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Chapter 2

Introduction toHydrocabons

Carbon Backbone,Nomenclature, Physical &

Chemical Properties

HYDROCARBONS• Compounds composed of only carbon and hydrogen atoms

(C, H). Each carbon has 4 bonds.

• They represent a “backbone” when other “heteroatoms”(O, N, S, .....) are substituted for H. (The heteroatoms givefunction to the molecule.)

• Acyclic (without rings); Cyclic (with rings); Saturated:only carbon-carbon single bonds; Unsaturated: containsone or more carbon-carbon double and/or triple bond

HYDROCARBONS• Alkanes contain only single (σ ) bonds and have the

generic molecular formula: [CnH2n+2]

• Alkenes also contain double (σ + π) bonds and have thegeneric molecular formula: [CnH2n]

• Alkynes contain triple (σ + 2π) bonds and have thegeneric molecular formula: [CnH2n-2]

• Aromatics are planar, ring structures with alternatingsingle and double bonds: eg. C6H6

Types of Hydrocarbons

Each C atom is trigonal planar with sp2 hybridized orbitals.There is no rotation about the C=C bond in alkenes.

Each C atom is tetrahedral with sp3 hybridized orbitals. They only have single bonds.

Question 2.1

• What is the hybridization of the starredcarbon in humulene (shown)?

• A) sp• B) sp2

• C) sp3

• D) 1s2 2s2 2p2

Question 2.2

• What is the hybridization of the starredcarbon of geraniol?

• A) sp• B) sp2

• C) sp3

• D) 1s2 2s2 2p2

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Types of Hydrocarbons

Each C atom is linear with sp hybridized orbitals.

Each C--C bond is the same length; shorter than a C-C bond: longer than a C=C bond.The concept of resonance is used to explain this phenomena.

Propane

It is easy to rotate about the C-C bond in alkanes.

Naming AlkanesC1 - C10 : the number of C atoms present in the chain.

Each member C3 - C10 differs by one CH2 unit. This is called a homologous series.Methane to butane are gases at normal pressures.Pentane to decane are liquids at normal pressures.

Nomenclature of Alkyl Substituents Examples of Alkyl Substituents

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Constitutional or structural isomers have the samemolecular formula, but their atoms are linkeddifferently. Naming has to account for them.

Question 2.3

• How many hydrogens are in a molecule ofisobutane?

• A) 6• B) 8• C) 10• D) 12

A compound can have more than one name, but a name must unambiguously specify only one compound

C7H16 can be any one of the following:

Question 2.4

• How many isomeric hexanes exist?• A) 2• B) 3• C) 5• D) 6

Question 2.5• The carbon skeleton shown at the bottom right

accounts for 9 carbon atoms. How many otherisomers of C10H22 that have 7 carbons in theirlongest continuous chain can be generatedby adding a single carbon to various positionson this skeleton?

• A) 2• B) 3• C) 4• D) 5

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Alkanes(Different types of sp3 carbon atoms)• Primary, 1o, a carbon atom with 3 hydrogen atoms: [R-

CH3]

• Secondary, 2o, a carbon atom with 2 hydrogen atoms: [R-CH2-R]

• Tertiary, 3o, a carbon atom with 1 hydrogen atom: [R-CH-R] R

• Quaternary, 4o, a carbon atom with 0 hydrogen atoms:CR4

Different Kinds of sp3 Carbons andHydrogens

Question 2.6

• In 3-ethyl-2-methylpentane, carbon #3(marked by a star) is classified as:

• A) primary (1°)• B) secondary (2°)• C) tertiary (3°)• D) quaternary (4°)

Question 2.7

• How many primary carbons are in themolecule shown at the bottom right?

• A) 2• B) 3• C) 4• D) 5

Nomenclature of Alkanes1. Determine the number of carbons in the parent hydrocarbon

2. Number the chain so that the substituent gets the lowest possible number

3. Number the substituents to yield the lowest possible number in the number of the compound

(substituents are listed in alphabetical order)

4. Assign the lowest possible numbers to all of the substituents

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5. When both directions lead to the same lowest number for oneof the substituents, the direction is chosen that gives the lowest possible number to one of the remaining substituents

6. If the same number is obtained in both directions, the firstgroup receives the lowest number

7. In the case of two hydrocarbon chains with the same number ofcarbons, choose the one with the most substituents

8. Certain common nomenclatures are used in the IUPAC system

Question 2.7

• The correct structure of 3-ethyl-2-methylpentane is:

• A) B)

• C) D)

CnH2n

Cycloalkane Nomenclature

Cycloalkanes• Cycloalkanes are alkanes that contain a ring

of three or more carbons.• Count the number of carbons in the ring,

and add the prefix cyclo to the IUPAC name ofthe unbranched alkane that has that number ofcarbons.

