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S  0 ``  X*  0H `   Z*  0 `   Z*z  bA޽h @ ?Parchment ̙33 Default Design 0 zr@ ( )   0Ĥ P    P*    0x     R*  d  c $ ?    0  @  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  6P `P   P*    6 `   R*  H  0޽h ? ̙3380___PPT10.l:P 0X(   $+ X X 0ì P    X*  X 08     Z*  X 6 `P   X*  X 6t `   Z* H X 0޽h ? ̙3380___PPT10.l:Pw 0L0 P*(    0- AAN INTRODUCTION TO ORGANIC CHEMISTRY A guide for A level studentsTB(2 f0f<ffJ  C "A KTRE: . " <̙? x @ 0 0 ( <̙? x @L 0 0 ) 0|0x m OKNOCKHARDY PUBLISHING(2 f * 0$5"`>B u2008 SPECIFICATIONS2(2$   B  s *޽h ? ̙33e  0L0  (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0I   0D INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the 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 available from 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 (2 2(2 2   75, dB   <D?     0t{0P( OKNOCKHARDY PUBLISHING(2    0X _ORGANIC CHEMISTRY2(2(f<fB  s *޽h ? ̙33  0L0   `- (    0  < CONTENTS Scope of organic chemistry Special nature of carbon Types of formulae Homologous series Functional groups Nomenclature Investigating molecules Revision check listV 202 2dB  <D)P0Q  0Pp@ 0 0dB  @ <D)PXP   0X0@ 0 0  0{d@V 0 0  6~dx !@ 0 0  6S\4 @ 0 0  6hdx @ 0 0  6. \ @ 0 0  6 \ @ 0 0  6 d> @ 0 0  6 d @ 0 0  0԰X _ORGANIC CHEMISTRY2(2(f<fB  s *޽h ? ̙33k 0L0 p(    0Hx ^Before you start it would be helpful to& Recall how covalent bonding arises Recall simple electron pair repulsion theoryH*(2T 2)TdB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  <̙? x @J 0 0  0|‹X _ORGANIC CHEMISTRY2(2(f<fB  s *޽h ? ̙33  0L0 y q   (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  04΋q~ _ORGANIC CHEMISTRY2(2ff   0ӋF0 Organic chemistry is the study of carbon compounds. It is such a complex branch of chemistry 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@7%%#f  B  s *޽h ? ̙33  0L0    |J (  |dB | <D)P0Q | 0Pp@ 0 0dB |@ <D)PXP | 0X0@ 0 0  | 0Hq~ %SPECIAL NATURE OF CARBON - CATENATIOND&(2f f | 0|g 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 it is able to FORM STRONG COVALENT BONDS WITH OTHER CARBON ATOMSn 8U 2ff   | 01g? 'Carbon forms a vast number of carbon compounds because of the strength of the C-C covalent bond. Other Group IV elements can do it but their chemistry is limited due to the 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 greater distance from the nucleus means that the shared electron pair is held less strongly.@TPPxPP&PPi>!fd.f" "   "!#@  k  b P  | C (Acaten1p= B | s *޽h ? ̙33 0L0 7/ ( w [  0t  yCHAINS AND RINGS CARBON ATOMS CAN BE ARRANGED IN STRAIGHT CHAINS BRANCHED CHAINS and RINGS R2H@PP,@Pf3f3f3  z dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0P   C (Acaten2pE>     0/q~ jTHE SPECIAL NATURE OF CARBON2(2ff   005- ^ 2You can also get a combination of rings and chains3X3 3 B  s *޽h ? ̙33 0L0 qi (  }  07Q cMULTIPLE BONDING AND SUBSTITUENTS CARBON-CARBON COVALENT BONDS CAN BE SINGLE, DOUBLE OR TRIPLE "HP=H@P"f3<f d dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0P   C (Acaten6p    0Cq~ jTHE SPECIAL NATURE OF CARBON2(2ffB  s *޽h ? ̙33  0L0 z r   ( w z  0Lq  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 FUNCTIONAL GROUP GROUP The chemistry of an organic compound is determined by its FUNCTIONAL GROUP"HP=H@P@HPFi@PxPngPxFPFLXF"f3<f?ff1ff3   K  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  <̙? x @~ 0 0P  C (Acaten6p   0Lwq~ jTHE SPECIAL NATURE OF CARBON2(2ffP   C (A caten5p  B  s *޽h ? ̙33  0L0   = (  [  0~X0 iMULTIPLE BONDING AND SUBSTITUENTS ATOMS/GROUPS CAN BE PLACED IN DIFFERENT POSITIONS ON A CARBON SKELETONb"HPG@"f3Ff j dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0   0dq~ jTHE SPECIAL NATURE OF CARBON2(2ff8  +    V   C &A1brbut v + V   C &A2brbutx % 8  gdST  C $A pent1 T  C $A pent2x   0، .THE C=C DOUBLE BOND IS IN A DIFFERENT POSITION,/H. /   0X <  ,THE CHLORINE ATOM IS IN A DIFFERENT POSITION,-H, -   0Ж  y PENT-1-ENE PENT-2-ENE* @f    0dhe/8 ' 1-CHLOROBUTANE 2-CHLOROBUTANE*(@'f ( B  s *޽h ? ̙33  0L0 P H   (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ cTYPES OF FORMULAE - 12(2ffh   0佗  MOLECULAR FORMULA C4H10 The exact number of atoms of each element present in the molecule EMPIRICAL FORMULA C2H5 The simplest whole number ratio of atoms in the molecule STRUCTURAL FORMULA CH3CH2CH2CH3 CH3CH(CH3)CH3 The minimal detail using conventional groups, for an unambiguous structure there are two possible structures DISPLAYED FORMULA Shows both the relative placing of atoms and the number of bonds between them @   1           J3ff!f3fM  R   C *Abutdisple    B`?0= #THE EXAMPLE BEING USED IS... BUTANE@$ 2B  s *޽h ? ̙33 0L0 # (     0 0  /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 6@   dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0zq~ cTYPES OF FORMULAE - 22(2ffN   C &AchskelW9 L  C $A  thalpu0 >G  0zX09 (  m CYCLOHEXANE2 H f    0$ zXx (  m THALIDOMIDE2 H f    0zD Cfor$ 2B  s *޽h ? ̙33> 0L0     l (   B   0z 0B 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... CnH2n+2 a homologous series possible formulae... CH4, C2H6 .... C99H200 The formula does not apply to cyclic compounds such as cyclohexane is C6H12 - by joining the atoms in a ring you need fewer H s@?Px@  G  3@     C dB   <D)P0Q   0Pp@ 0 0dB  @ <D)PXP   0X0@ 0 0   0\Gzq~ cTYPES OF FORMULAE - 22(2ffN   C &AchskelW9 L   C $A  thalpu0 >G   0PMzX09 (  m CYCLOHEXANE2 H f     0QzXx (  m THALIDOMIDE2 H f     0WzD Cfor$ 2B   s *޽h ? ̙33  0L0 > 6   ( w dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0czq~ _HOMOLOGOUS SERIES2(2ff   0hz 0 qA series of compounds of similar structure in which each member differs from the next by a common repeating unit, CH2. Series members are called homologues and... all share the same general formula. formula of a homologue differs from its neighbour by CH2. (e.g. CH4, C2H6, ... 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.`@=Pxt  f]ff fffffff  N   C &A homolgp D    0z 0P  |,ALCOHOLS - FIRST THREE MEMBERS OF THE SERIES -H- -   0z im :B CH3OH C2H5OH C3H7OH METHANOL ETHANOL PROPAN-1-OLC@      # C B  s *޽h ? ̙33 0L0 JB 29@( w dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0zq~ _FUNCTIONAL GROUPS2(2ff   0 z 0 qOrganic 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 have a particular atom, or group of atoms, FUNCTIONAL GROUP, in their structure. Functional groups can consist of one atom, a group of atoms or multiple bonds between carbon atoms. Each functional group has its own distinctive properties which means that the properties of a compound are governed by the functional group(s) in it. F@   L F .   # b _     0zF .  