Chemistry Lecture Notes 2010

General Chemistry - Dentistry1

Ajman UniversityOf Science And Technology Network

College Of PharmacyAnd Health Sciences

Notes on:GENERAL CHEMISTRY COURSEGENERAL CHEMISTRY COURSEFor Dentistry Students (700126)

2010/2011

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General:

Course Title: General Chemistry Course Code:(700126) Credit Hours: 3 Credit Hours

Lecture Hours: 2 Hrs. (2 C.H.)Lab Hours: 2 Hrs. (1

C.H.) Pre-requisite: None

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Course Description:

This course includes two major parts: Part I: General part will introduce the student

to basic aspects of general chemistry, i.e.: atomic structures, electronic configuration,

periodic table of elements, chemistry of metals, and the fundamentals of chemical bonds and chemical reactions.

Part II: Organic part will cover some important areas in organic chemistry, which include: aliphatic and aromatic hydrocarbons,

stereochemistry, as well as some functional groups, e.g: alcohols, phenols, carbonyl compounds.

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Course Objectives:

The course is aimed to:provide some fundamental

knowledge of general and organic chemistry.

provide the basis of more detailed understanding of that part of chemistry specifically related to living systems and drugs and on which biochemistry can be built.

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Learning and Information Resources:Student's Textbook: Masterton, Hurley: "Chemistry, Principles

and Reactions“Publisher: Saunders Golden Sunburst Series, Saunders College Publishing USA, ISBN: 0-03-005889-9

Further Readings: Hart, Craine and Hart: "Organic

Chemistry, A short Course" Publisher: Houghton Mifflin Company, Boston Toronto ISBN 0-395-70838-9

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Other References:

Computer Software: General Chemistry Interactive CD-ROM,

Ver 3.0

Recommended Web-Sites: http://www.lib.utexas.edu/Libs/Chem/ http://www.learnchem.net/tutorials/basc.shtml http://www.webelements.com/ http://homework.chem.uic.edu/IEMDL.HTM http://www.acdlabs.com/iupac/nomenclature/ http://www.chemhelper.com/tutorials.html

Interactive CDInteractive CD

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Assessment Of The Students: The evaluation is based on students’

performance in written examinations, quizzes and practical skills.

The 100 Marks are distributed as follows: 20 Marks for Quizzes 20 Marks for Mid-Term Examination 20 Marks for Practical Examination, and 40 Marks for Final Examination


PART – I:Introduction

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Introduction:(Ref.: Textbook, pp 2)

What is chemistry? What is matter?

Matter: is anything that has mass and occupies space

Classification of matter:Matter

Mixtures Pure substances

Elements Compoundsheterogeneous homogeneous

ionic molecular

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Elements: An element: is a type of matter that cannot be

broken down into two or more pure substances Symbols of elements

Compounds: A compound: is a pure substance that

contains more than one element. Fixed composition of elements Methods of separation:

• Heat• Electrolysis

Periodic TablePeriodic Tableof Elementsof Elements

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Mixtures: A mixture: is a type of matter that contains two or more

substances combined in such a way that each substance retains its chemical identity.

Types of mixtures:• Homogeneous mixtures (uniform)• Heterogeneous mixtures (nonuniform)

Methods of separation:• Filtration• Distillation• Crystallization• Extraction• Chromatography

Problems:-Compare between compounds and mixtures.-Solve exercise on screen 1.6 in


SSee ee DescriptionDescription on onscreen 1.7 inscreen 1.7 in


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Properties of Substances:(Ref.: Textbook, pp 15)

Identification of substances Intensive versus extensive properties Chemical properties Physical properties:

• m.p. & b.p.• Density• Solubility

Problems:- Textbook pages 21-25- Solve exercises and tutorials on screens 1.12

and 1.15 in Interactive CDInteractive CD


Atoms, molecules and ions

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Atoms, Molecules and Ions:(Ref.: Textbook, pp 26)

Atoms: Atomic structure:

(Source: interactive CD, screen 2.4)

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Dalton’s Atomic Theory: An element is a substance that cannot be broken

down by chemical means. Elements consist of atoms.

All the atoms in a particular element behave in the same way chemically, and they share the same ATOMIC NUMBER.

In ordinary chemical reactions atoms cannot be disappeared or changed to another atoms of another elements.

Elements combine chemically to form substances called COMPOUNDS.

The ATOM is the smallest particle of an element that can enter into a chemical reaction.

SSee screen 2.5 on ee screen 2.5 on Interactive CDInteractive CD

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Atoms: Components of the atom: (Rutherford

Hypothesis)• Electrons• Atomic Nucleus: protons and neutrons

Atomic Number (Z) Atomic Mass (A) Nuclear symbol Isotopes:

• Isotopes in medicine (see Textbook, page 510)

Problems:- Search for the radioactive isotope(s) used in dental

radiology- Further questions

SSee ee SummarySummary on onscreen 2.9 in screen 2.9 in Interactive CDInteractive CD

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Molecules: Definition Chemical bonds:

• Ionic chemical bonds• Covalent chemical bonds

Presentation of molecules: examples• Molecular formulas• Expanded structural formulas• Condensed structural formulas• Ball and stick molecular models• Space filling molecular models

Problems:Solve exercises and tutorials on screens 2.15 in Interactive CDInteractive CD

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Ions: Definition Ionization Classification of ions:

Problems:Solve tutorials on screens 2.16 in

ions

Monatomic(charged atoms)

Polyatomic(charged molecules)

cations anions cations anions


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Ionic compounds: Definition Ionic bonds Formulas of ionic compounds:

• Electrical neutrality• Ions with and without Noble-Gas

ConfigurationsPeriodic TablePeriodic Table

of Elementsof Elements

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Nomenclature of compounds: Naming cations:

• Monatomic• Polyatomic

Naming anions:• Monatomic• Polyatomic• acids

Naming ionic compounds Naming molecular compounds Naming hydrates

Problems:- Textbook pages 47-51- Solve tutorials on screens 2.17 -2.19 in - Online tutorial:http:gemini.tntech.edu/~snorthrup/chem111/tutorials/chap4c/prob1q.html



Electronic structureand periodic table of elements

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Electron configurations:(Ref.: Textbook, pp 140)

