Alkanes, Building Bridges to Knowledge

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Alkanes

Building Bridges to knowledge

Hydrocarbons contain hydrogen atoms and carbon atoms, and are classified as aliphatic or aromatic compounds. Diagram 2.1 is a schematic of the classification of hydrocarbons.

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Diagram 2.1 classifications of hydrocarbons

Alkanes have the general formula

,

and the hybridization about the alkane carbon atoms is 2sp33. Some

common alkanes are methane, ethane, propane, butane, and pentane. Higher molecular weight alkanes are formed from the distillation of crude oil. Gasoline, kerosene, petrolatum, paraffin wax, and lubricating oils can be obtained from the fractional distillation of crude oil.

Alkanes are among the least reactive of all organic compounds.

Hydrocarbons

Aliphatic Aromatic

Alkenes Alkynes Cyclic AliphaticAlkanes

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Methane, the simplest hydrocarbon, has a tetrahedral structure. “A simple view of the bonding in methane” is found at http://www.chemguide.co.uk/basicorg/bonding/methane.html.

The structural formula for methane can be inscribed in a cube where the hydrogen atoms are on the corners of a cube as indicated in Diagram 2.2

Diagram 2.2 methane inscribed in a cube

Each C-H bond is formed from the linear combination of an electron in the 1s atomic orbital of hydrogen with an electron in the 2sp3 3hybridized atomic orbital of C resulting in the formation of a (2sp3

3 + 1s) bonding molecular orbital. The H-C-H bond angle in methane is 109.5o

o.

Diagram 2.3 is a geometric arrangement of the methane carbon atom and two of its hydrogen atoms in a manner where algebraic and trigonometric calculations can be used to determine the H-C-H bond angle. In Diagram 2.3, a cube of x dimensions (similar to the one above) can be constructed around the carbon atom (centered in the middle of the cube) of methane where two of methane hydrogen atoms are positioned at “a” and “b.”

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Diagram 2.3: A geometric arrangement used to calculate the H-C-H bond angle in methane

If the side of the cube is x, using the Pythagorean Theorem, the line ab must be

.

Since the side of the cube is x, then the line running through C must also be x and the line running from c to the line ab, Co, must be

Therefore, the angle aCo in the triangle formed by Coa can be calculated in the following manner:

(ob)2 2 + (Co)22 = (aC)22

The hypotenuse is aC, and the value for aC can be calculated using Equation 2.1.

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Equation 2.1

x2

⎛⎝⎜

⎞⎠⎟

2

+ x 22

⎛⎝⎜

⎞⎠⎟

2

= (aC)2

Consequently, aC equals

The sine of angle aCo may be calculated using Equation 2.2

Equation 2.2

Angle aCb is the H-C-H bond angle

Methane can be represented as

CH44

or

6

or

or

Let’s review the bonding in methane. To understand the covalent bonds in methane, it is necessary to revisit the electron configurations of both carbon and hydrogen. The electron configurations of carbon is 1s2

2 2s22 2p2

2 and the electron configuration of hydrogen is 1s1

1. These electron configurations can be used to explain the bonding that exists in the molecule. To explain why the bond angle is 109.5o

o and that all the C-H bonds are equal, an electron in the 2s atomic orbital of carbon can be promoted to a 2p atomic orbital of carbon. The promotion of an electron from the 2s atomic orbital of C to a 2p atomic orbital is illustrated in Diagram 2.4.

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Diagram 2.4 Promotion of an electron from 2s atomic orbital of carbon to the 2p atomic orbital of carbon

The 2p atomic orbital that contains the promoted electron linearly combines with the remaining two 2p atomic orbitals of carbon (each containing an electron) and the 2s atomic orbital that contains one electron to form four degenerate (equal in energy) 2sp3 hybridized atomic orbitals. The resulting hybridized atomic orbitals are lower in energy than the 2p atomic orbitals from which they were formed and higher in energy than the 2s atomic orbital from which they were formed. Each 2sp3

3 hybridized atomic orbital has one electron. Diagram 2.5 illustrates this process.

Diagram 2.5 the four degenerate 2sp3 atomic orbitals of carbon

1s

2s

2px 2pz2py 2py 2pz2px

2s

1s

1s

2sp3 2sp3 2sp3 2sp3

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Each C-H covalent bond is formed from the linear combination of a 1s atomic orbital of H with a 2sp3 hybridized atomic orbital of carbon to form a bonding and an anti-bonding molecular orbital that describe the carbon-hydrogen bond in methane. Diagram 2.6 describes the bonding of each C-H bond in methane.

Diagram 2.6: the bonding and antibonding molecular orbitals formed from the linear combination of a 1s atomic orbital of hydrogen with a 2sp3 atomic orbital of carbon

Each bond between carbon and hydrogen is a single covalent bond (referred to as a sigma, σ, bond). The σ bond is formed from the linear combination of a 2sp3

3 hybridized atomic orbital of carbon and a 1s atomic orbital from hydrogen. In this case the single covalent bond is given the notation σ(2sp3 + 1s).

The result is illustrated in Diagram 2.7

2sp3

1s

σ(2sp 3 + 1s )

σ(2sp 3 - 1s )*

2sp3

1s

σ(2sp 3 + 1s )

σ(2sp 3 - 1s )*

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A tetrahedral molecular with a H-C-H bond angle equal to 109.5 o Diagram 2.7 the sigma C-H bonds in methane

The hybridization around any carbon atom of an alkane can be visualized in an analogous manner, i.e., the hybridization around any carbon atom of an alkane before it participates in the formation of a molecular orbital would be 2sp3.

Reactions of Methane/Alkanes

Methane and alkanes participate in two types of reactions:

(1) Combustion reactions; and

(2) Halogenation reactions

Oxidation reactions or combustion reactions involve reactions of hydrocarbons with oxygen.

The complete combustion of methane and alkanes to produce carbon dioxide and water can be represented by the following equations.

C

H H

H

H

σ(2sp3

+ 1s)

+ 1s)

σ(2sp3+ 1s)

σ(2sp3+ 1s)σ(2sp3

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Complete Combustion of Methane

General Equation for the Complete Combustion of Any Alkane

Incomplete Combustion of Methane

The incomplete combustion of methane forms acetylene, carbon monoxide and hydrogen, and can be illustrated by the following equation.

The structural formula for acetylene, C 2H22 is

The incomplete combustion of methane in the presence of water and a nickel catalyst produces carbon monoxide and hydrogen as illustrated by the following equation.

The second type of reaction alkanes can experience is halogenation reactions, i.e., the reaction of alkanes with fluorine, chlorine and bromine. The following sequences of chemical reactions represent

n (2n + 2) 2 2 23n + 1C H + O n CO + (n + 1) H O

2⎛ ⎞ →⎜ ⎟⎝ ⎠

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the halogenation of methane, the simplest alkane. X2 represents F

2,

Cl2, or Br

2,

Fluorine, chlorine, and bromine will undergo halogenation reactions, but iodine will not.

The order of reactivity of halogens in halogenation reactions is

Additional information concerning the halogenation of alkanes may be found at http://www.mhhe.com/physsci/chemistry/carey/student/olc/graphics/carey04oc/ref/ch04radical.html.

The series of elementary steps that rationalize the formation of product or products of a chemical reaction are referred to as the mechanism of the reaction. Mechanisms give insight into the reaction process and minimize perceived mysteries associated with the reaction. The mechanism of halogenation reactions can be explained by examining the following series of elementary steps.

CH4(g) + X2 (g) heat or light⎯ →⎯⎯⎯ CH3X (g) + HCl (g)

CH3X (g) + X2 (g) heat or light⎯ →⎯⎯⎯ CH2X 2 (g) + HCl (g)

CH2X2 (g) + X2 (g) heat or light⎯ →⎯⎯⎯ CHX 3 (g) + HCl (g)

CHX3 (g) + X2 (g) heat or light⎯ →⎯⎯⎯ CX 4 (g) + HCl (g)

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This illustration uses the simplest alkane to demonstrate the mechanism of halogenation. The illustration explains the formation of methyl halide from methane. X

2 represents F

2, Cl

2, or Br

2,

This mechanism includes three (3) steps. The first step of the mechanism is the dissociation of the halogen molecule to halogen atoms. This can be accomplished by light or heat. The second step of the mechanism is a propagation step. In this step, the halogen radical produced by the homolytic cleavage of the X-X bond reacts with methane to produce a methyl radical and a hydrogen halide gas molecule. The third step is another propagation step leading to the formation of methyl halide and a halogen atom. The final step, not included, but will be discussed momentarily, is a set of possible termination steps that quench the reaction.

