Chapter 4The Study of

Chemical Reactions

Jo BlackburnRichland College, Dallas, TX

Dallas County Community College District2003,Prentice Hall

Organic Chemistry, 5th EditionL. G. Wade, Jr.

Chapter 4 2

Tools for Study• To determine a reaction’s

mechanism, look at:Equilibrium constantFree energy changeEnthalpyEntropyBond dissociation energyKineticsActivation energy =>

Chapter 4 3

Chlorination of Methane

• Requires heat or light for initiation.• The most effective wavelength is blue, which

is absorbed by chlorine gas.• Lots of product formed from absorption of only

one photon of light (chain reaction). =>

C

H

H

H

H + Cl2heat or light

C

H

H

H

Cl + HCl

Chapter 4 4

Free-Radical Chain Reaction

• Initiation generates a reactive intermediate.

• Propagation: the intermediate reacts with a stable molecule to produce another reactive intermediate (and a product molecule).

• Termination: side reactions that destroy the reactive intermediate. =>

Chapter 4 5

Initiation Step

A chlorine molecule splits homolytically into chlorine atoms (free radicals)

=>

Cl Cl + photon (h) Cl + Cl

Chapter 4 6

Propagation Step (1)The chlorine atom collides with a methane

molecule and abstracts (removes) a H, forming another free radical and one of the products (HCl).

C

H

H

H

H Cl+ C

H

H

H

+ H Cl

=>

Chapter 4 7

Propagation Step (2)

The methyl free radical collides with another chlorine molecule, producing the other product (methyl chloride) and regenerating the chlorine radical.

C

H

H

H

+ Cl Cl C

H

H

H

Cl + Cl

=>

Chapter 4 8

Overall Reaction

C

H

H

H

H Cl+ C

H

H

H

+ H Cl

C

H

H

H

+ Cl Cl C

H

H

H

Cl + Cl

C

H

H

H

H + Cl Cl C

H

H

H

Cl + H Cl =>

Cl Cl + photon (h) Cl + Cl

Chapter 4 9

Termination Steps• Collision of any two free radicals

• Combination of free radical with contaminant or collision with wall.

C

H

H

H

Cl+ C

H

H

H

Cl

Can you suggest others? =>

Chapter 4 10

Equilibrium constant

• Keq = [products] [reactants]

• For chlorination Keq = 1.1 x 1019

• Large value indicates reaction “goes to completion.”

=>

Chapter 4 11

Free Energy Change• G = free energy of (products - reactants),

amount of energy available to do work.• Negative values indicate spontaneity.• Go = -RT(lnKeq)

where R = 1.987 cal/K-moland T = temperature in kelvins

• Since chlorination has a large Keq, the free energy change is large and negative. =>

Chapter 4 12

Problem

• Given that -X is -OH, the energy difference for the following reaction is -1.0 kcal/mol.

• What percentage of cyclohexanol molecules will be in the equatorial conformer at equilibrium at 25°C?

=>

Chapter 4 13

Factors Determining G• Free energy change depends on

enthalpyentropy

H = (enthalpy of products) - (enthalpy of reactants)

S = (entropy of products) - (entropy of reactants)

G = H - TS =>

Chapter 4 14

Enthalpy

• Ho = heat released or absorbed during a chemical reaction at standard conditions.

• Exothermic, (-H), heat is released.• Endothermic, (+H), heat is absorbed.• Reactions favor products with lowest

enthalpy (strongest bonds). =>

Chapter 4 15

Entropy• So = change in randomness, disorder,

freedom of movement.

• Increasing heat, volume, or number of particles increases entropy.

• Spontaneous reactions maximize disorder and minimize enthalpy.

• In the equation Go = Ho - TSo the entropy value is often small. =>

Chapter 4 16

Bond Dissociation Energy

• Bond breaking requires energy (+BDE)

• Bond formation releases energy (-BDE)

• Table 4.2 gives BDE for homolytic cleavage of bonds in a gaseous molecule.

A B A + B

We can use BDE to estimate H for a reaction. =>

Chapter 4 17

Which is more likely?

Estimate H for each step using BDE.

CH4 HCl+ +Cl CH3

CH3 + Cl2 CH3Cl + Cl

or

Cl+CH4 CH3Cl + H

H Cl2+ HCl Cl+

104 103

58 84

=>

104 84

58 103

Chapter 4 18

Kinetics

• Answers question, “How fast?”

• Rate is proportional to the concentration of reactants raised to a power.

• Rate law is experimentally determined.

=>

Chapter 4 19

Reaction Order• For A + B C + D, rate = k[A]a[B]b

a is the order with respect to Aa + b is the overall order

• Order is the number of molecules of that reactant which is present in the rate-determining step of the mechanism.

• The value of k depends on temperature as given by Arrhenius: ln k = -Ea + lnA RT =>

Chapter 4 20

Activation Energy• Minimum energy required to reach

the transition state.

• At higher temperatures, more molecules have the required energy.

=>

C

H

H

H

H Cl

Chapter 4 21

Reaction-Energy Diagrams• For a one-step reaction:

reactants transition state products

• A catalyst lowers the energy of the transition state.