Atomic structure: Shells / Energy levels Quantum Numbers

Atomic orbitals: Shapes of Atomic Orbitals (examples)

Electron configuration: Definition Pauli’s Exclusion Principle Filling of Atomic Orbitals Hund’s Rule

Problems:- Solve tutorials on screens 6.12, 6.13, 6.15 in - See online gallery of atomic orbitals at:http://www.shef.ac.uk/chemistry/orbitron/


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Periodic Table of Elements:(Ref.: Textbook, pp 33)

Classification of Elements, according to: Metallic and nonmetallic Acidic and basic Valency Atomic weight and atomic number

Periodic System Metals, nonmetals and metalloids

(See interactive CD, Tools: Periodic Table)


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Periodic Table of Elements:(Ref.: Textbook, pp 143)

Electron configuration and the Periodic Table Trends in the Periodic Table:

Atomic number and atomic radius Ionization Energy Electron Affinity Electronegativity:

• Electronegativity and types of chemical bonds Lewis Electron Dot Diagram (See interactive

CD, screen 7.4)

Problems:- Textbook pages 156, 158 pp- Solve tutorials on screens 6.17 -6.19 in Interactive CDInteractive CD

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Hybridization:(Ref.: Textbook, pp 183)

Hybrid Orbitals: (examples) sp-Hybridization sp2-Hybridization sp3-Hybridization

sigma (σ) bonds: s,s- sigma (σ) bonds s,px- sigma (σ) bonds

px,px- sigma (σ) bonds

pi (π) bonds: py,py-pi (π) bonds

pz,pz-pi (π) bonds


Solutions

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Reactions in aqueous solutions:

Aqueous Solutions:Solutes and solvents

Saturated, concentrated and diluted solutionsMethods used to express relative amount of

solute and solvent in solution. These methods include:

(1) Mass percent g %:A solution can be prepared to specify, the mass of solute in gram dissolved in 100 ml of a solvent.

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Reactions in aqueous solutions:

Example: An isotonic solution is prepared to contain both 0.9 % of

sodium chloride (NaCl) and 5% glucose (C6H12O6) in water. Calculate the mass of NaCl and C6H12O6 dissolved to prepare 500 mL.

Solution:A)0.9 % Sodium chloride means 0.9 grams per 100 mL so,

to prepare 500 mL, the amount needed is: 0.9 X 500/100 = 4.5 g

B)5 % Glucose means 5 grams per 100 mL so, to prepare 500 mL the amount needed is :

5 X 500/100 = 25 g Weigh 4.5 g of powdered NaCl and 25 g powdered

glucose dissolve in sufficient amount of water, then complete to 500 mL of the same solvent.

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Reactions in aqueous solutions:(Ref.: Textbook, pp 76)(2) Molarity (M):Molarity is the number of moles of solute per

liter of solution. Unit: M=moles/L

Mole: is a collection of 6.0122x1023 items. The mass in grams of one mole of a substance

is numerically equal to its formula mass (molecular weight).

1 Molar solution is prepared by dissolving one gram molecular weight (Mole) of any solute in 1 liter of solvent.

Molecular weight is the summation of atomic weights of the solute.

molarity (M)= moles of soluteliters of solution

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Reactions in aqueous solutions:(Ref.: Textbook, pp 76)Example: To prepare 1 mole of NaCl ( Na =23 and Cl =

35.5), 58.5 grams of NaCl dissolved in 1 liter

Problems:Show how you can prepare the following

solutions1. 0.1 M HCl from concentrated HCl (35%)2. 0.5 M copper(II) sulphate solution.3. 2 M glucose solution (C6H12O6)

4. 0.5 M of acetic acid solution from 10 % acetic acid solution (CH3COOH).

5. Further Problems: see Textbook pp 95, and interactive CD.

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Reactions in aqueous solutions:Problems:

Express the concentration of the following solution by M and g %:

1. 10 ml of 10 % acetic acid solution was diluted up to 500 mL with distilled water.

2. 5 grams of copper(II) sulphate penta hydrate in 1 liter distilled water

3. 10.8 grams of glucose in 100 mL distilled water


Chemistry of metals

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Chemistry of metals: (Ref.: Textbook, pp 530) Metals: are group of elements that share

certain properties: They conduct heat and electricity well, and They can be shaped easily.

88% of elements are metals Physical properties of metals: (used to

differentiate metals from nonmetals): Solids with high m.p. and b.p. (except Hg) High density Good conductors Shiny Malleable and ductile Formation of mixtures (alloys)

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Chemistry of metals: (Ref.: Textbook, pp 530) Classification of metals:

Main-group metals: (s-block metals)• Alkali metals (Group 1)• Alkaline earth metals (Group 2)

Transition metals:• d-block metals• f-block metals (Lanthanides and

Actinides) p-block metals metalloids

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Reactions of metals: (Ref.: Textbook, pp 537)Reactions of alkali and alkaline earth

metals: Alkali and alkaline earth metals are the

most reactive of all elements. (why?) Reaction with Hydrogen Reaction with nonmetals: (S, X2, O2, N2) Reaction with water

Problems:- Textbook pages 548 pp

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Reactions of metals: (Ref.: Textbook, pp 87)Redox Reactions of transition metals: Oxidation and reduction Oxidizing agents and reducing agents Oxidation number Calculation of oxidation number (page

89) Balancing Redox Equations (pages 90-

91)

(See Examples)

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Reactions of metals: (Ref.: Textbook, pp 79)Precipitation Reactions: Precipitation reaction is the reaction which

involves the formation of precipitates. A precipitate (ppt) is a substance which

separates as a solid phase out of the solution. Solubility is the molar concentration of the

saturated solution. It depends on: Temperature Pressure Concentration of other solutes Composition of the solvent

Practical uses of precipitation reactions

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Reactions of metals: (Ref.: Textbook, pp 402)Complexation Reactions: Complexation reaction is the reaction which

leads to the formation of complex ions. A complex ion is an ion containing a central

metal atom bonded to two or more ligands. Coordinate covalent bonds and coordination

number Charges of complexes Chelating Agents.

