Organic Chemistry year 1 2011-2012
Professor Martin Wills
Email: m.wills@warwick.ac.uk
Office: C504
CONTENT OF LECTURES
• Substitution reactions at saturated carbon atoms: ‘SN2 and SN1’.
Mechanisms of substitution reactions, intermediates, orbital structures, implications
for stereochemistry, inversion and racemisation. Clayden et al, Chapter 17.
• Formation and reactions of alkenes: ‘E2 and E1’.
Methods for making alkenes, structure and selectivity. Reactions of alkenes with
electrophiles and nucleophiles. Oxidation and reduction reactions. Clayden et al.
Chapters 19 and 20.
Prof M Wills
1
What you should know – what you will learn
What you should know by now: NOTE WE ARE GOING TO REVISE SOME OF THIS
-Hybridisation (sp, sp2, sp3) of carbon atoms.
-Orbital structure of C-C and C=C bonds.
-E/Z and R/S definition and how to assign configuration.
-Electronegativity and formal charge.
-Organic reaction mechanisms – ‘arrow pushing’.
-Factors which influence the stability of cations and anions.
-Free energy and reaction profiles.
What you will learn in this part of the course:
-The detailed mechanisms of substitution reactions at saturated carbon atoms
(SN2 and SN1 mechanisms).
-Factors which influence the mechanisms of substitution reactions.
-Methods for the formation of alkenes (E2 and E1 mechanisms).
-Reactions of alkenes with electrophiles and nucleophiles.
-Reduction and oxidation reactions of alkenes.
Prof M Wills
2
Nucleophilic substitution reactions – overview:
Why these reactions are important and some examples:
Substitution reaction: replacing one group with another one! Some examples:
3
Substitution at a saturated (sp ) carbon atom (M Wills section):
Br
O
O
+
Br
N
H
+
HBr
What about the
counterion?
Substitution at a saturated (sp3) carbon atom (M Wills section):
Br
H2N
Substitution at an unsaturated (sp2) carbon atom (discussed in G. Challis section, term 2):
O
Br
O
H2N
N
H
+
HBr
What mechanisms could there be for this reaction?
Can you define; the nucleophile, the electrophile, the leaving group.
Prof M Wills
3
Substitution reactions – some definitions.
What is the significance of the ‘saturated carbon atom’.
What ‘shape’ do the
groups around this
atom define?
sp3 hybridised carbon
Br
H
H
+
O
O
Br
Which leaving groups can be used – what ‘drives’ the reaction?
Leaving group
X
H
H
+
O
O
+
X
Key point: Good leaving groups are halides (Cl, Br, I), OSO2R, and other groups
which stabilise a negative charge.
Prof M Wills
4
Mechanisms of substitutions reactions: SN2
There are two major mechanisms of substitution reactions.
The first is called the SN2 mechanism – Substitution, Nucleophilic, Bimolecular:
What do these three terms mean?
It is a single step mechanism; the nucleophilic adds and the leaving group is simultaneously
displaced in the same step. A concerted mechanism.
Br
H
H
H
O
H
+
Br
O
Rate = k [nBuBr][nPrO-]
What happens if I double the concentration of bromide? What if I double the concentration
of bromide and of propoxide?
Transition state
Energy
Bromide
(starting material)
Ether (product)
Reaction co-ordinate.
Prof M Wills
5
The transition state for SN2 reactions:
The SN2 mechanism – structure of the transition state.
What does this symbol mean?
 Br
Br
H
H
H
H
H
H

O
What is the hybridisation at this C atom?
Nu:
Br
O
What ‘shape’ do the
groups around this
atom define?
O
+
C
C
* antibonding orbital (empty
C
C
 bonding orbital (full)
Note partial bonds to nucleophile and leaving group. Nucleophile adds electron density to * antibonding orbital.
*** Key point of nomenclature; it’s SN2 not SN2*** This is important ***
Prof M Wills
6
Stereochemical consquences of SN2 reactions:
The SN2 mechanism – What happens at chiral centres:
S configuration:
R configuration
 Br
Br
H
H
Me
H
Me
Me
+
Br
O
O

O
Key concept – inversion of configuration.