K H H H H H H C C C C C OH H H H H H:L 21fB  6DoM  fB  6DoM  fB  6DoE  fB  6DoE  fB  6Do-  fB  6DoE  T   {  #   { fB  6Do  { fB  6Do { fB  6Do s fB  6Do s fB  6Do s T   {  #  ;  fB  6Do  { fB  6Do { fB  6Do s fB  6Do s fB  6Do s lB   <DosL s  L  F x  !# d   " 0Dz F x  L H H H H H H C C C C C NH2 H H H H HNM 21fB # 6Do fB $ 6Do fB % 6Do  fB & 6Do  fB ' 6Dow  fB ( 6Do  T   {  )# W  7{ fB * 6Do  { fB + 6Do { fB , 6Do s fB - 6Do s fB . 6Do s T   {  /# W ; 7 fB 0 6Do  { fB 1 6Do { fB 2 6Do s fB 3 6Do s fB 4 6Do s lB 5 <DoL   6 6ܴz 9D ;Carbon Functional Carbon Functional skeleton Group = AMINE skeleton Group = ALCOHOLf@n f + f  f  B  s *޽h ? ̙33M 0L0 0 {(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0zq~ fCOMMON FUNCTIONAL GROUPS2(2ffM   06 01  GROUP ENDING GENERAL FORMULA EXAMPLE ALKANE - ane RH C2H6 ethane ALKENE - ene C2H4 ethene ALKYNE - yne C2H2 ethyne HALOALKANE halo - RX C2H5Cl chloroethane ALCOHOL - ol ROH C2H5OH ethanol ALDEHYDE -al RCHO CH3CHO ethanal KETONE - one RCOR CH3COCH3 propanone CARBOXYLIC ACID - oic acid RCOOH CH3COOH ethanoic acid ACYL CHLORIDE - oyl chloride RCOCl CH3COCl ethanoyl chloride AMIDE - amide RCONH2 CH3CONH2 ethanamide ESTER - yl - oate RCOOR CH3COOCH3 methyl ethanoate NITRILE - nitrile RCN CH3CN ethanenitrile AMINE - amine RNH2 CH3NH2 methylamine NITRO nitro- RNO2 CH3NO2 nitromethane SULPHONIC ACID - sulphonic acid RSO3H C6H5SO3H benzene sulphonic acid ETHER - oxy - ane ROR C2H5OC2H5 ethoxyethanex;PgP9                         +                             '       (                                                       %                ,               I  &        7     B  )              1                >        &     &    B  s *޽h ? ̙33 0L0 WO@ ( z dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0Eq~ fCOMMON FUNCTIONAL GROUPS2(2ff   0b] TALKANE ALKENE ALKYNE HALOALKANE AMINE NITRILE ALCOHOL ETHER ALDEHYDE KETONE<Fn ZF T  0`uIY KCARBOXYLIC ACID ESTER ACYL CHLORIDE AMIDE NITRO SULPHONIC ACIDFxPP2LN  C &A fgrpsp])B  s *޽h ? ̙33p  0L0   P  (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0x}q~ bHOW MANY STRUCTURES?2(2ff   0 0 <BDraw legitimate structures for each molecular formula and classify each one according to the functional group present. Not all the structures represent stable compounds. carbon atoms have 4 covalent bonds surrounding them oxygen atoms 2 nitrogen atoms 3 hydrogen 1 halogen atoms 1 C2H6 ONE C3H7Br TWO C4H8 FIVE - 3 with C=C and 2 ring compounds with all C-C s C2H6O TWO - 1 with C-O-C and 1 with C-O-H C3H6O SIX - 2 with C=O, 2 with C=C and 2 with rings C2H7N TWO C2H4O2 SEVERAL - Only 2 are stable C2H3N TWOPn|@P @P^1{      .    '        " P   C (A CHNOval x   PAParchmentFB  s *޽h ? ̙33  0L0   `  (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ bHOW MANY STRUCTURES?2(2ff   0 0 <BDraw legitimate structures for each molecular formula and classify each one according to the functional group present. Not all the structures represent stable compounds. carbon atoms have 4 covalent bonds surrounding them oxygen atoms 2 nitrogen atoms 3 hydrogen 1 halogen atoms 1 C2H6 ONE C3H7Br TWO C4H8 FIVE - 3 with C=C and 2 ring compounds with all C-C s C2H6O TWO - 1 with C-O-C and 1 with C-O-H C3H6O SIX - 2 with C=O, 2 with C=C and 2 with rings C2H7N TWO C2H4O2 SEVERAL - Only 2 are stable C2H3N TWOPn|@P @P^1{      .    '        " P   C (A CHNOval B  s *޽h ? ̙33  0L0 _ W p  (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ Z NOMENCLATURE2 (2 ff   0, 0} s]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, butane olefin alkene ethene, butene fatty acid alkanoic (carboxylic) acid ethanoic acid INDIVIDUAL COMPOUNDS trivial name derivation systematic name methane methu = wine (Gk.) methane (CH4) butane butyrum = butter (Lat.) butane (C4H10) acetic acid acetum = vinegar (Lat.) ethanoic acid (CH3COOH)@H@H@c&] kff=fD) -  8   "          l  '  F   B  s *޽h ? ̙33|  0L0     ( w dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0Cq~ eI.U.P.A.C. NOMENCLATURE2(2ff   0H0 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.&H@&*ffB&    q   0xU  Apart from the first four, which have trivial names, the number of carbons atoms is indicated by a prefix derived from the Greek numbering system. The list of alkanes demonstrate the use of prefixes. The ending -ane is the same as they are all alkanes.@$f0/9  Z  /        BLc/jJ A  qPrefix C atoms Alkane meth- 1 methane eth- 2 ethane prop- 3 propane but- 4 butane pent- 5 pentane hex- 6 hexane hept- 7 heptane oct- 8 octane non- 9 nonane dec- 10 decane`@f f f fl    V                0vjJ0 PWorking out which is the longest chain can pose a problem with larger molecules.Q@Qf Q B  s *޽h ? ̙33  0L0 L D 0@(  8   < lB  <Do]N]  6|+ SCH24 2   6Y SCH34 2 lB  <Do]]  6 SCH24 2 lB  <Do]]   6~` SCH24 2 lB   <DoW'W   6   SCH34 2 8 `   x  ' 6r. SCH24 2 lB ) <Do .  * 6(  SCH24 2  + 68 SCH24 2 lB , <Do6 - 6< SCH34 2 lB . <DoOTOfB  6Doa  0`  SCH34 2 8    ~k%   6X  SCH34 2 lB  <Do==   6ĺ4 SCH24 2 lB  <DoBB  6 SCH24 2   6Ł\ SCH24 2 lB  <Do<^<  6ʁY SCH34 2 lB  <Do8 sMN  g s  x 0TvN SCH24 2  { 0$ׁ&TN SCH24 2 fB | 6Do~ } 0܁#QK SCH24 2 fB ~ 6Do  0sQ/K SCH34 2 lB  <DoL  0tMG SCH34 2 fB  6Do77{  0lq~ eI.U.P.A.C. NOMENCLATURE2(2ff  0F iHow long is a chain? Because organic molecules are three dimensional and paper is two dimensional it can confusing when comparing molecules. This is because... 1. It is too complicated to draw molecules with the correct bond angles 2. Single covalent bonds are free to rotate All the following written structures are of the same molecule - PENTANE C5H12 H@JPZ.PxP@wfB  j   0  0 jA simple way to check is to run a finger along the chain and see how many carbon atoms 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.Z@n9  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0H  0޽h ? ̙33 0L0 *"3p(  Z8 E  9 lB   <DoS   68 SCH24 2    6V SCH34 2 lB   <Do   68 SCH24 2 lB  <Do  6l&e! SCH24 2 lB  <Do!  6$, BCH$ 2lB  <Do>    6/E  SCH34 2  * 0p1} SCH34 2 fB + 6Do|W|8 x OVlB  <Do  69| SCH24 2   6@? SCH34 2 lB  <Do|  6Ha SCH34 2  % 64ML BCH$ 2lB & <Do   ' 6QI SCH24 2 lB ( <Do^ M 0VIxr SCH34 2 fB N 6DoF  0\q~ eI.U.P.A.C. NOMENCLATURE2(2ffI  0Pb sHow long is the longest chain? Look at the structures and work out how many carbon atoms are in the longest chain.FHU@T t  F  * R     0i* r$  SCH34 2 fB  6Dox  0@o%VP SCH34 2   0XtVmP BCH$ 2fB  6DoO  0xvS jM SCH24 2 fB  6Do"!"fB  6Do" "i T >X ?d  #  F RfB  6Do    0x~j d  SCH24 2   0t>d ^  SCH34 2 fB  6DoD    0(j ?d  BCH$ 2  0̏X ?R  SCH34 2 fB  6Do.    0,1Bkh !THE ANSWERS ARE ON THE NEXT SLIDE."H! " dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0H  0޽h ? ̙33! 