Name: _______________________________

Unit 3 Packet: Activation Energy, Free Radical Chain Reactions,

Alkane Preparations, SN2, E

2

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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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Bonds Made = -85 - 103 = -188 kcal/mol

Bonds Broken = 105 + 59 = +164 kcal/mol

-24 kcal/mol

The reaction is exothermic!

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 + Cl2 hv CH3Cl + CH2Cl2 + CHCl3 + CCl4

Methane Chlorine Methyl chloride Methylene chloride Chloroform 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 CH3Cl + Cl2 hv

CH2Cl2 + Cl2 hv CHCl3 + Cl2 hv

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 + Cl──Cl hv H3C──Cl + H──Cl

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.

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.

105 kcal/mol 59 kcal/mol 85 kcal/mol 103 kcal/mol

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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.

.. .. ..

Initiation: :Cl──Cl: 2 :Cl .

˙˙ ˙˙ ˙˙

.. ..

H3C──H + . Cl: H3C. + H──Cl:

˙˙ ˙˙

Propagation:

.. .. .. ..

H3C. + :Cl──Cl: H3C──Cl: + . Cl:

˙˙ ˙˙ ˙˙ ˙˙

H3C. + H3C. H3C──CH3

or

.. .. Termination: H3C. + . Cl: H3C──Cl:

˙˙ ˙˙

or

.. .. .. .. :Cl . + . Cl: :Cl──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: Second Propagation Step:

.. .. .. .. .. ..

H3C──H . Cl: H3C. + H──Cl: H3C. :Cl──Cl: H3C──Cl: + .Cl:

˙˙ ˙˙ ˙˙ ˙˙ ˙˙ ˙˙

This reaction is endothermic by 2 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.

hv

105 kcal/mol 103 kcal/mol 59 kcal/mol 85 kcal/mol

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hv

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 CH3 CH3

| | |

CH3CHCH3 + Cl2 CH3CHCH2─Cl + CH3CCH3

|

Cl

isobutyl chloride tert-butyl chloride

(64%) (36%)

Isobutane has nine primary hydrogens and one tertiary hydrogen chlorination prefers 3º to 1º

by a factor of 5.1.

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Halogenation of Other Alkanes (cont.)

Bromination of isobutane:

CH3 CH3 CH3

| | |

CH3CHCH3 + Br2 CH3CHCH2─Br + 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

Energy Diagrams

Exothermic Reaction Endothermic Reaction

For Chlorination:

Halogen Primary Secondary Tertiary

F 1 1.2 1.4

Cl 1 3.9 5.1

Br 1 82 1600

hv

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Alkane Preparation Reactions

1. Formation of Alkyl Halide:

CH3CH3 + X2 CH3CH2—X + H—X

Alkane diatomic halide alkyl halide binary acid

Cl2 or Br2

2. Catalytic Hydrogenation of Alkenes:

CH3CH=CH2 + H2 CH3CH2CH3

Alkene Hydrogen Alkane

3. Reduction of Alkyl Halide:

2 CH3CHCH3 + 2 Zn + 2 HC2H3O2 2 CH3CH2CH3 + ZnX2 + Zn(C2H3O2)2

|

X

Alkyl halide Zinc 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 + 2 Na CH3CH2:Na + NaX

1º Alkyl Halide Sodium 1º Organosodium Sodium Salt

CH3CH2:Na + CH3CH2—X CH3CH2CH2CH3 + NaX

1º Organosodium 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 CH3CH2:Li + LiX

1º or 2 º Alkyl Halide Lithium 1º Organolithium Lithium Salt

6. Corey-House Reaction:

2 CH3CH2:Li + CuI [(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 **

hv

essureNiPdPt Pr,,

reflux

hexaneether,

hexaneether,

hexaneether,

hexaneether,

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7. Grignard Reaction:

CH3CH2Cl + Mg CH3CH2MgX

Alkyl halide Magnesium Alkane:Mg halide

8. Hydrolysis of Grignard:

H2O Mg(OH)X

CH3CH2MgX + or CH3CH3 + or

CH3OH Mg(CH3O)X

Grignard Reagent Acidic Solvent Alkane Magnesium salt

9. SN2 – Substitution Nucleophilic Bimolecular:

“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 .. H

CH3CH2:- + I CH3CH2:

- ------- ------ :I:

- and

˙˙ H ..

