Organic Chemistry A Levels A2

8/13/2019 Organic Chemistry A Levels A2

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AN INTRODUCTION TO

ORGANIC CHEMISTRY A guide for A level students

KNOCK HARDY PUBL ISHING 2008SPECIFICATIONS


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INTRODUCTION

This Powerpoint show is one of several produced to help students understandselected topics at AS and A2 level Chemistry. It is based on the requirements ofthe AQA and OCR specifications but is suitable for other examination boards.

Individual students may use the material at home for revision purposes or it may

be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are availablefrom the KNOCKHARDY SCIENCE WEBSITE at...

www.knockhardy.org.uk/sci.htm

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KNOCKHARDY PUBLISHING

ORGANIC CHEMISTRY


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CONTENTS Scope of organic chemistry

Special nature of carbon

Types of formulae Homologous series Functional groups Nomenclature Investigating molecules Revision check lis t

ORGANIC CHEMISTRY


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Before you start it would be helpful to

Recall how covalent bonding arises Recall simple electron pair repulsion theory

ORGANIC CHEMISTRY


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ORGANIC CHEMISTRY

Organic chemistry is the study of carbon compounds . It is such a complex branch ofchemistry because...

CARBON ATOMS FORM STRONG COVALENT BONDS TO EACH OTHER

THE CARBON -CARBON BONDS CAN BE SINGLE, DOUBLE OR TRIPLE

CARBON ATOMS CAN BE ARRANGED IN STRAIGHT CHAINS BRANCHED CHAINS

and RINGS

OTHER ATOMS/GROUPS OF ATOMS CAN BE PLACED ON THE CARBON ATOMS

GROUPS CAN BE PLACED IN DIFFERENT POSITIONS ON A CARBON SKELETON


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SPECIAL NATURE OF CARBON - CATENATION

CATENATION is the ability to form bonds between atoms of the same element .Carbon forms chains and rings, with single, double and triple covalent bonds, because itis able to FORM STRONG COVALENT BONDS WITH OTHER CARBON ATOMS

Carbon forms a vast number of carbon compounds because of the strength of the C-Ccovalent bond . Other Group IV elements can do it but their chemistry is limited due tothe weaker bond strength.

BOND ATOMIC RADIUS BOND ENTHALPY

C-C 0.077 nm +348 kJmol -1

Si-Si 0.117 nm +176 kJmol -1

The larger the atoms , the weaker the bond . Shielding due to filled inner orbitals and greaterdistance from the nucleus means that the shared electron pair is held less strongly .


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CHAINS AND RINGS

CARBON ATOMS CAN BE ARRANGED IN

STRAIGHT CHAINS

BRANCHED CHAINS

and RINGS

THE SPECIAL NATURE OF CARBON

You can also get a combination of rings and chains


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MULTIPLE BONDING AND SUBSTITUENTS

CARBON-CARBON COVALENT BONDS CAN BE SINGLE, DOUBLE OR TRIPLE



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CARBON-CARBON COVALENT BONDS CAN BE SINGLE, DOUBLE OR TRIPLE

DIFFERENT ATOMS / GROUPS OF ATOMS CAN BE PLACED ON THE CARBONS

The basic atom is HYDROGEN but groups containing OXYGEN , NITROGEN ,HALOGENS and SULPHUR are very common.

CARBON SKELETON FUNCTIONAL CARBON SKELETON FUNCTIONALGROUP GROUP

The chemistry of an organic compound is determined by its FUNCTIONAL GROUP



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ATOMS/GROUPS CAN BE PLACED IN DIFFERENT POSITIONS ON A CARBON SKELETON


THE C=C DOUBLE BOND IS IN A DIFFERENT POSITION

THE CHLORINE ATOM IS IN A DIFFERENT POSITION

PENT-1-ENE PENT-2-ENE

1-CHLOROBUTANE 2-CHLOROBUTANE


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TYPES OF FORMULAE - 1

MOLECULAR FORMULA C4H10The exact number of atoms of each

element present in the molecule

EMPIRICAL FORMULA C2H5The simplest whole number ratio of atoms in the molecule

STRUCTURAL FORMULA CH 3CH 2CH 2CH 3 CH 3CH(CH 3)CH 3 The minimal detail using conventionalgroups, for an unambiguous structure there are two possible structures

