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Organic reactions
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SERP course
A. Mickiewicz University,
Poznań 2016
Jan Milecki
Based on
Organic Chemistry
5th Edition
Paula Yurkanis Bruice
Thermodynamics and Kinetics
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Understanding energy changes…
Thermodynamics - description of a reaction at equilibrium
“How much product is formed?”
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“How fast is product formed?”
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Kinetics - explanation of the rates of chemical reactions
Thermodynamics - tells us relative amounts of reactants and products
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Kinetics - tells us how fast reactants are converted to products
Thermodynamics and Kinetics
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Visualizing energy changes…
The more stable the
species, the lower
the energy
x coordinate - progress of
the reaction
Reactants are plotted
on the left-side,
products on the right
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Reaction coordinate diagram - depiction of energy changes that take place
in the steps known to occur as reactants are converted into products
y coordinate - total (free)
energy of all species
Transition state (TS) - maximum energy state as reactants are converted to
products - height is proportion to likelihood of reaction occurring, molecular
structure is an intermediate between reactants and products (partial bonds)
Thermodynamics - How much product is formed?
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Thermodynamics and Kinetics
Thermodynamics - field of chemistry that describes the properties of a system at
equilibrium
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Equilibrium constant (Keq) - numerical expression of the relative
concentrations of reactants and products at equilibrium:
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What determines the relative concentrations of product and reactant at
equilibrium?
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Stability - the more stable the compound, the greater its concentration at
equilibrium
Thermodynamics - How much product is formed?
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Thermodynamics and Kinetics
What determines the relative concentrations of product/reactant at equilibrium?
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Stability - the more stable the compound, the greater its concentration at
equilibrium
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° = std. conditions
(1M, 25 °C, 1 atm)
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∆G° = Gibbs
free-energy
change =
difference
between the
free energy of
products and
reactants
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Now we understand why acid strength is dependent on conjugate base
stability...
RULE - as a species (i.e., base) becomes more stable, the equilibrium constant
(i.e., Ka) for its formation becomes larger
Thermodynamics - How much product is formed?
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Thermodynamics and Kinetics
∆G° = Gibbs free-energy change = difference between the free energy of
products and reactants:
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∆G° = Gproducts - Greactants
A negative ∆G means:
Products are more stable (lower in energy)
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More energy is released than consumed - exergonic
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A positive ∆G means:
Products are less stable (higher in energy)
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More energy is consumed than released - endergonic
RULE - a “successful” reaction is one in which products are favored at equilibrium
Thermodynamics - How much product is formed?
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Thermodynamics and Kinetics
Reaction “success” is measured by favorable product formation as indicated by:
Large equilibrium constant (Keq), or
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Large negative change in free energy (∆G°)
Are these two terms related (like pH and pKa?)
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∆G° = -RT ln Keq
R = gas constant (1.986 x 10-3 kcal mol-1 K-1, or 8.314 x 10-3 kJ mol-1 K-1)
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T = temperature in degrees Kelvin (K = °C + 273, 25 °C = 298 K)
RULE - small differences in
∆G° gives rise to a large
difference in Keq (and relative
concentrations of products
and reactants) – 10-fold
change for each 1.36
kcal/mol (5.7 kJ/mol)
Thermodynamics - How much product is formed?
How to influence the amount of product formed….
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Thermodynamics and Kinetics
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Le Châtelier’s principle - if an equilibrium is disturbed, the system will adjust
to offset the disturbance:
Decreases in [C] or [D] will force A and B to react to form more C and D to
maintain Keq
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Crystallization of a product out of solution
Product driven out of solution as a gas
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Increases in [A] or [B] will also force the production of more C and D
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Thermodynamics - How much product is formed?
Components of Gibbs standard free-energy change (∆G°)
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G  H  TS
Enthalpy (∆H°) - heat given off or consumed during a course of a reaction
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Atoms are held by bonds (= electrons = energy)
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RULE - Heat is given off when bonds are formed
RULE - Heat is consumed when bonds are broken
H  (energy of bonds being broken) - (energy of bonds being formed)
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Exothermic reaction - negative ∆H° - bonds formed in a reaction are
stronger than those consumed in a reaction - excess energy released
Endothermic reaction - positive ∆H° - bonds formed in a reaction are
weaker than those consumed in a reaction - additional energy required
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
Thermodynamics and Kinetics
Thermodynamics - How much product is formed?
