Synthetic planning (Retrosynthesis)

1
Synthetic planning (Retrosynthesis)
Work Backwards..
Trace the reactions sequence from the desired
product back to ultimate reactants.
Starting reactant
Target molecule.
Overall Sequence converts alkene ?alkyne
But typical of synthetic problems side reaction
occurs to some extent and must be taken into
account.
C
2
More Sythesis Nucleophilic Substitution

• Use the acidity of a terminal alkyne to create a
  nucleophile which then initiates a substitution
  reaction.

Note that we still have an acidic hydrogen and,
thus, can react with another alkyl group in this
way to make RCCR
Alkyl halides can be obtained from alcohols
3
Reactions alkyne with halogen
RCCR Br2? RBrCCBrR
No regioselectivity with Br2. Stereoselective for
trans addition.
4
Reactions Addition of HX
The expected reaction sequence occurs, formation
of the more stable carbocation.
Markovnikov orientation for both additions.
Now for the mechanism.
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Mechanism
The expected reaction sequence occurs, formation
of the more stable carbocation.
6
Addition of the second mole, another example of
resonance.
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Reactions Acid catalyzed Hydration (Markovnikov).
Markovnikov addition, followed by tautomerism to
yield, usually, a carbonyl compound.
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Reactions Anti Markovnikov Hydration of Alkynes,
Regioselectivity
Step 2
Step 1
Similar to formation of an anti-Markovnikov
alcohol from an alkene
Step 1, Internal Alkyne addition to the alkyne
with little or no regioselectivity issue.
Alternatively Asymmetric, terminal, alkyne if you
want to have strong regioselectivity then use a
borane with stronger selectivity for more open
site of attack.
Less exposed site.
Aldehyde not ketone.
More exposed site.
sia2BH
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Tautomerism, enol ? carbonyl
Step 2, Reaction of the alkenyl borane with H2O2,
NaOH would yield an enol. Enols are unstable and
rearrange (tautomerize) to yield either an
aldehyde or ketone.
Overall
internal alkyne ? ? ketone (possibly a mixture,
next slide)
Terminal alkyne ? ? aldehyde
10
Examples
Used to insure regioselectivity.
As before, for a terminal alkyne.
But for a non-terminal alkyne frequently will get
two different ketones
Get mixture of alkenyl boranes due to low
regioselectivity.
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Reduction, Alkyne ? Alkene
1. Catalytic Hydrogenation
If you use catalysts which are also effective
for alkene hydrogenation you will get alkane.
You can use a reduced activity catalyst
(Lindlar), Pd and Pb, which stops at the alkene.
You obtain a cis alkene.
Syn addition
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Reduction - 2
2. Treatment of alkenyl borane with a
carboxylic acid to yield cis alkene.
Instead of H2O2 / NaOH
Alkenyl borane
3. Reduction by sodium or lithium in liquid
ammonia to yield the trans alkene.
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Plan a Synthetic SequenceRetrosynthesis
YES!
Synthesize butan-1-ol from ethyne. Work backward
from the target molecule.
A big alkyne can be formed via nucleophilic
substitution. This is the chance to make the C-C
bond we need.
Is read as comes from.
Catalytic Lindlar reduction
Major problem make big from small. Be alert for
when the disconnect can be done.

• BH3
• H2O2, NaOH

Convert ethyne to anion and react with EtBr.
Do a disconnect here.
Target molecule
Catalytic reduction Lindlar
Addition of HBr.
Now, fill in the forward reaction details
Can we get an alkyne from smaller molecules?
Not yet! So how can we get it?
How about joining molecules to get an alkene?
Not yet!! So how can we get an alkene?
Ask yourself! Do we know how to join any two
molecules together to yield an alcohol?
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Alkyl Halides
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Boiling Points
The size of Br and CH3 about the same but bromo
compounds boil higher due to greater
polarizability more dispersion forces.
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Boiling Points
Branching in an alkane decreases the boiling
point. Likewise for haloalkanes.
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Boiling Points
Fluoroalkanes and alkanes of same MW have about
the same BP. In both cases the electrons are
tightly held, not very polarizable.
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Reaction of elemental halogen and alkanes
yielding haloalkanes
H3C H X X ? H3C X H - X

• Reaction Characteristics
• Requires heat or light to initiate.
• Fluorine is explosive. Reactivity decreases
  fluorine to chlorine to bromine to iodine which
  does not react.

Bond Dissociation Energies. Energy Required to
Break Bond.
Bonds being broken. Bonds being broken. Bonds being made. Bonds being made. Bonds being made. Bonds being made.
H3C - H 439 kJ
Cl - Cl 247 H3C - Cl 351 H Cl 431
Br - Br 192 H3C - Br 301 H - Br 368
Chlorination DH 439 247 -351 431 -96 kJ
exothermic
Bromination DH 439 192 301 368 -38 kJ
less exothermic
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Regioselectivity
Starting Point in Analysis Random Substitution.
Assume that all hydrogens are equally likely to
be replaced by X. There are 8 H in the molecule.
Equally likely to be replaced if random.