DISPLAYED FORMULA Shows both the relative placing of atomsand the number of bonds between them

H C C C C H

H H H H

H H H H

H C C C H

H H

H H H

H C H

H

THE EXAMPLE BEING

USED IS... BUTANE


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SKELETAL FORMULA A skeletal formula is used to show a simplified organic formula by removing hydrogen

atoms from alkyl chains, leaving just a carbon skeleton and associated functional groups


CH 2 CH 2

CH 2 CH 2

CH 2

CH 2

CYCLOHEXANE THALIDOMIDE

for


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SKELETAL FORMULA A skeletal formula is used to show a simplified organic formula by removing hydrogen

atoms from alkyl chains, leaving just a carbon skeleton and associated functional groups

GENERAL FORMULA Represents any member of for alkanes it is... C

nH

2n+2

a homologous series possible formulae... CH 4, C2H6 .... C99 H200

The formula does not apply to cyclic compounds such as cyclohexane is C 6H12- by joining the atoms in a ring you need fewer Hs


CH 2 CH 2

CH 2 CH 2

CH 2

CH 2

CYCLOHEXANE THALIDOMIDE

for


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HOMOLOGOUS SERIES

A series of compounds of similar structure in which each member differs from the nextby a common repeating unit, CH 2. Series members are called homologues and...

all share the same general formula. formula of a homologue differs from its neighbour by CH 2. (e.g. CH 4, C 2H6, ... etc ) contain the same functional group have similar chemical properties. show a gradual change in physical properties as molar mass increases. can usually be prepared by similar methods.

ALCOHOLS - FIRST THREE MEMBERS OF THE SERIES

CH 3OH C 2H5OH C 3H7OHMETHANOL ETHANOL PROPAN-1-OL


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FUNCTIONAL GROUPS

Organic chemistry is a vast subject so it is easier to split it into small sections for study.This is done by studying compounds which behave in a similar way because they havea particular atom, or group of atoms, FUNCTIONAL GROUP , in their structure.

Functional groups can consist of one atom, a group of atoms or multiple bondsbetween carbon atoms.

Each functional group has its own distinctive properties which means that theproperties of a compound are governed by the functional group(s) in it.

H H H H H

H C C C C C OH

H H H H H

H H H H H

H C C C C C NH2

H H H H H

Carbon Functional Carbon Functionalskeleton Group = AMINE skeleton Group = ALCOHOL


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COMMON FUNCTIONAL GROUPS

GROUP ENDING GENERAL FORMULA EXAMPLE

ALKANE - ane RH C 2H6 ethane

ALKENE - ene C 2H4 ethene

ALKYNE - yne C 2H2 ethyne

HALOALKANE halo - RX C 2H5Cl chloroethane

ALCOHOL - ol ROH C 2H5OH ethanol

ALDEHYDE -al RCHO CH 3CHO ethanal

KETONE - one RCOR CH 3COCH 3 propanone CARBOXYLIC ACID - oic acid RCOOH CH 3COOH ethanoic acid

ACYL CHLORIDE - oyl chloride RCOCl CH 3COCl ethanoyl chloride

AMIDE - amide RCONH 2 CH 3CONH 2 ethanamide

ESTER - yl - oate RCOOR CH 3COOCH 3 methyl ethanoate

NITRILE - nitrile RCN CH 3CN ethanenitrile AMINE - amine RNH 2 CH 3NH2 methylamine

NITRO nitro- RNO 2 CH 3NO 2 nitromethane

SULPHONIC ACID - sulphonic acid RSO 3H C 6H5SO 3H benzene sulphonic acid

ETHER - oxy - ane ROR C 2H5OC 2H5 ethoxyethane


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COMMON FUNCTIONAL GROUPS

ALKANE

ALKENEALKYNE

HALOALKANE

AMINE

NITRILE

ALCOHOL

ETHER

ALDEHYDE

KETONE

CARBOXYLIC ACID

ESTER

ACYL CHLORIDE

AMIDE

NITRO

SULPHONIC ACID


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HOW MANY STRUCTURES?