*** Key point of nomenclature; INVERSION *** This is important ***
Prof M Wills
7
Mechanisms of substitutions reactions: SN1
The second is called the SN1 mechanism – Substitution, Nucleophilic, Unimolecular:
What do these three terms mean?
It is a two step mechanism; the leaving group leaves in the first step to form a cationic
intermediate and then the nucleophile adds in the second step
INTERMEDIATE
cation
Br
Me
Me
first step
SLOW
Me
second step
FAST
Me
- Br
Me
Me
Me
Me
O
H
H
O
O
-H
H
The first step is the ratedetermining step (rds)
Rate = k [C6H13Br] i.e. [nPrOH (nucleophile)] is not featured
What happens if I double the concentration of bromide? What if I double the concentration
of bromide and of alcohol?
First transition state
Second transition state
Cation
(intermediate)
Energy
Prof M Wills
Bromide
(starting material)
Ether (product)
8
Stereochemical consquences of SN1 reactions:
Br
Me
Ph
Me
Ph
- Br
HO
HO
Me
Ph
or
S configuration:
Me
Ph
R configuration
50%
O
O
+
S configuration:
Me
Ph
A racemic product is formed.
(note - loss of proton not illustrated)
Why are there two products now?
What happened to the square brackets?
What is the ratio of the products?
50%
Two enantiomers are formed in a 1:1 ratio
The cations are intermediates, not transition states.
*** Key point of nomenclature; RACEMISATION *** This is important ***
Prof M Wills
9
Nucleophilic substitution reactions: Summary.
SN1 and SN2
The ‘1’ and ‘2’ refer to the molecularity of the reaction (the number of species in the rate
Expression).
SN1 is a two step reaction. SN2 is a one step reaction.
SN2 mechanisms go with inversion of configuration, SN1 with racemisation.
Make sure you understand the difference between an intermediate and a transition state.
Other substitution mechanisms include the SN2’ :
RNH
Br
H
Me
Prof M Wills
RHN
Me
- Br
10
Factors which influence SN2 and SN1 reactions:
‘If I do a substitution reaction, will it go through an SN2 or SN1 mechanism?’
i) Substrate structure. Steric hindrance and cation stability.
Unhindered substrate:
Nu: = nucleophile
-
Will not form a stable cation if Br leaves:
Br
H
H
more likely to undergo SN2
H
H
Nu:
Approach unhindered
Hindered (substituted) substrate
Will form very stable cation
Br
Me
Me
Nu:
more likely to undergo SN1
Me
Me
Approach hindered
by two large Me groups
Me
Cation stability:
Me
Me
tertiary
Me
>
Me
H
secondary
>
H
Me
H
primary
>
H
E2
Competes?
H
H
methyl cation
A primary halide is more likely to undergo SN2, a tertiary SN1. A secondary halide may do
both, although a good nucleophile would favour SN2 and a weak nucleophine SN1.
There are exceptions to all these guidelines.
Prof M Wills
11
Other factors that increase cation stability.
Cation stability can also be increased by an adjacent double bond or aromatic ring:
H
H
H
Charge is delocalised
into the aromatic ring
or double bond.
is more
stable than
H
benzyl cation
H
H
allyl cation
is more
stable than
H
H
However an adjacent benzyl or allyl group can also increase the rate of SN2 reactions, by
stabilising the transition state:

OMe
MeO
+
Br
OMe
H
H
via
+ Br
Br

Orbital overlap lowers the transition state
energy, and makes the reaction faster.
Prof M Wills
12
Factors which influence SN2 and SN1 reactions cont:
ii) Effect of the solvent. This section has been expanded from the original circulation.
Solvent effects can be complex and a good summary can be found on p428-429 of Clayden et al.
Solvents which stabilise cations will tend to increase the rate of S N1 (because a ions are formed in the rate
determining step). These include dipolar aprotic solvents such as dimethylformamide (DMF) and dimethylsulfoxide
(DMSO) and dipolar protic solvents such as water or carboxylic acids. For S N2 reactions the situation is more
complex. A nonpolar solvent may speed up the reaction in a situation where the transition state is less polar than the
localised anions, and the product is neutral. In situations where a charged product is formed by the reaction of neutral
substrates, then a polar solvent such as DMF will be better because the transition state is more polar. Dipolar aprotic
solvents such as DMF can also make anionic nucleophiles more reactive in S N2 reactions because they solvate the
cation and make the anion ‘freer’ to react. Polar protic solvents (e.g. water, alcohols, carboxylic acids) however can
retard the rate of SN2 reactions by solvating the anion and making it less reactive.
iii) Effect of the nucleophile.