Assignment:- Search for the medical uses of chelating agents in

Dentistry (Oral Chelating Therapy)

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Acid-Base Reactions: (Ref.: Textbook, pp 82)Definitions of acids and bases: Arrhenius Definition

Limitation of Arrhenius definition Lowry-Brønsted Definition (page: 351) Lewis Definition: (page: 404) Strong and weak acids and bases Conjugate acids and bases Amphiprotic solvents Monobasic and poly basic acids

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Acid-Base Reactions: (Ref.: Textbook, pp 350)pH Scale: The acidity or basicity (alkalinity) of a solution

can be described in terms of its hydrogen ion [H+] concentration.

It is defined by using the pH notation:

pH= -log10 [H+] Or

[H+] = 10 -pH

Most aqueous solutions have hydrogen ion concentrations between 1-10-14 M, and hence have pH between 0-14.

The pH of a natural solution is 7 If the pH <7, the solution is acidic If the pH >7, the solution is basic (alkaline)

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Acid-Base Reactions: (Ref.: Textbook, pp 350)

Relationship between pH and [H+]:

See Further Problems:

pH

[H+]

acidic alkaline

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

1 10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14

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Acid-Base Reactions: (Ref.: Textbook, pp 350)Ionization of water:

In any solution: pH+pOH = pKw

Kw is known as the Ion product constant of water

Kw = [H+] [OH-]

At 25oC, Kw=1.0x10-14 In pure water:

H2O H+ OH-+

[H+] = [OH-]

K = [H+]

[H+] = K = mol lit10 -7 -1w

w

2...

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Acid-Base Reactions: (Ref.: Textbook, pp 350)

Since water is neutral, a solution whose [H+] > 10-7 is acidic,

whereas a solution whose [H+] < 10-7 is alkaline.

Buffer Solutions: (page 379) Buffer solution is a solution that resists changes in pH upon

addition of small amounts of acids or bases. A buffer solution consists of a weak acid and its conjugate

base (e.g: CH3COOH and CH3COONa) or a weak base with its conjugate acid (e.g.: NH3 and NH4Cl)

The buffer action can be explained by the neutralization of the added acid by the base present in the buffer solution, and the neutralization of the added base by the acid present in the buffer solution.

Assignment:- Search for the medical uses of buffer solutions


PART – II:Organic Chemistry

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Introduction:(Ref.: Textbook, pp 577) What is Organic Chemistry? Uniqueness of carbon:

Carbon ranks only twelfth (12th) in abundance among the elements.

Carbon constitutes less than 0.1% of the earth's crust, oceans and atmosphere.

Structure of carbon atom Carbon in periodic table in Group IVA Carbon atom in tetravalent Carbon must form Four Bonds to satisfy its octet of

electrons Bonds may be with other carbons or with other elements

Types of carbon compounds Hybridization of carbon

See interactive CD #2: Screen 22.2

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Hydrocarbons:(Ref.: Textbook, pp 577) Definition Classification:

Hydrocarbons (HC)Hydrocarbons (HC)

aliphatic HCaliphatic HC Aromatic HCAromatic HC

Saturated HCSaturated HC Unsaturated HCUnsaturated HC homocyclichomocyclic heterocyclicheterocyclic

alkanesalkanes alkenesalkenes alkynesalkynes

See interactive CD #2: Screen 22.3


Saturated Hydrocarbons

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Saturated Hydrocarbons (Alkanes):(Ref.: Textbook, pp 577) Structure of alkanes:

sp3 hybridization tetrahedral structures:

methane ethane propane

General formula: CnH2n+2

Classification: Open-chain: straight and branched chains Cycloalkanes

homologous series

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Saturated Hydrocarbons (Alkanes):(Ref.: Textbook, pp 577) Isomeric structure of alkanes:

Isomers: are different compounds with identical molecular formulas.

Isomers are readily distinguishable from each other because they exhibit differences in their physical and chemical properties.

The first three members of alkane series have no isomers, because there is only one way in which the carbon atoms can join together.

All the higher alkanes have isomers. Nomenclature of alkanes: (Textbook page: 617)

Common Nomenclature IUPAC Nomenclature (International Union of Pure

and Applied Chemistry )

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Saturated Hydrocarbons (Alkanes):(Ref.: Textbook, pp 577) The IUPAC system of nomenclature follows the rules

indicated below:1. Select as the parent structure the longest

continuous chain.2. Number the parent carbon chain starting from

whichever end will give the lowest number for the point of attachment of the substituent.

3. If the same alkyl substituent occurs more than once on the parent carbon chain, the prefixes di-, tri-, tetra-, penta- etc.. are used to indicate whether there are two, three, four or five substituents.

4. If several different substituents are attached on the parent carbon chain, they are named either in order of increasing complexity or in alphabetical order.

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Saturated Hydrocarbons (Alkanes):(Ref.: Textbook, pp 577)

Problems: Give the IUPAC name for each of the following compounds:

CH3 CH

CH3

CH2 CH2 CH3 CH3 C

CH3

CH3

CH2 CH CH2

CH2

CH2

CH3

CH3

CH3 CH2 CH2 CH

CHCH3CH3

CH2 C

CH3

CH2 CH3

CH3

a) b)

c) d) CH3 CH

NO2

CH

CH2 NO2

CH

NH2

CH2 CH3

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Saturated Hydrocarbons (Alkanes):(Ref.: Textbook, pp 577)

Problems: Write the structural formula of a compound whose

IUPAC name is:a) 2,2,5-Trimethyl-4-ethyl heptaneb) 2-Bromo-3-chloro-4-isoprpyl octane

The following IUPAC names are incorrect, write the structural formula and correct the names:

a) 3,3-diethylbutaneb) 2-Ethyl-3-bromo butanec) 3-Chloro-2-bromo pentane

Draw all possible structures of the following general formulas:

a) C4H9Cl

b) C5H11Cl

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Cycloalkanes: Cycloalkanes are alkanes that exist in the form of a ring Nomenclature of cycloalkanes:

Cycloalkanes are also called:• cyclic aliphatic hydrocarbons, or• alicyclics

Cycloalkanes are named by simply attaching the prefix (cyclo-) to the name of the open chain hydrocarbon that has the same number of carbon atoms as in the ring:

Cyclopropane

CH2CH2

CH2

CC

H HC

H

H

H

H

CH

H

C

C

H

H

H

H

C

H

H

H2C CH2

H2C CH2

Cyclobutane

CHH

C C

H

H

H

H

CCH

H

H

H

CH2

H2C CH2

CH2C H2

Cyclopentane

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Cycloalkanes:

Problems:1. The general formula for an alkane is CnH2n+2. What

is the corresponding formula for a cycloalkane with one ring?

