Chapter 4

©2010,Prentice Hall

Organic Chemistry, 7th EditionL. G. Wade, Jr.

The Study of Chemical Reactions

Chapter 4 2

Introduction

• Overall reaction: reactants products

• Mechanism: Step-by-step pathway.

• To learn more about a reaction: Thermodynamics Kinetics

Chapter 4 3

Chlorination of Methane

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

absorbed by chlorine gas.• Many molecules of product are formed from

absorption of only one photon of light (chain reaction).

Chapter 4 4

The Free-Radical Chain Reaction

• Initiation generates a radical 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: Formation of Chlorine Atom

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

Chapter 4 6

Propagation Step: Carbon Radical

The chlorine atom collides with a methane molecule and abstracts (removes) an H, forming another free radical and one of the products (HCl).

Chapter 4 7

Propagation Step: Product Formation

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

Chapter 4 8

Overall Reaction

Chapter 4 9

Termination Steps

• A reaction is classified as a termination step when any two free radicals join together producing a nonradical compound.

• Combination of free radical with contaminant or collision with wall are also termination steps.

Chapter 4 10

More Termination Steps

Chapter 4 11

Lewis Structures of Free Radicals

• Free radicals are unpaired electrons.• Halogens have 7 valence electrons so one of them

will be unpaired (radical). We refer to the halides as atoms not radicals.

Chapter 4 12

Equilibrium Constant

• Keq = [products] [reactants]

• For CH4 + Cl2 CH3Cl + HCl

Keq = [CH3Cl][HCl] = 1.1 x 1019

[CH4][Cl2]

• Large value indicates reaction “goes to completion.”

Chapter 4 13

Free Energy Change

G = (energy of products) - (energy of reactants) G is the amount of energy available to do work. Negative values indicate spontaneity.

Go = -RT(lnKeq) = -2.303 RT(log10Keq)

where R = 8.314 J/K-mol and T = temperature in kelvins.

Chapter 4 14

Factors Determining G

Free energy change depends on: Enthalpy

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

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

G = H - TS

Chapter 4 15

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 16

Entropy

• So = change in randomness, disorder, or 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 17

Calculate the value of G° for the chlorination of methane.

G° = –2.303RT(log Keq)

Keq for the chlorination is 1.1 x 1019, and log Keq = 19.04

At 25 °C (about 298 ° Kelvin), the value of RT is RT = (8.314 J/kelvin-mol)(298 kelvins) = 2478 J/mol, or 2.48 kJ/mol

Substituting, we have

G° = (–2.303)(2.478 kJ/mol)(19.04) = –108.7 kJ/mol (–25.9 kcal>mol)This is a large negative value for G°, showing that this chlorination has a large driving force that pushes it toward completion.

Solved Problem 1

Solution

Chapter 4 18

Bond-Dissociation Enthalpies (BDE)

• Bond-dissociation requires energy (+BDE).• Bond formation releases energy (-BDE).• BDE can be used to estimate H for a reaction.• BDE for homolytic cleavage of bonds in a

gaseous molecule. Homolytic cleavage: When the bond breaks, each atom gets

one electron. Heterolytic cleavage: When the bond breaks, the most

electronegative atom gets both electrons.

Chapter 4 19

Homolytic and Heterolytic Cleavages

Chapter 4 20

Enthalpy Changes in Chlorination

CH3-H + Cl-Cl CH3-Cl + H-Cl

Bonds Broken H° (per Mole) Bonds Formed H° (per Mole)

Cl-Cl +242 kJ H-Cl -431 kJ

CH3-H +435 kJ CH3-Cl -351 kJ

TOTALS +677 kJ TOTAL -782 kJ

H° = +677 kJ + (-782 kJ) = -105 kJ/mol

Chapter 4 21

Kinetics

• Kinetics is the study of reaction rates.• Rate of the reaction is a measure of how the

concentration of the products increase while the concentration of the products decrease.

• A rate equation is also called the rate law and it gives the relationship between the concentration of the reactants and the reaction rate observed.

• Rate law is experimentally determined.

Chapter 4 22

Rate Law• For the reaction A + B C + D,

rate = kr[A]a[B]b

a is the order with respect to A b is the order with respect to B a + 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.

Chapter 4 23

Activation Energy

• The value of k depends on temperature as given by Arrhenius:

RTEaAek

/r

where A = constant (frequency factor)

Ea = activation energy R = gas constant, 8.314 J/kelvin-mole T = absolute temperature

Ea is the minimum kinetic energy needed to react.

Chapter 4 24

Activation Energy (Continued)

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

Chapter 4 25

Energy Diagram of an Exothermic Reaction

• The vertical axis in this graph represents the potential energy.