In general, more reactive nucleophiles favour the SN2 reaction. This is fairly logical.
Examples of 'reactive'
nucleophiles:
Alkoxides: RO
Amides: RNH
Alkyl anions (for example
Grignard (magnesium based)
reagents:
R (RMgBr)
Some relatively
stable anions are
good nucleophiles e.g.
Examples of 'less reactive'
nucleophiles:
Alcohols (not deprotonated) ROH
Anions of thiols R S
Anions of phenols
so are neutral
phosphines
R3P
Amines (not deprotonated) RNH2
O
Stabilised nucleophiles e.g. acetate:
O
O
iv) Effect of the leaving group; good leaving groups are needed for both mechanisms.
Prof M Wills
13
SN2 vs SN1 – all aspects must be considered:
MeO
K
(DMSO=Me2SO)
+
Br
H
H
OMe
INVERSION
i.e. good nucleophile, not very hindered substrate, good leaving group, hence SN2.
Ph
Br
Ph
H2O
Ph
+
Me
Me
OH
HO
50:50
RACEMIC
Me
i.e. weak nucleophile, more hindered substrate, potential for stable cation, hence SN1.
What would be your prediction of a mechanism for:
Ph
Ph
Ph
Ph
pyridine
Cl
+
R
OH
Ph
Ph
Ph
Ph
Ph
Prof M Wills
Ph
Ph
Ph
?
O
Ph
H
O
N
pyridine
(mild base)
N
Ph
14
Formation and use of the OTs leaving group (very common
in synthesis):
OH
H
H
+ Cl
O O
S
(TsCl)
O
O O
S
H
OTs
H
H
H
O
Et3N:
(base)
H
O O
S
O
H
O
+
O
O O
S
H
TsO
What is the mechanism, and why
go to all this trouble, i.e. why is OH
a poor leaving group? How else can it be
Made into a good leaving group?
Key point; Learn what a OTs (tosyl group) is – it will come up again!
Prof M Wills
15
Alkenes– reminder of structure and formation:
Already discussed in Prof M. Shipman’s lectures:
In the case of a carbon atom attached to three other groups (by two single bonds and one double bond)
the single 2s and two 2p orbitals mix (rehybridise) to form three sp 2 orbitals. These are all arranged at
mutual 120 degree angles to each other and define a trigonal shape, the remaining p orbital projects out
of the plane of the three sp2 orbitals and overlaps with an identical orbital on an adjacent atom
to form the double bond:
1 x 2s
2 x 2p
on a carbon atom
combine to form 3 x sp2 orbitals:
C
which lie at mutual 120 degrees
in a molecule such as etheneC2H6,
whilst the remaining p orbital
forms the double bond:
H
H
H
H
H
C
C
H
H
H
The resulting structure is rigid and cannot rotate about the C=C bond without breakage of the
bond between the p-orbitals (the p bond). The can be separated into E and Z configuration isomers.
The p bond is much more reactive than the  bond – the bonds are not equivalent to each other.
Prof M Wills
16
Alkynes – reminder of structure and formation:
In the case of a carbon atom attached to two other groups (by one single bonds and one triple bond)
the single 2s and one 2p orbitals mix (rehybridise) to form two sp orbitals. These are all arranged at
mutual 180 degree angles to each other and define a linear shape, the remaining p orbitals projecting out
from the sp orbital to overlap with identical orbitals on an adjacent atom to form the triple bond:
1 x 2s
1 x 2p
combine to form 2 x sp2 orbitals:
on a carbon atom
C
which lie at 180 degrees
in a molecule such as ethyneC2H4,
H
whilst the remaining p orbitals
form the triple bond:
H
H
C
C
H
Both p bonds in an alkyne are much more reactive than the  bond.
Prof M Wills
17
Alkenes – formation:
Most (but not all) alkene formation reactions involve an ELIMINATION reaction.