2. Draw the structural formulas for:• 1,3-dimethylcyclohexane• 1,2,3-trichlorocyclopropane

3. Give IUPAC names for:

Cl

Bra) b)

Br

Br

c) CH

CH3

CH

CH3

CH

Br

CH2 Br

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Physical Properties of Alkanes:1) Occurrence: At room temperature alkanes occur as gases,

liquids and solids: C1-C4 (methane to butane) are gases C5-C17 (pentane to heptadecane) are liquids C18 and larger alkanes are solid waxes

2) Solubility: Alkanes are nonpolar compounds

Their solubility characteristic may be predicted by "like-dissolves-like" rule.

This means:Nonpolar compounds are soluble in other nonpolar solvents and polar compounds are generally soluble in other polar solvents.

Thus, alkanes are soluble in the non-polar solvents carbon tetrachloride, CCl4, and benzene C6H6, but they are insoluble in polar solvents such as water, H2O.

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Physical Properties of Alkanes:

3) Boiling points: Generally the boiling point of n-hydrocarbon

increases with increasing molecular weight. 20-300C for each addition of carbon atom:

-200

-100

0

100

200

300

400

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

number of carbon atoms in n-alkanes

bo

ilin

g p

oin

ts

( oC

)

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Reactions of Alkanes: Alkanes are also called Paraffin hydrocarbons

(from the Latin: parum=little; affinis=affinity) because of their lack of reactivity.

Two important reactions that alkanes do undergo, however, are combustion and substitution.

1) COMBUSTION OF ALKANES: (OXIDATION) Alkanes combine with oxygen to form:

Carbon dioxide (CO2), water (H2O), and a large quantity of heat (the heat of combustion):

CnH2n+2 + O2 n CO2 + (n+1)H2O + HEAT3n+1

2

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Reactions of Alkanes:

2) SUBSTITUTION REACTIONS OF ALKANES: Halogenation:

Examples:

R H X2 R X H X+ +

Alkane Halogen

heat orU.V. light

Alkylhalide

a)

b)

CH4 Cl2+ CH3 Cl + HClU.V.

U.V.+ HCl+ Cl2CH3CH3 CH3 CH2 Cl

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MECHANISM OF SUBSTITUTION REACTIONS OF ALKANES:

a)

b)

CH4 Cl2+ CH3 Cl + HClU.V.

U.V.+ HCl+ Cl2CH3CH3 CH3 CH2 Cl

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Reactions of Alkanes: Orientation of Substitution

Ease of substitution of a hydrogen atom follows this sequence:

3oH > 2oH > 1oH

Ease of hydrogen substitution increases

U.V.+ Cl2CH3CH2CH3

CH3CH2CH2Cl

- HClCH3 CH CH3

Clpropane

n-propyl chloride

isopropyl chloride

a)

X

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Problem: Draw the structures and give the IUPAC names of

all possible monochloro and dichloro derivatives of:

n-butane Isobutane n-pentane Isopentane Neopentane

Determine the main products obtained by the chlorination reactions mentioned above


Unsaturated Hydrocarbons

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Unsaturated Hydrocarbons (Alkenes):(Ref.: Textbook, pp 581) Alkenes are characterized by the carbon-carbon

double bond. Structure of alkenes:

sp2 hybridization Trigonal planar structure:

General formula: CnH2n

Structure of ethylene:

o120

C CH

HH

H

-bond

-bond

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Unsaturated Hydrocarbons (Alkenes):

Nomenclature of alkenes: Common Nomenclature IUPAC Nomenclature

Rules of IUPAC Nomenclature:1. The longest continuos carbon chain containing the

double bond is selected as the parent chain.2. The name of the parent carbon chain is obtained by

replacing the ending (-ane) of the corresponding alkane by the suffix (-ene).

3. The parent chain is numbered so that the double bond is assigned the lowest number.

4. The position of the double bond is assigned the number of the first double-bonded carbon.

5. The entire compound is named by first listing the substituents either in alphabetical order or in order of complexity.

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Unsaturated Hydrocarbons (Alkenes):Problems: Give the IUPAC names of the following compounds:

Write the structural formula for the following compounds: 2-Methyl-2-butene 4,5-Dimethyl-3-isopropyl-2-hexene 2-Chloro-4-methyl-2-pentene

CH3 CH C

CH3

CH

CH CH3CH3

CH

CH3

CH2 CH2 CH3

CH3 CH

Cl

CH2 C

CH2

CH2 CH3

CH3 C

Br

C CH2 CH2 CH CH3

CH3CH3

a)

b)

c)

CH3

NH2

Cl

Br

Br

d)

e)

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Unsaturated Hydrocarbons (Alkenes):

Geometric Isomerism: Geometric isomers are compounds that differ only

in the arrangement of their atoms in space as result of restricted rotation about a bond:

cis-configuration:When alkenes have substituents on the same side of the double bond.

trans-configuration:When alkenes have substituents on opposite sides of the double bond.

For cis-trans isomerism in alkenes, each carbon of the double bond must have two different atoms or groups attached to it.

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Unsaturated Hydrocarbons (Alkenes):Example:

1,2-dichloroethene exists in two different forms:

Problem: Which of the following compounds can exist as cis-

trans isomers? Draw their structures. 1-butene 2-butene propene 3-hexene 2-hexene 2-methyl-2-butene

C CH

Cl Cl

HC C

Cl

Cl H

H

cis-1,2-dichloroethene(b.p 60 C, m.p -80 C)

o o ootrans-1,2-dichloroethene(b.p 47 C, m.p -50 C)

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Physical Properties of Alkenes:

In general, the physical properties of alkenes are much the same as those of corresponding alkanes.

At room temperature: the C2 to C4 alkenes are gases. the C5 to C18 alkenes are liquids. those above C18 are solids.

Solubility and boiling points (see alkanes) Like alkanes, the alkenes are insoluble in water

and soluble in organic solvents (explain why?)