Halogenation of alkanes follows a free radical mechanism. A free radical is an atom or group of atoms possessing an unpaired electron. The halogenation of methane is a chain reaction since the reaction involves a series of steps where each step generates a reactive free radical intermediate that propagates the reaction. If Cl

2

is the halogen in question that interacts with light, then the three (3) steps that explain the formation of the halomethane, chloromethane, can be described by the following series of elementary steps.

The first step is an initiation step where energy is absorbed and a

2 (g) (g)h

(g) 4 (g) (g) 3 (g)

3 (g) 2 (g) 3 (g) (g)

(1) X 2 X

(2) X + CH HX + CH

(3) CH + X CH X + X

ν⎯⎯→ ⋅

⋅ → ⋅

⋅ → ⋅

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free radical species is generated.

The second and third steps are chain-propagating steps where one reactive species is consumed and another is generated.

The fourth step is a series of chain termination steps where the reactive species are consumed and not generated, These quenching steps could include the collision of two chlorine radicals, the collision of a methyl radical with a chlorine atom, or the collision of two methyl radicals.

There is evidence to support the halogenation mechanism. Tetraethyl lead decomposes at 140oC in the following manner.

4 3Cl + CH CH + HCl⋅ → ⋅

2 3 3Cl + CH CH Cl + Cl⋅ → ⋅

2

3 3

3 3 3 3

Cl + Cl ClCH + Cl CH Cl CH + CH CH CH

⋅ ⋅ →⋅ ⋅→⋅ ⋅ →

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Since the extremely reactive ethyl radical,

,

is formed at140 oC, a mixture of methane and chlorine containing a small amount of tetraethyl lead should initiate the chlorination of methane at 140

oC instead of 250 oC by way of the following series

of elementary steps:

The overall reaction would be:

When tetraethyl lead is added to a reaction mixture of methane and chlorine, the reaction will occur at 140 oC rather 250o

oC. This is strong evidence for a free radical mechanism.

Bond Dissociation Energy

Bond dissociation energy is the amount of energy consumed when a

2 3CH CH⋅

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bond is broken. The magnitude of energy consumed when a bond is broken is identical to the energy liberated when a bond is formed. Bond dissociation energies are generally given in kilojoules (kJ) per mole, i.e., the number of kJ required to dissociate 6.022 x 1023

23 bonds of the same kind.

Table 2.1 lists some examples of bond dissociation energies.

Bond Dissociation Energy* Cl-Cl 2 Cl 243 CH4 CH3 + H 435 CH3 CH2 + H 444 CH2 CH + H 444 CH C + H 339

* Bond Dissociation Energy, kJ mol-1 Table 2.1 Some Dissociation Energies

For the reaction

∆H(bond dissociation) = 1662 kJ/mol

i.e., 435 kJ + 444 kJ + 444 kJ + 339 kJ = 1662 kJ

The average C-H bond dissociation energy would be 416 kJ/mol

1662 kJmole

4 = 416 kJ

mole

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Heat of A Reaction

During a chemical reaction, heat is either liberated or consumed. Chemical reactions involve bond breaking and bond making. Bond breaking requires the absorption of energy or heat from the surroundings, and bond making releases energy in the form of heat to the surroundings. The overall heat content change (enthalpy) represented by ∆H is referred to as heat of the reaction, and the heat of a reaction can be calculated using the following expression:

Heats or enthalpies of reactions are equal to the sum of the standard heats of the formation of the products minus the sum of the standard heats of formation of the reactants. If the heat of the reaction is negative, then the reaction releases heat to the surroundings, and the reaction is said to be exothermic. If the reaction has a positive value for the heat of the reaction, then the reaction absorbs heat from the surroundings, and the reaction is said to be endothermic.

The heat of the reaction can also be calculated from bond dissociation energies. For example, the heat of the reaction between methane and chlorine can be determined in the following manner.

Table 2.2 lists the bond energies (in kJ/mol) for making and breaking the bonds in the following reaction.

4 2 3CH + Cl CH Cl + HCl→

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Bond Breaking (Reactants)

Bond Making (Products)

Bond Energy (kJ mol-1)

H3C-H

+435

Cl-Cl +243 H3C-Cl

-351

H-Cl -431 Table 2.2 bond making and breaking energies for selected bonds

The bond breaking process absorbs +678 kJ mol-1 of energy (435 kJ + 243 kJ) from the surroundings, and the bond making process releases -782 kJ mol-1 (351 kJ + 431 kJ) to the surroundings.

or

One can use the bond dissociation energy and Hess’s Law to calculate the enthalpy of the reaction.

If chlorination of methane releases heat to the surroundings, then why is light or high heat required to initiate the reaction?

4 2 3CH + Cl CH Cl + HCl→

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This observation can be explained by examining the enthalpy values for the steps in the mechanism for the formation of methyl chloride from methane and chlorine. Let’s consider the elementary steps and the enthalpy of each step in the formation of methyl chloride from methane and chlorine.

(1) The Initiation Step

(2) A Propagation Step

(3) A Propagation Step

(4) A Termination Step

Although the chlorination of methane is an exothermic process, the initial step in the reaction involves an energy absorption process, the breaking of the Cl-Cl bond. This energy absorption step requires 243 kJ/mol of heat. Also, the next step of the process, i.e., the propagation of the methyl radical, is slightly endothermic. Consequently, the reaction requires some type of energy initiator, heat or light, hν, for the formation of the chlorinated product.

Cl2 → Cl ⋅ ΔH = + 243 kJ mol−1

Cl ⋅ + CH4 → HCl + ⋅CH3 ΔH = +4 kJ mol−1

⋅CH3 + Cl2 → CH3Cl + Cl ⋅ ΔH = -108 kJ mol−1

Cl ⋅ + Cl ⋅ → Cl2 ΔH = -243 kJ mol−1

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Using the data in table 2.3, suggest a reason why methane does not form iodomethane, CH

3I. If iodomethane could be produce, the

mechanism of the reaction pathway would be similar to the reaction pathway for the formation of chloromethane from chlorine and methane.

Bond Breaking (Reactants)

Bond Making (Products)

Bond Energy (kJ/mol)

H3C-H +435 I-I +151

H3C-I -234 H-I -297 Table 2.3 Bond making and bond breaking for

The following mechanism would rationalize the formation of iodomethane from iodine and methane.

Using the data from Table 2.3, it is obvious that the initiation, step 1, and the first propagation step, step 2, would be endothermic. Step 2 would require 138 kJ mol-1 of energy. This value is too high for the reaction to produce methyl iodide as a product; therefore, a mixture of iodine and methane in the presence of heat or light does not thermodynamically favor the production of iodomethane.

4 2 3CH + I CH I + HI→

2

4 3

3 2 3

2

(1) I 2 I(2) I + CH HI + CH(3) CH + I CH I + I(4) I + I I

→ ⋅⋅ → ⋅

⋅ → ⋅⋅ ⋅ →

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The Methyl Radical

Following is the structure of the methyl radical.

Experimental evidence suggests that the methyl radical is flat and the carbon atom, the central atom, is 2sp2

2 hybridized. The unpaired electron resides in a 2p atomic orbital perpendicular to the plane containing three (3) σ (2sp2 + 1s)

(bonding molecular orbitals. The 2p

orbital has a lobe that lies above the plane containing the three σ (2sp2 + 1s) bonding molecular orbitals and a lobe that lies below the plane containing three σ (2sp2 + 1s) molecular orbitals. A representation of the methyl radical is illustrated in figure 2.1.

