cl > H

C1: Cl > HC2: Cl > H

Ch. 7 - 13

●Other examples

(3)

Br

87

65

43

2

1

(Z )-3-Bromo-4-tert-butyl-3-octene

C3: Br > CC4: tBu > nBu

Ch. 7 - 14

3. Relative Stabilities of Alkenes

Cis and trans alkenes do not have the same stability

C C

H

R

H

R

C C

R

H

H

R

crowding

Less stable More stable

Ch. 7 - 15

3A.Heat of Reaction

C C + H HPt

C C

H H

Heat of hydrogenation●∆H° ≃ -120 kJ/mol

Ch. 7 - 16

En

thalp

y+ H2

≈

+ H2

≈

+ H2

≈

7 kJ/mol

5 kJ/mol

DH° = -115 kJ/mol

DH° = -127 kJ/mol

DH° = -120 kJ/mol

Ch. 7 - 17

3B.Overall Relative Stabilities ofAlkenes

The greater the number of attached alkyl groups (i.e., the more highly substituted the carbon atoms of the double bond), the greater the alkene’s stability.

Ch. 7 - 18

Relative Stabilities of Alkenes

R

R

R

R

tetra-substituted

R

R

H

R

R

R

H

H

H

R

R

H

H

R

H

R

H

R

H

H

H

H

H

H

> > > > > >

tri-substituted

di-substituted

mono-substituted

un-substituted

Ch. 7 - 19

Examples of stabilities of alkenes

(1) >

(2) >

Ch. 7 - 20

4. Cycloalkenes

Cycloalkenes containing 5 carbon atoms or fewer exist only in the cis form

cyclopropene cyclobutene cyclopentene

Ch. 7 - 21

Trans – cyclohexene and trans – cycloheptene have a very short lifetime and have not been isolated

cyclohexene Hypotheticaltrans - cyclohexene

(too strained to exist at r.t.)

Ch. 7 - 22

Trans – cyclooctene has been isolated and is chiral and exists as a pair of enantiomers

cis - cyclooctene trans - cyclooctenes

Ch. 7 - 23

5. Synthesis of Alkenes viaElimination Reactions

Dehydrohalogenation of Alkyl Halides

C CH

XH

H

H

H

H

H

H

Hbase

-HX

Dehydration of Alcohols

C CH

OHH

H

H

H

H

H

H

HH+, heat

-HOH

Ch. 7 - 24

6. Dehydrohalogenation of AlkylHalides

The best reaction conditions to use when synthesizing an alkene by dehydrohalogenation are those that promote an E2 mechanism

C C

H

X

B: C CE2

B:H + X+

Ch. 7 - 25

6A.How to Favor an E2 Mechanism

Use a secondary or tertiary alkyl halide if possible. (Because steric hinderance in the substrate will inhibit substitution)

When a synthesis must begin with a primary alkyl halide, use a bulky base. (Because the steric bulk of the base will inhibit substitution)

Ch. 7 - 26

Use a high concentration of a strong and nonpolarizable base, such as an alkoxide. (Because a weak and polarizable base would not drive the reaction toward a bimolecular reaction, thereby allowing unimolecular processes (such as SN1 or E1 reactions) to compete.

Ch. 7 - 27

Sodium ethoxide in ethanol (EtONa/EtOH) and potassium tert-butoxide in tertbutyl alcohol (t-BuOK/t-BuOH) are bases typically used to promote E2 reactions

Use elevated temperature because heat generally favors elimination over substitution. (Because elimination reactions are entropically favored over substitution reactions)

Ch. 7 - 28

Examples of dehydrohalogenations where only a single elimination product is possible

6B.Zaitsev’s Rule

Br(79%)(1)

EtONa

EtOH, 55oC

(2)Br

EtONa

EtOH, 55oC(91%)

(3) Br t -BuOK

t -BuOH, 40oC(85%)( )n

( )n

Ch. 7 - 29

Rate = H3CHC

Br

CH3 EtOk

Br

HbHaB2-methyl-2-butene

2-methyl-1-butene

WWW Ha

WWW Hb

(2nd order overall) bimolecular

Ch. 7 - 30

When a small base is used (e.g. EtO⊖ or HO⊖) the major product will be the more highly substituted alkene (the more stable alkene)

Examples:

Br

HbHaNaOEt

EtOH

70oC

+

69% 31%

(eliminate Ha) (eliminate Hb)

(1)

