1

Aliphatic Hydrocarbons

Jully TanSchool of Engineering

EP101 / EG101 2

Learning OutcomeAt the end of the lecture, students will be able to:

Understand the family under organic compounds.Explain the aliphatic hydrocarbon or alkanes specifically by its structure, physical and chemical properties.Naming the organic compound especially for alkanes according to IUPAC method.Explain the type of isomer in organic compound and draw the different isomer structures of the compound.Explain the different type of the synthesis of alkanes and reaction of alkanesspecifically:

HalogenationCombustionHydrogenation of alkenesReaction of Alkyl Halides with organometalsReduction of Alkyl Halides

2

EP101 / EG101 3

Classes of Organic CompoundsOrganic compound

Hydrocarbon

AlkaneAlkeneAlkyne

Hydrocarbon with O

Alcohol (-OH)

Ether/epoxideAldehyde/

Ketone

Carboxylic acid

Carboxylic Acid derivatives

Ester Halide acids anydride

Halogeneted HC HC with N

Amine Amide Nitrile

EP101 / EG101 4

Overview

HydrocarbonComposed entirely of C and H atoms.Has 3 main categories

Aliphatic - straight chain HCCyclic - ring shape HCAromatic – HC derived from benzene ring.

Each C can form a maximum of four single bonds, OR two single and one double bond, ORone single and triple bond.

3

EP101 / EG101 5

Some five-carbon skeletons

CC C C C

CC C C

C

CC C

C

C

C C C CC

CC C C C

CC C C

C

CC C C

C

CC C

C C

C

C C

CC

C C

C

C C

C

C

C

C C

C C

C

C C

CC C C

C

single bonds

double bond

ring

EP101 / EG101 6

Adding the H-atom skin to the C-atom skeleton

A C atom single-bonded to two other atoms gets two H atoms.

C C C

H

A C atom single-bonded to one other atom gets three H atoms.

C C

H

A C atom single-bonded to three other atoms gets one H atom.

C C C

C

H

H

H H

A C atom single-bonded to four other atom is already fully bonded (no H atoms).

C C C

C

C H

H

A double- and single-bonded C atom or a triple-bonded C atom is treated as if it were bonded to three other atoms.

CC CC C

H

HA double-bonded C atom is treated as if it were bonded to two other atoms.

CC

4

EP101 / EG101 7

EP101 / EG101 8

5

EP101 / EG101 9

Alkanes, CnH2n+2

CH

H

HH

carbon–carbon single bond: s (head–on) overlap of carbon sp3 orbitalssaturated hydrocarbons; only C-C and C-H bondsAlkane: a saturated hydrocarbon whose carbons are arranged in a chain. Saturated hydrocarbon: a hydrocarbon containing only single bonds.Aliphatic hydrocarbon: another name for an alkane: aliphatic, “fat”

EP101 / EG101 10

1. Molecular Formula of Alkanes

10Decane

9Nonane

8Octane

7Heptane

6Hexane

5Pentane

4Butane

3Propane

2Ethane

1Methane

Molecular FormulaNo of CName

6

EP101 / EG101 11

StructureShape

tetrahedral about carbonall bond angles are approximately 109.5°

EP101 / EG101 12

Isomer: same chemical formula, but different structure.There are 2 isomeric butanes

n-butaneC4H10

n means “normal” or in a straight chain

isobutaneC4H10

CCC

C

HH

HHH

H

H

H

H H

CCCC H

H

H

H

H

H

H

H

H

H

Branching vs. No Branching

7

EP101 / EG101 13

Draw the structures for C6H14 starting with the longest chain (6 in a row)

…next draw all structures with 5 carbons in the longest chain (substitute for the H s)

…then substitute for other H s

CCCCCC H

H

H

H

H

H

H

H

H

H

H

H

H

H

C C C C C

C

HH

H H

H

H

H

HH

H

H

HHH

Find the Isomers of C6H14

EP101 / EG101 17

Classification of C & H

Primary (1°) C: a carbon bonded to one other carbon

1° H: a hydrogen bonded to a 1° carbon

Secondary (2°): a carbon bonded to two other carbons

2°H: a hydrogen bonded to a 2° carbon

Tertiary (3°) C: a carbon bonded to three other carbons

3° H: a hydrogen bonded to a 3° carbon

Quaternary (4°) C: a carbon bonded to four other carbons

8

EP101 / EG101 18

C C

PRIMARY

C CC

C CC

C C CC

C

C

SECONDARY

TERTIARY QUATERNARY

CONNECTED TO ONEOTHER CARBON

CONNECTED TO TWOOTHER CARBONS

CONNECTED TO THREEOTHER CARBONS

CONNECTED TO FOUROTHER CARBONS

Degree of Substitution

EP101 / EG101 19

2. Chemical PropertiesNot reactive compound:

Alkanes are the least reactive among all other organic comp. They do not usually react with strong acids or bases, or with most oxidizing or reducing agents.

Highly combustible compoundAlkanes can burn easily in combustion reactions and releasing high energy.