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Reactions of Alkenes:(see interactive CD #2: screen 22.4)

ADDITION REACTIONS OF ALKENES A reagent (A-B) adds across the double bond to give a

saturated product that contains all atoms of both reactions:

General Equation:

1. Addition of Hydrogen: It is the catalytic hydrogenation of alkenes to give the

corresponding alkanes: General Equation:

C C A B+ C C

A B

C C + C C

H HH2

Pt / Nior Pd + Heat

alkanealkene

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Reactions of Alkenes:(see interactive CD #2: screen 22.4)2. Addition of Halogen: It is the addition of halogens (Chlorine or Bromine) to an

alkene to give an alkyl dihalide.

General Equation:

This reaction is known as the halogenation of alkenes.

C C X2 C C

X

X

alkene halogen dihalide

(X2= Cl2 or Br2)+

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Reactions of Alkenes:

3. Hydroxylation: The Baeyer Test: It is the addition of alkaline potassium permanganate,

KMnO4 to an alkene to give a dihydroxy alcohol or glycol and a brown precipitate of manganese dioxide MnO2:

General Equation:

This reaction is used to identify the presence of the carbon-carbon double bond.

It is called "the Baeyer Test".

C C + C C

OH OH

a glycolalkene

KMnO4

[OH-]MnO2+

(purple) (brown)ppt

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Reactions of Alkenes:(see interactive CD #2: screen 22.4)4. Addition of Hydrogen Halide(H-X): Alkenes react with hydrogen halides to form alkyl

halides:

General Equation:

This reaction is known as the hydrohalogenation of alkenes

C C + C C

H X

alkyl halidealkene

H X

O.

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Reactions of Alkenes:(see interactive CD #2: screen 22.4)Orientation of hydrohalogenation: MARKOVNIKOV'S RULE:

The Markovnikov's Rule states that:"In ionic addition of H-X to unsymmetrical alkenes, the positive hydrogen of the hydrogen halide adds to the double-bond carbon that bears the greater number of hydrogen atoms, and the negative halide ion adds to the double-bond carbon that bears the least number of hydrogen atoms."

CH3 C

CH3

CH2 + HCl CH3 C CH3

CH3

Cl

isobutylene(2-methylpropene)

t-butylchloride(2-chloro-2-methyl-propane)

isobutylchloride(1-chloro-2-methyl-propane)


CH3 CH CH2

CH3

Cl+ HClCH3 C

CH3

CH2

and NOT:

CH3 C

CH3

CH2 + HCl CH3 C CH3

CH3

Cl


t-butylchloride(2-chloro-2-methyl-propane)

isobutylchloride(1-chloro-2-methyl-propane)


CH3 CH CH2

CH3

Cl+ HClCH3 C

CH3

CH2

and NOT:

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Unsaturated Hydrocarbons (Alkynes):(Ref.: Textbook, pp 582) Alkynes are characterized by the carbon-carbon

triple bond. Structure of alkynes:

sp hybridization linear structure:

General formula: CnH2n-2

Structure of acetylene:

o180

C CH H

σ bond

bond

bond

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Unsaturated Hydrocarbons (Alkynes):

Nomenclature of alkynes: Common names are used for low-molecular-

weight alkynes, and they named as substituted acetylenes

For more complicated alkynes the IUPAC names are used.

The IUPAC rules for naming alkynes are similar to those for alkanes and alkenes.

The ending (-yne) is used for a triple bond, and (-diyne) for two triple bonds.

Compounds with a double and a triple bond are (-enynes), the double bond receives the lowest numbers.

Numbers as low as possible are given to double and triple bonds, even though this may at times give (-yne) a lower number than (-ene).

When there is a choice in numbering, the double bonds are given the lowest numbers.

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Nomenclature of alkynes: Examples:

CH3 CH

Br

C C CH3

4-Bromo-2-pentyne

5 4 3 2 1a)

c)

b)

d)

CH C CH

CH3

CH2 C CH

3-Methyl-1,5-hexadiyne

1 2 3 4 5 6

CH2 CH C CH

1-Buten-3-yne

1 2 3 4CH3 CH CH C CH

3-Penten-1-yne

5 4 3 2 1

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Problem: Give the IUPAC name of the following

compounds:

CH3 CH

CH3

CH

Cl

C C CH2 CH

CH3

CH2 CH3a)

b) CH2 C C CH

CH3

c)Br

O.

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Physical Properties of Alkynes:

In general, the physical properties of alkynes are much the same as those of corresponding alkanes and alkenes.

At room temperature: the C2 to C4 alkynes are gases. the C6 to C18 alkynes are liquids. those above C18 are solids.

Solubility and boiling points (see alkanes and alkenes)

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Reactions of Alkynes:(see interactive CD #2: screen 22.4)

ADDITION REACTIONS OF ALKYNES

1. Addition of Hydrogen: Alkynes add hydrogen in the presence of catalysts (Pt,

Ni or Pd) to form the corresponding alkanes: General Equation:

This reaction is called catalytic hydrogenation of alkynes.

C C+H2

Pd or NiC C

H H Pd or Ni

H2+C C

H H

HHalkyne alkane

O.

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Reactions of Alkynes:(see interactive CD #2: screen 22.4)2. Addition of Halogen: Alkynes react with two molecules of halogen to give

tetrahalides: General Equation:

3. Addition of Hydrogen Halide: Hydrohalogenation of alkynes:

The addition of hydrogen chloride, bromide or iodide to alkynes follows the Markovnikov's Rule.

General Equation:

C C + X2 C

X

C

X

X2+ C

X

C

X

X X

a gem-dihalide

C

H

C

X

XH

+ HXC

H

C

X

haloalkenealkyne

C

H

C

X+ HXC C

a gem-dihalide

C

H

C

X

XH

+ HXC

H

C

X

haloalkenealkyne

C

H

C

X+ HXC C

+HX

O.

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Nomenclature of Further Functional Groups in Organic

Chemistry

A functional group is a specific group of atoms responsible for the characteristic

reactions of a compound.