Draw legitimate structures for each molecular formula and classify each one accordingto the functional group present. Not all the structures represent stable compounds .

carbon atoms have 4 covalent bonds surrounding themoxygen atoms 2nitrogen atoms 3hydrogen 1halogen atoms 1

C 2H6 ONE

C 3H7Br TWO

C 4H8 FIVE - 3 with C=C and 2 ring compounds with all C- Cs

C 2H6O TWO - 1 with C-O-C and 1 with C-O-H

C 3H6O SIX - 2 with C=O, 2 with C=C and 2 with rings

C 2H7N TWO

C 2H4O 2 SEVERAL - Only 2 are stable

C 2H3N TWO


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HOW MANY STRUCTURES?

Draw legitimate structures for each molecular formula and classify each one accordingto the functional group present. Not all the structures represent stable compounds .

carbon atoms have 4 covalent bonds surrounding themoxygen atoms 2nitrogen atoms 3hydrogen 1halogen atoms 1

C 2H6 ONE

C 3H7Br TWO

C 4H8 FIVE - 3 with C=C and 2 ring compounds with all C- Cs

C 2H6O TWO - 1 with C-O-C and 1 with C-O-H

C 3H6O SIX - 2 with C=O, 2 with C=C and 2 with rings

C 2H7N TWO

C 2H4O 2 SEVERAL - Only 2 are stable

C 2H3N TWO


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NOMENCLATURE

Ideally a naming system should tell you everything about a structure without ambiguity.There are two types of naming system commonly found in organic chemistry;

Trivial : based on some property or historical aspect;the name tells you little about the structure

Systematic : based on an agreed set of rules (I.U.P.A.C);exact structure can be found from the name (and vice-versa).

HOMOLOGOUS SERIES trivial name systematic name example(s)paraffin alkane methane, butaneolefin alkene ethene, butenefatty acid alkanoic (carboxylic) acid ethanoic acid

INDIVIDUAL COMPOUNDStrivial name derivation systematic namemethane methu = wine (Gk.) methane (CH 4)butane butyrum = butter (Lat.) butane (C 4H10 )acetic acid acetum = vinegar (Lat.) ethanoic acid (CH 3COOH)


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I.U.P.A.C. NOMENCLATURE

A systematic name has two main parts.

STEM number of carbon atoms in longest chain bearing the functional group +a prefix showing the position and identity of any side-chain substituents.

Apart from the first four, which have trivialnames , the number of carbons atoms is indicatedby a prefix derived from the Greek numberingsystem.

The list of alkanes demonstrate the use of prefixes.

The ending -ane is the same as they are all alkanes.

Prefix C atoms Alkane

meth- 1 methane

eth- 2 ethaneprop- 3 propanebut- 4 butanepent- 5 pentanehex- 6 hexanehept- 7 heptaneoct- 8 octane

non- 9 nonanedec- 10 decane

Working out which is the longest chain can pose a problem with larger molecules.


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CH 2 CH 3 CH 2 CH 2 CH 3 CH 2 CH 2 CH 2

CH 3

CH 3

CH 3 CH

2 CH

2 CH

2

CH 3

CH 2 CH 2 CH

2 CH

3 CH

3


How long is a chain?

Because organic molecules are three dimensional and paper is two dimensional itcan confusing when comparing molecules. This is because...

1. It is too complicated to draw molecules with the correct bond angles2. Single covalent bonds are free to rotate

All the following written structures are of the same molecule - PENTANE C 5H12

A simple way to check is to run a finger along the chain and see how many carbonatoms can be covered without reversing direction or taking the finger off the page.In all the above there are... FIVE CARBON ATOMS IN A LINE .


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CH 2 CH 3 CH 2 CH 2 CH CH 3 CH 3

CH 2 CH 3 CH 3 CH CH 2 CH 3


How long is the longest chain?

Look at the structures and work out how many carbon atoms are in the longest chain.

CH 3

CH 3 CH CH 2

CH 2 CH 3 CH CH 3

THE ANSWERS AREON THE NEXT SLIDE


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CH2 CH3 CH2 CH2 CH CH3 CH 3

CH2 CH 3 CH3 CH CH2 CH3


How long is the longest chain?

Look at the structures and work out how many carbon atoms are in the longest chain.