0L0 R!J!5  (  F E    fB  6DoS  08 eCH2F 2   0ܥV eCH3F 2 fB  6Do  0  eCH2F 2 fB  6Do   0e! eCH2F 2 fB   6Do!   0 TCH6 2fB   6Do>     0E  eCH3F 2   0} SCH34 2 fB  6Do|W|OF x " VfB # 6Do $ 0Ą| eCH2F 2  % 0ʄ SCH34 2 fB & 6Do| ' 0τa eCH3F 2  ( 0 քL TCH6 2fB ) 6Do   * 0LۄI eCH2F 2 fB + 6Do^ , 0Ixr eCH3F 2 fB - 6DoF i 0q~ eI.U.P.A.C. NOMENCLATURE2(2ffI j 0 sHow long is the longest chain? Look at the structures and work out how many carbon atoms are in the longest chain.FHU@T t 8 8  * R   l 6* r$  eCH3F 2 lB n <Dox o 6%VP SCH34 2  r 6$VmP TCH6 2lB s <DoO t 6S jM eCH2F 2 lB u <Do"!"lB v <Do" "i T >X ?d  w#  F RfB x 6Do   y 0 j d  eCH2F 2  z 0>d ^  eCH3F 2 fB { 6DoD   | 0j ?d  TCH6 2 } 0xX ?R  SCH34 2 fB ~ 6Do.    0" l QLONGEST CHAIN = 5$ 2  0\&    QLONGEST CHAIN = 6$ 2  0(+$    QLONGEST CHAIN = 6$ 2dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0H  0޽h ? ̙33H  0L0    v (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  006q~ eI.U.P.A.C. NOMENCLATURE2(2ffw   0; 0 SUBSTITUENTS Many compounds have substituents (additional atoms, or groups) attached to the chain. Their position is numbered.L@ tff&!   U z   0CF hA systematic name has two main parts. SUFFIX An ending that tells you which functional group is present&HC@$: i r   0,Mf B See if any functional groups are present. Add relevant ending to the basic stem. In many cases the position of the functional group must be given to avoid any ambiguity8@QfX  C   BX/jJ   }Functional group Suffix ALKANE - ANE ALKENE - ENE ALKYNE - YNE ALCOHOL - OL ALDEHYDE - AL KETONE - ONE ACID - OIC ACID$`e@fffffff  ~ N   C &A1brbut Ar N  C &A2brbut @ l   0g  w'1-CHLOROBUTANE 2-CHLOROBUTANE (H(f ( B  s *޽h ? ̙33  0L0   +0  ( \ dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0   0v0d  =KSIDE-CHAIN carbon based substituents are named before the chain name. they have the prefix -yl added to the basic stem (e.g. CH3 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 &@PPPPP@PPZ T f!  B-";. .    C    g       7   0袆jJ gAlkyl radicals methyl CH3 - CH3 ethyl CH3- CH2- C2H5 propyl CH3- CH2- CH2- C3H7&(`@`f f f &G    + 0xuq~ eI.U.P.A.C. NOMENCLATURE2(2ffB  s *޽h ? ̙33 0L0 QI$&$(  $dB $ <D)P0Q $ 0Pp@ 0 0dB $@ <D)PXP $ 0X0@ 0 0 $ 00 $SIDE-CHAIN carbon based substituents are named before the chain name. they have the prefix -yl added to the basic stem (e.g. CH3 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 3,5,6 order is ethyl, methyl, propyl 3-ethyl-5-methyl-4-propyloctane`@PPPPP@PPZP<@ T f!  B-";. . f   C    g        & 7  $ 0jJ gAlkyl radicals methyl CH3 - CH3 ethyl CH3- CH2- C2H5 propyl CH3- CH2- CH2- C3H7&(`@`f f f &G   F ! o  $  !   $ 0T q  aCH3B 2f3f3f3fB  $ 6Do  $ 0D]m ]CH2> 2fB $ 6Do $ 0m ]CH3> 2 $ 0m LCH. 2fB $ 6Do! $ 0 z ]CH2> 2f3f3fB $ 6DojQjfB $ 6DojQ j fB $ 6Dod   $ 0X% o ]CH2> 2f3f3 $ 0|*!  i _CH3@ 2f3f3f3fB $ 6Dog $ 0/Ro HCH* 2 $ 04 Ru HCH* 2fB $ 6Do L  $ 09 [ o aCH2B 2f3f3f3fB $ 6Do E fB $ 6Dod   $ 0X?  ]CH2> 2  $ 0|D! y _CH3@ 2 !$ 0I [  aCH2B 2fB "$ 6Do E fB #$ 6Do\  fB $$ 6Dog %$ 0O Ks  ]CH3> 2f3f3 &$ 0Tq~ eI.U.P.A.C. NOMENCLATURE2(2ffB $ s *޽h ? ̙33 0L0 QI35 (   &F E    e k fB   6DoS   0x\8 SCH24 2    0aV SCH34 2 fB   6Do   0g SCH24 2 fB   6Do   0ce! SCH24 2 fB   6Do!   0q BCH$ 2fB   6Do>     0LvE  SCH34 2    0{} SCH34 2 fB   6Do|W|F x   V#fB   6Do   0| SCH24 2    0І SCH34 2 fB   6Do|   0a SCH34 2    0䓈L BCH$ 2fB   6Do     0I SCH24 2 fB   6Do^   0pIxr SCH34 2 fB   6DoF   00q~ eI.U.P.A.C. NOMENCLATURE2(2ff   00 &Apply the rules and name these alkanes 'H'&    F  * R       02* r$  SCH34 2 fB   6Dox !  