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Thermodynamics and Kinetics
Components of Gibbs standard free-energy change (∆G°)
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G  H  TS
Entropy (∆S°) - measure of the freedom of movement in a system
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Restricting the movement of a molecule decreases entropy
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- Entropy increases when more molecules are moving more
RULE
(temperature dependence)
S  (freedom of motion of the products) - (freedom of motion of the reactants)
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Putting it all together - a reaction is favored when…
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∆S° is positive (greater freedom of motion in larger number of molecules)
∆H° is negative (products are formed with stronger (more stable) bonds)
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Overall - ∆G° is made negative by the contributions of these two terms
Notice that only entropy has a temperature dependence
Thermodynamics - How much product is formed?
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Thermodynamics and Kinetics
Experimental consideration of Gibbs standard free-energy change (∆G°)
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G  H  TS
∆H° is easy to determine experimentally - chemists frequently ignore entropy
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Only valid if reactions involve a small change in entropy, or occur at low T:
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 is small then we can assume G  H
If TS
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RULE - Prediction of the favorability of an equilibrium constant can be estimated
from enthalpy only when entropy does not contribute significantly
Thermodynamics - How much product is formed?
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Thermodynamics and Kinetics
Predicting ∆G° from enthalpy - addition of HBr to an alkene:
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Enthalpy can be calculated from bond dissociation energies (DH, Table 3.2)
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This examination tells us that the addition of HBr to ethene is an exothermic
reaction (∆H° = -24 kcal/mol) - can we infer, then, that ∆G° is also negative?
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RULE - ∆H° will only approximate ∆G° if the entropy change is small
RULE - Entropy changes can only be assumed to be small for reactions in the
gas phase (true here)
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Describing the electrophilic addition reaction of HBr to 2-butene:
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Reaction coordinate diagrams can be drawn for each step of the reaction:
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Step 1 (a) - the alkene is converted to a carbocation - product is less stable
than reactants - endergonic reaction (positive ∆G°)
Step 2 (b) - the carbocation reacts with a nucleophile to form a product that is
more stable than the carbocation reactant - exergonic reaction (negative ∆G°)
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Thermodynamics and Kinetics
Kinetics - How fast is the product formed?
Kinetics - field of chemistry that studies the rates of chemical reactions and the
factors that affect those rates
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Free energy of activation (∆G‡)- energetic barrier (“hill”) to convert
reactants to products,
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Remember - ∆G°
only describes the
difference between
the stability of the
reactants and
products - it does
not indicate
anything about the
energy barrier of
the reaction
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∆G‡ = (free energy of the transition state) - (free energy of the reactants)
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Describing the electrophilic addition reaction of HBr to 2-butene:
Reactions where products of one step are reactants for another can be linked:
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∆G° for the overall reaction becomes the difference between the free
energy of the final products and the free energy of the initial reactants
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Intermediate - species that is a product of one step and reactant for next
(ability to isolate is proportional to species stability - carbocation unstable)
CAUTION - do not
confuse transition states
and intermediates:
Transition states partially formed bonds,
species too unstable to be
isolated
Intermediates - fully
formed bonds, species
can be isolated if low
enough in energy (stable)
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Chemical reaction
O


:NH3
Dipol and free
electron pair
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CN
Dipol and ion
O

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
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Occurs when molecules come into very close contact. Usually it has the form of collision,
which most often ends in changing of the direction of movement, without any reaction.
For the reaction to occur, energy of the collision has to be high enough to overcome repulsion
force bettween valence orbitals of the colliding molecules.
Sometimes also attraction between orbitals occurs:
F

F
B
:NH3
F
Electron- deficient
orbital and free pair
Br
Br
Br
Br
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+
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In the last case there can be no repulsion because p orbital of the boron atom is vacant
(no electrons present). Another case is the attraction between occupied bonding orbitals
of one molecule and and unoccupied antibonding orbitals of the other:
Bonding orbitals of the alkene interact
with antibonding * orbitals of the bromine
molecule. The empty * orbitals accept
 electrons and form bonds
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CN

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O
CN
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
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Molecule (or molecular individuum) which donates electrons is a NUCLEOPHILE and the one
which accepts electrons is called an ELECTROPHILE. Electron movement is pictured by the
bent arrow directed from the nucleophile to the electrophile.