Cyclopentane Cyclohexane

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Ethylcyclopentane

CH2CH3

• Name any alkyl groups on the ring in theusual way. A number is not needed for a singlesubstituent.

Cycloalkanes• Name any alkyl groups on the ring in the

usual way. A number is not needed for a singlesubstituent.

• List substituents in alphabetical order andcount in the direction that gives the lowestnumerical locant at the first point of difference.

3-Ethyl-1,1-dimethylcyclohexane

CH2CH3

H3C CH3

Cycloalkanes

For more than two substituents, Question 2.8

• Which one contains the greatest numberof tertiary carbons?

• A) 2,2-dimethylpropane• B) 3-ethylpentane• C) sec-butylcyclohexane• D) 2,2,5-trimethylhexane

2.17Physical Properties of Alkanes

and Cycloalkanes Crude oil

Refinery gas

C1-C4

Light gasoline(bp: 25-95 °C)

C5-C12

Naphtha(bp 95-150 °C)

Kerosene(bp: 150-230 °C)

C12-C15

Gas oil(bp: 230-340 °C)

C15-C25

Residue

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Fig. 2.15

Question 2.9

• Arrange octane, 2,2,3,3-tetramethylbutaneand 2-methylheptane in order of increasingboiling point.

• A) 2,2,3,3-tetramethylbutane < octane< 2- methylheptane

• B) octane < 2-methylheptane <2,2,3,3- tetramethylbutane

• C) 2,2,3,3-tetramethylbutane < 2-methylheptane < octane

• D) 2-methylheptane < 2,2,3,3-tetramethylbutane < octane

Example of Intramolecular Forces:Protein Folding

10-40kJ/mol

700-4,000kJ/mol

150-1000kJ/mol

0.05-40kJ/mol

Ion-dipole(Dissolving)40-600kJ/mol

van der Waals ForcesWeak Intermolecular Attractive Forces

The boiling point of a compound increases withthe increase in van der Waals force…and a

Gecko uses them to walk!

Ion-Dipole Forces (40-600 kJ/mol)• Interaction between an ion and a dipole (e.g. NaOH and

water = 44 kJ/mol)• Strongest of all intermolecular forces.

Intermolecular ForcesIon-Dipole & Dipole-Dipole Interactions:

like dissolves like• Polar compounds dissolve in polar solvents & non-polar in non-polar

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Dipole-Dipole Forces(permanent dipoles)

Intermolecular Forces

5-25 kJ/mol

Dipole-Dipole Forces

Intermolecular Forces

Boiling Points &Hydrogen Bonding Hydrogen Bonding

• Hydrogen bonds, aunique dipole-dipole (10-40 kJ/mol).

Hydrogen Bonding

Intermolecular Forces

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DNA: Size, Shape & Self Assemblyhttp://www.umass.edu/microbio/chime/beta/pe_alpha/atlas/atlas.htm

Views & Algorithms

10.85 Å10.85 Å

London or Dispersion Forces• An instantaneous dipole can induce another dipole in an

adjacent molecule (or atom).• The forces between instantaneous dipoles are called

London or Dispersion forces ( 0.05-40 kJ/mol).

Intermolecular Forces

Gecko: toe, setae, spatulae6000x Magnification

http://micro.magnet.fsu.edu/primer/java/electronmicroscopy/magnify1/index.html

Geim, Nature Materials(2003)Glue-free Adhesive100 x 10 6 hairs/cm2

Full et. al., Nature (2000)5,000 setae / mm2

600x frictional force; 10-7

Newtons per seta

Boiling Points of Alkanes

• governed by strength of intermolecularattractive forces

• alkanes are nonpolar, so dipole-dipole anddipole-induced dipole forces are absent

• only forces of intermolecular attraction areinduced dipole-induced dipole forces

Induced dipole-Induced dipoleAttractive Forces

+–+ –

• two nonpolar molecules• center of positive charge and center of

negative charge coincide in each

+–+ –

• movement of electrons creates aninstantaneous dipole in one molecule (left)

Induced dipole-Induced dipoleAttractive Forces

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+–+ –

• temporary dipole in one molecule (left)induces a complementary dipole in othermolecule (right)

Induced dipole-Induced dipoleAttractive Forces

+–+ –

• temporary dipole in one molecule (left)induces a complementary dipole in othermolecule (right)

Induced dipole-Induced dipoleAttractive Forces

+–+ –

• the result is a small attractive forcebetween the two molecules

Induced dipole-Induced dipoleAttractive Forces

+– + –

• the result is a small attractive forcebetween the two molecules

Induced dipole-Induced dipoleAttractive Forces

Boiling Points•Increase with increasing number of carbons

• more atoms, more electrons, moreopportunities for induced dipole-induceddipole forces

•Decrease with chain branching

• branched molecules are more compact withsmaller surface area—fewer points of contactwith other molecules

London DispersionForces

Intermolecular Forces

Which has the higherattractive force?