CH3

CH 3 CH CH2

CH2 CH3 CH CH 3

LONGEST CHAIN = 5

LONGEST CHAIN = 6

LONGEST CHAIN = 6


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SUBSTITUENTS Many compounds have substituents (additional atoms, or groups)attached to the chain. Their position is numbered.

A systematic name has two main parts .

SUFFIX An ending that tells you which functional group is present

See if any functional groups are present.Add relevant ending to the basic stem.

In many cases the position of the functional

group must be given to avoid any ambiguity

Functional group Suffix

ALKANE - ANEALKENE - ENEALKYNE - YNEALCOHOL - OLALDEHYDE - ALKETONE - ONEACID - OIC ACID

1-CHLOROBUTANE 2-CHLOROBUTANE


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SIDE-CHAIN carbon based substituents are named before the chain name.they have the prefix -yl added to the basic stem (e.g. CH 3 is methyl).

Number the principal chain from one end to give the lowest numbers .

Side-chain names appear in alphabetical order butyl, ethyl, methyl, propyl Each side-chain is given its own number .

If identical side-chains appear more than once, prefix with di, tri, tetra, penta, hexa

Numbers are separated from names by a HYPHEN e.g. 2-methylheptane

Numbers are separated from numbers by a COMMA e.g. 2,3-dimethylbutane

Alkyl radicals methyl CH 3 - CH 3 ethyl CH 3- CH 2- C 2H5 propyl CH 3- CH 2- CH 2- C 3H7



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SIDE-CHAIN carbon based substituents are named before the chain name.they have the prefix -yl added to the basic stem (e.g. CH 3 is methyl).

Number the principal chain from one end to give the lowest numbers .

Side-chain names appear in alphabetical order butyl, ethyl, methyl, propyl Each side-chain is given its own number .

If identical side-chains appear more than once, prefix with di, tri, tetra, penta, hexa

Numbers are separated from names by a HYPHEN e.g. 2-methylheptane

Numbers are separated from numbers by a COMMA e.g. 2,3-dimethylbutane

Example longest chain 8 (it is an octane)3,4,6 are the numbers NOT 3,5,6order is ethyl, methyl, propyl

3-ethyl-5-methyl-4-propyloctane

Alkyl radicals methyl CH 3 - CH 3 ethyl CH 3- CH 2- C 2H5 propyl CH 3- CH 2- CH 2- C 3H7

CH 3

CH 2 CH 3 CH CH 2

CH 2 CH 3 CH CH

CH 2 CH 2 CH 3 CH 2

CH 3



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Apply the rules and name these alkanes

CH 3

CH 3 CH CH 2

CH 2 CH 3 CH CH 3

THE ANSWERS ARE ON THE NEXT SLIDE


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CH 3

CH 3 CH CH 2

CH 2 CH 3 CH CH 3




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CH2 CH 3 CH3 CH CH2 CH3


CH 3

CH 3 CH CH 2

CH 2 CH 3 CH CH 3

Longest chain = 5 so it is a pentane

A CH 3, methyl, group is attached to thethird carbon from one end...

3-methylpentane




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CH2 CH3 CH2 CH2 CH CH3 CH 3



CH 3

CH 3 CH CH 2

CH 2 CH 3 CH CH 3



3-methylpentane



Longest chain = 6 so it is a hexane

A CH 3, methyl, group is attached to thesecond carbon from one end...

2-methylhexane


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CH3

CH 3 CH CH2

CH2 CH3 CH CH 3



3-methylpentane




A CH 3, methyl, group is attached to thesecond carbon from one end...

2-methylhexane


CH 3, methyl, groups are attached to thethird and fourth carbon atoms(whichever end you count from).

3,4-dimethylhexane


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NAMING ALKENES

Length In alkenes the principal chain is not always the longest chainIt must contain the double bond the name ends in -ENE

Position Count from one end as with alkanes.Indicated by the lower numbered carbon atom on one end of the C=C bond

5 4 3 2 1

CH 3CH 2CH=CH CH 3 is pent-2-ene (NOT pent-3-ene)

Side-chain Similar to alkanes position is based on the number allocated to the double bond

1 2 3 4 1 2 3 4CH 2 = CH (CH 3)CH 2CH 3 CH 2 = CHCH (CH 3)CH 3

2-methylbut-1-ene 3-methylbut-1-ene


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WHICH COMPOUND IS IT?