Figure 2.1

Figure 2.1 suggests that the C-H bonds in the methyl radical lie in the same plane, and that the 2p orbital containing one electron is perpendicular to the plane containing the three C-H bonds.

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The Transition State

When molecules react, bonds are broken and bonds are formed. The following factors must be satisfied during the breaking of bonds and formation of bonds:

• The reactant molecules collide. • The molecules must collide with the proper orientation so that

bond breaking and bond making may occur. • Collisions between the reactant molecules must be

accompanied by sufficient kinetic energy so that bond breaking and making can be initiated; otherwise, the molecules will “bounce” off one another and no product or products will form. The energy that the molecules possess upon collision must be sufficient to overcome any energy barrier to the reaction.

When these three factors are satisfied, bond breaking and bond making can occur, and the molecules react. However, prior to bond formation and bond disruption, there is an intermediate state that cannot be isolated where a bond is not quite broken and a bond is not quite formed. This quasi intermediate state is referred to as the activated complex, and its potential energy is higher than the potential energies of the reactants and the products. For example, in the propagation step where methane reacts with a halogen radical to form the methyl radical and hydrogen halide, the reaction, including the formation of the activated complex, can be illustrated in the following manner.

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This reaction may be illustrated in the following manner where the geometry of the molecule is taken into consideration.

An energy profile diagram may be used to illustrate the progress of a reaction (the reaction coordinates versus the potential energy) for the above reaction. The energy of the activated complex is represented by the “top of a hill.” The “hill top” position along the reaction

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coordinates is the transition state. The transition state is where the activated complex is formed. The activated complex involves the simultaneous breaking and making of bonds. The activated complex cannot be isolated.

Figure 2.1 illustrates the formation of the activated complex at the transition state along the reaction coordinates.

Figure 2.1 The Energy Profile Diagram for the reaction between methane and a halogen radical forming a methyl radical and hydrogen halide molecule

The energy of activation, E act is the difference in energy between the

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reactants and the transition state. For this reaction to occur, the energy of the colliding molecules must exceed the Eact . The enthalpy of the reaction, ΔHrxn , is the energy required to homolytically break one C-H bond.

Determination of Molecular Masses of Volatile Organic Compounds,

Composition of Organic Compounds, and Empirical Formulas

The molecular masses of volatile organic compounds can be obtained by using the ideal gas equation, PV = nRT. For example, if one is able to obtain the mass and volume of a volatile organic compound, then knowing the temperature and the pressure, the molecular mass can be obtained by solving the ideal gas equation. The experimental method for determining the molecular mass of a volatile organic compound is called the Dumas method.

The Dumas method, http://web.centre.edu/miles/che135/che135labs/Molecular%20Weight%20by%20the%20Dumas%20Method.pdf, for determining the molecular mass of a volatile organic compound can be illustrated by calculating the molecular mass of a volatile unknown compound given the following data:

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Physical Parameters of the Volatile

Organic Compound Data

mass 0.309 g volume 488 mL

temperature 23.0oC pressure 737 torrs

The molecular mass can be determined by using the ideal gas equation.

Therefore, the molecular mass of the volatile organic compound would be 15.9 grams per mole, i.e., one mole (6.022 x 1023

23 molecules of the volatile compound) would weigh 15.9 grams.

Molecular Mass of Organic Compounds that May not be Volatile.

The molecular mass of organic compounds that may not be volatile can be determined using the colligative properties of solvents. Conducting boiling point elevation experiments; freezing point depression experiments; osmotic pressure experiments; or vapor

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pressure lowering experiments may be used to determine the molecular masses of non-volatile organic compounds.

The selection of a specific colligative property for determining the molecular mass of the organic compound depends on whether the organic compound is a gas, a liquid or a solid.

If boiling point elevation is used, then the organic compound must not dissociate; be non-reactive with the selected solvent; and be completely soluble in the selected solvent.

The solution, a mixture of the solvent and the organic compound, would have a higher boiling point than the pure solvent. The extent to which the boiling point of the solution is elevated is directly proportional to the number of moles of the solute, the organic compound, dissolved in a given mass of solvent.

The equation for calculating molecular mass by the elevation of the boiling point is:

where ∆Tb = boiling point of the solution minus the boiling point of the

pure solvent

Kb = molal boiling point constant whose numerical value depends

upon the solvent and is independent of the solute

b bT = K mΔ

2(boiling point of solvent)

bvaporization

molecular mass of solvent x R x TK =

1000 HΔ

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m = molality of the solution

The molality equals

R = 8.314 J/K-mole

∆Hvap is in kJ/mol

and T is in K

Equation 2.1, the equation for the elevation of the boiling point, can be derived from the Clausius- Clapeyron Equation, Equation 2.2.

Equation 2.1

Equation 2.2

Equation 2.3 can be obtained from the Clausius-Clapeyron equation.

Equation 2.3

To avoid calculus and for the sake of simplicity, let’s say that the

number of moles of solute dissolved in the solventnumber of kilograms of solvent in the solution

b b T = K mΔ

vap2

PdP = dT RT

ΔΗ

VAPHln P = - + C'RT

Δ

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expression

of the Clausius-Clapeyron is a small change in the pressure divided by a small change in temperature (in Kelvin). Therefore, a larger change in the pressure divided by a larger change in the temperature would be

and equation 2.4, a modified version of the Clausius-Clapeyron equation, would be:

Equation 2.4

With this modification of the Clausius-Clapeyron equation, ΔT would give equation 2.5

Equation 2.5

Equation 2.6 is a statement of Raoult’s Law.

dPdT

PT

ΔΔ

vap2

PP = T RT

ΔΗΔΔ

2RT PT = P

vap

ΔΔΔΗ

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Equation 2.6

or

since

then

and

Substituting

o oA A AP = P - PχΒ

o oA A AP - P = - PχΒ

o oA A AP - P = PχΒ

oA AP - P = PΔ

oAP = PχΒΔ

oA

P =

PχΒ

Δ

BoA A B

nP =

P n nΔ

+

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for

In Equation 2.5

would give equation 2.7

Equation 2.7

for the elevation of the boiling point when a nonvolatile nonelectrolyte is added to a solvent

Since only a small amount of the nonvolatile nonelectrolyte is generally added to a solvent, an assumption may be made that n

A >> n

B

then

B

A B

nn n+

oA

PPΔ

2RT PT = P

vap

ΔΔΔΗ

2B

A B

nRTT = x

n nvap

ΔΔΗ +

31

is approximately equal to

and an approximation of equation 2.7 would be Equation 2.8.

Equation 2.8

since

then equation 2.8 can be written as:

the molality is

B

A B

nn n+

B

A

nn

2B

A

nRTT = x

nvap

ΔΔΗ

A (in grams)A

A (in g/mol)

massn =

MW

2B

A (in g)

A

nRTT = x

massMW

vap

ΔΔΗ

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Consequently,

since

then,

where Tb = boiling point of water

Table 2.4 lists molal boiling point constants for some solvents used in determining molecular masses of organic compounds.

B

A (in g)

nmass

1000

2ART MWT = m

1000vap

ΔΔΗ

2b A

bRT MWK = 1000

vapΔΗ

b bT = K mΔ

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Solvent Boiling Point, K Kb, K kg mol-1

water 373 0.512 benzene 353.3 3.07 carbon

tetrachloride 349.7 5.03

chloroform 334.4 3.63 ethanol 351.7 1.22

methanol 337.7 0.83 n-octane 400 4.02

Table 2.4 molal boiling point constants

The value of Kb means that 1 mole of a non-dissociated solute in one

kilogram of a solvent will change the boiling point of the solvent by the value of K

b. For example, benzene has a K

b equal to

Therefore, 1 mole of a pure non-dissociated solute in one kilogram of benzene would raise the boiling point of benzene from 353.3 K to

.

If 0.500 mole of solute is added to benzene, then the resulting mixture would have a boiling point of

K-kg mol353.3K + (3.07 x 1.00 ) = 356.4K

mol kg

1 K-kg mol353.3K + ( x 3.07 x 1.00 ) = 354.8K

2 mol kg

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Following is an application of colligative property technique used to determine the molecular mass of an unknown organic compound.