CH4 O2 CO2 OH2+ +

9

EP101 / EG101 20

a. Both bp and mp increase with increasing carbon number for straight-chain alkanes with formula CnH2n+2

b. Branching tends to raise the melting point and lower the boiling point. Why??

3. Physical Properties, Density & SolubilityAlkanes physical forms as gas, liquid and solid.

C1-C4 -GasC5-C17 -LiquidC18 and above -Solid

Only dissolves in organic solvent, such as benzene, CCl4, but do not dissolve in water (hydrophobic). Alkane is less density than water. (near 0.7 g/mL)

Why do alkane do not dissolve in

water?

4. Boiling Point

Hint: surface area of alkane.

EP101 / EG101 21

b. Branching tends to lower the boiling point and raise the melting point

BP 60oC 58oC 50oC

MP -154oC -135oC -98oC

Explanation:

MP Branching reduces the flexibility of the molecule which reduces the entropy term ∆S in the equation Tmp = ∆H/∆S. Since ∆S is in the denominator, Tmp increases. The structure of the alkane packed better into a compact 3D structure which is said the atom are bonded more tightly and need higher T to transform the alkane from solid to liquid phase.

BP Branching reduces surface area (more compact structure), and therefore Van der Waals dispersion forces which control boiling point for these molecules been reduces. Less energy to overcome these reactions to transform liquid to gas phase.

Answer: Solubility – alkanes are nonpolar molecules and therefore insoluble in water, whichis polar. Alkanes are hydrophobic.

10

EP101 / EG101 22

Boiling points of the first 10 unbranchedalkanes

EP101 / EG101 23

Boiling Points of Alkanes

11

EP101 / EG101 24

EP101 / EG101 25

Sources2 main sources are petroleum & natural gas

12

EP101 / EG101 26

International

Union of

Pure and

Applied

Chemistry

“eye-you-pac”colloquially:

Nomenclature

International Union of Pure and Applied Chemistry (IUPAC)

EP101 / EG101 27

13

EP101 / EG101 28

1. Numerical Roots for Carbon Chains and Branches

Number of C atomsRoots/Parent

1

2

3

4

5

6

8

7

9

10

meth-

eth-

prop-

but-

hex-

pent-

hept-

oct-

non-

dec-

PREFIX + ROOT (Parent) + SUFFIX

EP101 / EG101 29

STEP 1 : Find the parent hydrocarbon1. Find longest continuous chain of carbons and use as parent name.

2. If 2 chains have same number of carbons, choose the one with the more branch points.

STEP 2 : Number the atoms in parent chain1. Begin at the end nearer to the first branch point.

Which one is the correct way of longest chain?

Which one is the correct way of numbering the

atom of the chain?

14

EP101 / EG101 30

2. If branching occurs at equal distance from both ends, begin numbering at the end nearer the second branch point.

Which one is the correct way of numbering the

atom of the chain?

EP101 / EG101 31

2. Naming the Compound Types of Suffix

Suffix is the type of the family of that specifics organic compound.For example, alkanes, alkene and etc.

PREFIX + ROOT (Parent) + SUFFIX

15

EP101 / EG101 32

3. Identify Prefix of the Chain

CH 3

CH 3

CH 3

-CH 2 CH 2 CH 2 CH 2 CH 3

-CH 2 CH 2 CHCH 3

-CH 2 CCH 3neopentyl

isopentyl

pentyl

EP101 / EG101 33

STEP 1 : Identify and number the substituents1. Give each substituent a number that corresponds to its position on the parent chain.

2. Two substituents on the same carbon gets the same number

SubstituentsC3 CH2CH3 3-ethylC4 CH3 4-methylC6 CH2CH3 6-ethyl

Substituents :C2 CH3 2-methylC4 CH3 4-methylC4 CH2CH3 4-ethyl

16

EP101 / EG101 34

STEP 2: Write the name as a single word (follow alphabetical order)1. Use hyphens to separate prefixes and commas to separate numbers or more identical substituents are present, use multipliers (di, tri, tetra)

Note : the multipliers are not use for alphabetizing.

Name a complex substituent as if it were a compound and put them in parentheses

EP101 / EG101 35

Naming & Drawing Cyclic Alkanes

Name of branching group or substitute (if any)

Parent name

+ +Use Prefix “cyclo”

Parent Name = longest continuous chain of carbon atomsCyclo = parent chain forms a ring

STEP 1: Find parentIf number of carbon atoms in the ring is larger than the number in the largest substituent, the compound is an alkyl-substituted cycloalkane.If the number of carbon atoms in the ring is smaller than the number in the largest substituent, the compound is an cycloalkyl-substituted alkane.

17

EP101 / EG101 36

STEP 2: Number the substituents and write the name1. Start at a point of attachment and number the substituents so that the second substituent has

the lowest possible number.2. If necessary, proceed to the next substituent until a point of difference is found.

3. If 2 or more substituents might potentially receive the same number, number them alphabetical priority. Treat halogens as if they are alkyl groups.

Which one is the correct way of

numbering the atom of the chain?

Which one is the correct way of

numbering the atom of the chain?