Random substitution 2/8 25 6/8 75
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Regioselectivity
Now include experimental results for X Cl, Br.
Secondary Primary
Random substitution 2/8 25 6/8 75
X Cl, experimental 57 43
X Br, experimental 92 8

• Replacement of secondary H is favored over
  primary H. Generally order of reactivity is
  tertiary gt secondary gt primary gt methyl.
• Bromination displays greater selectivity than
  does chlorination.

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Reactivities of Hydrogens
Quantitative Analysis Deviations from random
replacement quantified by assigning a Reactivity
to each kind of Hydrogen (primary, secondary, or
primary).
Substitution at a carbon proportional to (
hydrogens at C) x (Reactivity of the Hs)

• For chlorination
• tertiarysecondaryprimary 541
• For bromination
• tertiarysecondaryprimary 1600801

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Predict product mixture
For chlorination of 2,2,4,4-tetramethylpentane
predict the product mixture.
Cl2

UV light
18 x 1 18
2 x 4 8
Number of hydrogens leading to this product
Expected ratio 9 4 Expected
fraction 9/13 4/13
Reactivity of those hydrogens
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Chain Reaction Mechanism
Weak Cl-Cl bond may be broken by heat or light.
Initiation
Heat or light

• 1. Cl Cl 2 Cl.

Chlorine atom. Highly reactive, only seven
electrons in valence shell
2. R H Cl. R. H - Cl
Chain steps.
Alkyl radical, only seven electrons around the C,
highly reactive alkyl radical.
3. R. Cl Cl R Cl Cl.
Repeat 2, 3, 2, 3,.
Hydrogen to be abstracted. Trade bonds R-H for
H-Cl
Regenerates the Cl atom used in step 2
Termination steps.
Trade bonds weak Cl-Cl for a stronger C-Cl
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Energetics of the Chain Steps
Chlorination of ethane
Step 2, abstraction of the H, controls the
regioselectivity of the reaction. Isothermic or
slightly exothermic for Cl endothermic for Br.
Step 3, attachment of the halogen, controls the
stereochemistry, which side the halogen attaches.
Exothermic
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Step 2 and Bond Dissociation Energies, breaking
bonds
More highly substituted radicals are easier to
make. This gives rise to regioselectivity
non-random replacement.
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Compare chlorination and bromination of 2-methyl
propane. Bromination is more Regioselective.
Examine Step 2 only for regioselectivity.
First chlorination. Two kinds of H.
BDE, reflecting different radical stabilities
Slightly exothermic
Now bromination.
H-Cl is a more stable bond than H-Br.
Step 2 is exothermic for chlorination.
Endothermic for Br
Endothermic
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Step 2 Transition State Energetics, Cl vs Br
Early transition states, little difference in
energies of activation, rates of abstraction and
regioselectivity
chlorination
RH.Cl
Halogenation of 2-methylpropane yields two
differerent radicals, primary and tertiary.
Exothermic, tertiary radical more stable
R.H..Br
Late transition states, larger difference in
energies of activation, rates of abstraction and
regioselectivity.
bromination
Endothermic, but tertiary radical still more
stable by same amount.
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Now Step 3 Stereochemistry
Mirror objects. If a chiral carbon has been
produced we get both configurations.
Alkyl radical, sp2 hybridization, one electron in
the p orbital.

• Step 3 has these characteristics
• Determines stereochemistry
• Is exothermic
• Is fast, not rate determining

