Name: _______________________________
Unit 3 Packet:
Activation Energy, Free Radical Chain Reactions,
Alkane Preparations, SN2, E2
Honors Organic Chemistry
Key Terms For Unit 3
Free Radical Chain Reaction
Homolytic Cleavage
Free Radical
Initiation
Propagation
Termination
Endothermic
Exothermic
Activation Energy
Bond Energies
Selectivity
Alkane Preparation
Hydrogenation
Organometallics
Nucleophile
Electrophile
Grignard Reaction
Wurtz Reaction
Corey-House Reaction
SN2
E2
Saytzeff Product
Hoffmann Product
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Honors Organic Chemistry
Making an Alkane More Useful: Free Radical Chain Reaction
As you’ve already learned, alkanes are not very useful in the world of organic chemistry. They
are involved in relatively few reactions of interest, capable of doing very little other than
burning. There is only really one simple way of taking an alkane, and converting it to something
more useful. The pathway used is called a ______ __________ chain reaction.
Consider methane, the simplest alkane, with diatomic chlorine. While methane and chlorine are
indefinitely stable when combined in the dark, upon irradiation with light, a mixture of the
following products are formed.
Chlorination:
CH4
+
Methane
Cl2
hv

Chlorine
CH3Cl
Methyl chloride
+
CH2Cl2
+
Methylene chloride
CHCl3
Chloroform
+
CCl4
Carbon Tetrachloride
The composition (relative abundances) of the products for this chlorination reaction depends
upon the amount of chlorine available and the amount of time the reaction is allowed to run. If
sufficient chlorine and sufficient time were allowed, all of the methane would be converted to
carbon tetrachloride. Through a composition vs. time analysis, it has been determined that
methyl chloride is formed first, and the products are made sequentially from there as shown
below.
CH4 + Cl2
hv
hv
hv
hv

CH3Cl + Cl2  CH2Cl2 + Cl2  CHCl3 + Cl2  CCl4
Reaction 1
Reaction 2
Reaction 3
Reaction 4
Let’s focus in for a minute on the initial reaction between methane and chlorine to form methyl
chloride.
H3C──H
105 kcal/mol
+
Cl──Cl
hv

59 kcal/mol
H3C──Cl
85 kcal/mol
Breaking bonds costs energy (endothermic),
but forming new bonds gives off energy
(exothermic). So to calculate the overall
energy of reaction, the following calculation
is performed.
+
H──Cl
103 kcal/mol
Bonds Made = -85 - 103 = -188 kcal/mol
Bonds Broken = 105 + 59 = +164 kcal/mol
-24 kcal/mol
The reaction is exothermic!
This reaction, like many others in organic chemistry, is not explained with a single step, but
rather a combination of several steps. The set of specific steps in a reaction, that display which
bonds are being formed and broken, is called a reaction mechanism.
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Honors Organic Chemistry
Mechanism – Free Radical Chlorination of Methane
The mechanism for the free radical chlorination of methane (or any alkane, for that matter)
involves an initiation step, two propagation steps, and one of any number of termination steps.
..
..
:Cl──Cl:
˙˙
˙˙
Initiation:
..
2 :Cl .
˙˙

hv
..
+ . Cl:
˙˙


H3C.
..
..
:Cl──Cl:
˙˙
˙˙


..
H3C──Cl:
˙˙


H3C──CH3


..
H3C──Cl:
˙˙


..
..
:Cl──Cl:
˙˙
˙˙
H3C──H
+
..
H──Cl:
˙˙
Propagation:
H3C.
+
H3C.
+
H3C.
+
..
. Cl:
˙˙
+
..
. Cl:
˙˙
..
+ . Cl:
˙˙
or
Termination:
H3C.
or
..
:Cl .
˙˙
Let’s take a closer look at the thermochemistry of each of the propagation steps. Although the
second step is exothermic by 26 kcal/mol, and thus is very favorable, the first propagation step is
endothermic by about 2 kcal/mol.
First Propagation Step:
H3C──H
..
. Cl:
˙˙
105 kcal/mol
Second Propagation Step:
..