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Chemical Group

Functional Group

ExamplesCommon NamesIUPAC Names

Alcohols

Methyl alcoholMethanol

Ethyl alcoholEthanol

n-Propyl alcohol1-Propanol

Isopropyl alcohol2-Propanol

Cyclohexyl alcoholCyclohexanol

C OH

CH3 OH

CH3 CH2 OH

CH3 CH2 CH2 OH

CH3 CH

CH3

OH

OH

O.

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Aldehyde (carbonyl):

C O

H

H C O

H

C O

H

CH3

C O

H

CH2CH3

Aldehydes

FormaldehydeMethanal

AcetaldehydeEthanal

PropionaldehydePropanal

Keto (carbonyl):

C O

C O

CH3

CH3

CH3 CH2 CH2 C O

CH3

O

Ketones

Dimethyl ketone (Acetone)

Propanone

Methyl n-propyl ketone

2-Pentanone

Cyclohexyl ketoneCyclohexanone

Chemical Group

Functional Group


O.

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Carboxyl:

C O

OH

C OH

OH

C OCH3

OH

C O

OH

CH2CH3

Ester:

C

O

O R

C

O

OCH3 CH3

C

O

OCH2 CH3CH3

Ether (Alkoxy):

C O C

O CH3CH3

O CH3CH2CH3

O CH2CH2CH3 CH3

Carboxylic Acids:

Formic acidMethanoic acid

Acetic acidEthanoic acid

Propionic acidPropanoic acid

Esters:

Methyl acetateMethyl ethanoate

Methyl propionateMethyl propanoate

Ethers:

Dimethyl etherMethoxymethane

Methyl ethyl etherMethoxyethane

Diethyl etherEthoxyethane

Chemical Group

Functional Group


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Amino:

N

CH2CH3 NH2

(1o amine)

N

H

CH3CH3(2o a mine)

N

CH3

CH3CH2CH3(3o amine)

Carboxamide:

C

O

N

C

O

NH2CH3

C

O

NHCH3 CH3

Amines:

Ethyl amineEthanamine

Dimethyl amineN-Methyl methanamine

Dimethyl ethyl amineN,N-Dimethylethanamine

Amides:

AcetamideEthanamide

N-Methyl acetamideMethyl ethanamide

Chemical Group

Functional Group


O.

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I) Name the following compounds:

(1)(2)(3)

(4)(5)(6)

(7)(8)(9)

(10)(11)(12)

CH3 CH

OH

C CH2 CH3

Cl

Br

OH

CH2 CH3CH2

NO2

C

O

CH2 CH CH3

CH2 CH3

CH3 CH

CH3

CH

NH2

CH2 C

O

H

CH3 C

NH2

COOH

CH2 CH3CH3 CH

CH3

O CH3

CH3 CH

Cl

C

Cl

CH2 OHCH

CHCH3 CH3

C

O

CH3 CH3

C

O

NH CH3

CH3 C

O

NH CH3 CH

CH3

C

O

O CH2 CH3 C

O

O CH3

O.

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II) Draw structures of the following compounds:

(1) 2,5-diamino cyclopentanone(2) N,N-dimethyl benzamide

(3) Ethyl butanoate(4) Cyclobutyl acetaldehyde

(5) diisopropyl ether(6) Nitro acetate

(7) Cyclopentyl methyl ketone(8)3-bromo-4-chloro-5-nitro-2-hexanol

(9) Triethyl amine(10) 2-amino-4-methyl-pentanoic acid


Aromatic Hydrocarbons

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Aromatic Hydrocarbons:

The term aromatic was originally used to designate compounds with spicy or sweet smelling odours derived from plants.

A compound is said to be aromatic, if it is benzenelike in its properties

Structure of Benzene: (Kekule's Proposal) F.A. Kekule's proposed in 1865 a structure for

benzene: Benzene consists of a cyclic, hexagonal, planar

structure of six carbon atoms with alternating single and double bonds:

C

CC

C

CC

H

H

H

HH

H

benzene (1,3,5-cyclohexatriene)The Kekule's structure

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Aromatic Hydrocarbons:

Structure of Benzene: Theory of Resonance( The modern proposal )

The Resonance Theory:This theory may be stated as follows:

1. Whenever a molecule can be represented by two or more equivalent structures that differ only in the arrangement of electrons, there is resonance. The actual structure of the molecule is a hybrid that combines the characteristics of all these structures. The individual structures are called contributing structures and do not exist independently.

2. The resonance hybrid is more stable (contains less internal energy) than any of the contributing structures. This increase in stability is referred to as resonance energy.

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Nomenclature of Aromatic Compounds:

A) Monosubstituted Benzenes: Because all six positions in benzene are

equivalent there is no need to specify by a number the position of a substituent for monosubstituted benzenes.

Monosubstited benzenes can be named by placing the name of the substituent in front of the word : "benzene":

nitrobenzenet-butylbenzeneethylbenzenechlorobenzene

NO2C CH3CH3

CH3CH2CH3Cl

O.

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A) Monosubstituted Benzenes: Some monosubstituted benzenes are

considered parent compounds because they occur so frequently:

CH3 CH CH2 OH COH

COOH NH2 SO3H

toluene styrene phenol benzaldehyde

benzoic acid aniline benzene sulfonic acid

O.

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B) Disubstituted Benzenes: There are three possible isomers of disubstituted

benzenes which are differentiated by the use of the prefixes ortho (o-), meta (m-) and para (p-) :

Note that when the two substituents are different, they are listed in alphabetical order:

Br

Br

Br

Br

Br

Bro-dibromobenzene m-dibromobenzene p-dibromobenzene

NO2

NO2

C2H5

Cl

F

I

m-dinitrobenzene o-chloroethylbenzene p-fluoroiodobenzene

O.

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B) Disubstituted Benzenes: If one of the substituent is part of a parent

compound, then the disubstituted benzene is named as a derivative of the parent compound:

Like mono substituted benzenes, certain disubstituted benzenes are referred to by their common names

OH

Cl

CH3

Br

NO2

COOHo-chlorophenol m-bromotoluene p-nitrobenzoic acid

CH3

CH3

CH3

CH3

CH3

CH3o-xylene m-xylene p-xylene

O.

