chapter 5 mcmurry

7/27/2019 Chapter 5 Mcmurry

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Chapter 5: Overview of Chemical ReactionsCoverage:

1. Types of Organic Reactions

2. Mechanisms and Curved Arrows

3. Radical and Polar Reactions

4. Reactive Intermediates and Transition States5. Thermodynmics

6. Kinetics

End of Chapter Problems: 5.23-5.32, 5.39, 5.40

Goals:

1. Know the types of organic reactions and be able to recognize when you see them.

2. Know how to used curved arrows when writing mechanisms.

3. Understand the difference between radical and polar reactions.

4. Know the nature of reactive intermediates in organic chemistry.5. Understand how a transition state differs from a reactive intermediate.

6. Understand what controls the equilibrium position of a reaction (thermodynamics).

7. Understand what controls the rate of a reaction (kinetics).



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.



Curved Arrows

1. Electrons move from a nucleophile to a electrophile

N: + E-X Æ N-E+

+ X –

HO:- CH3-I:+

..

..

Nucleophile Electrophile

HO-CH3 :I:-+

..

..

..

......

The above reacton is an example of a substitution reaction. The OH substitutes for the I atomin the reaction.

δ+ δ-



2. The nucleophile can either be negatively charged or neutral. However, it must

possess a nonbonded pair of electrons.

3. The electrophile can either be positively charged or neutral.

N: + E Æ N-E Neutral product

N: – + E Æ N-E– Negative product

N: + E+ Æ N-E+ Positive Product

N: – + E+ Æ N-E Neutral Product

Important: Remember that charge must be conserved in going from

left to right in the reaction.



Kinds of Organic Reactions

1. Addition reactions - two reactants are added together to form a single new product.

N: + E Æ N-E

C CH

H

H

H

H-Br C C

Br

H

HH

H

H

+

2. Elimination Reactions - the reverse of Addition, a molecule splits into two products

C CH

H

H

H

H-BrC C

Br

H

HH

H

H

+



3. Substitution Reactions – two reactants exchange parts to give a two new products

CH4

Cl Cl CH3Cl H Cl+ +

light

4. Rearrangement Reactions - a molecule undergoes a reorganization of bonds to yield an isomeric product.

C7H16 C7H16



Reactive Intermediates

1. 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.

CHHH

.



• Stabilized by the presence of alkyl groups.

C R

R

R

C R

R

H

C R

H

H

C H

H

H

Most Stable Least Stable

3o 2o 1o methyl



Radical Reactions

Cl2→ 2 Cl.

Cl. + CH4 → HCl + CH3.

CH3. + Cl2 → CH3Cl + Cl

Chlorination of Methane

Initiation – light or heat is required to produce the initial radicals.

Propagation – these two steps repeat until one or more reactant is consumed.

CH3. + CH3

. Æ CH3CH3

CH3. + Cl. Æ CH3Cl

Termination – this step causes the reaction to stop.

Overall Reaction:

CH4 + Cl2 Æ CH3Cl + HCl



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.

CH

HH

+

empty p orbital



• Order of Stability same as Radical Stability

C+

R

R

R

C+

R

R

H

C+

R

H

H

C+

H

H

H

3o 2o 1o methyl

Most Stable Least Stable



Polar Reaction involving Carbenium ion intermediate: Addition of HBr to ethylene.

C CH

H

H

H

H-Br C C

Br

H

HH

H

H

+

Nucleophile

Electrophile



Why is the double bond so reactive?

Answer: The Pi electrons are more loosely held.



3. Carbanion Intermediates

• Carbanions are negatively charged and electron-rich.

CH3

-CH3CH2

-


• Destabilized by the presence of alkyl groups.

C HH

H:

C

R

R

R

C

R

R

H

C

R

H

H

C

H

H

H

3o 2o 1o methyl

Least Stable Most Stable

-sp3 orbital



4. Carbenes Intermediates

• Uncharged reactive intermediates containing a divalent carbon atom.

• Simplest carbene has the formula :CH2, called methylene


• The carbene can act either as an electrophile or a nucleophile.

Why?

Answer: It has an empty p orbital – electrophile

It has a nonbonding pair of electrons - nucleophile

CBr

Br:

Empty p orbital

Nonbonding electrons paired

in this sp2 orbital

Dibromocarbene



Thermodynamics

Thermodynamics is the branch of chemistry that deals with energy changes that

accompanies chemical and physical reactions.

A + B C + D

K eq = [products]/[reactants] = [C][D]/[A]/[B]

Example: CH4 + Cl2 CH3Cl + HCl Radical Chlorination

K eq = 1.1 x 1019 i.e. this reaction goes to completion!

The standard Gibbs free energy change, ∆Go, can be calculated from K eq ..

∆G = -RT (ln K eq )

R = 1.987 cal/K mol

T = absolute temperature in Kelvin

e = 2.718, the base of the natural logarithms



If ∆Go is negative, then forward reaction is favored at equilibrium (K eq > 1)

If ∆Go is positive, then the forward reaction is unfavorable (K eq < 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



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 asthe 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!!!!



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 A. + B. ∆Ho Bond Dissociation

Energy

Homolytic Cleavage only!!

CH3

H Cl Cl CH3

Cl H Cllight

+ +

Cl Cl

CH3

HCH

3Cl

H Cl

Bonds Broken Bonds Made

+58 kcal/mol

+104 kcal/mol

+162 kcal/mol

-103 kcal/mol

-84 kcal/mol

-187 kcal/mol

∆Ho = +162 + (-187) = -25 kcal/mol



Focus on Propagation since it is the major sequence of reactions that produces

products of CH3Cl and HCl.

a.

CH3

Cl2

CH3Cl Cl

Cl CH4 ClH CH

3

+ +

+ +

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

∆Ho = +1 kcal/mol

∆Ho = -26 kcal/mol

∆Ho = -25 kcal/mol

What do the transition states for these reactions look like? Use dotted lines to show bonds breaking

and making.



CH4 + Cl.

CH3. + Cl2

CH3Cl + Cl.

-------------------

------------------------------

∆Hooverall = -25

Ea1 = 4 kcal/mol

Ea2

= 1 kcal/mol

∆Ho

Kcal/mol

Reaction Coordinate

Propagation Steps Rate-determining Step

Highest energy step of multistep

reaction. It represents the bottle-neck

of the reaction.

The first step is rate-determining due to its

highest activation energy.



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.

k r = Ae-Ea/RT

k r rate constant

A frequency factor

Ea energy of activation

R gas constant

T absolute temperature

The higher the temperature, the faster the reaction (larger k r )The higher the Energy of Activation, the slower the reaction (smaller k r )

chapter 5 mcmurry

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