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RULES FOR DRAWING THE ARROW
1. The arrow starts at the source of the electron pair (bond, lone pair, negative charge)
2. End of the arrow should indicate approximately the middle of the newly formed bond
3. If the bond is broken and no new bond is formed, the arrow ends on the atom gaining
lone pair or charge
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REMEMBER Summary charge cannot change
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Direction of the new bond is the consequence of the symmetry of the orbitals. It means that the
direction of the collision is important. Orbitals of the nucleophile and electrophile have to match
in order to ensure their good overlapping. Energy of activation stems from the fact that energy of
unoccupied orbital is higher than the energy of the occupied one. This energy difference has to
be covered by the collision energy. Therefore the easiest are the reactions where the energy
difference between occupied and unoccupied orbitals is the lowest. After overlapping of these
orbitals two new arise: bonding one of lower energy and antibonding of higher energy. The most
often cases are:
Energy
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Antibonding
Reactants
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Nu
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Bonding
E+
E+
E+
Nu
Nu
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The most close in energy are the highest occupied molecular orbital (HOMO) of the
nucleophile and the lowest unoccupied molecular orbital (LUMO) of the electophile.
The reaction result is the consequence of their energy nad symmetry.
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N
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C
Cyanide ion: two electron pairs (of sp hybridization) on carbon atom
and nitrogen atom have different energies. The HOMO at the carbon
atom is higher in energy and the cyanide nucleophile attacs with its
carbon end! It occurs despite the fact that nitrogen is more electronegative
than carbon.
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Electron–accepting orbitals are usually antibonding orbitals. The most often case are
the * orbitals, especially * orbitals of the carbonyl group
HOMO  orbital is polarized and most of
the electron density is localized at oxygen
atom (electronegative). This is reversed in
the LUMO * orbital, where carbon end is
of lower energy and accepts electron pair
of nucleophile, forming new bond at carbon
atom.


Cl
H
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Even * orbital can be electrophile if one of atoms forming the  bond is strongly electronegative
H
Cl
:B
:
B
Antibonding 
orbital of the acid
Bonds between carbon and other elements – difference
in electronegativity:
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C-Cl
C-Br
C-I
=0.9
=0.3
~0
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Second factor, except electronegativity is the accesability of LUMO. The bond between carbon
and iodine is very slightly polarized but iodine is easily substituted, LUMO of carbon atom has
low energy level. Also bromine (Br2 molecule) is electrophilic (reaction with carbon – carbon double
bonds). The Br – Br bond is weak, which means that energy difference between LUMO and HOMO
is small
CH3
Br
:S
Br
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Br
CH3
CH3
S
+
Br
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CH3
Br
Br
+
H3 C
CH3
Just a mixture
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Energy
 H3C
 Br
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CH3
Br
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easy
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 Br
difficult
Br
 H3 C
Nu
CH3
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Fragmentation
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Even strong bonds can split relatively easily if they are polarized. Uneven distribution of
electron density initiates charge shift and eventually separation of full charge (deficit of one
electron on one side and one electron excess on the other).
Example: C-O bond is stronger than C-C or C-H, but is splits much more often - it is a
strongly polarized bond.
If the molecule contains electrodonor group (X) on one side of the bond system and
elctroacceptor group (Y) on the other side, the bond system undergoes fragmentation heterolytic bond cleavage
H
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OH
H
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HO
Y
Y
X
nt
X
O
OH2
H
H2 O
O
Polarization initiated by protonation of OH group
OTs
OTs
H
OH
O
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O
retro-aldol
OH
H
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O
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Base
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More examples:
OH
O
H
O
H
O
Cl
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Mechanism:
Cl
N
H
O
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Cl
Cl O
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P
P
O
N
nt
Cl
O
N
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N
H
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O
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POCl3
Cl
H
Cl
Cl
N
H
Cl
N
H3C N CH3
O
CH3
O
O
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N
R
CH3
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O
H3 C N CH3
H3C N CH3
N
H3 C
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OCH3, NH2, NR2 etc. activate pyridine ring
N
O
H
O
CH3
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Activation by N-oxides
E
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ri
N
nt
RCO3H
H
E
E
N
N
N
N
O
O
O
O
N-oxides activate position 2- and 4- of the pyridine ring.