• The transition state is the highest point on the graph, and the activation energy is the energy difference between the reactants and the transition state.

Chapter 4 26

Rate-Limiting 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 27

Energy Diagram for the Chlorination of Methane

Chapter 4 28

Rate, Ea, and Temperature

X + CH4 HX + CH3

X Ea(per Mole) Rate at 27 °C Rate at 227 °C

F 5 140,000 300,000

Cl 17 1300 18,000

Br 75 9 x 10-8 0.015

I 140 2 x 10-19 2 x 10-9

Chapter 4 29

Consider the following reaction:

This reaction has an activation energy (Ea) of +17 kJ/mol (+4 kcal/mol) and a H° of +4 kJ/mol (+1 kcal/mol). Draw a reaction-energy diagram for this reaction.

We draw a diagram that shows the products to be 4 kJ higher in energy than the reactants. The barrier is made to be 17 kJ higher in energy than the reactants.

Solved Problem 2

Solution

Chapter 4 30

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 31

Primary, Secondary, and Tertiary Hydrogens

Chapter 4 32

Chlorination Mechanism

Chapter 4 33

Bond Dissociation Energies for the Formation of Free Radicals

Chapter 4 34

Tertiary hydrogen atoms react with Cl• about 5.5 times as fast as primary ones. Predict the product ratios for chlorination of isobutane.

There are nine primary hydrogens and one tertiary hydrogen in isobutane.

(9 primary hydrogens) x (reactivity 1.0) = 9.0 relative amount of reaction(1 tertiary hydrogen) x (reactivity 5.5) = 5.5 relative amount of reaction

Solved Problem 3

Solution

Chapter 4 35

Even though the primary hydrogens are less reactive, there are so many of them that the primary product is the major product. The product ratio will be 9.0:5.5, or about 1.6:1.

Solved Problem 3 (Continued)

Solution

Chapter 4 36

Stability of Free Radicals

• Free radicals are more stable if they are highly substituted.

Chapter 4 37

Chlorination Energy Diagram

• Lower Ea, faster rate, so more stable

intermediate is formed faster.

Chapter 4 38

Rate of Substitution in the Bromination of Propane

Chapter 4 39

Energy Diagram for the Bromination of Propane

Chapter 4 40

Hammond Postulate

• Related species that are similar in energy are also similar in structure.

• The structure of the transition state resembles the structure of the closest stable species.

• Endothermic reaction: Transition state is product-like.

• Exothermic reaction: Transition state is reactant-like.

Chapter 4 41

Energy Diagrams: Chlorination Versus Bromination

Chapter 4 42

Endothermic and Exothermic Diagrams

Chapter 4 43

Radical Inhibitors

• Often added to food to retard spoilage by radical chain reactions.

• 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 44

Radical Inhibitors (Continued)

• A radical chain reaction is fast and has many exothermic steps that create more reactive radicals.

• When an inhibitor reacts with the radical, it creates a stable intermediate, and any further reactions will be endothermic and slow.

Chapter 4 45

Carbon Reactive Intermediates

Chapter 4 46

Carbocation Structure

• Carbon has 6 electrons, positively charged.• Carbon is sp2 hybridized with vacant p orbital.

Chapter 4 47

Carbocation Stability

Chapter 4 48

Carbocation Stability (Continued)

• Stabilized by alkyl substituents in two 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 49

Free Radicals

• Also electron-deficient.• Stabilized by alkyl substituents.• Order of stability:

3 > 2 > 1 > methyl

Chapter 4 50

Stability of Carbon Radicals

Chapter 4 51

Carbanions

• Eight electrons on carbon: 6 bonding plus one lone pair.

• Carbon has a negative charge.

• Destabilized by alkyl substituents.

• Methyl >1 > 2 > 3

Chapter 4 52

Carbenes

• Carbon is neutral.• Vacant p orbital, so can be electrophilic.• Lone pair of electrons, so can be nucleophilic.

Chapter 4 53

Basicity of Carbanions

• A carbanion has a negative charge on its carbon atom, making it a more powerful base and a stronger nucleophile than an amine.

• A carbanion is sufficiently basic to remove a proton from ammonia.

Chapter 4 54

Carbenes as Reaction Intermediates

• A strong base can abstract a proton from tribromomethane (CHBr3) to give an inductively stabilized carbanion.

• This carbanion expels bromide ion to give dibromocarbene. The carbon atom is sp2 hybridized with trigonal geometry.

• A carbene has both a lone pair of electrons and an empty p orbital, so it can react as a nucleophile or as an electrophile.