 H3C. + H──Cl:
H3C.
˙˙
103 kcal/mol
..
..
..
..
:Cl──Cl: 
 H3C──Cl: + .Cl:
˙˙
˙˙
˙˙
˙˙
59 kcal/mol
This reaction is endothermic by 2 kcal/mol
85 kcal/mol
This reaction is exothermic by 26 kcal/mol
The carbon-hydrogen bond in methyl chloride is weaker than the one in methane, so methyl
chloride that is formed converts quickly to form methylene chloride. Methylene chloride and
chloroform are more reactive still, and the reaction will speed up as it progresses towards carbon
tetrachloride.
64
Honors Organic Chemistry
Halogenation of Other Alkanes
Chlorination of n-butane:
In methane, all of the hydrogens are equivalent and primary. In n-butane, there are ___________
hydrogens on the terminal carbons and ___________ hydrogens on carbons in the middle of the
chain. By analyzing the products of monochlorination of n-butane, we can see that chlorine has
a preference towards adding to the 2º position to form sec-butyl chloride.
CH3CH2CH2CH3
+ Cl2
hv

CH3CH2CHCH3
+
CH3CH2CH2CH2─Cl
|
Cl
sec-butyl chloride
n-butyl chloride
(72.2%)
(27.8%)
This analysis would seem to show that chlorination prefers secondary to primary by a ratio of
72.2 : 27.8, or a factor of 2.6. This however, does not take into consideration that there are 6
primary hydrogens and 4 secondary hydrogens, so to correct for the statistical advantage we
must multiply the factor by 6/4 to produce the true factor of 3.9. Thus, chlorination prefers 2º
to 1º by a factor of 3.9.
Bromination of n-butane:
Bromine is significantly more _____________ than chlorine in its free radical halogenation:
CH3CH2CH2CH3
+ Br2
hv

CH3CH2CHCH3
+
CH3CH2CH2CH2─Br
|
Br
sec-butyl bromide
n-butyl bromide
(98.2%)
(1.8%)
The above statistical adjustment would be the same for bromine, and would indicate that
bromination prefers 2º to 1º by a factor of 82.
Chlorination of isobutane:
CH3
|
CH3CHCH3
+
Cl2
CH3
|
CH3CHCH2─Cl
hv

isobutyl chloride
(64%)
+
CH3
|
CH3CCH3
|
Cl
tert-butyl chloride
(36%)
Isobutane has nine primary hydrogens and one tertiary hydrogen  chlorination prefers 3º to 1º
by a factor of 5.1.
65
Honors Organic Chemistry
Halogenation of Other Alkanes (cont.)
Bromination of isobutane:
CH3
|
CH3CHCH3
+
CH3
|
CH3CHCH2─Br
hv

Br2
+
CH3
|
CH3CCH3
|
Br
isobutyl bromide
tert-butyl bromide
(~0%)
(~100%)
***Bromination, being much more selective than chlorination, prefers 3º to 1º by a factor
of 1600.***
Selectivities of Halogen atoms in Carbon-Hydrogen Bond Abstraction
Halogen
Primary
Secondary
Tertiary
F
1
1.2
1.4
Cl
1
3.9
5.1
Br
1
82
1600
Energy Diagrams
Exothermic Reaction
Endothermic Reaction
For Chlorination:
66
Honors Organic Chemistry
Alkane Preparation Reactions
1.
Formation of Alkyl Halide:
CH3CH3
Alkane
2.
CH3CH2—X + H—X
diatomic halide
Cl2 or Br2
alkyl halide
binary acid
Catalytic Hydrogenation of Alkenes:
CH3CH=CH2
Pt , Pd , NiPr essure

 CH3CH2CH3
+ H2
Alkene
3.
hv

+ X2
Hydrogen
Alkane
Reduction of Alkyl Halide:
2 CH3CHCH3 + 2 Zn + 2 HC2H3O2
|
X
Alkyl halide
Zinc
reflux

 2 CH3CH2CH3 + ZnX2 + Zn(C2H3O2)2
Acetic acid
Alkane
Zinc halide
Zinc acetate
Alkane Preparation Reactions #4 - #8 - Organometallics
***Organic compound with Carbon directly attached to a metal; highly reactive
and extremely usefel. Strong bases, good nucleophiles. Mg, Li, Na, Cu.***
4.
Wurtz Reaction: Organosodiums – very reactive, cannot stop
CH3CH2—X
ether,
hexane
CH3CH2:Na + NaX
+ 2 Na
1º Alkyl Halide
Sodium
1º Organosodium
CH3CH2CH2CH3 + NaX
CH3CH2:Na + CH3CH2—X
1º Organosodium
Sodium Salt
ether,
hexane
1º Alkyl Halide
Alkane
Sodium Salt
**Doubles chain length, good for symmetrical alkanes, Alkyl Halide must be primary**
5.
Organolithiums – 1st step to Corey-House:
CH3CH2—X
+ 2 Li
1º or 2 º Alkyl Halide
6.
ether,
hexane
 CH3CH2:Li + LiX
Lithium
1º Organolithium
Lithium Salt
Corey-House Reaction:
2 CH3CH2:Li
+
CuI
ether,
hexane
 [(CH3CH2:)2Cu]Li + LiI
1º 2 º Organolithium Cuprous Iodide
Lithium dialkyl cuprate
Lithium Iodide
**The lithium dialkyl cuprate can then be used to perform SN2 on a different 1º alkyl halide **
67
Honors Organic Chemistry
7.
Grignard Reaction:
CH3CH2Cl
+
Alkyl halide
8.
Mg
ether