Alkyne reduction is also important (see later).
Ph
H
H
Ph
H
Ph
H
Br
Ph
Ph
- HBr
H
E-alkene
and/or
Ph
H
H
Z-alkene
Recap on alkene structure.
Can you use the Cahn-Ingold-Prelog rules to determine the configuration?
Key point – H from one carbon atom and a leaving group (typically a halide) from the adjacent
carbon atom.
Prof M Wills
18
Mechanisms for the formation of alkenes: E1 and E2:
Most elimination reactions, to form alkenes, involve an E2 or E1 elimination.
E2 = Elimination, bimolecular. It is a one-step reaction.
A strong base is needed - why is this?
MeO
H
Br
H
H
H
H
Rate = k [Cyclohexylbromide][MeO-]
(This is a base)
Transition state
Energy
Bromide
(starting material)
Alkene (product)
Reaction co-ordinate.
Prof M Wills
19
E2 elimination – stereochemical implications: The ‘anti
periplanar’ requirement.
E2 reactions require correct orbital alignment in order to work. The optimal
arrangement is ‘anti periplanar’, where the ‘H’ and ‘Br’ (in an alkyl bromide) are anti to
each other.
Ph
Ph
can form a E- or Z-alkene upon elimination:
Br
Base:
H
H
Ph
Ph
H
Br
Base:
H
Ph
Ph
H
H
Ph
E- alkene.
H
H
Ph
Ph
H
H
Ph
Br
H
Ph
H
Ph
H
Ph
H
Ph
Z- alkene.
Which base would you use?
EtO-, HO-, alkoxide. Etc.
Prof M Wills
20
E2 elimination – stereochemical implications: orbital
alignment:
Orbital alignment in E2 elimination reactions:
But O H
Substrate
H
H
Ph
Ph
H
tBuOK
(base)
Ph
H
tBuOH
H
Ph
Br -
Br
But O H
H
H
Ph
H
Ph
Br
tBuOK
(base)
H
Ph
H
Ph
Ph
H
Ph
H
H
Ph
E- alkene.
Br
tBuOH
H
Ph
Br -
H
Ph
H
Ph
Br
H
Ph
H
Ph
Z- alkene.
The alignment of  and * orbitals in the substrate leads to a smooth transition to a p bond in
the product. Which is the most likely product? The E, or trans, product.
Prof M Wills
21
The E1 elimination mechanism:
E1 = Elimination, unimolecular. It is a two-step reaction.
It proceeds via a cationic intermediate.
Me
Br
H
-Br
H
Me
Me
H
H
H
Et3N:
Rate = k [Methylcyclohexylbromide]
Triethylamine (Et3N)
is a weak base.
Why does the substrate now have an extra methyl group?
Why was a weak base used in this reaction?
First transition state
Second transition state
Cation
(intermediate)
Energy
Bromide
(starting material)
Alkene (product)
Reaction co-ordinate.
Prof M Wills
22
Nature of the intermediate in an E1 reaction:
E1 reactions proceed through a ‘flat’, i.e. trigonal, cation (like SN1 reactions).
Me
Br
H
Me
Me
H
-Br
H
H
H
2
sp carbon
H
3
sp carbon
Me
H
Sometimes multiple products are formed (irrespective of mechanism)–
What products would you predict from this reaction (more than one is possible)?
Me
Br
Me
Me
Me
Et3N (mild base)
Me
Me
Me
and
and
E1
Prof M Wills
23
Formation of alkenes by elimination of alcohols:
See if you can write the mechanism for the following (E1) reaction:
Me
HCl (strong acid) H Me
O
MeOH (solvent)
H
Me
HO
Me
H
MeOH
H
H
Why is acid needed when the alcohol is the leaving group – why can’t we rely on a base?
What other alkene product can be formed, and how?
What would be the effect of using a base and an acid in the following? – write mechanisms.
HO
H2SO4
Br
KOtBu
?
?
K O
To be completed in lectures.
E1cb ‘conjugate base’ is less common mechanism, but important.
Prof M Wills
24
Some alternatives to ‘simple’ elimination – the Wittig reaction:
The Wittig reaction is one of a number of reaction that provide a means for controlling where
the double bond ends up.
