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Transcript of Created by Professor William Tam & Dr. Phillis Chang Ch. 7 - 1 Chapter 7 Alkenes and Alkynes I:...
![Page 1: Created by Professor William Tam & Dr. Phillis Chang Ch. 7 - 1 Chapter 7 Alkenes and Alkynes I: Properties and Synthesis. Elimination Reactions of Alkyl.](https://reader035.fdocuments.in/reader035/viewer/2022081417/56649c7d5503460f94932b84/html5/thumbnails/1.jpg)
Created byProfessor William Tam & Dr. Phillis
Chang Ch. 7 - 1
Chapter 7
Alkenes and Alkynes I:Properties and
Synthesis.Elimination Reactions
of Alkyl Halides
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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.
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Ch. 7 - 3
1. Introduction
Alkenes●Hydrocarbons containing C=C●Old name: olefins
CH2OH
Vitamin A
HO
H3C
H3C
H HCholesterol
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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)
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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
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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
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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
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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
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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
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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
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Ch. 7 - 11
ClBr
CH3
H (E )-2-Bromo-1-chloropropene
●Examples
ClCH3
Br
H (Z )-2-Bromo-1-chloropropene
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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
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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
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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
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Ch. 7 - 15
3A.Heat of Reaction
C C + H HPt
C C
H H
Heat of hydrogenation●∆H° ≃ -120 kJ/mol
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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
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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.
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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
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Ch. 7 - 19
Examples of stabilities of alkenes
(1) >
(2) >
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Ch. 7 - 20
4. Cycloalkenes
Cycloalkenes containing 5 carbon atoms or fewer exist only in the cis form
cyclopropene cyclobutene cyclopentene
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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.)
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Ch. 7 - 22
Trans – cyclooctene has been isolated and is chiral and exists as a pair of enantiomers
cis - cyclooctene trans - cyclooctenes
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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
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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+
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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)
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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.
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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)
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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
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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
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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%
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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
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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⊖
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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‡
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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
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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)
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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
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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%
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Ch. 7 - 38
●Examples
NaOEt, EtOH, 70oC
+
91% 9%
(eliminate Ha) (eliminate Hb)
(2)
KOtBu, tBuOH, 75oC 7% 93%
BrHb
Ha
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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
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Ch. 7 - 40
C CH
LG
B
Anti coplanartransition state
(preferred)
C CH LG
Syn coplanartransition state
(only with certainrigid molecules)
B
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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
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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
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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
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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%)
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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
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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)
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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%)
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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%)
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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
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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
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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
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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
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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
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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
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Ch. 7 - 55
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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
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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
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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
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Ch. 7 - 59
Step 2
CH3
C
H3C
H3C CH CH3
OH2
+ H O
H
a 2o carbocation
CH3
C
CH3
H3C CH CH3
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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.
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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
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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
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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
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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
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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
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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< < < < <
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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
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Ch. 7 - 68
Mechanism
Br
R C
H
C R
Br
H NH2
R R
NH2
H
RR
BrE2
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Ch. 7 - 69
Examples
Br
Br H
H (78%)
NaNH2
heat(1)
NaNH2heat
Ph
Ph
Br2
CCl4 Ph
Br H
BrH
PhPh
Ph(2)
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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
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Ch. 7 - 71
11.Replacement of the AcetylenicHydrogen Atom of TerminalAlkynes
The acetylide anion can be prepared by
R HNaNH2
liq. NH3R Na + NH3
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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
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Ch. 7 - 73
Examples Ph H
Ph Na
NaNH2liq. NH3
CH3 I
I
H
SN2 E2
Ph CH3
+
NaI
Ph H
+
+
I
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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
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Ch. 7 - 75
Examples
H2
Rh(PPh3)3Cl
H
H
H
H
H2
Pd/C
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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
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Ch. 7 - 77
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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.
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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
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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
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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)
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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%)
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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
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Ch. 7 - 84
Example
2. NH4Cl
1. Li, liq. EtNH2, -78oC
H
H
anti addition
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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
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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
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Ch. 7 - 87
TAXOL
Isolated from Pacific Yew treeLeaves
usually appear on separatemale and female trees
Cones and Fruit
seed pollen cones
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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
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Ch. 7 - 89
16B. Retrosynthetic Analysis
targetmolecule
1stprecursor
2ndprecursor
startingcompound
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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
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Ch. 7 - 91
16C. Identifying Precursors
C C
(target molecule)
Synthesis of
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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
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Ch. 7 - 93
C C NaC C HNaNH2
liq. NH3
Synthesis
I(SN2)
NaI + C C
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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)
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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
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Ch. 7 - 96
END OF CHAPTER 7