Elucidation of the structures of organic compounds - a brief summary

Organic chemistry is so vast that the identification of a compound can be involved. Thecharacterisation takes place in a series of stages (see below). Relatively large amountsof substance were required to elucidate the structure but, with modern technology andthe use of electronic instrumentation, very small amounts are now required.

Elemental composition

One assumes that organic compounds contain carbon and hydrogen but it can beproved by letting the compound undergo combustion. Carbon is converted to carbondioxide and hydrogen is converted to water.

Percentage composition by massFound by dividing the mass of an element present by the mass of the compoundpresent, then multiplying by 100 . Elemental mass of C and H can be found by allowingthe substance to undergo complete combustion. From this one can find...

mass of carbon = 12/44 of the mass of CO 2 producedmass of hydrogen = 2/18 of the mass of H 2O produced


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INVESTIGATING MOLECULES

Empirical formula The simplest ratio of elements present in the substance . It is calculated by dividing themass or percentage mass of each element by its molar mass and finding the simplestratio between the answers. Empirical formula is converted to the molecular formulausing molecular mass.


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Empirical formulaThe simplest ratio of elements present in the substance. It is calculated by dividing themass or percentage mass of each element by its molar mass and finding the simplestratio between the answers. Empirical formula is converted to the molecular formulausing molecular mass.

Molecular mass Traditionally found out using a variety of techniques such as ... volumetric analysis ormolar volume methods (Dumas, Victor-Meyer or gas syringe experiments) . Mass

spectrometry is now used. The m/z value of the molecular ion and gives the molecularmass. The fragmentation pattern gives information about the compound.


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Molecular massTraditionally found out using a variety of techniques such as ... volumetric analysis ormolar volume methods (Dumas, Victor-Meyer or gas syringe experiments). Mass


Molecular formula The molecular formula is an exact multiple of the empirical formula . Comparing themolecular mass with the empirical mass allows one to find the true formula. e.g.

if the empirical formula is CH (relative mass = 13) and the molecular mass is 78the molecular formula will be 78/13 or 6 times the empirical formula i.e. C 6H6 .


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Molecular massTraditionally found out using a variety of techniques such as ... volumetric analysis ormolar volume methods (Dumas, Victor-Meyer or gas syringe experiments). Mass


Molecular formulaThe molecular formula is an exact multiple of the empirical formula. Comparing themolecular mass with the empirical mass allows one to find the true formula. e.g.


Structural formula Because of the complexity of organic molecules, there can be more than one structurefor a given molecular formula. To work out the structure, different tests are carried out.


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Empirical formula The simplest ratio of elements present in the substance . It is calculated by dividing themass or percentage mass of each element by its molar mass and finding the simplestratio between the answers. Empirical formula is converted to the molecular formulausing molecular mass.

Molecular mass Traditionally found out using a variety of techniques such as ... volumetric analysis ormolar volume methods (Dumas, Victor-Meyer or gas syringe experiments) . Mass


Molecular formula The molecular formula is an exact multiple of the empirical formula . Comparing themolecular mass with the empirical mass allows one to find the true formula. e.g.


Structural formula Because of the complexity of organic molecules, there can be more than one structurefor a given molecular formula. To work out the structure, different tests are carried out.


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Chemical Chemical reactions can identify the functional group(s) present .

Spectroscopy IR detects bond types due to absorbance of i.r. radiation

NMR gives information about the position and relativenumbers of hydrogen atoms present in a molecule

Confirmation By comparison of IR or NMR spectra andmass spectrometry


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REVISION CHECK

What should you be able to do?

Recall and explain the reasons for the large number of carbon based compoundsBe able to write out possible structures for a given molecular formula

Recognize the presence of a particular functional group in a structure

Know the IUPAC rules for naming alkanes and alkenes

Be able to name given alkanes and alkenes when given the structure

Be able to write out the structure of an alkane or alkene when given its name

Recall the methods used to characterise organic molecules

CAN YOU DO ALL OF THESE? YES NO


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WELL DONE!Try some past paper questions


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AN INTRODUCTION TO

ORGANIC CHEMISTRY THE END

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Organic Chemistry A Levels A2