A solution containing 9.30 grams of a pure unknown organic compound and 105 grams of benzene exhibited a boiling point of 82.5oC. Calculate the molecular mass of the unknown organic compound to two significant figures.

Parameters Data

boiling point of pure benzene 353.3 K boiling point of the solution 355.7 K

kilograms of solvent 0.105 kg ΔTb 355.7 K - 353.3 K = 2.4 K

Kb for benzene

moles of solute = molality of solute x mass of solvent

The table is incoplete

Therefore, using this technique, the approximate molecular mass (with two significant figures) of the unknown organic compound is 110 g/mol.

The freezing point depression technique can also be employed to determine the molecular mass of an unknown organic compound. The freezing point of a solution is lower than the freezing point of the pure solvent comprising the solution. In a dilute solution, the freezing point depression depends on the number of moles of the dissolved solute.

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Equation 2.9, the equation for the depression of the freezing point, can also be derived from the Clausius- Clapeyron equation.

Equation 2.9

Where ∆Tf is equal to freezing point of the pure solvent minus the

freezing point of the solution; Kf is the molal freezing point

depression constant, and m is the molality of the solution.

Table 2.5 lists the molal freezing point constants for several solvents commonly used to determine the molecular masses of unknown organic compounds.

solvent freezing or melting point, K Kf, K kg mol-1

water 273 1.86 benzene 278.7 5.12

benzophenone 321.3 9.8

biphenyl 343 8.0 camphor 451.6 40.0

acetic acid 289.8 3.90 naphthalene 353.4 6.8

Table 2.5 molal freezing point constants

In an analogous fashion as Kb, the numerical value for K

f is the

property of the solvent. The values are for 1molal concentrations, i.e., 1 mole of pure non-dissociated solute in 1 kg of solvent.

For example, one mole of non-dissociated solute dissolved in 1000

f fT = K mΔ

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grams of water would have a freezing point of 271.29K

0.500 mole of non-dissociated solute dissolved in 1000 grams of water would have a freezing point of 272.23K

Following is a second application of colligative properties that can be used to determine the molecular mass of an unknown organic compound.

A solution containing 2.10 grams of a pure unknown organic compound and 45.0 grams of camphor exhibited a melting point of 170.2oC. Calculate the molecular mass of the unknown organic compound to two significant figures.

Parameters Data

melting point of pure camphor 451.6 K melting point of the mixture 443.4 K

kilograms of solvent 0.0450 kg ΔTf 451.6 K -443.4 K = 8.2 K

Kf for camphor

moles of solute = molarity of solute x mass of solvent

K kg mol273.15K + (-1.86 x 1.00 ) = 271.29K

mol kg

1 K kg mol273.15K + x (-1.86 x 1.00 ) = 272.23K

2 mol kg

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The Table is incomplete

Using this technique, the approximate molecular mass of the unknown organic compound is 230 g/mol.

An example of this experimental process can be examined by looking at the following Website.

http://home.millsaps.edu/lewisll/Learning%20Modality,%20Lab%20Report%20Aids-Genchem%20and%20pchem/Format%20example.pdf

Composition and Empirical Formulas

The empirical formula of an organic compound may be determined by knowing the percent composition of the organic compound. The following Website may be used to obtain an understanding of how percent composition can be used to determine the empirical formula of an organic compound.

http://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm.

For instance, if a student analyzes 100 grams of a compound, and discovers that the compound contains 75.0 % carbon and 25.0% hydrogen, then the percent composition of the carbon and the hydrogen can be calculated by determining the mole ratio of the constituent elements.

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Since the percentages equal 100, one may assume that there are one hundred grams of material. Consequently, the compound would contain 75.0 grams of carbon and 25.0 grams of oxygen.

The number of moles of carbon and the number of moles of hydrogen in the compound can be calculated by multiplying each mass by the reciprocal of its respective atomic mass (in grams per mole).

Now, obtain the mole ratio by dividing the moles of C and the moles of H by the smaller number of moles of the component elements (in this case the smaller number would be 6,25 moles).

Therefore, the empirical formula for this compound is CH4.

Following is another example for determining the percent composition of a compound.

Calculate the percent composition for an unknown organic compound if it is subjected to the following conventional wet

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analysis.

When 5.95 milligrams of an unknown organic compound were heated in a special device, 13.97 milligrams of CO

2 were trapped in a

specially prepared substance and 2.39 milligrams of H2O were

trapped in a specially prepared substance. In another experiment 7.55 milligrams of AgCl precipitated.

The masses of chorine, hydrogen, and carbon can be calculated in the following manner.

Total mass of C; H; and Cl would be:

3.81 mg + 0.266 mg + 1.87 mg = 5.95 mg, the total mass of the unknown compound; therefore, the percent composition can be calculated using the following process.

weight of carbon = 13.97 mg CO2 x 12.02 g/mol C44.0 g/mol CO2

= 3.81 mg of C

40

Total percent = 64.0% + 4.48% + 31.4% = 99.9% (essentially 100%)

Ethane

http://www.chemguide.co.uk/basicorg/bonding/methane.html describes the hybridization and bonding in methane and ethane.

Ethane has the molecular formula C2H

6 and the structural formula

As expected for alkanes, the hybridization about the carbon atoms is 2sp3. The C-C bond is formed from the linear combination of a 2sp3 hybridized atomic orbital of one carbon with a 2sp3 hybridized atomic orbital of the other carbon to produce a σ

(2sp

3

+ 2sp

3

) bonding molecular

orbital. The H-C-H bond angle is about 109.5o, and the C-C-H bond angle is approximately 109.5o. The C-C bond length is 1.54 Ǻ. The bond angle and the bond length in ethane can be illustrated in the following manner:

41

There is free rotation about the C-C single bond in ethane, and the various rotational patterns are referred to as rotational isomers or conformations of ethane. The rotational isomers can be converted into one another by rotating about the C-C single bond. The resulting isomers are called conformational isomers. The various conformational isomers can be ascertained by using a Newman Projection formula, (I) and a Sawhorse model (II) for the staggered conformation of ethane, and a Newman Projection formula (III) and a Sawhorse model (IV) for the eclipsed conformation of ethane.

42

Clearly, there is less steric interaction from the hydrogen atoms in the staggered conformation of ethane; therefore, this is the most stable conformation. The energy barrier to rotation about the C-C single bond is small; consequently, none of the conformational isomers can be isolated. There are an infinite number of skewed

43

conformations between the eclipsed conformation and the staggered conformation. The relative stability of the various conformations can be illustrated using energy Diagram 2.8.

Diagram 2.8 Potential energy diagram for ethane conformational isomers

44

Butane, C4H

10

The structure of butane is CH3CH

2CH

2CH

3. Butane can be viewed by

using the following sawhorse structure,

where carbon atoms 2 and 3 form the middle C-C bond in butane. Carbon atoms 2 and 3 can rotate to form various conformational or rotational isomers of the molecule. In reality, one can select any of the connecting C-C single bonds for this purpose, but the conformational isomers resulting from the rotation about carbon atoms 2 and 3 are symmetrical and can easily illustrate the potential energy of the various conformational isomers as the system rotates from the staggered conformation to the eclipse conformation. Perhaps an easier way of understanding this is to look at the Newman projection formula for butane represented by the following structure.

45

Even though the conformers in butane differ in potential energy, the energy barrier to rotation is small and the rotational isomers cannot be isolated. All the possible conformers are in rapid equilibrium with each other. However, in some molecules where the energy barrier to rotation is large, separation of the conformers is possible. Also, it is possible to freeze the rotation in certain molecules and study a particular conformation.

Diagram 2.9 represents an energy diagram showing rotation from 0o to 240o for four conformational isomers of butane in relationship to their potential energies.

46

Diagram 2.9 Two-thirds of the potential energy diagram for rotating to the different rotational conformations of n-butane

The remaining rotational conformations, 300o and 360o , an eclipsed conformation and the anti-staggered conformation respectively, are not observed in this potential energy versus rotation diagram.