 CH3CH2MgX
Magnesium
Alkane:Mg halide
Hydrolysis of Grignard:
CH3CH2MgX
H2O
+ or
CH3OH
Grignard Reagent
Acidic Solvent


CH3CH3
+
Alkane
Mg(OH)X
or
Mg(CH3O)X
Magnesium salt
SN2 – Substitution Nucleophilic Bimolecular:
9.
“Backside Attack” – Involves only one step – the rate of reaction is dependant on the
concentration of the two molecules in the rate determining step. Substrate must be 1º
or 2º. Competes with E2.
**Inversion of Configuration**
H
H
H
H
H
Et
H
-
..
------ :I: - 
˙˙
-
I  CH3CH2: -------
CH3CH2: +
H
H
and
..
:I: ˙˙
H
**Transition State**
10.
E2 – Elimination Bimolecular:
One Step – Competes with SN2 – Preferred when the carbon to be attacked is
sterically hindered (2º or 3º).
CH3
|
CH3CH2:- + CH3CCH2CH3
|
X
CH3
|
CH2 = C—CH2—CH3


Hoffmann
Product
(Minor)
and
CH3
|
CH3—C = CH—CH3
**Saytzeff’s Rule** – major product is determined
by the greater # of alkyl groups attached to double bonded carbons
68
Saytzeff
Product
(Major)
Honors Organic Chemistry
2
2
SN vs. E - Continued
11.
Ethers:
**Ethers are chosen as the solvent for these reactions because they are un-reactive in
both substitution and elimination**
..
CH3CH2:- + CH3—O—CH3 
 No Reaction
˙˙
When determining whether SN2, E2 or hydrolysis will occur in a reaction, consider the
following:
1.
2.
3.
4.
5.
6.
Look for a metal – Is one of your reactants a base (anion)? Recognize acid/base rxns
Look for an easy (acidic) proton for a base to grab
Can one of your reactants act as a nuc:?
1º  SN2
3º  E2
2º  SN2 + E2
Organometallics Used as Nucleophiles:
CH3CH2MgX
CH3CH2:Li
+
CH3CH2CH2—Br
SN 2

CH3CH2CH2CH2CH3
[(CH3CH2:)2Cu]Li
Organometallics Used as Bases:
CH3CH2MgX
CH3CH2:Li
CH3CH3
+
CH3CH2—O—H
+ Mg(OCH3)Cl
Hydrolysis

 CH3CH3 + Li(OCH3)
[(CH3CH2:)2Cu]Li
CH3CH3 + Li(OCH3) + CH3CH2Cl
More Uses as Bases:
[(CH3CH2:)2Cu]Li +
CH3
|
CH3CCH2CH3
|
X
CH3
|
CH2 = C—CH2—CH3


and
CH3
|
CH3—C = CH—CH3
69
Hoffmann
Product
(Minor)
Saytzeff
Product
(Major)
Honors Organic Chemistry
2
2
SN and E Mechanisms – A Summary
**Correlation of structure and reactivity for substitution and elimination reactions**
Halide Type
Methyl
Primary
Secondary
Tertiary
SN2
Highly favored with
most nuc:
Highly favored with
strong nuc:
Favored by good nuc: in
polar, aprotic solvents
Does not occur
SN2 – Substitution, Nucleophilic, Bimolecular
- Rate = k[substrate][:nuc]
- One-step, concerted mechanism
- 100% Stereochemical Inversion
- Can compete with E2
E2 – Elimination, Bimolecular
- Rate = k[substrate][base]
- One-step, concerted mechanism
- Competes with SN2
70
E2
Does not occur
Favored with strong,
hindered bases
Favored when strong
bases are used
Highly favored when
bases are used
Honors Organic Chemistry
Alkane Reactions and Properties – Worksheet #1
1.
Which of the following is most soluble in water?
a. Pentan – 1 – ol
b. Hexane
c. Diethyl ether
d. Ethan – 1,2 – diol
2.
Complete and balance the following (assume complete oxidation):
a. Isohexane + oxygen 
b. Hentriacontane + oxygen 
c. Tetradecane + oxygen 
3.
Determine the oxidation number of the indicated atoms in each of the following
molecules
a. CH3—CH2—CH3
e.
Br—CH—CH2—CH3
|
Br
b. CH3—CH—CH3
|
OH
f.
CH3—CH = CH2
c. CH3—C—CH3
||
:O:
g. CH3—CH—CH3
|
:NH2
d.
CH3—CH2—C—OH
||
:O:
h. CH3—C—NH2
||
:O:
i.
71
CH3—CH2:Mg:Cl
Honors Organic Chemistry
Alkane Properties and Preparation – Worksheet #2
1.
What isomer of hexane has two (and only two) monobromo derivatives?
2.
Without referring to tables, list the following hydrocarbons in order of decreasing
boiling points:
a. 3,3 – dimethylpentane
b. n – heptane
c. 2 – methylheptane
d. n – pentane
e. 2 – methylhexane
3.
Write balanced equations, naming all organic products, for the following reactions:
a. Isobutyl bromide + Mg ether