(2)Br KOEt

EtOH

51% 18% 31%

+ +

69%

Ch. 7 - 31

Zaitsev’s Rule●In elimination reactions, the

more highly substituted alkene product predominates

Stability of alkenes

C C

Me

Me

Me

Me

> C C

Me

Me

H

Me

> C C

H

Me

Me

H

> C C

H

Me

H

Me

> C C

H

Me

H

H

Ch. 7 - 32

Mechanism for an E2 Reaction

Et O

C CBr

H

HH3C

CH3CH3

Et O

C CBr

H

HH3C

CH3CH3

C C

H

H3C

CH3

CH3

Et OH +Br

+β

α

EtO⊖ removes a b proton; C−H breaks; new p bond forms and Br begins to depart

Partial bonds in the transition state: C−H and C−Br bonds break, new p C−C bond forms

C=C is fully formed and the other products are EtOH and

Br⊖

Ch. 7 - 33

Free

Ene

rgy

Reaction Coordinate

DG2‡

EtO- +

CH3

CCH3CH2

Br

CH3

+ EtOH + Br-CH3

CCH3CH2 CH2

+ EtOH + Br-CH3

CCH3CH CH3

Et O

C CBr

H

HH3C

CH3CH3

EtO

CCBr

H

HH

H3CCH3CH2

DG1‡

Ch. 7 - 34

Hofmann’s Rule●Most elimination reactions

follow Zaitsev’s rule in which the most stable alkenes are the major products. However, under some circumstances, the major elimination product is the less substituted, less stable alkene

6C. Formation of the Less SubstitutedAlkene Using a Bulky Base

Ch. 7 - 35

●Case 1: using a bulky base

CH3CH2CHCH3

EtO

(small)

CH3CH CHCH3

CH3CH2CH CH2

CH3CH CHCH3

CH3CH2CH CH2

+(80%)

(20%)

+(30%)

(70%)

tBuO

(bulky)

Br

H C C C C

H

H

H

H

H

Br

H

H

H

EtO⊖

(small base)tBuO⊖

(bulky base)

Ch. 7 - 36

●Case 2: with a bulky group next

to the leaving halide

C C C C

Me

H3C

Me H

H

Me

Br H

H

H C C C CH2

Me

H3C

Me H

H

Me

(mainly)

EtO

more crowded β-H

less crowded β-H

Ch. 7 - 37

Zaitsev Rule vs. Hofmann Rule

●Examples

Br

HbHa

NaOEt, EtOH, 70oC

+

69% 31%

(eliminate Ha) (eliminate Hb)

(1)

KOtBu, tBuOH, 75oC 28% 72%

Ch. 7 - 38

●Examples

NaOEt, EtOH, 70oC

+

91% 9%

(eliminate Ha) (eliminate Hb)

(2)

KOtBu, tBuOH, 75oC 7% 93%

BrHb

Ha

Ch. 7 - 39

The 5 atoms involved in the transition state of an E2 reaction (including the base) must lie in the same plane

The anti coplanar conformation is the preferred transition state geometry●The anti coplanar transition

state is staggered (and therefore of lower energy), while the syn coplanar transition state is eclipsed

6D.The Stereochemistry of E2Reactions

Ch. 7 - 40

C CH

LG

B

Anti coplanartransition state

(preferred)

C CH LG

Syn coplanartransition state

(only with certainrigid molecules)

B

Ch. 7 - 41

Orientation Requirement●H and Br have to be anti

periplanar (trans-coplanar)●Examples

+CH3CH2

Br CH3

CH3CH2

CH3

EtO

since:

H

CH3CH2

CH3

H

Br

H

EtO

Only H isanti periplanarto Br

Ch. 7 - 42

E2 Elimination where there are two axial β hydrogens

H

H3C

Cl

H

Hb

H

EtO

EtO

Ha

CH(CH3)21

23

4

H3C CH(CH3)21

23

4

H3C CH(CH3)21

23

4

1-Menthene (78%)(more stablealkene)

2-Menthene (22%)(less stable alkene)

(a)

(b)Both Ha and Hb hydrogens are anti to the chlorine in this, the more stable conformation

Ch. 7 - 43

E2 elimination where the only axial β hydrogen is from a less stable Conformer

H

H3C

H

Cl

H

H

H

CH(CH3)2123

4

CH(CH3)2

H

H

H

Cl

H

CH3

Menthyl chloride(more stable conformer)Elimination is not possible for this conformation because no hydrogen is anti to the leaving group

Menthyl chloride(less stable conformer)

Elimination is possible for this conformation because the green hydrogen is anti to the chlorine

Ch. 7 - 44

CH(CH3)2

H

H

H

Cl

H

CH3

H3C CH(CH3)2

CH(CH3)2

H

H

H

Cl

H

CH3

OEt

The transition state for the E2 elimination is anti coplanar

2-Menthene (100%)