O
Ph3P
CH2
Phosphonium ylid
H2C
+
Ph3P
O
the isomer, H3C
is not formed
This is how the ylid is formed:
PPh3 +
base
CH3Br
Ph3P
CH3
Ph3P
CH2
Ph3P
CH2
Key point: this is important – learn it
Prof M Wills
25
Some alternatives to ‘simple’ elimination – the Wittig reaction
(this is important – learn it)
Here is the mechanism of the Wittig reaction (you need to complete it).
O
O
Ph3P
C
H2
Ph3P
C
H2
H2C
+
Prof M Wills
Ph3P
O
26
Substitution vs elimination, base vs nucleophile:
Sometimes a particular substrate can undergo a substitution or an elimination reaction.
The outcome depends on all the factors involved in the reaction;
The ‘is an alkoxide a nucleophile or a base?’ question. Answer - depends what it does:.
Br
MeO Na
MeO
Alkoxide is a nucleophile.
Br
MeO Na
Alkoxide is a base.
The most important factor is probably the substrate structure – deprotonation may outpace
nucleophilic addition when a substrate is very hindered. Certain substrates cannot undergo
elimination reactions.
Prof M Wills
27
Reactions of alkenes with electrophilic reagents - bromine:
Alkenes are electron rich (in the p system) and react with electrophilic reagents:
Br
R
Br2
R
R
R
Br
The mechanism is as follows, the intermediate is a bromonium ion:
R
Br
R
R
R
Br
Br
Br
R
R
Br
Br
Br
what about
Br
trans
racemic
Prof M Wills
28
Further additions of electrophiles to double bonds - HBr:
Hydrogen halides (HCl, HBr) also add across double bonds.
Br
R
R
R
R
HBr
R
R
R
H
R
The mechanism involves the addition of a proton first (with the electron-rich alkene), then the
bromide. This is logical, because the alkene is electron rich.
R
R
R
R
R
R
R
Prof M Wills
Br
H
R H
Br
R
R
R
R
H
29
Regioselectivity of electrophilic additions to alkenes:
Addition of HCl and HBr (and other acids) across unsymmetric alkenes results in formation of the
more substituted halide (via the more substituted cation).
H
H
H3C
H
HBr
H3C
H
Br
H
H
Major
Br
H
H
H3C
H
H
Minor
H
The mechanism involves the addition of a proton first, as before, but in this case the unsymmetrical
intermediate has a larger density of positive charge at one end.
more stable
cation
BrH
H3C
H
Br
H
H3C
H
H
What about
H
H
H
Br
H
H
H
H3C
H
H
H
H
H3C
H
less stable
cation
major
Br
H3C
H
H
H
H
There are two
options.
The reaction goes via
the most stable (most
substituted) cation.
minor
?
Ph
Prof M Wills
30
Acid catalysed hydration (addition of water) to alkenes:
Acid catalysed hydration (addition of water) is a very important reaction of alkenes:
H
H
H3C
H
H2O
H3C
H
OH
H
H
+
H catalysis
H
OH
H
H3C
H
+
Major
H
Minor
H
The mechanism involves the addition of a proton first, as before, followed by addition of water, the
regioselectivity is the same as for addition of HBr:
OH2
acid H3C
H3C
H
H
H
(H )
H
H
H
H
H
H
H2O
H3C H
H
O
H
H
H
H
H3C H
regeneration of H
O
H
H
H
This mode of addition of H2O is referred to as ‘Markonikov’ selectivity (i.e. formation of the
MOST substituted alcohol via the MOST substituted cation.
Prof M Wills
31
Radical reactions of alkenes: HBr in diethylether containing
peroxides.
HBr, peroxides in ether
Br
anti Markovnikov
addition.
H
Mechanism (Clayden et al p 1033-1035)
RO
HO.
H Br
Br .
Br
Prof M Wills
2 RO . (radical)
OR
HO.
Br .
Br
.
.
.
Br
H Br
H
+ Br .
32
Nucleophilic additions to C=C bonds – require nearby
electron withdrawing groups (‘ewg’).
The polar effects of C=O bonds can be transmitted through adjacent C=C bonds, e.g.