b. Tert-butyl bromide + Mg
ether


c. Product of “a” + water
d. Product of “b” + water
e. Product of “a” + CH3OH
f. Sec-butyl chloride + Li, then CuI
g. Product of “f” + ethyl bromide
4.
Write Equations for the preparation of n-butane from:
a. n – butyl bromide
b. sec – butyl bromide
c. ethyl chloride
d. but – 1 – ene (CH3CH2CH = CH2)
72
Honors Organic Chemistry
Alkane Reactions Worksheet #3
1.
If the following names are correct, say so. If not, give the correct name.
a. 3 – isopropylhexane
b. 3 – n-propylhexane
2.
Arrange the following in order of increasing boiling point. Explain your choices:
a. Pentane
b. 2,2 – dimethylpropane
c. Hexane
d. Ethane
3.
Complete the following with proper equations:
a. The production of tert-butylmagnesium bromide from 2 – bromo – 2 –
methylpropane
b. Product of “a” plus water
c. Product of “a” plus aqueous HCl
d. Isopentyl chloride + zinc + acetic acid
e. Complete combustion of hexadecane with oxygen
f. Isobutane + chlorine
hv

g. Show all steps in the free-radical chain reaction for the formation of one of the
products in “f”
h. Starting with propane, outline a synthesis of 2,3 – dimethylbutane
i. Starting with isobutane and propane, how could you prepare 2,2 –
dimethylpentane?
j. Consider the following picture:
i. What is the activation energy?
ii. What is the sign and magnitude of ΔH?
iii. Endothermic or Exothermic?
73
Honors Organic Chemistry
Alkane Reactions Worksheet #4
I.
Outline all steps in the free-radical chain reaction representing the monobromination of isobutane. Show 3 possible termination steps.
II.
Give the IUPAC names for:
a. CH3(CH2)55CH3
c.
b.
III.
Give the common names for:
a. 1 – bromo – 2,2 – dimethylpropane
b. 2 – methylbutane
c. 2 – bromo – 2 – methylbutane
d. 1 – bromo – 4 – methylpentane
IV.
Complete and balance (name the products):
hv,heat
a. Ethane and I2


b. 2 – chloropentane
Zn, HAc


c. 4 – methylpent– 2 – ene + H2
d. Propane + Cl2
hv

e. CH3CH2CH2CH2CH2Cl + Na
f. Isopropyl chloride + Mg
V.
Pd

ether


ether


Starting with only ethane and any needed inorganic reagents, outline a potential
synthesis of 3 – methylpentane
74
Honors Organic Chemistry
Material Covered on the Unit 3 Exam
1. Write and balance combustion reactions for alkanes.
2. Be able to assign oxidation numbers to carbon and other elements in organic compounds.
3. Define oxidation and reduction, carbanion and free radical.
4. Write the mechanism for free-radical chain reaction, including termination steps.
5. Define Activation Energy.
6. Draw and label a reaction energy profile for exothermic and endothermic steps in a free
radical chain reaction.
7. Use bond energy tables to find ΔHrxn.
8. Explain why chlorine is more reactive than bromine with methane.
9. Explain why Iodine does not react with methane.
10. Use probability and success rate factors to predict % yields of isomers, assuming
monohalogenation.
11. Explain what is meant by the statement “Bromine is more selective than Chlorine”.
12. Explain the stabilities of 3°, 2° and 1° free radicals.
13. Define hyperconjugation and the role it plays in the stability of a charged/radical species.
14. Diagram mechanisms for SN2 and E2 reactions.
15. Describe the “competition” between SN2 and E2, and identify the factors that influence
which mechanism is preferred in a given situation.
16. Predict the products of various alkane preparation reactions.
17. Use alkane preparation reactions in synthesis problems to make more complicated
products from simpler reactants.
75