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Simple Example monochlorination of 2-methylbutane
From a and a.
a
a
b
c
Chiral carbon
d
From b
First, look carefully at molecule
Observations Optically inactive molecule (can
show reflection plane) and products will be
optically inactive.
From c
Four optically inactive fractions if distilled.
From d
Chiral carbon
Next, organize approach, label the carbons.
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Example
Can get relative amounts made of each using
reactivities of 145
First get stereochemical relationships between
carbons.
a
enantiomers
a
a
a
c
b
b
3 x 1
3 x 1
b
b
enantiomers
b
b
enantiomers
Diastereomers both sides used.
Diastereomers both sides used.
1 x 5 / 2
1 x 5 / 2
1 x 5 / 2
1 x 5 / 2
c
c
Distillation would yield 5 optically inactive
fractions.
meso
meso
2 x 4 / 2
2 x 4 / 2
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Allylic Systems
CH3CH2 H 411 kJ/mol (101 kCal/mol)
Vinyl C H bonds, difficult to break.
Allylic C H bonds, weak easily broken. Removal
of H produces the allylic radical.
Now the allylic radical
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Resonance in Allylic Radical
Resonance provides the stabilization.
The pi system is delocalized.
Odd electron located on alternate carbons, C1 and
C3, not C2.
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Allylic Substitution
Allylic C-H, 372 kJ
H-Br 368 kJ
Br-Br 192 kJ
Allylic C-Br 247 kJ
DH -247 - 386 (-372 192) -51kJ
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Mechanism
initiation
Weakest C-H bond selected, highest reactivity
Chain steps
368 kJ
372 kJ
Termination usual combining of radicals
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We have a Problem seem to have two possible
reactions for an alkene with Br2.
1. Addition to the double bond yielding a
dibromide.
And/or
2. Substitution at allylic position.
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Competing Reactions Addition vs Substitution
Substitution
Addition
Produced from Br2 at high temperature
Br2
Br., not Br2
Alkene reacts preferentially with Br atoms if
present. Favored by high Br atom
concentrations. High temperature favors Br2 ? 2Br
and thus substitution.
Alkene reacts directly with Br2 Happens at low
Br atom concentrations. Low temperature keeps Br
concentration low and thus favors addition.
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Convenient allylic bromination
For allylic substitution to occur we need both
bromine atoms and Br2 Br R-H Br. ? R.
H-Br Br2 R. Br2 ? R-Br Br.
Both Br and Br2 can be supplied from Br2 at high
temperature or from NBS (N-bromosuccinimide).
NBS
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Allylic Rearrangement
Expect bromination of but-2-ene to yield 1-bromo
but-2-ene by replacing allylic hydrogen.
But get rearranged product as well.
Major product, more stable with subsituted double
bond.
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Mechanism of Rearrangement
Two different sites of reactivity
More highly substituted alkene (more stable,
recall hydrogenation data) is the major product
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An interesting competition is occurring. Consider
the allylic radical
The actual radical is a blending of these two
structures. Secondary radicals are more stable
than primary. This predicts most of the radical
character at the secondary carbon, favoring this
structure. But But more highly substituted
alkenes are more stable. This predicts most
radical character at primary carbon favoring this
structure. This appears to be the dominant
factor leading to dominant product. But also.
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Results of Calculation of Spin Densities in
radical formed from 1-butene
Blue is unpaired electron density. More at
primary than secondary.
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Anti Markovnikov addition of HBr
Only with HBr, not HCl, HI
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Mechanism of anti-Markovnikov Radical Addition of
HBr
Contrast radical and ionic addition of HBr
Ionic
Radical
Not a Chain Process
Chain Steps
Common Concept More stable intermediate formed,
secondary radical or secondary carbocation
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Autoxidation, reaction of allylic sites with
oxygen.
Double allylic
Lower energy since double bonds are in
conjugation with each other
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Contd
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Sample Problem
4. A mixture of 1.6 g of methane and 1.5 g of
ethane are chlorinated for a short time. The
moles of methyl chloride produced is equal to the
number of moles of ethyl chloride. What is the
reactivity of the hydrogens in ethane relative to
those in methane? Show your work.
Solution
Recall The amount of product is proportional to
the number of hydrogens that can produce it
multiplied by their reactivity.
Number of hydrogens leading to methyl chloride
1.6g (1 mol/16 g) (4 mol H/1 mol methane)
0.40 mol H
Number of hydrogens leading to ethyl chloride
1.5 g (1 mol/30 g) (6 mol H/ 1 mol ethane)
0.30 mol H
0.40 mol H Rmethane 0.30 mol H
Rethane Rethane/Rmethane 0.4/0.3 1.3
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How do we form the orbitals of the pi system

• First count up how many p orbitals contribute to
  the pi system. We will get the same number of pi
  molecular orbitals.

Three overlapping p orbitals. We will get three
molecular orbitals.
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If atomic orbitals overlap with each other they
are bonding, nonbonding or antibonding
Anti-bonding, destabilizing. Higher Energy
But now a particular, simple case distant atomic
orbitals, on atoms not directly attached to each
other. Their interaction is weak and does not
affect the energy of the system. Non bonding
If atoms are directly attached to each other the
interactions is strongly bonding or antibonding.
Bonding, stabilizing the system. Lower energy.
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Molecular orbitals are combinations of atomic
orbitals. They may be bonding, antibonding or
nonbonding molecular orbitals depending on how
the atomic orbitals in them interact.
Example Allylic radical
Two antibonding interactions.
Only one weak, antibonding (non-bonding)
interaction.
All bonding interactions.
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Allylic Radical Molecular Orbital vs Resonance
Molecular Orbital. We have three pi electrons
(two in the pi bond and the unpaired electron).
Put them into the molecular orbitals.
Note that the odd electron is located on the
terminal carbons.
Resonance Result
Again the odd, unpaired electron is only on the
terminal carbon atoms.
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But how do we construct the molecular orbitals of
the pi system? How do we know what the molecular
orbitals look like?
Key Ideas
For our linear pi systems different molecular
orbitals are formed by introducing additional
antibonding interactions. Lowest energy orbital
has no antibonding, next higher has one, etc.
2 antibonding interactions
1 weak antibonding Interaction, non-bonding
Antibonding interactions are symmetrically
placed.
0 antibonding interactions
This would be wrong.
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Another example hexa-1,3,5-triene
Three pi bonds, six pi electrons. Each atom is
sp2 hybridized.
Have to form bonding and antibonding combinations
of the atomic orbitals to get the pi molecular
orbitals.
Expect six molecular orbitals. molecular
orbitals atomic orbitals
Start with all the orbitals bonding and create
additional orbitals. The number of antibonding
interactions increases as we generate a new
higher energy molecular orbital.