Chapter 5: Overview of Chemical Reactions

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Chapter 5 Overview of Chemical Reactions
Coverage 1. Types of Organic Reactions 2.
Mechanisms and Curved Arrows 3. Radical and Polar
Reactions 4. Reactive Intermediates and
Transition States 5. Thermodynmics 6. Kinetics

• Goals
• Know the types of organic reactions and be able
  to recognize when you see them.
• Know how to used curved arrows when writing
  mechanisms.
• Understand the difference between radical and
  polar reactions.
• Know the nature of reactive intermediates in
  organic chemistry.
• Understand how a transition state differs from a
  reactive intermediate.
• Understand what controls the equilibrium position
  of a reaction (thermodynamics).
• Understand what controls the rate of a reaction
  (kinetics).

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Mechanism step-by-step description of
bond-breaking and bond-making processes
to give product.
Bond-breaking
Bond-making
Curved arrows are used to indicate flow of
electrons in these processes.
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Curved Arrows
1. Electrons move from a nucleophile to a
electrophile
? ?-
N E-X ? N-E
X
..
..
..
..
..
..
..
..
Nucleophile
Electrophile
The above reacton is an example of a substitution
reaction. The OH substitutes for the I atom in
the reaction.
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2. The nucleophile can either be negatively
charged or neutral. However, it must
possess a nonbonded pair of electrons.
Neutral product
N E ? N-E
N E ? N-E
Negative product
3. The electrophile can either be positively
charged or neutral.
N E ? N-E
Positive Product
N E ? N-E
Neutral Product
Important Remember that charge must be conserved
in going from left to right in
the reaction.
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Kinds of Organic Reactions
1. Addition reactions - two reactants are added
together to form a single new product.
N E ? N-E
2. Elimination Reactions - the reverse of
Addition, a molecule splits into two products
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3. Substitution Reactions two reactants
exchange parts to give a two new products
light
4. Rearrangement Reactions - a molecule
undergoes a reorganization of bonds to yield
an
isomeric product.
C7H16 C7H16
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Reactive Intermediates

• Free Radicals
• Neutral intermediates possessing unpaired
  electrons
• CH3. CH3CH2.
• Carbon is sp2 hybridized.
• Radical is electron deficient and therefore acts
  as an electrophile.
• Stabilized by the presence of alkyl group.

.
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• Stabilized by the presence of alkyl groups.

3o 2o
1o
methyl
Most Stable Least
Stable
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Radical Reactions
Chlorination of Methane
Overall Reaction CH4 Cl2 ? CH3Cl
HCl
Initiation light or heat is required to produce
the initial radicals.
Cl2 2 Cl.    
Propagation these two steps repeat until one or
more reactant is consumed.
Cl. CH4 HCl CH3. CH3 . Cl2 CH3Cl Cl
Termination this step causes the reaction to
stop.
CH3. CH3. ? CH3CH3 CH3.
Cl. ? CH3Cl
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2. Carbenium (Carbocation) Intermediates

• Carbenium are positively charged and
  electron-deficient.
• Strong Lewis acids and electrophiles
• CH3 CH3CH2
• Central Carbon is sp2 hybridized
• Stabilized by the presence of alkyl groups due
  to hyperconjugative effect
• or donation of sigma electrons of C-H bond to
  empty p orbital.

empty p orbital

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• Order of Stability same as Radical Stability

3o 2o
1o
methyl
Most Stable Least
Stable
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Polar Reaction involving Carbenium ion
intermediate Addition of HBr to ethylene.
Electrophile
Nucleophile
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Why is the double bond so reactive? Answer The
Pi electrons are more loosely held.
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3. Carbanion Intermediates

• Carbanions are negatively charged and
  electron-rich.
• CH3- CH3CH2-
• Central Carbon is sp3 hybridized
• Destabilized by the presence of alkyl groups.
  

-
sp3 orbital
3o 2o
1o
methyl
Least Stable Most
Stable
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• 4. Carbenes Intermediates
• Uncharged reactive intermediates containing a
  divalent carbon atom.
• Simplest carbene has the formula CH2, called
  methylene
• Central Carbon is sp2 hybridized
• The carbene can act either as an electrophile or
  a nucleophile.