An enone: (a compound with a directly linked C=C and C=O double bond) can react with a
nucleophile at either the C of the C=O bond or at the C at the end of the C=C bond. This is called
conjugate addition, 1,4-addition and/or ‘Michael’ addition.
H
C
H
O
C
O
H
C
H
R
H
C
H
C
R
C
H
H
H
(a nucleophile)
Add acid at end
of reaction.
H
The oxygen atom drives the reactionit is more likely to gain a negative
charge because it is more
electronegative than adjacent atoms.
C
O
H
R
C
H
C
H
H
More on this from Prof Challis later in course.
Prof M Wills
33
Polymerisation of alkenes/reactions of alkynes.
Alkenes/Alkynes
Another important reaction of alkenes is polymerisation, which is often radical-initiated:
CH3
CH3
CH3
CH3
H3C
n
polymerisation
Alkynes are capable of many of the same reactions as alkenes, but twice if enough
reagent is used, e.g. addition of bromine:
Br
Br
Br2
Br
Br2
Br
Br
Br
More on polymerisation later in the course (Professor Haddleton).
Prof M Wills
34
Cycloaddition reactions of alkenes: The Diels-Alder reaction.
Complete the diagram below:
note error corrected here
O
O
+
a conjugated
diene
O
O
O
Alkene
(dienophile)
O
Reaction is concerted i.e. all bonds are formed and broken at the same time:
O
O
O
O
O
O
O
O
O
Stereochemistry: Bonds are formed on one face of the alkene, hence there is a high degree of stereocontrol.
Prof M Wills
35
Alkene hydroboration reaction (important)!
Overall reaction (reagents to be added):
i) BH3
OH
ii) H2O2, NaOH
What happens if you carry out acid-catalysed hydration (addition of water).
OH
H
, H2O
You get the secondary alcohol!
Prof M Wills
36
Hydroboration mechanism – to be completed in lecture:
H
Concerted addition
of B-H bond across
the alkene boron adds to least
hindered end:
BH2
i) BH3
OH
+
B(OH)3
ii) H2O2, HO
BH3
H2O
BH2
O
B
3
HO-O
BH
O
2
B
2
This is a migration
reaction
All the B-H bonds
are utilised.
Three alkenes add
to one borane.
B
3
HO-O
HO-O
- OH
B
2
O
B
O
2
OH
note there was an error on the handout should not be 'H' on 'B' at this stage.
Prof M Wills
37
Reduction reactions of alkenes: and alkynes, stereochemistry
– formation of cis alkenes by hydrogenation.
First you need to make a substituted alkyne:
H
H
NH2
NaNH2
EtI
H
i) NaNH2
H
ii)
I
Br
Why do you need to use NH2? Can you use EtO or Et3N?
These would not be powerful enough as bases.
This is the key reduction reaction:
1
R
R2
H2 (gas), Pd/C, quinoline
H
1
or H2 (gas), Lindlar catalyst
R
H
2
R1
R2
H
H
R
catalyst surface
Why is the quinoline added to the catalyst? What happens if you do not add it?
You get reduction right down to the alkane.
The reduction takes place on a surface, and the hydrogen is transferred to one side
of the alkyne.
Prof M Wills
38
trans Alkenes can also be formed from alkynes:
R1
Na, NH3, then H
R2
tBuOH is often used as a source of protons.
This is commonly known as ‘reducing metal’ reduction. It works by a mechanism in which
‘electrons’ are generated from the metal. Li, K and Mg are also sometimes used.
Here is the mechanism:
Na
Na
1
R
2
R
+
e
R2
e
NH3
2
H
R
.
R1
.
R1
Note this is a radical anion.
e
H
R1
Prof M Wills
R2
H
NH3
H
R1
R2
H
R1
R2
39
Reduction of alkenes to alkanes – hydrogenation most commonly used:
Commonly used
Metals: Pd, Rh, Ru, Ir,
Commonly used
Supports: Carbon (graphite), silica
H H
H
H
Stereochemistry:
note a
cis alkene is
formed
Relative rate of hydrogenation:
Me
Me
Me
H
Me
H
Me
H
Me
H
H
H
H
H
H
H
in general, more stable (more substituted) alkenes react more slowly.
slow reduction
Prof M Wills
fast reduction
Why use a support to aid product separation, how does the reaction work?