The rotational isomers at 300o and 360o are represented by the Diagram 2.9b.

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Diagram 2.9b

The energy barrier to rotation is greater in butane than in propane or ethane, because, in butane, there is steric interaction or repulsion between the two methyl , CH

3, groups. In propane, the interaction is

between a methyl group and a smaller H atom. In ethane, the interaction is between two H atoms.

Table 2.6 lists the energy barrier to rotation to the conformational isomers of butane along the C2 and C3 carbon single bond.

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Table 2.6 Energy barrier to rotation along the C2-C3 carbon single bond

The energy barriers to rotation to the most eclipsed rotation conformation of butane, propane, and ethane are listed in Table 2.7.

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Table 2.7 energy barrier to the most eclipsed conformations for butane, propane, and ethane

http://wetche.cmbi.ru.nl/organic/nalkanesconf/butane/jmindex.html illustrates the dynamics of butane as the two carbon atoms labeled 1 and 4 with hydrogen atoms attached rotate about carbon atoms 2 and 3.

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Higher Alkanes

Compounds where each member differs from the next by a constant amount is referred to as an homologous series. The alkanes are members of a homologous series where each member differs from the next by a –CH

2- , methylene, group. Each member of a

homologous series is called a homolog.

Alkanes can assume a straight chain structure, e.g., compound (V) and a branched structures, e.g., compound VI.

Table 2.8 list examples of some alkanes with their respective number of carbon atoms and boiling points.

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Number of Carbon

Atoms Boiling Point, K Nomenclature

1 112 Methane 2 184.6 Ethane 3 228.7 Propane 4 272.7 Butane

261.5 isobutane or 2-methylpropane

5 309 Pentane 310.1 isopentane

2-methylbutane 282.7 2,2-dimethylpropane

6 342 Hexane 333.5 Isohexane or

2-methylpentane 331 2,3-dimethylbutane 322 2,2-dimethylbutane

336.5 3-methylpentane 7 371.6 Heptanes 8 398 Octane

10 447.3 Decane 11 468 Undecane 15 543.7 Pentadecane 20 616 Eicosane

Table 2.8 Names and boiling points of representative alkanes

The International Union of Pure and Applied Chemistry, IUPAC, organization provides some basic rules for naming organic compounds. In Table 2.8, the IUPAC names appear in blue. C1 to C10 compounds are common; therefore, familiarization with these structures is essential since they are frequently encountered.

Table 2.9 lists some structural units that are frequently encountered

52

in a study of organic chemistry; therefore, it is important to be familiar with them.

Table 2.9 some common alkyl groups

When naming organic compounds containing these groups, the “ane” of the alkane is dropped and replaced with “yl.” For

53

example,CH3CH

2CH

2Cl is named n-propyl chloride. CH

3CH

2I is referred

to as ethyl iodide.

is called isopropyl chloride

is called tert-butyl bromide

The following rules should be followed when naming alkanes (using the IUPAC system of nomenclature).

(1) Select the longest continuous carbon-carbon chain as the parent structure. For example, using the carbon skeleton of structures I and II, the carbon atoms included in the ellipses would represent the longest continuous chain for each structure.

54

The parent chain for I is butane and the parent chain for II is hexane

(1) Indicate by using numbers where the alkyl side group or groups is or are attached to the parent structure. The numbering should commence from the carbon atom of the parent structure that gives the lowest number for the attached alkyl group.

55

The numbering process is indicated in compound I. Carbon atom 2 has a methyl group attached. For compound II, carbon atoms 3 and 4 have methyl groups attached. The name of compound I would be 2-methylbutane. There is a dash, “-“, after the 2, and the dash is a part of the nomenclature.

If the same alkyl group appears more than once as a side chain attachment, then indicate the presence of that alkyl group with a Greek prefix di (2), tri (3), tetra (4), penta (5) etc. depending on how many groups of the same structure are attached.

Compound II has two methyl groups attached- one on carbon atom 3 and the other on carbon atom 4; therefore, the attached methyl groups would carry the name dimethyl. The name of the compound would be 3,4-dimethylhexane. There are no spaces between the 3,4, dash, and the dimethylhexane. This is standard procedure for

56

the IUPAC system of nomenclature for organic compounds. Let’s consider the nomenclature of compounds III and IV.

The longest continuous chain in compound III has five carbon atoms, and two methyl groups are attached to carbon atom 2; therefore, the name of compound III is 2,2-dimethylpentane. The longest continuous chain in compound IV has six carbon atoms, and an isopropyl group is attached to carbon atom 3; therefore, the name for compound IV is 3-isopropylhexane. Compound IV could also be called 3-ethyl-2-methylhexane since the carbon atoms may be numbered in the following manner.

57

Also, compound IV is a segue to rule (3).

(1) When multiple alkyl groups are attached to the parent structure, the groups are cited in alphabetical order; therefore, the ethyl group is written first, followed by the isopropyl group, followed by methyl, etc.

Let’s use rules 1-3 to name compounds V and VI.

The Greek prefixes, di, tri, tetra, etc, are ignored in the alphabetical arrangement, i.e., triethyl- would precede dimethyl, because the Greek prefixes would not be considered in alphabetizing triethyl and dimethyl; therefore, ethyl would precede methyl. Iso, tert, and neo are part of the alphabetized nomenclature, e.g., neopentyl or isopropyl or tert-butyl would be considered in alphabetizing the attached groups.

58

Using the IUPAC nomenclature rules, the name for Compound V is 2,2,4-trimethylpentane, and analogous to compound IV, compound VI can have two names: 3,5-diethyl-2,3,4-trimethylheptane or 5-ethyl-3-isopropyl-3,4-dimethylheptane.

Isomers

As the number of alkane carbons increases, the number of possible isomers also increases.

For example, C4H

10 has the following two isomers:

CH3CH

2CH

2CH

3

n-butane

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b.p. = 272.6K

and

Isobutane

b.p. = 260.0K

C5H

12 has the following three (3) isomers:

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C6H

14 has five (5) isomers. Following are the five isomers of C

6H

14 and

their boiling points. The observation that these five isomers have different physical properties contributes to the fact that they are different compounds.

CH3CH

2CH

2CH

2CH

2CH

3

n-hexane

boiling point = 342 K

61

C7H

16 has nine (9) isomers; C

10H

22 has seventy-five (75) isomers; and C

20H

42

has 366,319 isomers.

62

The nine isomers for C7H

16 are

(1)

n-heptane

(2)

2-methyhexane

(3)

3-methylhexane

(4)

CH3CH2CH2CH2CH2CH2CH3

CH3CH CH2CH2CH2CH3

CH3

CH3CH2CH CH2CH2CH3

CH3

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2,2-dimethylpentane

(5)

3,3-dimethylpentane

(6)

2,3-dimethylpentane

(7)

CH3C CH2CH2CH3

CH3

CH3

CH3CH2 CCH2CH3

CH3

CH3

CH3CH CHCH2CH3

CH3

CH3

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2,4-dimethylpentane

(8)

3-ethylpentane

(9)

2,2,3-trimethylbutane

Classification of Carbon and Hydrogen Atoms

Carbon atoms in alkanes and related compounds can be classified as primary, secondary or tertiary. A primary, 1o, carbon atom is a carbon atom that is bonded to one other carbon atom. Not all primary carbon atoms are equivalent. A secondary, 2o, carbon atom

CH3CHCH2CHCH3

CH3 CH3

CH3CH2CH CH2CH3

CH2CH3

CH3C CHCH3

CH3

CH3

CH3

65

is a carbon atom that is bonded to two other carbon atoms. Not all secondary carbon atoms are equivalent. A tertiary, 3o, carbon atom is a carbon atom that is bonded to three other carbon atoms. Not all tertiary carbon atoms are equivalent.

Figure 2.2 highlights the primary, secondary, and tertiary carbon atoms of 2,3-dimethylhexane.

Figure 2.2 1o, 2o and 3o carbon atoms in 2,3-dimethylhexane

The hydrogen atoms on a primary carbon atom are referred to as primary hydrogen atoms. Secondary hydrogen atoms are bonded to secondary carbons and tertiary hydrogen atoms are bonded to tertiary carbons.

