Basic Organic ChemistryCourse code : CHEM 12162

(Pre-requisites : CHEM 11122)

Chapter – 02Chapter – 02

Chemistry of Alkanes

Dr. Dinesh R. Pandithavidana

Office: B1 222/3

Phone: (+94)777-745-720 (Mobile)

Email: dinesh@kln.ac.lk

Introduction:

• Recall that alkanes are aliphatic hydrocarbons having C—C

and C—H σ bonds. They can be categorized as acyclic or

cyclic.

• Acyclic alkanes have the molecular formula CnH2n+2 (where n

= an integer) and contain only linear and branched chains of

carbon atoms. They are also called saturated hydrocarbons

Alkanes

carbon atoms. They are also called saturated hydrocarbons

because they have the maximum number of hydrogen atoms

per carbon.

• Cycloalkanes contain carbons joined in one or more rings.

Because their general formula is CnH2n, they have two fewer

H atoms than an acyclic alkane with the same number of

carbons.

Introduction:

• All C atoms in an alkane are surrounded by four groups,

making them sp3 hybridized and tetrahedral, and all bond

angles are 109.5°.

• The 3-D representations and ball-and-stick models for these

Alkanes

• The 3-D representations and ball-and-stick models for these

alkanes indicate the tetrahedral geometry around each C

atom. In contrast, the Lewis structures or structural formulas

are not meant to imply any 3-D arrangement.

• Additionally, in propane and higher molecular weight

alkanes, the carbon skeleton can be drawn in a variety of

ways and still represent the same molecule.

• The three-carbon alkane CH3CH2CH3, called propane, has a

Alkanes

• The three-carbon alkane CH3CH2CH3, called propane, has a

molecular formula C3H8. Note in the 3-D drawing that each

C atom has two bonds in the plane (solid lines), one bond in

front (on a wedge) and one bond behind the plane (on a

dashed line).

Introduction:

Alkanes

Alkanes

Number Molecular Name

of C atoms formula (n-alkane) .

11 C11H24 undecane

12 C12H26 dodecane

Names of straight chain alkanes:

12 26

13 C13H28 tridecane

14 C14H30 tetradecane

15 C15H32 pentadecane

16 C16H34 hexadecane

17 C17H36 heptadecane

18 C18H38 octadecane

19 C19H40 nonadecane

Cycloalkanes:

Cycloalkanes have molecular formula CnH2n and contain

carbon atoms arranged in a ring. Simple cycloalkanes are

named by adding the prefix cyclo- to the name of the acyclic

alkane having the same number of carbons.

Alkanes

Nomenclature:

The name of every organic molecule has 3 parts:

1. The parent name indicates the number of carbons in the

longest continuous chain.

2. The suffix indicates what functional group is present.

3. The prefix tells us the identity, location, and number of

Alkanes

3. The prefix tells us the identity, location, and number of

substituents attached to the carbon chain.

• The carbon substituents bonded to a long carbon chain are

called alkyl groups.

• An alkyl group is formed by removing one H atom from an

alkane.

• To name an alkyl group, change the –ane ending of the

Alkanes

Nomenclature:

• To name an alkyl group, change the –ane ending of the

parent alkane to –yl. Thus, methane (CH4) becomes

methyl (CH3-) and ethane (CH3CH3) becomes ethyl

(CH3CH2-).

Naming three- or four-carbon alkyl groups is more

complicated because the substituent can have isomeric

forms. For example, propane has both 1° and 2° H atoms,

and removal of each of these H atoms forms a different

alkyl group with a different name, propyl or isopropyl.

Alkanes

Nomenclature:

10

1. Find the parent carbon chain and add the suffix.

Alkanes

Rule 1:

The parent is the longest continuous chain of carbon atoms.

Note that if there are two chains of equal length, pick the chain

with more substituents to simplify naming. In the following

example, two different chains in the same alkane have seven C

atoms. We circle the longest continuous chain as shown in the

diagram on the left, since this results in the greater number of

substituents.

Alkanes

2. Number the atoms in the carbon chain from the end

that gives the first substituent the lowest number.

Alkanes

Rule 2:

If the first substituent is the same distance from both ends,

number the chain to give the second substituent the lower

number.

Alkanes

When numbering a carbon chain results in the same numbers

from either end of the chain, assign the lower number

alphabetically to the first substituent.

Alkanes

3. Name and number the substituents.

• Name the substituents as alkyl groups.

• Every carbon is part of the longest chain or a part of a

substituent, not both.

• Each substituent needs its own number.

• If two or more identical substituents are bonded to the longest

AlkanesRule 3:

16

• If two or more identical substituents are bonded to the longest

chain, use prefixes to indicate how many: di- for two groups,

tri- for three groups, tetra- for four groups, etc.

4. Finally, write the name.• Substituent names and numbers come before the parent.

• List substituents in alphabetical order, ignoring all prefixes except iso, as in

isopropyl and isobutyl.