Ch. 7 - 45

7. Acid-Catalyzed Dehydration ofAlcohols

Most alcohols undergo dehydration (lose a molecule of water) to form an alkene when heated with a strong acid

C

H

C

OH

HA

heatC C + H2O

Ch. 7 - 46

The temperature and concentration of acid required to dehydrate an alcohol depend on the structure of the alcohol substrate● Primary alcohols are the most

difficult to dehydrate. Dehydration of ethanol, for example, requires concentrated sulfuric acid and a temperature of 180°C

H

CH

H

C

H

OH

Hconc. H2SO4

180oCC C

H

H

H

H

+ H2O

Ethanol (a 1o alcohol)

Ch. 7 - 47

● Secondary alcohols usually dehydrate under milder conditions. Cyclohexanol, for example, dehydrates in 85% phosphoric acid at 165–170°C

OH85% H3PO4

165-170oC+ H2O

Cyclohexanol Cyclohexene(80%)

Ch. 7 - 48

● Tertiary alcohols are usually so easily dehydrated that extremely mild conditions can be used. tert-Butyl alcohol, for example, dehydrates in 20% aqueous sulfuric acid at a temperature of 85°C

H3C C

CH3

CH3

OH20% H2SO4

85oC+ H2O

CH2

H3C CH3

tert-Butyl alcohol 2-Methylpropene(84%)

Ch. 7 - 49

●The relative ease with which alcohols will undergo dehydration is in the following order:

C

R

R OH

R

> C

R

R OH

H

> C

H

R OH

H

3o alcohol 2o alcohol 1o alcohol

Ch. 7 - 50

Some primary and secondary alcohols also undergo rearrangements of their carbon skeletons during dehydration

C CH

OH

CH3H3C

3,3-Dimethyl-2-butanol

85% H3PO4

80oC

C C

H3C

H3C

CH3

CH3

2,3-Dimethyl-2-butene(80%)

+ C CHCH3

H2C

H3C CH3

2,3-Dimethyl-1-butene(20%)

CH3

CH3

Ch. 7 - 51

●Notice that the carbon skeleton of the reactant is

C CC

C

C

C

C C

C

C

C

Cwhile that of the product is

Ch. 7 - 52

7A.Mechanism for Dehydration of 2o & 3o Alcohols: An E1 Reaction

Consider the dehydration of tert-butyl alcohol●Step 1

CH3

C

CH3

H3C O H

H

O

H

H+

H3C

C

CH3

H3C O H

H

+ H O

H

protonatedalcohol

Ch. 7 - 53

●Step 2H3C

C

CH3

H3C O H

H

C

CH3

H3C CH3

+ H O

Ha carbocation

●Step 3

C

C

H3C CH3

+ H O

H

H

H

HH

O

H

H+CH2

CH3C CH3

2-Methylpropene

Ch. 7 - 54

7B.Carbocation Stability & theTransition State

Recall

R

CR

R

H

CR

R

H

CH

R

H

CH

H

> >>

3o 2o 1o methyl> >>

moststable

leaststable

Ch. 7 - 55

Ch. 7 - 56

7C. A Mechanism for Dehydration of Primary Alcohols: An E2 Reaction

C C

H

H

H

O H + H Afast

C C

H

H

H

O H

H

+ A

C C

H

H

+HA+

H

OH

slowr.d.s

alkene

1o alcohol

acidcatalyst

protonatedalcohol

conjugatebase

Ch. 7 - 57

8. Carbocation Stability & Occurrenceof Molecular Rearrangements

8A.Rearrangements duringDehydration of Secondary Alcohols

C CH

OH

CH3H3C

3,3-Dimethyl-2-butanol

85% H3PO4heat

C C

H3C

H3C

CH3

CH3

2,3-Dimethyl-2-butenol(major product)

+ C CHCH3

H2C

H3C CH3

2,3-Dimethyl-1-butene(minor product)

CH3

CH3

Ch. 7 - 58

Step 1

CH3

C

CH3

H3C CH CH3

HOH

H

+

O H

+ H O

H

protonatedalcohol

CH3

C

CH3

H3C CH CH3

OH2

Ch. 7 - 59

Step 2

CH3

C

H3C

H3C CH CH3

OH2

+ H O

H

a 2o carbocation

CH3

C

CH3

H3C CH CH3

Ch. 7 - 60

Step 3

CH3

C

CH3

H3C CH CH3

2o carbocation

(less stable)

transition state

CH3

C

CH3

H3C CH CH3

C

CH3

H3C CH CH3

CH3

3o carbocation

(more stable)The less stable 2o carbocation rearranges to a more stable 3o carbocation.