H :I: -

˙˙

10. E2 – Elimination Bimolecular:

One Step – Competes with SN2 – Preferred when the carbon to be attacked is

sterically hindered (2º or 3º).

CH3

|

CH3 CH2 = C—CH2—CH3

|

CH3CH2:- + CH3CCH2CH3 and

|

X CH3

|

CH3—C = CH—CH3

**Saytzeff’s Rule** – major product is determined

by the greater # of alkyl groups attached to double bonded carbons

ether

**Transition State**

Hoffmann

Product

(Minor)

Saytzeff

Product

(Major)

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SN2 vs. E

2 - 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, E

2 or hydrolysis will occur in a reaction, consider the

following:

1. Look for a metal – Is one of your reactants a base (anion)? Recognize acid/base rxns

2. Look for an easy (acidic) proton for a base to grab

3. Can one of your reactants act as a nuc:?

4. 1º SN2

5. 3º E2

6. 2º SN2 + E

2

Organometallics Used as Nucleophiles:

CH3CH2MgX

CH3CH2:Li + CH3CH2CH2—Br CH3CH2CH2CH2CH3

[(CH3CH2:)2Cu]Li

Organometallics Used as Bases:

CH3CH2MgX CH3CH3 + Mg(OCH3)Cl

CH3CH2:Li + CH3CH2—O—H CH3CH3 + Li(OCH3)

[(CH3CH2:)2Cu]Li CH3CH3 + Li(OCH3) + CH3CH2Cl

More Uses as Bases:

CH3

|

CH3 CH2 = C—CH2—CH3

|

[(CH3CH2:)2Cu]Li + CH3CCH2CH3 and

|

X CH3

|

CH3—C = CH—CH3

2SN

Hydrolysis

Saytzeff

Product

(Major)

Hoffmann

Product

(Minor)

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SN2 and E

2 Mechanisms – A Summary

**Correlation of structure and reactivity for substitution and elimination reactions**

SN2 – Substitution, Nucleophilic, Bimolecular

- Rate = k[substrate][:nuc]

- One-step, concerted mechanism

- 100% Stereochemical Inversion

- Can compete with E2

E

2 – Elimination, Bimolecular

- Rate = k[substrate][base]

- One-step, concerted mechanism

- Competes with SN2

Halide Type SN2 E

2

Methyl Highly favored with

most nuc: Does not occur

Primary Highly favored with

strong nuc:

Favored with strong,

hindered bases

Secondary Favored by good nuc: in

polar, aprotic solvents

Favored when strong

bases are used

Tertiary Does not occur Highly favored when

bases are used

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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

b. CH3—CH—CH3

|

OH

c. CH3—C—CH3

||

:O:

d. CH3—CH2—C—OH

||

:O:

e. Br—CH—CH2—CH3

|

Br

f. CH3—CH = CH2

g. CH3—CH—CH3

|

:NH2

h. CH3—C—NH2

||

:O:

i. CH3—CH2:Mg:Cl

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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

b. Tert-butyl bromide + Mg

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)

ether

ether

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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

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?

hv

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Alkane Reactions Worksheet #4

I. Outline all steps in the free-radical chain reaction representing the mono-

bromination of isobutane. Show 3 possible termination steps.

II. Give the IUPAC names for:

a. CH3(CH2)55CH3

b.

c.

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):

a. Ethane and I2

b. 2 – chloropentane

c. 4 – methylpent– 2 – ene + H2

d. Propane + Cl2

e. CH3CH2CH2CH2CH2Cl + Na

f. Isopropyl chloride + Mg

V. Starting with only ethane and any needed inorganic reagents, outline a potential

synthesis of 3 – methylpentane

heathv,

HAcZn,

Pd

hv

ether

ether

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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 E

2 reactions.

15. Describe the “competition” between SN2 and E

2, 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.