• Separate numbers by commas and separate numbers from letters by

hyphens. The name of an alkane is a single word, with no spaces after

hyphens and commas.

Alkanes

17

Cycloalkanes are named by using similar rules, but the prefix

cyclo- immediately precedes the name of the parent.

CycloalkanesCycloalkane Nomenclature:

1. Find the parent cycloalkane.

2. Name and number the substituents. No number is

needed to indicate the location of a single substituent.

For rings with more than one substituent, begin numbering at

one substituent and proceed around the ring to give the second

Cycloalkanes

19

one substituent and proceed around the ring to give the second

substituent the lowest number.

With two different substituents, assign numbers to the

substituents alphabetical order.

Cycloalkanes

In the case of an alkane having a long chain substituent, if the number of

carbons in the ring is greater than or equal to the number of carbons in the

longest chain, the compound is named as a cycloalkane.

Naming compounds containing both a ring and a long

chain of carbon atoms

Examples of cycloalkane nomenclature

Cis-Trans Isomerism

• Cis: like groups on same side of ring

• Trans: like groups on opposite sides of ring

Major Uses of Alkanes

• C1-C2: gases (natural gas)

• C3-C4: liquified petroleum (LPG)

• C5-C8: gasoline

• C9-C16: diesel, kerosene, jet fuel

• C17-up: lubricating oils, heating oil

• Origin: petroleum refining

Synthesis of Alkanes

1. Reduction of an alkyl halide

a) hydrolysis of a Grignard reagent

b) with an active metal and an acid

2. Corey-House synthesis

(coupling of an alkyl halide with lithium

dialkylcopper)

1.Reduction of an alkyl halide

a) hydrolysis of a Grignard reagent (two steps)

i) R—X + Mg � RMgX (Grignard reagent)

ii) RMgX + H2O � RH + Mg(OH)X

CH3CH2CH2-Br + Mg � CH3CH2CH2-MgBr

n-propyl bromide n-propyl magnesium bromide

CH3CH2CH2-MgBr + H2O � CH3CH2CH3 + Mg(OH)Br

propane

CH3 CH3

CH3CH-Br + Mg � CH3CH-MgBrisopropyl bromide isopropyl magnesium bromide

CH3

CH3CH-MgBr + H2O � CH3CH2CH3

propane

CH3CH2CH2-MgBr + D2O � CH3CH2CH2D

heavy water

CH3 CH3

CH3CH-MgBr + D2O � CH3CHD

b) with an active metal and an acid

R—X + metal/acid � RH

active metals = Sn, Zn, Fe, etc.

acid = HCl, etc. (H+)

CH CH CHCH + Sn/HCl � CH CH CH CH + ClCH3CH2CHCH3 + Sn/HCl � CH3CH2CH2CH3 + Cl2Cl

sec-butyl chloride n-butane

CH3 CH3

CH3CCH3 + Zn/H+ � CH3CHCH3 + ZnBr2

Brtert-butyl bromide isobutane

2. Corey-House synthesis

R-X + Li � R-Li + CuI � R2CuLi

R2CuLi + R´-X � R—R´ (alkane)

(R´-X should be 1o or methyl)

This synthesis is important because it affords a synthesis

of a larger alkane from two smaller alkyl halides.

note: the previous equations are not balanced:

R-X + 2 Li � R-Li + LiX

2 R-Li + CuI � R2CuLi + LiX

R

R2CuLi = R-Cu-, Li+

R2CuLi + R´X � R-R´ + RCu + LiX

CH3 CH3 CH3

CH3CH-Br + Li � CH3CH-Li + CuI � (CH3CH)2-CuLiisopropyl bromide

CH3 CH3

(CH3CH)2-CuLi + CH3CH2CH2-Br � CH3CH-CH2CH2CH3(CH3CH)2-CuLi + CH3CH2CH2-Br � CH3CH-CH2CH2CH3

2-methylpentane

(isohexane)

Note: the R´X should be a 1o or methyl halide for the best yields

of the final product.

Reactions of Alkanes

1. Combustion (oxidation)

2. Pyrolysis (cracking)

3. Halogenation

1. Combustion

CnH2n+2 + O2, flame � n CO2 + (n+1) H2O + heat

gasoline, diesel, heating oilG

2. Pyrolyis (cracking)2. Pyrolyis (cracking)

alkane, 400-600oC � smaller alkanes + alkenes + H2

Used to increase the yield of gasoline from petroleum. Higher

boiling fractions are “cracked” into lower boiling fractions that

are added to the raw gasoline. The alkenes can be

separated and used in to make plastics.