Figure 2.3 highlights the primary, secondary, and tertiary hydrogen

66

atoms in 2,3-dimethylhexane.

Figure 2.3 1o, 2o and 3o hydrogen atoms in 2,3-dimethylhexane

Alkyl groups are classified as primary, secondary and tertiary.

An alkyl group with a carbon attached to a functional group (e.g., halogen and OH), another carbon atom and two hydrogen atoms is referred to as a primary alkyl group. For example,

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is referred to as n-propyl, a primary alkyl group. The compound CH

3CH

2CH

2Cl is n-propyl chloride or 1-chloropropane, and

CH3CH

2CH

2OH is

n-propyl alcohol or 1-propanol.

is referred to as a sec-butyl group. The compound

is sec-butyl chloride or 2-chlorobutane, and the compound

is sec-butyl alcohol or 2-butanol.

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.

is referred to as a tert-butyl group. The compound

is tert-butyl chloride of 2-chloro-2-methylpropane, and the compound

is tert-butyl alcohol or 2-methyl-2-propanol

http://www.sciencegeek.net/APchemistry/APtaters/alkanes.htm

is a practice Website for naming alkanes. Take a few moments to

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access http://www.sciencegeek.net/APchemistry/APtaters/alkanes.htm and practice naming alkanes.

Some Physical Properties of Alkanes

The unique carbon atoms of alkanes are connected by covalent bonds. Alkanes are non-polar molecules.

The boiling points and melting points of alkanes increase with increasing molecular mass. The boiling points of alkanes increase 20 to 30 degrees with the addition of each carbon atom.

Branching, i.e., side chain alkyl groups, decreases the boiling points of alkanes.

Alkane Synthesis

Alkanes can be prepared industrially or in a small laboratory setting. Industrial preparations provide alkanes in large quantities at minimum costs. Laboratory preparations only require a small amount perhaps a few hundred grams maximally. Several alkanes, e.g., methane, ethane, propane and butane, are obtained by the fractional distillation of petroleum and crude oil.

Alkanes not found in nature can be prepared in the laboratory. Some of these laboratory preparations include:

Reduction of unsaturated hydrocarbons

1,4-Pentadiene can undergo hydrogenation in the presence of a

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metal catalyst to produce pentane.

3,3-Dimethyl-1-butyne can undergo hydrogenation in the presence of a metal catalyst to produce 2,2-dimethylbutane.

2-Pentene can undergo hydrogenation in the presence of palladium or nickel to produce n-pentane

The reduction of unsaturated hydrocarbons generally results in a clean quantitative production of a desired alkane.

Hydrolysis of a Grignard reagent

Alkanes can be produced by the hydrolysis of a Grignard reagent. The Grignard reagent can be obtained by treating an alkyl halide with magnesium under dry conditions in an aprotic solvent. An aprotic solvent is a solvent that does not have a hydrogen atom attached to an electronegative atom. Diethyl ether is an aprotic solvent;

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therefore, diethyl ether would be an appropriate solvent for preparing Grignard reagents. Once the Grignard reagent, RMgX (where R represents an alkyl group), has been prepared, it can be treated with water to produce an alkane. Following are examples of preparations of Grignard reagents followed by treatment with water to obtain an alkane.

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Reacting methylmagnesium bromide with water can be used to quantitatively determine the amount of water in some inert compounds since the stoichiometry of the reaction shows a clear relationship between the number of moles of methane and the number of moles of water present in the sample. The moles of methane formed from the hydrolysis is equivalent to the moles of water present in the sample.

The Grignard reagent is a versatile reagent, because it will react with compounds other than water to produce a variety of organic compounds. The Grignard reagent will be studied in greater detail in Chapter 12.

The Corey-House Reaction

The Corey-House Reaction can be used to synthesize symmetrical and unsymmetrical alkanes. The Corey-House reaction involves the coupling of an alkyl halide with an organometallic compound, a lithium diakylcuprate reagent (the Gilman reagent), to produce an alkane. The reaction occurs between a primary or secondary alkyl halide (works best for a primary alkyl halide) and a lithium dialkylcuprate reagent,

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The overall reaction can be represented by the following equation:

The reaction is used to prepare unsymmetrical alkanes. The preparation of R

2CuLi is from the corresponding alkyl halide.

(1)

R’X + 2 Li → R’Li + LiX

(2)

The lithium dialkylcuprate reagent reacts with a primary alkyl halide to form the symmetrical or unsymmetrical alkane.

R‘2CuLi + 2RX → 2 R’R + R’Cu + LiX

This is not a simple coupling reaction. The alkyl group in the lithium dialkylcuprate reagent can be primary, secondary, or tertiary; however, as indicated, the reagent reacts best with a primary alkyl halide. Nevertheless, the lithium diaklycuprate reagent will work fairly well on an unhindered secondary alkyl halide.

The synthesis of 2-methylpentane can be accomplished using the Corey-House synthesis via the following sequence of reactions.

' '2RX + R CuLi RR + RCu + LiX→

2 R'Li + CuX → R2' CuLi + LiX

C

CH3

CH3

H

I

+ Li2

H

CH3

CH3

C

Li

+ LiI

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The Wurtz Reaction

The Wurtz Reaction, not a very useful or practical reaction, may be used to synthesize symmetrical alkanes.

RX + 2 Na + XR → R-R + 2 NaX

2 CH3CH

2Br + 2 Na → CH

3CH

2CH

2CH

3 + 2 NaBr

2(CH3)

3CHCl + 2 Na → (CH

3)

2CHCH(CH

3)

2 + 2NaCl

CH

CH3

CH3

+LiCu2

H

CH3

CH3

C

Li

+LiI

CuI

2

CH3

CH

CH3

2

LiCu + CH3CH2CH2I

CH3

CH

CH3

CH2CH2CH3 +

CH3

CH

CH3

Cu + LiI

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http://en.wikipedia.org/wiki/Wurtz_reaction gives additional information about the Wurtx reaction.

The Kolbe Synthesis

A more effective synthesis for symmetrical alkanes than the Wurtz reaction is the Kolbe Electrolysis Synthesis for Alkanes. Kolbe published his synthesis during the mid nineteenth century. The method involves the electrolysis of an aqueous solution of the salt of a carboxylic acid, and leads to the synthesis of symmetrical alkanes. Following is a general equation for the Kolbe synthesis of symmetrical alkanes.

The synthesis of n-hexane from butanoic acid is an illustration of the the Kolbe Electrolysis reaction for the synthesis of symmetrical alkanes.

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The mechanism of the reaction involves five steps. Steps (1) and (2) involve the loss of electron from the carboxylate anion to form a free radical. Steps (3) and (4) results in the loss of carbon dioxide. The final step is the combination of the two alkyl radicals to form the symmetrical alkane.

(1)

(2)

(3)

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(4)

(5)

Tertiary radical are more stable than secondary and primary radicals; therefore, the Kolbe Electrolysis Synthesis is effective for the synthesis of branched alkanes like 2,2,3,3-tetramethylbutane.

78

This is the first opportunity to begin a discussion about mechanisms. A mechanism is a series of elementary steps that rationalize the formation of the product or products. The elementary steps are unimolecular or bimolecular. Molecularity is the number of species interacting in a single step. A unimolecular reaction involves only one species, and a bimolecular reaction involves the interaction of two species. For example, steps (1) through (4) are unimolecular steps in the mechanism of the Kolbe Synthesis for n-hexane, and step (5) is a bimolecular step. The sum of all the elementary steps of the mechanism gives the reactants and the products. Following is an explanation of the steps in the mechanism for the Kolbe synthesis of n-hexane from the electrolysis of sodium butanoate.

Steps (1) and (2) are unimolecular.

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This step occurs twice, because two alkyl radicals are required to form the symmetrical alkane.

Steps (3) and (4) are unimolecular.

The half arrow, , represents the movement of one electron. Therefore, an electron moves from the oxygen and an electron moves from a C-C bond to form a C-O double bond.