Ch. 7 - 61

Step 4

H

C

CH3

C CH3CH2

CH3

H

A

(a) or (b)

(a)

(b)

(a) (b)

H

C

CH3

C CH3

H3C

H2C+ HAC

CH3

C

CH3

H3C

H3CHA +

(major) (minor)

less stable alkene more stable alkene

Ch. 7 - 62

a 2o carbocation

CH3

C

CH3

H3C CH CH3

Other common examples of carbocation rearrangements

●Migration of an allyl group

3o carbocation

C

CH3

H3C C CH3

methanide

migration

CH3

Ch. 7 - 63

●Migration of a hydride

a 2o carbocation

H

C

CH3

H3C CH CH3

3o carbocation

C

CH3

H3C C CH3

hydride

migration

H

Ch. 7 - 64

8B.Rearrangement after Dehydrationof a Primary Alcohol

+ H O

H

C C C O H

R

H

H

R

H

H

H A+E2

C C

R

C

H

H

R

HH A+

H A+protonation

C C

R

C

H

H

R

HA+C C

R

C

H

H

R

HH

H A+deprotonation

C C

R

C

H

HR

A + C C

R

C

H

H

R

HH

H

Ch. 7 - 65

sp sp2 sp3

9. The Acidity of Terminal Alkynes

pKa = 25 pKa = 44 pKa = 50

Acetylenic hydrogen

C CH H C C

H

H

H

H

C C

H

H

H H

H

H

Relative basicity of the conjugate base

CH3CH2 >CH2 CH >CH CH

Ch. 7 - 66

Comparison of acidity and basicity of 1st row elements of the Periodic Table●Relative acidity

●Relative basicity

H OH H OR C CRH H NH2 H CH CH2 H CH2CH3> > > > >

pKa 15.7 16-17 25 38 44 50

OH OR C CR NH2 CH CH2 CH2CH3< < < < <

Ch. 7 - 67

10.Synthesis of Alkynes by Elimination Reactions

Synthesis of Alkynes by Dehydrohalogenation of Vicinal Dihalides

Br

C

H

C

Br

H

C CNaNH2

heat

Ch. 7 - 68

Mechanism

Br

R C

H

C R

Br

H NH2

R R

NH2

H

RR

BrE2

Ch. 7 - 69

Examples

Br

Br H

H (78%)

NaNH2

heat(1)

NaNH2heat

Ph

Ph

Br2

CCl4 Ph

Br H

BrH

PhPh

Ph(2)

Ch. 7 - 70

Synthesis of Alkynes by Dehydrohalogenation of Geminal DihalidesO

R CH3

PCl5

0oC R CH3

Cl Cl

gem-dichloride

1. NaNH2 (3 equiv.), heat2. HA

Ph H

Ch. 7 - 71

11.Replacement of the AcetylenicHydrogen Atom of TerminalAlkynes

The acetylide anion can be prepared by

R HNaNH2

liq. NH3R Na + NH3

Ch. 7 - 72

Acetylide anions are useful intermediates for the synthesis of other alkynes

R R' X R R' X+

∵ 2nd step is an SN2 reaction, usually only good for 1o R’

2o and 3o R’ usually undergo E2 elimination

Ch. 7 - 73

Examples Ph H

Ph Na

NaNH2liq. NH3

CH3 I

I

H

SN2 E2

Ph CH3

+

NaI

Ph H

+

+

I

Ch. 7 - 74

13.Hydrogenation of Alkenes

Hydrogenation is an example of addition reaction

C C

H

C C

HH2

Pt, Pd or Ni

solventheat and pressure

C C

H

C C

H

H

H

H2Pt, Pd or Ni

solventheat and pressure

Ch. 7 - 75

Examples

H2

Rh(PPh3)3Cl

H

H

H

H

H2

Pd/C

Ch. 7 - 76

14.Hydrogenation: The Functionof the Catalyst

Hydrogenation of an alkene is an exothermic reaction●∆H° ≃ -120 kJ/mol

R CH CH R

+ H2

hydrogenationR CH2 CH2 R

+ heat

Ch. 7 - 77

Ch. 7 - 78

14A. Syn and Anti Additions An addition that places the parts

of the reagent on the same side (or face) of the reactant is called syn addition

C C + X Y C C

X Y

synaddition

C C + H H C C

H H

Pt

Catalytic hydrogenation is a syn addition.

