12_radicals_post

Sec. 12 - free radicals 1

Free Radical Reactions

Free Radical any atoms or groups of atoms that has one or more unpaired electrons. They usually has do not have a positive or negative charge, but are highly reactive and short lived reaction intermediates

Production of Radicals

Homolysis of relatively weak bonds such as O-O or X-X bonds can occur with addition of energy in the form of heat or light

homolytic bond cleavage of bond


Homolytic Bond Dissociation Energies and Heats of ReactionThe homolytic bond dissociation energy is abbreviated DHo

Homolytic Bond Dissociation energies can be used to calculate the enthalpy change (DHo) for a reaction

DHo is positive for bond breaking and negative for bond forming

The more stable the bond the more energy required to cleave the covalent bond

Example This reaction below is highly exothermic since DHo is a large and negativeDHo is not dependant on the mechanism; only the initial and final states of the molecules are considered in determining DHo


Free Radical Reactions

Initiation

Cl Cl + 246 kJ mol-1 2 Cl

Propagation: The initial formation of a few radicals results in the propagation of new radicals in a self perpetuating reaction called a chain reaction.

Cl + H CH3

CH3 + Cl Cl

Cl CH2Cl+ H

CH2Cl Cl Cl+

Termination: The formation of stablemolecules

CH3Cl +

CH3+CH3

etc.

etc.


Hydrogen Abstraction

Hydrogen abstraction is usually the rate determining step. The reaction does not follow 1st or 2nd order kinetics because of cyclical process.

CH4 + Cl CH3 + HCl 12 times faster

CD4 + Cl CD3 + DCl C-D is a stronger bond than C-H bond

Stability of RadicalsWhich hydrogen is extracted?

Look at the stability of the transition state. The abstraction of a hydrogen that produces a more stable radical (lower G‡) is more likely to occur.

Remember the transition state for an endergonic reaction looks very much like the product. Any bonds that are forming are more than half formed, and any bonds that are breaking are more than half broken


Radical stability parallels cation stability

Methyl 1 2 3 allylic and benzylic

Increasing rate of reaction towards Br2

1°

3°

benzylic

allylic

2°

Halogen ReactivityI2 < Br2 < Cl2 < F2

I2 Forms stable radicals F2

Explosive

Not due to halogen cleavage

I2 Br2 Cl2 F2

Bond 151 193 243 159 dissociation (kJmol-

1)


Order of reactivity depends on Eact for the propagation steps.First lets define Activation Energy Eact

The Overall Free-Energy Change: Go = Ho - T (So)

In radical reactions such as the chlorination of methane the overall entropy change (So) in the reaction is small and thus it is appropriate to

use Ho values to approximate Go values (Go ~ Ho)

Go = -102 kJ mol-1 and Ho = -101 kJ mol-1 for this reaction

Activation Energies

When using enthalpy values (Ho) the term for the difference in energy between starting material and the transition state is the energy of activation (Eact)

Recall when free energy of activation (Go) values are used this difference is G‡


Energy of activation values can be predicted

A reaction in which bonds are broken will have Eact > 0 even if a stronger bond is formed and the reaction is highly exothermic

Bond forming always lags behind bond breaking

An endothermic reaction which involves bond breaking and bond forming will always have Eact > Ho


A gas phase reaction in which only bond homolysis occurs has Ho = Eact

A gas phase reaction in which small radicals combine to form a new bond usually has Eact = 0


Back to the Order of reactivity of the Halogens which depends on Eact for the rate determining step in the propagation steps.

Initiation

F2 + 159 kJ mol-1 2F • H = 159 kJ mol-1 Eact = 159 kJ mol-1

Propagation

F• + CH4 HF + •CH3 + 130 kJ mol-1 H = - 130 kJ mol-1 Eact = 5 kJ mol-1

•CH3 + F2 CH3F + F• + 302 kJ mol-1 H = - 302 kJ mol-1 Eact = small

CH4 + F2 CH3F + HF H = - 432 kJ mol-1

Explosive because Eact is low and the heat produced by the exothermic reaction

increases the rate of reaction.Initiation

I2 + 151 kJ mol-1 2I • H = 151 kJ mol-1 Eact = 151 kJ mol-1

Propagation

I• + CH4 + 142 kJ mol-1 HI + •CH3 H = 140 kJ mol-1 Eact = 142 kJ mol-1

•CH3 + I2 CH3I + I• + 89 kJ mol-1 H = - 89 kJ mol-1 Eact = small

CH4 + I2 CH3I + HI H = 53 kJ mol-1

The propagation steps are overall endothermic.The I radical is stable and does not abstract hydrogens.

rate determining step

rate determining

step


CH3 + HBr

(+Br2)

CH3Br + Br

CH4 + Br

Reaction Coordinate

Po

tent

ial E

ner

gy

CH3 + HCl

(+Cl2)

CH3Cl + Cl

CH4 + Cl

Reaction Coordinate

Pot

en

tial E

ne

rgy

H = 8 kJ mol-1

H = -109 kJ mol-1

H = -101 kJ mol-1

H = 74 kJ mol-1

H = -100 kJ mol-1

H = -26 kJ mol-1

Chlorination is more reactive than Bromination


Mono Halogenation of higher alkenes and the selectivity differencesbetween Chlorination and Bromination

CH3CH2CH3 + Br2hv CH3CH2CH2 CH3CHCH3+

2 Br

+ HBrBr

CH3CH2CH2Br CH3CHCH3+

Br

98 %

CH3CH2CH3 + Cl2hv CH3CH2CH2 CH3CHCH3+

2 Cl

+ HClCl

CH3CH2CH2Cl CH3CHCH3+

Cl

55 %

Bromination is more selective than chlorination because it is less reactive.There are only a few bromine radicals that have enough energy to abstract a hydrogen that will produce the less stable molecule. Thus a greater number of the more stable intermediates are produced in the reaction. This creates the selectivity in the reaction.

The chlorine radical being more reactive will produce more of the unstable intermediate, thus producing more of a mixture of products. Fluorine being much more reactive than chlorine is even less selective and there is little difference between the rate of abstracting a 1°, 2° or 3° radicals.

1°

1°

2°

2°


C

CH3

H3C CH3

H

What are the major products formed in the following reactions

hv, 127°

Br2

C

CH3

H3C CH3

Hhv, 25°

Cl2

hv

Br2

Note: Radicals do not rearrange.


Stereochemistry of an alkyl radical

(-)-1-chloro-2-methylbutane

C

H3C

H3CH2C

CH2Cl

H

CH3C

H3CH2C CH2Cl

()-1,2-dichloro-2-butane (a racemic form)

Cl

radical intermediate

Cl2

hv

Cl

Halogenation of butane

Br

1 equiv.HBr +

Br

Br

C(S)

C(R)


Now chlorinate at C3

C

CH3

HC

Br

H3C

HCl

1 equiv. 2(S)

CC

CC

CH3

H Br

H H

CH3

Cl2+

DiastereomersMay not be

equally produced

If we did the same thing with (2R)- CH3CH2CHBrCH3 the results would be:

steric hindrance from bottom less product


(2R)-2-bromobutane → (2R,3R)-2,3-bromochlorobutane (60%) + (2R,3S)-2,3-bromochlorobutane (40%)

If we did the same thing with (2R) - CH3CH2CHBrCH3 the results would be:

+ +

Which compounds are enantiomers?

Would this solution be optically active?


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