This step occurs twice to produce two alkyl radicals.

Step 5 is a quenching step that combines the two alkyl radicals to form the symmetrical alkane.

The sum of the steps of the mechanism should give the reactants and the products with their stoichiometric quantities. The mechanism for the Kolbe Electrolysis Synthesis of 2,2,3,3-tetramethylbutane is an example of this.

1.

80

2.

3.

4.

5.

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The sum of 1-5 is

Reactions of Higher Alkanes

Alkanes are inert toward most reagents. The major reactions of alkanes are oxidation and free radical halogenation. However, there are a few other reactions that alkanes will undergo including carbene insertion. A carbene is :CH

2, and the following are examples of two

carbene insertion reactions:

(1)

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(2)

Notice that in both cases, there are two products obtained because there are two types of carbon atoms and two types of hydrogen atoms in the reactant. Therefore, the products of the reactions would probably have different percentages due to statistical factors and the rates of insertions. The fractions x, and y would equal 1.0.

Based on these observations, what would be the product(s), if any, when diazomethane is added to hexane? What would be the product(s), if any, when ketene is added to n-hexane?

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Answers

The products of the reactions would probably have different percentages due to statistical factors and the rates of insertions.

The fractions x, y, and z would equal 1.0.

The following website gives some syntheses of alkanes that we may encounter in later chapters after obtaining an understanding of functional groups. The site also includes some recent syntheses of alkanes.

http://www.organic-chemistry.org/synthesis/C1C/chains/alkanes.shtm

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Halogenation of Alkanes Revisited

Free radical chlorination is an excellent example of the formation of multiple products from a single starting material. The mechanism for the halogenation of higher molecular weight alkanes is the same as the mechanism discussed for methane. However, for alkanes beyond ethane, more than one monohalogenated product is possible. Table 2.10 illustrates this point.

For example, chlorination of propane would produce two products. The fraction of products produced depends on statistical factors and the rate of abstraction of hydrogen atoms. The two products formed are 1-chloropropane and 2-chloropropane. The fraction of 1-Chloropropane produced is 0.44, and the fraction of 2-chloropropane produced is 0.56.

Table 2.10 is a list of the chlorination products of selected alkanes with their respective percentages of isomers. Apparently, no isomer is dominant In the chlorination of alkanes.

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Table 2.10

In the case of bromination, one isomer clearly dominates. The chlorine radical is more reactive than the bromine radical; therefore, it is less selective about the type of hydrogen it abstracts. The isomer ratio is statistically related to the ratio of the types of

86

hydrogen atoms available in the alkane. For example, n-butane has four secondary hydrogen atoms and six primary hydrogen atoms. Consequently, from a statistical perspective, 60% of the chlorinated alkane should be n-butylchloride and 40% should be sec-butyl chloride. The actual percentages are 28% n-butyl chloride and 72% sec-butyl chloride. This observation suggests that the reactive chlorine radical prefers the abstraction of secondary hydrogen atoms over primary hydrogen atoms. The abstraction of secondary hydrogen atom of butane would result in the formation of a secondary radical, and the abstraction of a primary hydrogen atom would result in the formation of a primary radical. The secondary hydrogen atoms are more easily abstracted than the primary hydrogen atoms; therefore, the secondary radical will be formed at a faster rate resulting in the dominance of the secondary alkyl chloride. The relative rates in which a chlorine radical will abstract various classes of hydrogen atoms at room temperature has been studied, and found to follow the following observations:

A chlorine radical will abstract a secondary hydrogen atom approximately 3.80 times faster than a primary hydrogen atom.

Tertiary hydrogen atoms are abstracted approximately 5 times faster than a primary hydrogen.

The ratio of abstraction is:

These data can be used to predict the percentage of isomers

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obtained from the chlorination of alkanes.

Figure 2.4 highlights the hydrogen atoms that can be replaced to produce the monochlorinated isomers possible for 2-methylbutane.

Figure 2.4 hydrogen atoms that can be replace to form the monochlorinated isomers from 2-methylbutane

Abstraction of one of the six equivalent primary hydrogen atoms would result in the production of 1-chloro-2-methylbutane (I).

Abstraction of the tertiary hydrogen would result in the production of 2-chloro-2-methylbutane (II):

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Abstraction of one of the two equivalent secondary hydrogen atoms would result in the production of 2-chloro-3-methylbutane (III):

Abstraction of one of the three equivalent primary hydrogen atoms would result in the production of 1-chloro-3-methylbutane (IV):

Table 2.11 gives the results for predicting the percentages for the monochlorinated structural isomers I, II, III and IV. This result is based on the reactivity ratio for the formation of a primary radical versus a secondary radical versus tertiary alkyl free radicals and the statistical data for the abstraction of a primary, secondary and tertiary hydrogen in 2-methylbutane.

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Monochlorinated isomer

Statistical Factor x

Reactivity Factor

Percent of Isomer

I 6 x 1.00 = 6.00

II 1 x 5.00 = 5.00

III 2 x 3.80= 7.60

IV 3 x 1.00 = 3.00

Table is incomplete

Table 2.11 the percentage of monoclorinated isomers from the chlorination of 2-methylbutane

The bromine radical is not as reactive as the chlorine radical; consequently, bromination of alkanes is more selective. Where possible, bromination of an alkane will result in a much larger percentage of the tertiary isomer than chlorination of the same alkane. If the alkane contains only primary and secondary hydrogen atoms, abstraction of the secondary hydrogen proceeds at a faster rate leading almost exclusively to the secondary alkyl bromide.

The Ease of Abstraction of Alkane Hydrogen Atoms by Halogen Radicals

Alkane primary hydrogen atoms are the more difficult to abstract by halogen radicals than secondary or tertiary hydrogen atoms. Secondary hydrogen atoms are more difficult to abstract than tertiary hydrogen atoms. The ease of abstraction is related to the stability of the resulting alkyl radical. A tertiary hydrogen atom is

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relatively easy to abstract because the resulting tertiary radical is stable. The methyl radical is relatively unstable; therefore, the abstraction of hydrogen from methane is difficult. The stability of a radical is related to its dissociation energy. The higher the dissociation energy, the more difficult will be the abstraction process. Table 2.12 lists the dissociation energy for several compounds.

Alkanes Formation of Radical Energy, kJ mol-1

methane ·CH3 427 ethane ·CH2CH3 406

propane CH3CH2CH2·

406

propane 393

2-methylpropane

381

Table is incomplete

Table 2.12 dissociation energies for the formation of selected alkyl radicals

The values of the dissociation energies in Table 2.12 clearly show that primary radicals are more difficult to form than secondary or tertiary radicals, and that secondary radicals are more difficult to form than tertiary radicals.

The Stability of Alkyl Radicals

The stability of alkyl radicals can be explained by hyperconjugation. Hyperconjugation can best be explained by assuming that the unpaired electron delocalizes through orbital overlap. The orbital

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overlap is between the unpaired electron in the 2p atomic orbital and the alkane’s sigma, σ, bonds. The following is an illustration of hyperconjugation of an alkyl radical:

The “etc” means that the other two hydrogen atoms, in an analogous manner, can participate in the hyperconjugation stabilization process. The unpaired electron is delocalizes throughout the alkyl radical.

This process could also be written in the following manner:

Therefore, delocalization of the electron results in the C-C single bond acquiring some double bond character, and the C-H sigma bond losing some of its sigma bond character. The greater the

92

extent of delocalization, as ascertained by the number of possible hyperconjugated structures, the more stable the intermediate radical. Consequently, an unpaired electron residing on a carbon atom with more alkyl groups attached would be more stable than an unpaired electron on a carbon with fewer alkyl groups attached. This means that an unpaired electron on a tertiary carbon would have greater stability than an unpaired electron on a secondary carbon atom, and an unpaired electron on a secondary carbon would have greater stability than an unpaired electron on a primary carbon atom.

Following is the representation of hyperconjugation in the isopropyl radical.

The tert-butyl radical is more stable than the isopropyl radical. Why? Try drawing the hyperconjugation structures for the tert-butyl radical.

