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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
Navigation is achieved by...
either clicking on the grey arrows at the foot of each page
or using the left and right arrow keys on the keyboard
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
THE SPECIAL NATURE OF CARBON
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MULTIPLE BONDING AND SUBSTITUENTS
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
THE SPECIAL NATURE OF CARBON
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MULTIPLE BONDING AND SUBSTITUENTS
ATOMS/GROUPS CAN BE PLACED IN DIFFERENT POSITIONS ON A CARBON SKELETON
THE SPECIAL NATURE OF CARBON
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
TYPES OF FORMULAE - 2
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
TYPES OF FORMULAE - 2
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
I.U.P.A.C. NOMENCLATURE
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
I.U.P.A.C. NOMENCLATURE
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
I.U.P.A.C. NOMENCLATURE
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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I.U.P.A.C. NOMENCLATURE
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
I.U.P.A.C. NOMENCLATURE
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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
I.U.P.A.C. NOMENCLATURE
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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
I.U.P.A.C. NOMENCLATURE
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 2 CH 3 CH 2 CH 2 CH CH 3 CH 3
CH 2 CH 3 CH 3 CH CH 2 CH 3
I.U.P.A.C. NOMENCLATURE
CH 3
CH 3 CH CH 2
CH 2 CH 3 CH CH 3
I.U.P.A.C. NOMENCLATURE
Apply the rules and name these alkanes
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CH 2 CH 3 CH 2 CH 2 CH CH 3 CH 3
CH2 CH 3 CH3 CH CH2 CH3
I.U.P.A.C. NOMENCLATURE
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
I.U.P.A.C. NOMENCLATURE
Apply the rules and name these alkanes
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CH2 CH3 CH2 CH2 CH CH3 CH 3
CH 2 CH 3 CH 3 CH CH 2 CH 3
I.U.P.A.C. NOMENCLATURE
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
I.U.P.A.C. NOMENCLATURE
Apply the rules and name these alkanes
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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CH 2 CH 3 CH 2 CH 2 CH CH 3 CH 3
CH 2 CH 3 CH 3 CH CH 2 CH 3
I.U.P.A.C. NOMENCLATURE
CH3
CH 3 CH CH2
CH2 CH3 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
I.U.P.A.C. NOMENCLATURE
Apply the rules and name these alkanes
Longest chain = 6 so it is a hexane
A CH 3, methyl, group is attached to thesecond carbon from one end...
2-methylhexane
Longest chain = 6 so it is a hexane
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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INVESTIGATING MOLECULES
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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INVESTIGATING MOLECULES
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 massTraditionally 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.
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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INVESTIGATING MOLECULES
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 massTraditionally 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.
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.
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 .
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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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.
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.
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 .
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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INVESTIGATING MOLECULES
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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You need to go over therelevant topic(s) again
Click on the button toreturn to the menu
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WELL DONE!Try some past paper questions
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AN INTRODUCTION TO
ORGANIC CHEMISTRY THE END