0%VP SCH34 2  "  0ܸVmP BCH$ 2fB #  6DoO $  0`S jM SCH24 2 fB %  6Do"!"fB &  6Do" "i T >X ?d  ' #  F RfB (  6Do   )  0Ĉj d  SCH24 2  *  0Ȉ>d ^  SCH34 2 fB +  6DoD   ,  0LΈj ?d  BCH$ 2 -  0҈X ?R  SCH34 2 fB .  6Do.   /  08؈1" !THE ANSWERS ARE ON THE NEXT SLIDE."H!   " dB 0  <D)P0Q 1  0Pp@ 0 0dB 2 @ <D)PXP 3  0X0@ 0 0H   0޽h ? ̙33 0L0 7/3<(  &F E   - 3 fB  6DoS  0p8 SCH24 2   0V SCH34 2 fB  6Do  0 SCH24 2 fB  6Do   0xވe! SCH24 2 fB   6Do!   0 BCH$ 2fB   6Do>     0E  SCH34 2   0} SCH34 2 fB  6Do|W|F x  VfB  6Do  0 | SCH24 2   0 SCH34 2 fB  6Do|  0a SCH34 2   0L BCH$ 2fB  6Do    0P I SCH24 2 fB  6Do^  0&Ixr SCH34 2 fB  6DoF  0t+q~ eI.U.P.A.C. NOMENCLATURE2(2ff F  * R  * R  00* r$  SCH34 2 fB   6Dox ! 0x5%VP SCH34 2  " 0P#VmP BCH$ 2fB # 6DoO $ 0?S jM SCH24 2 fB % 6Do"!"fB & 6Do" "i T >X ?d  '#  F RfB ( 6Do   ) 0tEj d  SCH24 2  * 0J>d ^  SCH34 2 fB + 6DoD   , 0Oj ?d  BCH$ 2 - 0lX ?R  SCH34 2 fB . 6Do.   2 0|[q~ eI.U.P.A.C. NOMENCLATURE2(2ff 4 0_0 &Apply the rules and name these alkanes 'H'&   dB 7 <D)P0Q 8 0Pp@ 0 0dB 9@ <D)PXP : 0X0@ 0 0H  0޽h ? ̙33a  0L0 480( w 0&F E  0 - 3 fB 0 6DoS 0 0Dm8 SCH24 2  0 0hV SCH34 2 fB 0 6Do 0 00x SCH24 2 fB 0 6Do  0 0}e! SCH24 2 fB  0 6Do!  0 0l BCH$ 2fB  0 6Do>    0 0|E  SCH34 2  0 0i} SCH34 2 fB 0 6Do|W|WF x 0 VfB 0 6Do 0 0| eCH2F 2  0 0x [CH3< 2fB 0 6Do| 0 0a eCH3F 2  0 0L TCH6 2fB 0 6Do   0 0<I eCH2F 2 fB 0 6Do^ 0 0ثIxr eCH3F 2 fB 0 6DoF 0 0q~ eI.U.P.A.C. NOMENCLATURE2(2ff F  * R 0 * R 0 0* r$  SCH34 2 fB 0 6Dox  0 0%VP SCH34 2  !0 0hVmP BCH$ 2fB "0 6DoO #0 0ŊS jM SCH24 2 fB $0 6Do"!"fB %0 6Do" "i T >X ?d  &0#  F RfB '0 6Do   (0 0ˊj d  SCH24 2  )0 0Њ>d ^  SCH34 2 fB *0 6DoD   +0 00֊j ?d  BCH$ 2 ,0 0ڊX ?R  SCH34 2 fB -0 6Do.  X .0 0Dr  zLongest chain = 5 so it is a pentane A CH3, methyl, group is attached to the third carbon from one end... 3-methylpentaneb{ 2*ff@f /0 0<q~ eI.U.P.A.C. NOMENCLATURE2(2ff 00 0,0 &Apply the rules and name these alkanes 'H'&   dB 30 <D)P0Q 40 0Pp@ 0 0dB 50@ <D)PXP 60 0X0@ 0 0H 0 0޽h ? ̙33! 0L0 \!T!58, (  ,F E  , - 3 fB , 6DoS , 0T8 eCH2F 2  , 0DV eCH3F 2 fB , 6Do , 0x eCH2F 2 fB , 6Do  , 0` e! eCH2F 2 fB  , 6Do!  , 0 TCH6 2fB  , 6Do>    , 0E  eCH3F 2  , 0L} [CH3< 2fB , 6Do|W|F x , VfB , 6Do , 0#| SCH24 2  , 0T) SCH34 2 fB , 6Do| , 00.a SCH34 2  , 02L BCH$ 2fB , 6Do   , 07I SCH24 2 fB , 6Do^ , 0p>Ixr SCH34 2 fB , 6DoF , 0Bq~ eI.U.P.A.C. NOMENCLATURE2(2ff F  * R , * R , 0H* r$  SCH34 2 fB , 6Dox  , 00M%VP SCH34 2  !, 0hEVmP BCH$ 2fB ", 6DoO #, 0VS jM SCH24 2 fB $, 6Do"!"fB %, 6Do" "i T >X ?d  &,#  F RfB ', 6Do   (, 0,]j d  SCH24 2  ), 0Pb>d ^  SCH34 2 fB *, 6DoD   +, 0gj ?d  BCH$ 2 ,, 0hlX ?R  SCH34 2 fB -, 6Do.  J ., 0rr  zLongest chain = 5 so it is a pentane A CH3, methyl, group is attached to the third carbon from one end... 3-methylpentaneT{ 2*O /, 0xq~ eI.U.P.A.C. NOMENCLATURE2(2ff 0, 0}0 &Apply the rules and name these alkanes 'H'&   W 1, 0Xr   yLongest chain = 6 so it is a hexane A CH3, methyl, group is attached to the second carbon from one end... 2-methylhexanebz 2)ffAfdB 3, <D)P0Q 4, 0Pp@ 0 0dB 5,@ <D)PXP 6, 0X0@ 0 0H , 0޽h ? ̙33N# 0L0 ""68(v"(  (.F E  ( - 3 fB ( 6DoS ( 0h8 SCH24 2  ( 0HV SCH34 2 fB ( 6Do ( 0H SCH24 2 fB ( 6Do  ( 0`e! SCH24 2 fB  ( 6Do!  ( 0 BCH$ 2fB  ( 6Do>    ( 0,E  SCH34 2  ( 0د} [CH3< 2fB ( 6Do|W|F x ( VfB ( 6Do ( 0,| SCH24 2  ( 0غ [CH3< 2fB ( 6Do| ( 0a SCH34 2  ( 0ČL BCH$ 2fB ( 6Do   ( 0ɌI SCH24 2 fB ( 6Do^ ( 0HˌIxr SCH34 2 fB ( 6DoF ( 0Ԍq~ eI.U.P.A.C. NOMENCLATURE2(2ff& F  * R ( * R ( 0،* r$  eCH3F 2 fB ( 6Dox  ( 0ߌ%VP [CH3< 2 !