Answer

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There are more hyperconjugation structures for the tert-butyl radical than for the isopropyl radical; therefore, the tert-butyl radical is more stable than the isopropyl radical.

Hyperconjugation at one time was a controversial subject; however, electron spin resonance spectroscopy studies suggest that hyperconjuagation contribute to the overall stability of radicals.

After studying Chapters 7 and 8, you may want to read http://books.google.com/books?id=RlY7AAAAIAAJ&pg=PA37&lpg=PA37&dq=alkyl+radicals+and+esr&source=bl&ots=Y3oSyB-

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vp4&sig=jYUiCCoXOPK0x-76VGPcxrLIajc&hl=en&ei=r-U_StqfLpD-sgOB0pn5Cg&sa=X&oi=book_result&ct=result&resnum=7 which discusses the stereochemistry of free radicals.

Hyperconjugation offers a rationale that connects the stability of alkyl radicals with the delocalization of electrons. This type of delocalization argument is not unique to free radicals, and will be used in later discussions to rationalize the stability of other intermediates.

Cyclopropane

Cycloalkanes will be discussed in greater detail in Chapter 8; however, in this Chapter, the syntheses of the small ring structure of cyclopropane will be considered.

Cyclopropane is an unusual molecule, because it exhibits a considerable amount of angle strain. The bond angle for sp3

hybridized carbon atoms is ideally 109.5o , but the three C-C-C bond

angles in cyclopropane are approximately 60o . Therefore, cyclopropane exhibits 49.5o bond angle strain ( 109.5o - 60.0o ) for each C-C-C bond angle. Nevertheless, 1,3-dihalopropane can be synthesized via the following reaction scheme.

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cyclopropane

Also, cyclopropane can be synthesized by methylene insertion reactions with diazomethane or ketene.

96

Carbene generation can produce a singlet state carbene or a triplet state carbene. The singlet state carbene has an unshared pair of electrons, i.e., electrons with opposite spins.

The C-H bond length is 1.12Ǻ

The pairing of the electrons can be

Therefore, the term singlet is applied to this carbene. The singlet carbene can be synthesized from diazomethane.

The triplet state carbene has unshared electrons which are not paired, i.e., electrons with or without opposite spins.

97

Therefore, the possible spin arrangements for the two electrons could be:

1:2:1, hence the term triplet is applied to this carbene system

In the triplet state, the C-H bond length is 1.03 Ǻ.

Photochemical reaction of ketene produces carbon monoxide and the triplet state of carbene.

The singlet carbenes add to alkenes stereospecifically (i.e., produces a particular spatial arrangement of the product) to form cyclopropanes. For example, the singlet carbene would insert into a cis alkene via an activated complete that would specifically lead to the cis isomer.

98

cis-alkene

cis-cyclopropane

Carbenes in the triplet state add to alkenes in a nonstereospecific manner to form cyclopropane.

(1)

The dirdical intermediate can rotate in such a manner that both the cis and the trans isomers can be formed.

99

(2)

As previously stated, diazomethane generates a singlet carbene, and ketene generates a triplet carbene.

Also, carbenes can be generated by treating chloroform with a strong base.

(CH3)

3CO- K+ + H:CCl

3 → K+ - :CCl

3

K+ - :CCl3 → KCl + :CCl

2

The Simmons-Smith reaction uses an organozinc carbenoid that can instantaneously inserted into the double bond. The carbenoid inserts in a manner that results in the formation of an activated complex at the transition state that leads to a stereospecific product. The Simmons-Smith reaction will be revisited in Chapter 3.

(1)

100

(2)

(3)

cis-cyclopropane

101

Important Organic Chemistry Functional Organic Function Groups

As organic chemistry discussions evolve, students may encounter functional groups before they are actually introduced. Table 2.13 lists a number of functional groups encountered in an elementary organic chemistry course. These functional groups should be familiarized since they are frequently encountered.

Table 2.13 Common Functional Groups in Organic Chemistry

102

Continuation of Table 2.13, Common Functional Groups in Organic Chemistry

103

Continuation of Table 2.13, Common Functional Groups in Organic Chemistry

104

Problems

Alkanes

1. Name the following compounds using the IUPAC system of nomenclature:

(a)

(b)

CH3 C

CH2CH3

CH2CH2 CH

H CH3

CH3

CH3 C CH2CH2 CHCH2

H CH3

CH3

CH

CH3 CH3

105

(c)

(d)

(e)

(f)

CH2CH3CH

CH3

CH2

CH3

CH3

CH3CH2CH

CH2CH2CH3

CH2CH3

CH3

CH3

CH2CH2CH3

CHCH2CH

CH2CH3

CH3

CH3

CH3CHCH2CCH3

106

(g)

(h)

(i)

2. Write a balanced chemical equation for the following reactions:

(a)

CH2CH3

CH3

CH3

CHCH

CH3CH2

CH3

CH3

CHCH

CH3(CH2)4CH2

CH3CH2

CH2CH3

CH3 CH3

CH3(CH2)4CH2

CH3CH2

C CH2CH

n (2n + 2) 2C H + O →

107

(b)

(c)

(d)

(e)

h3 2 3 2CH CH CH + Br ν⎯⎯→

CH3CH2C

Br

CH3

H + Mg(1) dry ether

H2O(2)

CH3C C CH2CH3 + H22Pt

CH3CH2C

Br

CH3 +

H

Na

108

(f)

(g)

(h)

3. Suggest mechanisms for the following conversions

(a)

(b) 3 CH3CH

2Br + 4 Li + CuBr → CH

3CH

2CH

2CH

3 + CH

3CH

2Cu +

4 LiBr

(c)

1. Calculate ∆H for

CH3CH2CH3 + CH2N2hν

hνCH3

CH3

CH3

CH3C + C C OH

H

CH3CH2CH2Br + [(CH3)2CH]2CuLi

h4 2 3CH + Cl CH Cl + HClν⎯⎯→

3 2 3 2 2 32 CH CH Br + 2 Na CH CH CH CH + 2NaBr→

109

If the bond dissociation energies in kJ/mol are

bond bond dissociation energy, kJ/mol

C-H 410

Br-Br 192

C-Br 280

H-Br 364

1. Sketch a potential energy versus reaction coordinates digram for the following reaction:

On the graph, indicate the:

(a) Intermediates

(b) enthalpy of reaction

(c) energy of activation

(d) transition states

(a) the activated complex

Use the following data to approximate the potential energy diagram.

3 3 2 3 2CH CH + Br CH CH Br + HBr→

CH3CH2CH3 + Br2hν

(CH3)2CHBr + HBr

110

bond bond dissociation energy, kJ/mol

C-H 413

Br-Br 192

C-Br 280

H-Br 364

6. Benzo[a]pyrene is a carcinogenic compound which has been detected in grilled steaks. Combustion analysis of 3.22 mg of benzo[a]pyrene produced 11.25 mg of CO

2 and 1.38 mg of H

2O.

(a) Calculate the empirical formula of benzo[a]pyrene.

(b) In an experiment, 2.523 g of benzo[a]pyrene lowers the melting point of 50.0 grams of camphor to 170.6oC. Calculate the molecular mass of the carcinogen.

7. Draw the Newman projection formulas for CH3CHBrCH

2CH

3 and

sketch the potential energy versus rotation diagram for a 360o

rotation of 2-bromobutane along C2 and C3 carbon atoms.

8. Suggest syntheses for the following compounds from organic compounds containing no more than four or fewer carbon atoms and any other necessary inorganic reagents.

(a) (CH3)

2CHCH

3

111

(c) CH3CH

2CH

2CH

2CH

2CH

2CH

2CH

3

(d) CH3CHDCH

3 where D is deuterium, an isotope of H

(e) CH3CH

2CH

2CH

2T where T is tritium, an isotope of H

9. Calculate the percentage of monochlorinated products anticipated in the monochlorination of 2,4-dimethylpentane.

10. The freezing point of 3.5 grams of an unknown organic compound in 100 mL of benzene is 5.0oC. Calculate the molecular weight of the unknown organic compound.