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Hydrogenation of alkenes to make margarine:
Polyunsaturated fats are regarded as healthier than saturated ones but tend to be liquids so
They are partially hydrogenated to make margarine – solid but still with double bonds in..
Polyunsaturated vegetable oil.
CO2H
Linolenic acid mp -11 oC
H2 / Ni
CO2H
Linoleic acid mp -5 oC
CO2H
Oleic acid mp 16 oC
H2 / Ni
H2 / Ni
CO2H
Stearic acid mp 71 oC
Fully saturated fatty acid.
Saturated fats have high melting points because they pack more efficiently.
Prof M Wills
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Alkene oxidation reactions:
Alkene oxidation reactions can give epoxides, diols, or even ketones from complete
cleavage of the alkene.
The p bond in alkenes
is very reactive.
2
H
R1
R
H
peracid
O
R
H
R1
O
2
R
O OH
H
H
R1
HO
R2
OsO 4
H
H
osmium
tetroxide
R1
Epoxide
OH 2
R
H
diol
O3
H
O
1
R
+
R2
O
H
What is the structure of ozone? Look it up Why might these products be useful?
Prof M Wills
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Epoxidation of alkenes using peracids:
O
This is mCPBA,
(meta-chloroperbenzoic acid)
which is a commonly used peracid
O
H
Ar
H
O O
O O
Cl
mCPBA
O
H
2
R
Ar
H
H
R1
trans (E) alkene
Learn the mechanism:
O
H
O O
R2
R1
H
H
H
mCPBA
O
H
Ar
O O
H O
R1
R1
R2
cis (Z) alkene
H
R2
Ar
O
O
O
H
H
R2
R1
H
H
R1
O
R2
H
This is one of the best mechanisms!
Prof M Wills
43
Ozonolysis of alkenes cleaves the double bond:
R2
H
i) O3
H
O
1
ii) reducing agent
(e.g. Zn metal, Ph3P, Me2S
H
R
1
R2
+
O
R
H
reducing agent
O
O
O
H
2
R
1
R
H
H
O
O
R1
O
H O O R2
R2
R1
H
O
H
ozonide
Predict the products from:
O
i) O3
ii) reducing agent
(e.g. Zn metal, Ph3P, Me2S
OMe
i) O3
ii) reducing agent
(e.g. Zn metal, Ph3P, Me2S
Prof M Wills
O
O
OMe
O
44
Alkene dihydroxylation:
Potassium permanganate (KMnO4)
O
O
Mn
O
O
O
O
Mn
O
O
O
O
Os
O
O
O
O
Os
O
O
2 H2O
OH
OH
Mechanism should be
fairly easy to work out.
2 H2O
OH
OH
What products are formed using
the following alkenes and OsO4?
i) OsO 4
ii) 2 H2O
i) OsO 4
ii) 2 H2O
HO
OH
OH
OH
Note – OsO4 is expensive and very toxic. Better to use it catalytically (how would you do this?).
Prof M Wills
45
The Wacker Oxidation:
ClPd
Cl
O
catalytic PdCl2, H2O, O2, CuCl2
R1
1
R
keto-enol
tautomerism
Cl
Pd
HO
H
PdCl
1
R
R1
HO
+ HPdCl
R1
OH2
Pd
+ HCl
This is a commercial reaction used on a large scale in industry.
The CuCl2 and O2 reoxidise the PdCl2 (Pd is expensive).
Prof M Wills
46
What you should understand and know:
• MECHANISM OF REACTIONS AT SATURATED CARBON ATOMS
The mechanisms and stereochemical implications of SN2 and SN1 reactions. Inversion
with SN2, racemisation with SN1. Factors which determine the likely pathway.
Alternative leaving groups. Examples of applications in synthesis.
• SYNTHESIS AND REACTIONS OF ALKENES (C=C)
E1 and E2 elimination mechanisms. Stereoselective alkene synthesis from alkynes
by hydrogenation and dissolving metal reduction; Wittig reaction including ylid
formation and mechanism; hydrogenation of alkenes; epoxidation, dihydroxylation and
ozonolysis of alkenes; hydration and hydroboration of alkenes; Wacker oxidation;
Diels-Alder cycloaddition.
Prof M Wills
47