Ch. 7 - 79

An anti addition places parts of the adding reagent on opposite faces of the reactant

C C + X Y C C

Y

X

antiaddition

Ch. 7 - 80

15.Hydrogenation of Alkynes

Using the reaction conditions, alkynes are usually converted to alkanes and are difficult to stop at the alkene stage

H2

Pt or Pd

H H

H2

H H

HH

Ch. 7 - 81

15A. Syn Addition of Hydrogen: Synthesis of cis-Alkenes

Semi-hydrogenation of alkynes to alkenes can be achieved using either the Ni2B (P-2) catalyst or the Lindlar’s catalyst● Nickel boride compound (P-2

catalyst)

● Lindlar’s catalyst Pd/CaCO3, quinoline

NiO CH3

O

2

NaBH4

EtOHNi2B

(P-2)

Ch. 7 - 82

Semi-hydrogenation of alkynes using Ni2B (P-2) or Lindlar’s catalyst causes syn addition of hydrogen●Examples

Ni2B(P-2)

H2 H H

(cis)

(97%)

Pd/CaCO3quinoline

H2Ph CH3

Ph CH3

H H(86%)

Ch. 7 - 83

15B. Anti Addition of Hydrogen: Synthesis of trans-Alkenes

Alkynes can be converted to trans-alkenes by dissolving metal reduction

Anti addition of dihydrogen to the alkyne

R'R2. aqueous work up

1. Li, liq. NH3, -78oCH

RR'

H

Ch. 7 - 84

Example

2. NH4Cl

1. Li, liq. EtNH2, -78oC

H

H

anti addition

Ch. 7 - 85

C C

R

R

H

Mechanism

C C

R

R

radical anion

C C RR

Li

H NHEt

C C

R

R

H

Li

C C

H

R

R

H HEtHN

vinyl radical

vinyl aniontrans alkene

Ch. 7 - 86

16. An Introduction to Organic Synthesis

16A. Why Do Organic Synthesis? To make naturally occurring

compounds which are biologically active but difficult (or impossible) to obtain

BzN

H

Ph

OH

O

O

OAcH

OHHO

AcO O OH

TAXOL

Anti-tumor, anti-cancer agent

Ch. 7 - 87

TAXOL

Isolated from Pacific Yew treeLeaves

usually appear on separatemale and female trees

Cones and Fruit

seed pollen cones

Ch. 7 - 88

Approved by the U.S. Food & Drug Administration in 1992 for treatment of several types of cancer, including breast cancer, lung cancer, and melanoma

An estimation: a 100-year old yew tree must be sacrificed in order to obtain 300 mg of Taxol, just enough for one single dose for a cancer patient

Obviously, synthetic organic chemistry methods that would lead to the synthesis of Taxol would be extremely useful

TAXOL

Ch. 7 - 89

16B. Retrosynthetic Analysis

targetmolecule

1stprecursor

2ndprecursor

startingcompound

Ch. 7 - 90

When doing retrosynthetic analysis, it is necessary to generate as many possible precursors, hence different synthetic routes, as possible

targetmolecule

1st precursor A2nd precursor a

1st precursor B

1st precursor C

2nd precursor b

2nd precursor c

2nd precursor d

2nd precursor e

2nd precursor f

Ch. 7 - 91

16C. Identifying Precursors

C C

(target molecule)

Synthesis of

Ch. 7 - 92

C C

Retrosynthetic Analysis

disconnection 1

disconnection 2

C C

X

+

X +

SN2 on 1o alkyl halide: good

SN2 on 2o alkyl halide: poor will get E2 as major pathway

Ch. 7 - 93

C C NaC C HNaNH2

liq. NH3

Synthesis

I(SN2)

NaI + C C

Ch. 7 - 94

16D. Raison d’Etre

Summary of Methods for the Preparation of Alkenes

C C

C C

H OHH+

heat

C C

Li, liq. NH3(give (E)-alkenes)

C C

H2, Ni2B (P-2)or Lindlar's catalyst(give (Z)-alkenes)

C C

H X base, heat

(Dehydration of alcohols)

(Dissolvingmetal reduction of alkynes)

(Semi-hydrogenation of alkynes)

(Dehydrohalogenationof alkyl halides)

Ch. 7 - 95

Summary of Methods for the Preparation of Alkynes

C C R'R

(Dehydrohalogenation of geminal dihalide)

(Deprotonation of terminal alkynes and SN2 reaction of the acetylide anion)

(Dehydrohalogenation of vicinal dihalide)

R H

ClCl

H

R'

NaNH2heat

C C HR

1. NaNH2, liq. NH3

2. R'-X (R' = 1o alkyl group)

R H

HX

X

R'

NaNH2heat

Ch. 7 - 96

END OF CHAPTER 7