The life of the chain depends on the ongoing presence of the highly reactive Cl atoms and alkyl radicals. Eliminating these species ends chains.

4. 2 Cl. Cl – Cl

5. 2 R. R – R

6. R. + Cl. R - Cl

Heat

or light

Chain Reaction Mechanism

1. Cl – Cl 2 Cl.

2. R – H + Cl. R. + H - Cl

3. R. + Cl – Cl R – Cl + Cl.

Repeat 2, 3, 2, 3,….

Cha

in s

teps

.

Chlorine atom. Highly reactive, only seven electrons

in valence shell

Weak Cl-Cl bond may be broken by heat or light.

Hydrogen to be abstracted. Trade bonds: R-H for H-Cl

Alkyl radical, only seven electrons around the C,

highly reactive alkyl radical.

Trade bonds: weak Cl-Cl for a

stronger C-Cl

Regenerates the Cl atom used in step 2

Ter

min

atio

n st

eps.

Initi

atio

n

Energetics of the Chain Steps

Chlorination of ethane

Step 2, abstraction of the H, controls the regioselectivity of the reaction. Isothermic or slightly exothermic for Cl; endothermic for Br.

Step 3, attachment of the halogen, controls the stereochemistry, which side the halogen attaches. Exothermic

Step 2 and Bond Dissociation Energies, breaking bonds…

More highly substituted radicals are easier to make.

This gives rise to regioselectivity = non-random replacement.

Now bromination.

Compare chlorination and bromination of 2-methyl propane. Bromination is more Regioselective. Examine Step 2 only for

regioselectivity.

BDE, reflecting different radical stabilities

Slightly exothermic

Endothermic

H-Cl is a more stable bond than H-Br.

Step 2 is exothermic for chlorination. Endothermic for Br

First chlorination. Two kinds of H.

Step 2 Transition State Energetics, Cl vs Br

chlorination

Exothermic, tertiary radical more stable

Early transition states, little difference in energies of

activation, rates of abstraction and regioselectivity

bromination

Endothermic, but tertiary radical still more

stable by same amount.

Late transition states, larger difference in

energies of activation, rates of abstraction and

regioselectivity.

R…H……….Cl

R……….H..Br

Halogenation of 2-methylpropane yields two differerent radicals, primary and tertiary.

Now Step 3: Stereochemistry

HStep 2X

Alkyl radical, sp2 hybridization, one electron in the p orbital.

X

XX

Step 3

X

X X

Mirror objects. If a chiral carbon has been produced we get both configurations.

Step 3 has these characteristics

•Determines stereochemistry

•Is exothermic

•Is fast, not rate determining

CH3

H CH3

CH3

H H

C6H14

Cl2

C6H13Cl

Simple Example: monochlorination of 2-methylbutane

Observations:

Optically inactive molecule (can show reflection plane) and products will be optically inactive.

First, look carefully at molecule

Next, organize approach, label the carbons.

aa’

b

c

From a and a’.

From b

d

H

H3C CH3

CH3

H H

CH2Cl

H CH3

CH3

H H

CH3

H CH2Cl

CH3

H H

CH3

Cl CH3

CH3

H H

From cCH3

H CH3

CH3

Cl H

CH3

H CH3

CH3

H Cl

From dCH3

H CH3

CH2Cl

H H

Four optically inactive fractions if distilled.

Chiral carbon

Chiral carbon

Example

H(S) (R)

H3C CH3 Cl2

H H

Ha a’

b b’

c

H(S) (R)

ClH2C CH3

H H

H

a

H(S) (R)

H3C CH2Cl

H H

H

a’

H(R) (R)

H3C CH3

Cl H

H

H(S) (R)

Cl CH3

H3C H

Hb

Diastereomers both sides used.

H(S) (S)

H3C CH3

H Cl

H b’

H(S) (R)

H3C Cl

H CH3

H

Diastereomers both sides used.

(s)

H(R) (S)

H3C CH3

H H

Cl

(r)

Cl(R) (S)

H3C CH3

H H

H

meso meso

b b’enantiomers

enantiomers

First get stereochemical relationships between carbons. enantiomers

Can get relative amounts made of each using reactivities of 1:4:5

3 x 1 3 x 1

1 x 5 / 2 1 x 5 / 2 1 x 5 / 2 1 x 5 / 2

2 x 4 / 2 2 x 4 / 2

Distillation would yield 5 optically inactive fractions.

c c

Allylic Systems

Allylic C – H bonds, weak easily broken.

Removal of H produces the allylic radical.

Vinyl C – H bonds, difficult to

break.

CH3CH2 – H: 411 kJ/mol (101 kCal/mol)

Now the allylic radical…

Resonance in Allylic RadicalResonance provides the stabilization.

The pi system is delocalized.

Odd electron located on alternate carbons, C1 and C3, not C2.

Allylic Substitution

H2C

H

HH

H2C CH2 +

Br

H BrBr Br

heat

Allylic C-H,

372 kJBr-Br

192 kJ Allylic C-Br

247 kJ

H-Br 368 kJ

H = -247 - 386 – (-372 – 192) = -51kJ

Mechanism

initiation

Br2 2Brheat

Chain steps

Weakest C-H bond selected, highest

reactivity

372 kJ368 kJ

H2C CH2

H2C CH2

Br2

H2C CH2 + H2C CH2

BrBr

+ Br

H2C

H

HH Br H2C CH2

+ H Br

H2C CH2

Termination: usual combining of radicals

We have a Problem: seem to have two possible reactions for an alkene with Br2.