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Nomenclature of Aromatic Compounds:Problems:(1) Name the following compounds:

(2) Draw the structures of the following compounds:a) 2-Phenyl pentaneb) p-Nitroethyl benzenec) o-Ethyl anilined) m-Bromobenzene sulfonic acide) 2,2-Dimethyl-1-phenyl butanef) m-Nitrobenzoic acidg) p-bromotolueneh) m-chloro benzaldehyde

FF

CH CH2

Cl

CH3

NO2

HO

C2H5

C CHCH2 Cl

O2N

CHCH3 CH CH3

Cl

CH2Br

a) b) c) d)

e) f) g) h)

O.

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C) Polysubstituted Benzenes: Polysubstituted benzenes contain three or more

substituents. In polysubstituted benzenes the ring must be

numbered and the position of the substituents must be specified by numbers.

The numbering starts with one substituent and continues around the ring so as to use the lowest possible numbers for the other substituents.

O

BrBr

Br

6

5

4

3

2

16

54

3

2

1

6

5

4

3

21

2,4,6-trinitro-toluene

2,4,6-tribromo-phenol

1,2,3-trichloro-benzene

2

2NO

NO

NO

C

2

H3H

l

l

lC C

C

O.

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Reactions of Aromatic Compounds: Aromatic compounds are stable molecules. The stability is attributed to their resonance energy,

which due to the extensive delocalization of the electrons in the ring.

Benzenes and aromatic compounds in general tend to undergo substitution rather than addition reactions in order to preserve the delocalized system.

Because the aromatic ring is extremely stable, a highly reactive agent is required to react with the benzene ring.

The aromatic ring is also an electron-rich system. Attack on the ring takes place by means of electron-

deficient agent, an Electrophile (E+).

The role of catalyst in electrophilic aromatic substitution:

In almost all electrophilic aromatic substitution a catalyst is needed for a reaction to take place.

The catalysts are usually Lewis acids or a protonic acid. The role of these catalysts is to generate powerful

electrophiles.

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Reactions of Aromatic Compounds: General equation:

Examples of substitutions of benzenes:1. Halogenation:

2. Alkylation (Friedel-Crafts Alkylation):

+ catalyst

E

+ H - BE+ B

-

+ Br2FeBr3

Br

+ H Br

+ H Cl

Cl

FeCl3Cl2+

+ R Cl AlCl3

R

+ H Cl

alkylchloride

alkylbenzene

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Reactions of Aromatic Compounds: Examples of substitutions of benzenes:

3. Acylation (Friedel-Crafts Acylation):

4. Nitration:

R C Cl

O+

C R

O

+ HCl

alkyl phenylketone

acyl chloride

HNO3+H2SO4

NO2

H2O+

O.

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Reactions of Aromatic Compounds:Other Reactions Of Aromatic Compounds:1. Halogenation of the alkyl side chain:

2. Conversion of the alkyl side chain into carboxyl group (COOH):

Reaction of alkylbenzenes with hot potassium permanganate, a strong oxidizing agent, forms benzoic acid: (Regardless of the length of alkyl chain):

CH3

Br2+

CH2 Br

+ HBr

benzylbromidetoluene

sun-light

CH3

hot KMnO4

COOH

H2O+

toluene benzoic acid

a)

CH3

CH3

hot KMnO4

COOH

COOH

H2O+ 2

p-xylene terephthalic acid

c)

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Reactions of Aromatic Compounds:Other Reactions Of Aromatic Compounds:3. Conversion of Nitro compounds to amines:

Aromatic nitro compounds can be converted to aromatic amines by treatment with (Fe) and (HCl):

Problems:

Starting from benzene show how can you prepare the following aromatic compounds:

1. Benzyl chloride2. Benzoic acid3. Aniline4. Cyclohexyl benzene

NO2

Fe, HCl

NH2

nitrobenzene aniline


Stereochemistry

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Stereoisomerism Stereochemistry is a branch of chemistry that deals

with description of the spatial arrangements of atoms in molecules and with the consequences of molecular shape of the physical and chemical properties of molecules.

It is also defined as: The study of the three-dimensional orientations of

the atoms in molecules as they undergo a chemical reaction.

Stereoisomerism: is one aspect of Stereochemistry.

Isomers are compounds that have the same molecular formula, but are not identical.

There are two main classes of isomers:a) Structural isomers (or constitutional isomers)b) Stereoisomers

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Stereoisomerism

Isomers

Structural Isomers Stereoisomers

Geometric(cis - trans)

OpticalIsomers

There are two main classes of isomers:

Structural isomers (or constitutional isomers) Stereoisomers

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Stereoisomerisma) Structural Isomers: are compounds that have the same molecular formulas

but different structural formulas.

b) Stereoisomers: are compounds that have not only the same molecular

formulas but also identical sequence of bonding of the atoms in the molecule.

They differ in one respect only: the spatial orientation or configuration of their

constituent atoms.

Types of Stereoisomerism: 1. Geometric isomerism:

The restricted rotation about the carbon-carbon double bond and the resulting planar geometry is responsible for geometric isomerism.

2. Optical isomerism: It manifests itself by its effect on plane-polarized light.

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Optical Activity Optical isomers are optical active substances. An optical active substance is any compound that has the

ability to change the direction of plane-polarized light or to rotate it.

The rotation itself is called Optical Activity.

Measurement of optical activity Optical activity can be measured by placing a solution of the

compound suspected of being optically active in front of the plane-polarized light and seeing what happens to the light transmitted by the solution.

Polarimeters are used to measure the optical activities of substances.

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Optical Activity Plane-polarized light passing through an optically

active solution is rotated by a certain number of degrees alpha () called the observed rotation.