( 0XVmP TCH6 2fB "( 6DoO #( 0S jM eCH2F 2 fB $( 6Do"!"fB %( 6Do" "i T >X ?d  &(#  F RfB '( 6Do   (( 0j d  eCH2F 2  )( 0>d ^  eCH3F 2 fB *( 6DoD   +( 0j ?d  TCH6 2 ,( 0PX ?R  [CH3< 2fB -( 6Do.  J .( 0`r  zLongest chain = 5 so it is a pentane A CH3, methyl, group is attached to the third carbon from one end... 3-methylpentaneT{ 2*O /( 0 q~ eI.U.P.A.C. NOMENCLATURE2(2ff 0( 0t0 &Apply the rules and name these alkanes 'H'&   E 1( 0xr   yLongest chain = 6 so it is a hexane A CH3, methyl, group is attached to the second carbon from one end... 2-methylhexanePz 2)O| 2( 0& r  Longest chain = 6 so it is a hexane CH3, methyl, groups are attached to the third and fourth carbon atoms (whichever end you count from). 3,4-dimethylhexaneb 2'ffdfdB 3( <D)P0Q 4( 0Pp@ 0 0dB 5(@ <D)PXP 6( 0X0@ 0 0H ( 0޽h ? ̙33m  0L0     ( Ow@ dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0l9q~ \NAMING ALKENES2(2ffw   0= 0  Length In alkenes the principal chain is not always the longest chain It 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 CH3CH2CH=CHCH3 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 4 CH2 = CH(CH3)CH2CH3 CH2 = CHCH(CH3)CH3 2-methylbut-1-ene 3-methylbut-1-ene@(F7PZ@4PF.P);  "      =s        @     B  s *޽h ? ̙33  0L0    . (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0hq~ cWHICH COMPOUND IS IT?2(2ff   0P 0 Elucidation of the structures of organic compounds - a brief summary Organic chemistry is so vast that the identification of a compound can be involved. The characterisation takes place in a series of stages (see below). Relatively large amounts of substance were required to elucidate the structure but, with modern technology and the use of electronic instrumentation, very small amounts are now required. Elemental composition One assumes that organic compounds contain carbon and hydrogen but it can be proved by letting the compound undergo combustion. Carbon is converted to carbon dioxide and hydrogen is converted to water. Percentage composition by mass Found by dividing the mass of an element present by the mass of the compound present, then multiplying by 100. Elemental mass of C and H can be found by allowing the substance to undergo complete combustion. From this one can find... mass of carbon = 12/44 of the mass of CO2 produced mass of hydrogen = 2/18 of the mass of H2O produced zEHW@@Pz@EUffJf d)ff5ff f  B  s *޽h ? ̙33 0L0  @%( w @dB @ <D)P0Q @ 0Pp@ 0 0dB @@ <D)PXP @ 0X0@ 0 0 @ 0pq~ eINVESTIGATING MOLECULES2(2ff  @ 0t0 *Empirical formula The simplest ratio of elements present in the substance. It is calculated by dividing the mass or percentage mass of each element by its molar mass and finding the simplest ratio between the answers. Empirical formula is converted to the molecular formula using molecular mass. >+@f8 + B @ s *޽h ? ̙33 0L0 B: <( w <dB < <D)P0Q < 0Pp@ 0 0dB <@ <D)PXP < 0X0@ 0 0 < 0q~ eINVESTIGATING MOLECULES2(2ff  < 0Î0  K}Empirical formula The simplest ratio of elements present in the substance. It is calculated by dividing the mass or percentage mass of each element by its molar mass and finding the simplest ratio between the answers. Empirical formula is converted to the molecular formula using molecular mass. Molecular mass Traditionally found out using a variety of techniques