=>

Chapter 4 22

Energy Diagram for a Two-Step Reaction

• Reactants transition state intermediate• Intermediate transition state product

=>

Chapter 4 23

Rate-Determining Step

• Reaction intermediates are stable as long as they don’t collide with another molecule or atom, but they are very reactive.

• Transition states are at energy maximums.

• Intermediates are at energy minimums.

• The reaction step with highest Ea will be the slowest, therefore rate-determining for the entire reaction. =>

Chapter 4 24

Rate, Ea, and Temperature

X + CH4 HX + CH3

X E a Rate @ 300K Rate @ 500KF 1.2 kcal 140,000 300,000Cl 4 kcal 1300 18,000

Br 18 kcal 9 x 10-8 0.015

I 34 kcal 2 x 10-19 2 x 10-9

=>

Chapter 4 25

Conclusions• With increasing Ea, rate decreases.

• With increasing temperature, rate increases.

• Fluorine reacts explosively.

• Chlorine reacts at a moderate rate.

• Bromine must be heated to react.• Iodine does not react (detectably).

=>

Chapter 4 26

Chlorination of Propane

• There are six 1 H’s and two 2’s. We expect 3:1 product mix, or 75% 1-chloropropane and 25% 2-chloropropane.

• Typical product mix: 40% 1-chloropropane and 60% 2-chloropropane.

• Therefore, not all H’s are equally reactive.

=>

1 C

2 CCH3 CH2 CH3 + Cl2

h CH2

Cl

CH2 CH3 + CH3 CH

Cl

CH3

Chapter 4 27

Reactivity of Hydrogens• To compare hydrogen reactivity, find

amount of product formed per hydrogen: 40% 1-chloropropane from 6 hydrogens and 60% 2-chloropropane from 2 hydrogens.

• 40% 6 = 6.67% per primary H and60% 2 = 30% per secondary H

• Secondary H’s are 30% 6.67% = 4.5 times more reactive toward chlorination than primary H’s. =>

Chapter 4 28

Predict the Product Mix

Given that secondary H’s are 4.5 times as reactive as primary H’s, predict the percentage of each monochlorinated product of n-butane + chlorine.

=>

Chapter 4 29

Free Radical Stabilities

• Energy required to break a C-H bond decreases as substitution on the carbon increases.

• Stability: 3 > 2 > 1 > methylH(kcal) 91, 95, 98, 104

=>

Chapter 4 30

Chlorination Energy DiagramLower Ea, faster rate, so more stable

intermediate is formed faster.

=>

Chapter 4 31

• There are six 1 H’s and two 2’s. We expect 3:1 product mix, or 75% 1-bromopropane and 25% 2-bromopropane.

• Typical product mix: 3% 1-bromopropane and 97% 2-bromopropane !!!

• Bromination is more selective than chlorination. =>

1 C

2 CCH3 CH2 CH3 + CH2

Br

CH2 CH3 +Br2heat

CH3 CH

Br

CH3

Bromination of Propane

Chapter 4 32

• To compare hydrogen reactivity, find amount of product formed per hydrogen: 3% 1-bromopropane from 6 hydrogens and 97% 2-bromopropane from 2 hydrogens.

• 3% 6 = 0.5% per primary H and97% 2 = 48.5% per secondary H

• Secondary H’s are 48.5% 0.5% = 97 times more reactive toward bromination than primary H’s. =>

Reactivity of Hydrogens

Chapter 4 33

Bromination Energy Diagram• Note larger difference in Ea

• Why endothermic?

=>

Chapter 4 34

Bromination vs. Chlorination

=>

Chapter 4 35

Endothermic and Exothermic Diagrams

=>

Chapter 4 36

Hammond Postulate• Related species that are similar in energy are

also similar in structure. The structure of a transition state resembles the structure of the closest stable species.

• Transition state structure for endothermic reactions resemble the product.

• Transition state structure for exothermic reactions resemble the reactants. =>

Chapter 4 37

Radical Inhibitors

• Often added to food to retard spoilage.• Without an inhibitor, each initiation step

will cause a chain reaction so that many molecules will react.

• An inhibitor combines with the free radical to form a stable molecule.

• Vitamin E and vitamin C are thought to protect living cells from free radicals. =>

Chapter 4 38

Reactive Intermediates

• Carbocations (or carbonium ions)

• Free radicals

• Carbanions• Carbene

=>

Chapter 4 39

Carbocation Structure

• Carbon has 6 electrons, positive charge.

• Carbon is sp2 hybridized with vacant p orbital. =>

Chapter 4 40

Carbocation Stability• Stabilized by alkyl

substituents 2 ways:• (1) Inductive effect:

donation of electron density along the sigma bonds.

• (2) Hyperconjugation: overlap of sigma bonding orbitals with empty p orbital. =>

Chapter 4 41

Free Radicals

• Also electron-deficient

• Stabilized by alkyl substituents

• Order of stability:3 > 2 > 1 > methyl

=>

Chapter 4 42

Carbanions

• Eight electrons on C:6 bonding + lone pair

• Carbon has a negative charge.

• Destabilized by alkyl substituents.

• Methyl >1 > 2 > 3

=>

Chapter 4 43

Carbenes

• Carbon is neutral.• Vacant p orbital, so

can be electrophilic.• Lone pair of

electrons, so can be nucleophilic.

=>

Chapter 4 44

End of Chapter 4