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Question 2.10

• Which alkane has the highest boilingpoint?

• A) hexane• B) 2,2-dimethylbutane• C) 2-methylpentane• D) 2,3-dimethylbutane

•Increase with increasing number of carbons

• more atoms, more electrons, more opportunities for induced dipole-induceddipole forces

Heptanebp 98°C

Octanebp 125°C

Nonanebp 150°C

Boiling Points

•Decrease with chain branching• branched molecules are more compact with

smaller surface area—fewer points of contactwith other molecules

Octane: bp 125°C

2-Methylheptane: bp 118°C

2,2,3,3-Tetramethylbutane: bp 107°C

Boiling Points

•All alkanes burn in air to givecarbon dioxide and water.

2.18Chemical Properties:

Combustion of Alkanes

4817 kJ/mol

5471 kJ/mol

6125 kJ/mol

654 kJ/mol

654 kJ/mol

Heptane

Octane

Nonane

Heats of Combustion

What pattern is noticed in this case?

•Increase with increasing number of carbons

• more moles of O2 consumed, more molesof CO2 and H2O formed

Heats of Combustion

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5471 kJ/mol

5466 kJ/mol

5458 kJ/mol

5452 kJ/mol

5 kJ/mol

8 kJ/mol

6 kJ/mol

Heats of Combustion

What pattern is noticed in this case?8CO2 + 9H2O

5452 kJ/mol5458 kJ/mol

5471 kJ/mol

5466 kJ/molO2+ 25

2

O2+ 252 O2+ 25

2 O2+ 252

Figure 2.17

•Increase with increasing number of carbons

• more moles of O2 consumed, more molesof CO2 and H2O formed

•Decrease with chain branching• branched molecules are more stable

(have less potential energy) than theirunbranched isomers

Heat of CombustionPatterns

•Isomers can differ in respect to their stability.

•Equivalent statement:

–Isomers differ in respect to their potential energy.

Important Point

Differences in potential energy canbe measured by comparing heats ofcombustion. (Worksheet problems)

2.19Oxidation-Reduction in Organic Chemistry

Oxidation of a carbon atom correspondsto an increase in the number of bonds tothe carbon atom and/or a decrease in thenumber of hydrogens bonded to thecarbon atom. See examples on the board.

increasing oxidation state of carbon

-4 -2 0 +2 +4

H

H

H

C H

H

H

H

C OH

O

CHH

O

COHH

O

COHHO

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increasing oxidation state of carbon

-3 -2 -1

HC CH

C C

H

H H

H

C C H

HH

H H

H

• How to calculate the oxidation stateof each carbon in a molecule that containscarbons in different oxidation states?

CH3CH2OH C2H6O

Table 2.5 How to CalculateOxidation Numbers

• 1. Write theLewis structureand includeunshared electronpairs.

H

C

H

H

H

O

H

C H••

••

Table 2.5 How to CalculateOxidation Numbers

• 2. Assign theelectrons in acovalent bondbetween twoatoms to the moreelectronegativepartner.

H

O

H

C

H

H

H

C H••••

••••

••

••

••

••••

• 3. For a bondbetween twoatoms of thesame element,assign theelectrons in thebond equally.

H

O

H

C

H

H

H

C H••••

••••

••

••

••

••••

Table 2.5 How to CalculateOxidation Numbers

• 3. For a bondbetween twoatoms of thesame element,assign theelectrons in thebond equally.

H

O

H

C

H

H

H

C H••••

••••

••

••

••

•••• ••

Table 2.5 How to CalculateOxidation Numbers

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• 4. Count the numberof electronsassigned to eachatom and subtractthat number fromthe number ofvalence electrons inthe neutral atom;the result is theoxidation number.

H

O

H

C

H

H

H

C H••••

••••

••

••

••

•••• ••

Each H = +1C of CH3 = -3

C of CH2O = -1O = -2

Table 2.5 How to CalculateOxidation Numbers

X Y

X less electronegative than carbonY more electronegative than carbon

oxidation

reductionC C

GeneralizationOxidation of carbon occurs when a bond betweencarbon and an atom which is less electronegativethan carbon is replaced by a bond to an atom thatis more electronegative than carbon. The reverseprocess is reduction.

CH3Cl HClCH4 Cl2+ +

Oxidation

+ 2Li LiClCH3Cl CH3Li +Reduction

Examples Question 2.11

• To carry out the reaction shown below weneed:

•• CH3OH → H2C=O•• A) an oxidizing agent• B) a reducing agent