If the observed rotation found to be to the right (clockwise rotation), the optically active compound is designated as dextrorotatory with the symbol (+):

If the optically active compound rotates plane-polarized light to the left (counter clockwise rotation) is termed levorotatory, with the symbol (-)

Example: 2-chlorobutane has two optical isomers:

1. (+) 2-chlorobutane: this isomer rotates plane-polarized light to the right (clockwise), and

2. (-) 2-chlorobutane: It rotates plane-polarized light to the left (counter clockwise)

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Optical Activity The observed rotation () depends upon:

1. The concentration of the solution (c)2. the length of the polarimeter tube (l)3. the temperature (T)4. the wavelength of the light ()

Example: observed rotation of (+)-2-butanol:

Concentration Polarimeter tube 1) 2.0 g/ml 10 cm +27.8o

2) 2.0 g/ml 5 cm +13.9o

3) 4.0 g/ml 10 cm +55.6o

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Optical Activity Specific Rotation The value of the optical rotation of a compound under

standard conditions is called the specific rotation [] , and is defined by the equation:

Where: []=Specific rotation (in degrees) T=Temperature (in 0C) =Wavelength of light*) used (in nm**), usually sodium D-line

589 nm) α=Experimental observed rotation (in degrees) l=Length of polarimeter tube (in dm, decimeters: 1 dm = 10

cm) c=Concentration of the solution (in g/ml)

*) Wavelength is the linear distance between successive maximum or minimum of a wave

**) nm (nanometer) = 10 -9 m

α[ ]T α

λ=

l x c

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Optical Activity

Problems:

1. Calculate the specific rotation for a solution of 2.0 g/ml of (+)-2-butanol in a 10 cm polarimeter tube and the observed rotation was +27.800

2. Calculate the specific rotation of cholesterol if the concentration of the solution is 5.20 g/ml per 100 ml, the length of the polarimeter tube is 5 cm and the observed rotation was -2.500

3. Search the following web site in the internet for further problems: http://www.chem.uic.edu/web1/OCOL-II/WIN/STEREO/31/FRAMES.HTM

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Optical Activity andStructure of Compounds We can predict whether or not a compound will be

optically active based on its structure. In order to exhibit optical activity, a compound must

be chiral, that is, it cannot be superimposed on its mirror image.

Example: The mirror image (the "left" hand is not

superimposable of the "right" hand):

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Optical Activity andStructure of Compounds Compounds that are mirror images of each other

and are not superimposable, are called enantiomers. (from Greek: Enantio = opposite, Meros = part)

Enantiomers are optical isomers. They have the same properties except that they rotate plane-polarized light in opposite directions (but in equal amounts)

C

CH3

CH3CH2

HCl Cl

C

CH3

CH2CH3

H

mirror

the two optical isomers (enantiomers)of 2-chlorobutane

* *

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Optical Activity andStructure of Compounds Chiral compounds have a chiral center, which is an

atom (usually carbon, called: stereogenic carbon atom) bonded to four different atoms or groups.

The chiral center is usually indicated by an asterisk(*)

Compounds without chiral center are called achiral (not chiral).

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Problems

1. The formulas for a series of compounds are given below. Indicate the chiral carbon(s) in each molecule with an asterisk (*):

A) 1-chloropentaneB) 2-chloropentaneC) 3-chloropentaneD) 1-chloro-2-methyl pentaneE) 2-chloro-2-methyl pentaneF) 3-chloro-2-methyl pentaneG) 4-chloro-2-2methyl pentaneH) 1-chloro-2-bromo butaneI) (Amphetamine)

J) (Penicillamine)

2. Search the following web site in the internet for further problems: http://www.chem.uic.edu/web1/OCOL-II/WIN/STEREO/21/FRAMES.HTM

CH2 CH

NH2

CH3

CH3 C

SH

CH3

CH

NH2

COOH

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Racemic Mixture

o When enantiomers are mixed together, the rotation caused by a molecule of one isomer is exactly cancelled by an equal and opposite rotation caused by a molecule of its enantiomer.

An equal mixture of enantiomers called a racemic mixture with prefix (±)

Example: if a given amount of (+) amphetamine causes the observed

rotation to be +6.250 then an equal amount of (-) amphetamine under the same

conditions will result in an observed rotation of -6.250 Thus, in mixing equal weights of the two enantiomers (a

racemic mixture) we find that the net observed rotation is zero.

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Diastereomers and Van't Hoff's Rule

The Van't Hoff's Rule stated that:"the maximum number of stereoisomers for a structure

with (n) different chiral centres is equal to 2n“Examples: 2-methyl-1-butanol has one chiral center (n=1) and two

enantiomers (2n =21 =2)

Enantiomers of 2-methyl-1-butanol

CH2

C CH3H

CH2 CH3

OH

*

CH2

C CH3H

CH2 CH3

OH

* *CH2

C HCH3

CH2 CH3

OH

(I) (II)

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Diastereomers and Van't Hoff's Rule 2,3,4-Trihydroxy-butanal has two unlike chiral centers

(n=2) and can therefore exist as four possible isomers (2n =22 =4)

These four stereoisomers are

COH

C OHH

C OHH

CH2OH

*

*

*

*C OHH

C OHH

CH2OH

CHO CHO

C HHO

C HHO

CH2OH

*

*

(I) (II) (III) (IV)

*

*C HHO

C OHH

CH2OH

CHOCHO

C OHH

C HHO

CH2OH

*

*

enantiomers diastereomers enantiomers

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Diastereomers and Van't Hoff's Rule Diastereomers: are stereoisomers that are not

mirror images of each other. Unlike enantiomers, which have the same physical

properties, Diastereomers have different boiling points and melting points, refractive indices, solubilities etc

Problem: Give all stereoisomers of the following compounds:1. 2-aminobutane2. 3-nitro-2-butanol3. 2-amino-4-bromo-3-pentanol

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Meso Compounds:

A meso compound is an optically inactive compound even though it possesses more than one chiral center.

Example Tartaric acid

Tartaric acid has two chiral centers. According to the van't Hoff's rule, we could expect that tartaric acid

forms a maximum of four (22) stereoisomers. In fact there are only three possible stereoisomers for tartaric acid

as shown below:

COOH

CHOH

CHOH

COOH

*

*

*

*C OHH

C HHO

COOH

COOH COOH

C HHO

C OHH

COOH

*

*

(I) (II) (III) (IV)

*

*C HHO

C HHO

COOH

COOHCOOH

C OHH

C OHH

COOH

*

*

enantiomers diastereomers identical (MESO)

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Problems

1. Draw ALL stereoisomers of 3,4-dinitrobutane, and indicate which isomers are enantiomers and which are diastereomers or meso compounds.

2. Search the following web site in the internet for further problems:

http://www.chem.uic.edu/web1/OCOL-II/WIN/STEREO/11/FRAMES.HTM http://www.chem.uic.edu/web1/OCOL-II/WIN/STEREO/101/FRAMES.HTM

Documents

Chemistry Lecture Notes 2010