Products of electrophilic substitution can easily be reducedPCl
(
3)
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Radical reactions are another story!
H Br
H
+
Br
H Br
H
+
Br homolytic (bond energy=366 kJ / mole)
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heterolytic (high energy)
.
Initiator
.
Addition
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O
O
O
O
O
Ph
fo
O
Reacting radical
Br
Br
Ph
RO + HBr → ROH + Br
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RO-OR →
.
2RO
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Radicals are formed by abstraction of an electron from atom or molecule. The starting specie is
an initiator (chain reactions). There is also possibility of addition of the radical or eliminaton
of stable molecule.
Ph
Ph
O
+
C
O
Elimination
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Radical electron („bachelor electron”) is very reactive.
Exceptions:
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O
N
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O
TEMPO
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Usually radical stability reflects that of the corresponding carbocation or anion. Stabilization
occurs when there is conjugation with groups donating or accepting electrons (Ph3C.
radical). . Another stabilizing factor is the steric crowding
.
.
R +R
paired-spin molecule
.
.
R + paired-spin molecule
.
new R + new paired-spin molecule
.
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new R + paired-spin molecule
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R
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Different ways of reacting
ROOR
H
Br
.
Br
H
Br
.
H
fo
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nt
.
2RO
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na
Chain reaction
Br
Br
Br
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Acyloin Condensation
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The bimolecular reductive coupling of carboxylic esters by reaction with metallic sodium in an
inert solvent under reflux gives an α-hydroxyketone, which is known as an acyloin. This
reaction is favoured when R is an alkyl. With longer alkyl chains, higher boiling solvents can
be used. The intramolecular version of this reaction has been used extensively to close rings
of different sizes, e.g. paracyclophanes or catenanes. Sometimes the reaction is hampered
by competing Claisen reaction
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Mechanism of Acyloin Condensation
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Pinacol Coupling Reaction
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CO 
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This reaction involves the reductive homo-coupling of a carbonyl compound to produce a
symmetrically substituted 1,2-diol. The first step is single electron transfer of the carbonyl bond,
which generates radical ion intermediates that couple via carbon-carbon bond formation to give
a 1,2-diol. The example depicted above shows the preparation of pinacol itself.
CO 
O
fo
CO 
ri
+e
O
+e
CO 
O
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These reactions are the exception among radical reactions – they proceed through stable
(hence low reactive) radicals. Usually reactive radicals do not „wait” for another unpaired electron
of partner radical, but grab the electron from the chemical bond of randomly chosen molecule.
It is the process of chain reaction. Selectivity is associated with less rective radicals.
lu
.
Br
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Selective bromination – allylic position
H
.
.
Br
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na
Br2
.
Competing reaction
fo
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nt
.
Br
Br
.
Br
Br
.
Br
Br
Br
Br
Br
Bromine radical is less reactive (more selective) than chlorine one
Br
Sn – H
C – Br
Bond strenghth
308 kJ/mole
280 kJ/mole
vs
vs
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Reducton with Bu3SnH
C–H
Sn – Br
418 kJ/mole
552 kJ/mole
N
2
CN
.
NC
+
N2
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na
N
NC
.
H
. Sn
Br
Bu
Bu
Bu
Bu
Sn
Bu
.
Br
Bu
Sn
Bu
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nt
Bu
.
Bu
Bu
H
Sn
Bu
+
Bu
H
. Sn
Bu
. Sn
Bu
Bu
Bu
fo
NC
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AIBN is employed as an initiator since it produces less reactive radicals than peroxides. Only
weak Sn–H bond is broken. Bu3Sn. radical is formed and it reacts to form strong Sn–Br bond
(Bu3SnBr).
Bu
Bu