EP101 / EG101 37

H2C CH2

CH2

1. What is the parent name?Propane (Longest chain is 3 carbons)

2. Is the parent chain in a ring? YesCorrect name is: cyclopropane

18

EP101 / EG101 38

H2C CH

CH2

CH3

Parent name is cyclopropane.

What is the branching group? Methyl –CH3

Correct name is: methylcyclopropaneNote: The number (1) is not needed to identify the place of the substituent if only one is present in the molecule.

EP101 / EG101 39

H2C CH

H2C

CH3

CH CH3

Parent name is cyclobutane.What is the branching group? Methyl –CH3

Correct name is: 1, 2-dimethylcyclobutaneNote: The first named substituent is arbitrarily given the number one (1) position

1

23

4

19

EP101 / EG101 40

Reaction of Alkanes & Synthesis of Alkanes

ReactionHalogenationCombustionCracking & hydrocracking

SynthesisHydrogenation of AlkenesReduction of Alkyl HalidesReaction of Alkyl Halides with organometals

20

EP101 / EG101 42

Synthesis of Alkanes: 1. Hydrogenation of Alkenes

Catalytic hydrogenation of alkene to form alkanes.General rxn eqn:

Its take place under heterogenous ( more than 1 phase) system, the alkane & H2 are adsorbed on the metallic surface. Therefore they are weakened (both multiple bond and H-H bond).The H2 will add to the same side of the multiple bond: cis-condition.

)()(

22/,2

AlkaneAlkene

HCHC nnNiPdPtnn +⎯⎯⎯ →⎯H2

−−−⎯→⎯−+−=− CCHHCC Pt

What is the function of

the catalyst??

EP101 / EG101 43

A: Hydrolysis by Grignard Reagent

Gridgnard reagent: prepared by treating an alkyl/aryl halide with magnesium mether in dry ether. General eqn. of the preparing of Grignard reagent

Grignard reagent is a reactive reagent. It can react with H2O to form Alkane.IBrClXgrouparylalkylR

XMgRRMgXMgRXreagentGrignard

Ether

,,:/:

+−

−⎯⎯ →⎯+δδ

Synthesis of Alkanes: 2. Reduction of Alkyl Halides

XOHMgHROHRMgXreagentGrignard

)(2 +−⎯→⎯+

21

EP101 / EG101 44

B: Reduction by metal & acid

General eqn.

Eg. Reduction of Cloro-cyclohexane with Zn & acetic acid able to produce 83% of cyclohexane.

Alkyl halide as well react with Litium aluminium hydride (LiAlH4)

Tetrahydroaluminate ion will react with alkyl halide to produce alkaneRHRX LiAlH⎯⎯ →⎯ 4

RHRXCOOHCH

Zn

3

⎯→⎯

−− +−−+−⎯→⎯−−+−− XHCHAlHXCHHAlH

tetrahydroaluminate Alkyl Halide Alkane

EP101 / EG101 45

Alkane can be synthesis via nucleophilic substitution reaction between alkyl halide and a negative nucleophile. The carbon in the polar carbon-halogen bond is positive, therefore the negative nucleophile is attracted to it.

LiXRCuRRXRCuLiR ++−⎯→⎯+ ''

Cumprum diaalkyl Litium2

3. Reaction of Alkyl Halides with organometals

RRuLiCRLiRRX XRCuXLi −+⎯⎯→⎯+⎯⎯→⎯−+⎯→⎯ '2

22

EP101 / EG101 46

Reaction of Alkane1. Halogenation

EP101 / EG101 47

• Radical halogenation has three distinct parts.

Halogenation of Alkanes

• A mechanism such as radical halogenation that involves two or more repeating steps is called a chain mechanism.

• The most important steps of radical halogenation are those that lead to product formation—the propagation steps.

23

EP101 / EG101 48

• When a single hydrogen atom on a carbon has been replaced by a halogen atom, monohalogenation has taken place.

• When excess halogen is used, it is possible to replace more than one hydrogen atom on a single carbon with halogen atoms.

• Monohalogenation can be achieved experimentally by adding halogen X2 to an excess of alkene.

• When asked to draw the products of halogenation of an alkane, draw the products of monohalogenation only, unless specifically directed to do otherwise.

Halogenation of Alkanes

EP101 / EG101 49

• In the presence of heat or light, alkanes react with halogens to form alkyl halides.

• Halogenation of alkanes is a radical substitution reaction.

• Halogenation of alkanes is only useful with Cl2 or Br2. Reaction with F2 is too violent, and reaction with I2 is too slow to be useful.

• With an alkane that has more than one type of hydrogen atom, a mixture of alkyl halides may result.

Halogenation of Alkanes

24

EP101 / EG101 50

EP101 / EG101 51

2. Combustion

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

gasoline, diesel, heating oil…

3. 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 andused in to make plastics.

25

EP101 / EG101 52

REVISION CHECKREVISION CHECK

What should you be able to do?

Recall and explain the physical properties of alkanes

Recall and explain the types of isomerism found in alkanes

Recall and explain why alkanes undergo chlorination free radical reaction

Write balanced equations representing the reactions taking place in this section

CAN YOU DO ALL OF THESE? CAN YOU DO ALL OF THESE? YES YES NONO