1. Addition to the double bond yielding a dibromide.

Br2

Br

Br

2. Substitution at allylic position.

Br2

Br

+ HBr

And/or

Competing Reactions: Addition vs Substitution

Alkene reacts preferentially with Br atoms if present.

Favored by high Br atom concentrations.

High temperature favors Br2 2Br and thus substitution.

Alkene reacts directly with Br2

Happens at low Br atom concentrations.

Low temperature keeps Br concentration low and thus favors addition.

BrBr

Br Br2 Br., not Br2

Addition Substitution

Produced from Br2 at

high temperature

Convenient allylic bromination

For allylic substitution to occur we need both bromine atoms and Br2

Br: R-H + Br. R. + H-Br

Br2: R. + Br2 R-Br + Br.

Both Br and Br2 can be supplied from Br2 at high temperature or from NBS (N-bromosuccinimide).

NBS

Allylic Rearrangement

NBS

Br

Expect bromination of but-2-ene to yield 1-bromo but-2-ene by replacing allylic hydrogen.

But get rearranged product as well….

+

Br

Major product, more stable with subsituted double bond.

Mechanism of Rearrangement

H

HH Br

CH2

+ H Br

CH2

Two different sites of reactivity

CH2

CH2

Br2

CH2 + CH2

BrBr

+ Br

More highly substituted alkene (more stable, recall hydrogenation data) is

the major product

H

HH Br

CH2

+ H Br

CH2

The actual radical is a blending of these two structures.

Secondary radicals are more stable than primary. This predicts most of the radical character at the secondary carbon, favoring this structure. But…

But more highly substituted alkenes are more stable. This predicts most radical character at primary carbon favoring this structure. This appears to be the dominant factor leading to dominant product. But also….

An interesting competition is occurring. Consider the allylic radical…

Results of Calculation of Spin Densities in radical formed from 1-butene

Blue is unpaired electron density. More at primary than secondary.

Anti Markovnikov addition of HBr

Only with HBr, not HCl, HI

HBr

ionic condition, polar solvents

H

Br

Markovnikov

Br

H

anti-Markovnikovradical conditions, peroxides, light,heat

HBr

Mechanism of anti-Markovnikov Radical Addition of HBr

.BrBr

RO-OR 2RO .

peroxide

RO . + H-Br ROH + Br .

Initiation via peroxide to generate Br .

Br

H

Br H

Br

.Br

Chain Steps

Contrast radical and ionic addition of HBr

Radical Ionic

HH

BrBr

Br

Br

HBr

Not a Chain Process

Common Concept: More stable intermediate formed, secondary radical or secondary carbocation

4. A mixture of 1.6 g of methane and 1.5 g of ethane are chlorinated for a short time. The moles of methyl chloride produced is equal to the number of moles of ethyl chloride. What is the reactivity of the hydrogens in ethane relative to those in methane? Show your work.

Sample Problem

Solution:

Recall: The amount of product is proportional to the number of hydrogens that can produce it multiplied by their reactivity.

Number of hydrogens leading to methyl chloride =

1.6g * (1 mol/16 g) * (4 mol H/1 mol methane) = 0.40 mol H

Number of hydrogens leading to ethyl chloride =

1.5 g * (1 mol/30 g) * (6 mol H/ 1 mol ethane) = 0.30 mol H

0.40 mol H * Rmethane = 0.30 mol H * Rethane

Rethane/Rmethane = 0.4/0.3 = 1.3

How do we form the orbitals of the pi system…

First count up how many p orbitals contribute to the pi system. We will get the same number of pi molecular orbitals.

Three overlapping p orbitals. We will get three molecular orbitals.

If atomic orbitals overlap with each other they are bonding, nonbonding or antibonding

Anti-bonding, destabilizing.Higher Energy

pi type anti-bond sigma type anti-bonding

If atoms are directly attached to each other the interactions is strongly bonding or antibonding. Bonding, stabilizing the system. Lower energy.

But now a particular, simple case: distant atomic orbitals, on atoms not directly attached to each other. Their interaction is weak and does not affect the energy of the system. Non bonding

non-bonded

pi type bond sigma type bonding

or

or

or or

Molecular orbitals are combinations of atomic orbitals.

They may be bonding, antibonding or nonbonding molecular orbitals depending on how the atomic orbitals in them interact.

All bonding interactions.

Only one weak, antibonding (non-bonding) interaction.

Two antibonding interactions.

Example: Allylic radical

Allylic Radical: Molecular Orbital vs Resonance

Note that the odd electron is located

on the terminal carbons.

Molecular Orbital. We have three pi electrons (two in the pi bond and the unpaired electron). Put them into the molecular orbitals.

Resonance Result

Again the odd, unpaired electron is only on the terminal carbon atoms.

But how do we construct the molecular orbitals of the pi system? How do we know what the molecular orbitals look like?

Key Ideas:

For our linear pi systems different molecular orbitals are formed by introducing additional antibonding interactions. Lowest energy orbital has no antibonding, next higher has one, etc.

0 antibonding interactions

1 weak antibonding Interaction, “non-bonding”

2 antibonding interactions

Antibonding interactions are symmetrically placed.

This would be wrong.

Another example: hexa-1,3,5-triene

Three pi bonds, six pi electrons.Each atom is sp2 hybridized.

Have to form bonding and antibonding combinations of the atomic orbitals to get the pi molecular orbitals.

Expect six molecular orbitals.

# molecular orbitals = # atomic orbitals

Start with all the orbitals bonding and create additional orbitals. The number of antibonding interactions increases as we generate a new higher energy molecular orbital.