3. Halogenation

R-H + X2, heat or hv � R-X + HX

a) heat or light required for reaction.a) heat or light required for reaction.

b) yields mixtures �

c) H: 3o > 2o > 1o > CH4

d) bromine is more selective

CH3CH3 + Cl2, hv � CH3CH2-Cl + HClethane ethyl chloride

CH3CH2CH3 + Cl2, hv � CH3CH2CH2-Cl + CH3CHCH33 2 3 2 3 2 2 3 3

propane n-propyl chloride Cl

isopropyl chloride45%

55%

gives a mixture of both the possible

alkyl halides! �

CH3CH2CH2CH3 + Cl2, hv � CH3CH2CH2CH2-Cl

28%n-butane n-butyl chloride

+

CH3CH2CHCH3 72%

Cl

sec-butyl chloride

CH3 CH3

CH3CHCH3 + Cl2, hv � CH3CHCH2-Cl 64%

isobutane isobutyl chloride

+

CH3

CH3CCH3 36%

Cltert-butyl chloride

CH3CH3 + Br2, hv � CH3CH2-Br + HBrethane ethyl bromide

CH3CH2CH3 + Br2, hv � CH3CH2CH2-Br + CH3CHCH33 2 3 2 3 2 2 3 3

propane n-propyl bromide Br

isopropyl bromide3%

97%

In the reaction of alkanes with halogens, bromine is less

reactive but more selective. Why? How?

CH3CH2CH2CH3 + Br2, hv � CH3CH2CH2CH2-Br

2%n-butane n-butyl bromide

+

CH3CH2CHCH3 98%

Br

sec-butyl bromide

CH3 CH3

CH3CHCH3 + Br2, hv � CH3CHCH2-Br <1%

isobutane isobutyl bromide

+

CH3

CH3CCH3 99%

Brtert-butyl bromide

Mechanism of Halogenation

1) Initiating step:

X—X � 2 X•

2) Propagating steps:

X• + R—H � H—X + R•

R• + X—X � R—X + X•R• + X—X � R—X + X•

3) Terminating steps:

2 X• � X—X

R• + X• � R—X

2 R• � R—R

Mechanism of Chlorination of Propane

1) Cl—Cl � 2 Cl•

2) abstraction of 1o hydrogen:

Cl• + CH3CH2CH3 � CH3CH2CH2• + HCl

or abstraction of 2o hydrogen:

Cl• + CH3CH2CH3 � CH3CHCH3 + HCl Cl• + CH3CH2CH3 � CH3CHCH3 + HCl •

3) CH3CH2CH2• + Cl2 � CH3CH2CH2Cl + Cl•

or CH3CHCH3 + Cl2 � CH3CHCH3 + Cl•• Cl

plus terminating steps

Relative Reactivity in Halogenation

Stability of free radicals:

Ease of formation of free radicals:

Ease of abstraction of H’s:Ease of abstraction of H’s:

3o > 2o > 1o > CH4

Factors Affected on Hydrogen Abstraction

Cl• + CH3CH2CH3 � CH3CH2CH2• + HCl

or abstraction of 2o hydrogen:

Cl• + CH3CH2CH3 � CH3CHCH3 + HCl •

The chloride that is produced depends on which hydrogen

is abstracted by the chlorine free radical in step 2. The n-

propyl free radical gives the n-propyl chloride while the

isopropyl free radical yields the isopropyl chloride.

The relative reactivity in chlorination:

H: 3o : 2o : 1o = 5.0 : 3.8 : 1.0

The number of hydrogens (probability factor) may also be

important.

CH3CH2CH2CH3 + Cl2, hv � CH3CH2CH2CH2-Cl

n-butane

+ CH3CH2CHCH3

Cl

n-butyl chloride = (# of 1o hydrogens) x (reactivity of 1o)n-butyl chloride = (# of 1o hydrogens) x (reactivity of 1o)

= 6 x 1.0 = 6.0

sec-butyl chloride = (# of 2o hydrogens) x (reactivity of 2o)

= 4 x 3.8 = 15.2

% n-butyl chloride = 6.0/(6.0 + 15.2) x 100% = 28%

% sec-butyl chloride = 15.2/(6.0 + 15.2) x 100% = 72%

CH3 CH3 CH3

CH3CHCH3 + Cl2, hv � CH3CHCH2-Cl + CH3CCH3

isobutane Cl

isobutyl chloride = (# of 1o H’s) x (reactivity of 1o)

= 9 x 1.0 = 9.0

tert-butyl chloride = (# of 3o H’s) x (reactivity of 3o)

= 1 x 5.0 = 5.0= 1 x 5.0 = 5.0

% isobutyl = (9.0/(9.0 + 5.0)) x 100% = 64%

In this case the probability factor outweighs the difference in

relative reactivity of 1o and 3o hydrogens.

Relative Reactivity in Bromination

3o : 2o : 1o = 1600 : 82 : 1

In bromination the relative reactivity differences are much

greater than any probability differences.

isobutane + Br2, hv � isobutyl bromide + tert-butyl isobutane + Br2, hv � isobutyl bromide + tert-butyl

bromide

isobutyl bromide = 9 H x 1 = 9

tert-butyl bromide = 1 H x 1600 = 1600

% tert-butyl bromide = (1600/1601) x 100% = >99%