Empty p orbital
Nonbonding electrons paired in this sp2 orbital
Dibromocarbene
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Thermodynamics
Thermodynamics is the branch of chemistry that
deals with energy changes that accompanies
chemical and physical reactions.
A B C D
Keq products/reactants CD/A/B
Example CH4 Cl2 CH3Cl
HCl Radical Chlorination
Keq 1.1 x 1019 i.e. this reaction goes to
completion!
The standard Gibbs free energy change, ?Go, can
be calculated from Keq..
?G -RT (ln Keq) R 1.987 cal/K mol T
absolute temperature in Kelvin e 2.718, the
base of the natural logarithms
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If ?Go is negative, then forward reaction is
favored at equilibrium (Keq gt 1) If ?Go is
positive, then the forward reaction is
unfavorable (Keq lt 1)
What factors contribute to ?Go?
?Go ?Ho - T ?So ?Go change in free
energy ?Ho change in enthalpy (heat) ?So
change in entropy (amount of disorder)
At low temperatures, the T?So term is small
compared to ?Ho, so chemists often make the
approximation that ?Go ? ?Ho
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• Change in Enthalpy, ?Ho, the amount of heat
  absorbed or released by the reaction.
• Measure of the relative strengths of bonds in
  the products and reactants.
• If weaker bonds are broken and stronger bonds
  are formed, heat is released and
• the reaction is termed exothermic (negative
  ?Ho).
• If stronger bonds are broken and weaker bonds
  are formed, heat is absorbed and
• the reaction is termed endothermic (positive
  ?Ho).
• Example Free radical chlorination of methane
• ?Ho -26.0 kcal/mol exothermic reaction

• Change in Entropy, ?So, is the change in the
  amount of randomness in the reaction.
• A positive value means that the products have
  more randomness (degrees of freedom).
• Value reflects the number of molecules in the
  reactant versus the product as well as
• the flexibility of the molecules.
• Example Free radical chlorination of methane
• ?So 2.9 cal/K mol
• T ?So 298 (2.9) -860
  cal/mol or 0.860 kcal/mol Very small!!!!

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Conclusion Enthalpy changes dominate in the
free radical chlorination of methane!!!
How can we calculate the enthalpy change for a
reaction? Answer Utilize Bond Dissociation
Energies (BDE) obtained from Tables in the text.
A. B.
?Ho Bond Dissociation Energy
Homolytic Cleavage only!!
Bonds Broken
Bonds Made
-103 kcal/mol -84 kcal/mol -187 kcal/mol
58 kcal/mol 104 kcal/mol 162 kcal/mol
?Ho 162 (-187) -25 kcal/mol
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Focus on Propagation since it is the major
sequence of reactions that produces products of
CH3Cl and HCl.
a.
?Ho 1 kcal/mol ?Ho -26 kcal/mol ?Ho
-25 kcal/mol
b.
Calculate ?Ho for each step a. Bonds
Broken
Bonds Made H3C-H 104
H-Cl -103 b. Bonds Broken
Bonds Made
Cl-Cl 58
H3C-Cl -84
What do the transition states for these reactions
look like? Use dotted lines to show bonds
breaking and making.
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Propagation Steps
Rate-determining Step Highest energy step of
multistep reaction. It represents the
bottle-neck of the reaction.
Ea1 4 kcal/mol
Ea2 1 kcal/mol
----
---------------
?Ho Kcal/mol
CH3. Cl2
The first step is rate-determining due to
its highest activation energy.
CH4 Cl.
?Hooverall -25
------------------------------
CH3Cl Cl.
Reaction Coordinate
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Kinetics and Energy of Activation
Energy of Activation is the minimum kinetic
energy the molecules must possess to overcome
the repulsion between their electron clouds when
they collide. If they do not have sufficient
energy, no reaction will occur. kr
Ae-Ea/RT kr rate constant A frequency
factor Ea energy of activation R gas constant T
absolute temperature The higher the
temperature, the faster the reaction (larger
kr) The higher the Energy of Activation, the
slower the reaction (smaller kr)