such as ... volumetric analysis or molar 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 molecular mass. The fragmentation pattern gives information about the compound. ~@)fBE ~ B < s *޽h ? ̙33  0L0  81( w 8dB 8 <D)P0Q 8 0Pp@ 0 0dB 8@ <D)PXP 8 0X0@ 0 0 8 0Ԏq~ eINVESTIGATING MOLECULES2(2ff  8 0َ0  Empirical formula The simplest ratio of elements present in the substance. It is calculated by dividing the mass or percentage mass of each element by its molar mass and finding the simplest ratio between the answers. Empirical formula is converted to the molecular formula using molecular mass. Molecular mass Traditionally found out using a variety of techniques such as ... volumetric analysis or molar 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 molecular mass. The fragmentation pattern gives information about the compound. Molecular formula The molecular formula is an exact multiple of the empirical formula. Comparing the molecular 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 78 the molecular formula will be 78/13 or 6 times the empirical formula i.e. C6H6 . @}fD    B 8 s *޽h ? ̙33  0L0 { s  4 (  4dB 4 <D)P0Q 4 0Pp@ 0 0dB 4@ <D)PXP 4 0X0@ 0 0 4 0q~ eINVESTIGATING MOLECULES2(2ff  4 0H0 Empirical formula The simplest ratio of elements present in the substance. It is calculated by dividing the mass or percentage mass of each element by its molar mass and finding the simplest ratio between the answers. Empirical formula is converted to the molecular formula using molecular mass. Molecular mass Traditionally found out using a variety of techniques such as ... volumetric analysis or molar 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 molecular mass. The fragmentation pattern gives information about the compound. Molecular formula The molecular formula is an exact multiple of the empirical formula. Comparing the molecular 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 78 the molecular formula will be 78/13 or 6 times the empirical formula i.e. C6H6 . Structural formula Because of the complexity of organic molecules, there can be more than one structure for a given molecular formula. To work out the structure, different tests are carried out.@f  B 4 s *޽h ? ̙33  0L0 !    ( w dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ eINVESTIGATING MOLECULES2(2ff   00 *Empirical formula The simplest ratio of elements present in the substance. It is calculated by dividing the mass or percentage mass of each element by its molar mass and finding the simplest ratio between the answers. Empirical formula is converted to the molecular formula using molecular mass. Molecular mass Traditionally found out using a variety of techniques such as ... volumetric analysis or molar 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 molecular mass. The fragmentation pattern gives information about the compound. Molecular formula The molecular formula is an exact multiple of the empirical formula. Comparing the molecular 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 78 the molecular formula will be 78/13 or 6 times the empirical formula i.e. C6H6 . Structural formula Because of the complexity of organic molecules, there can be more than one structure for a given molecular formula. To work out the structure, different tests are carried out.\@f8fBEfD  f  B  s *޽h ? ̙33} 0L0  0`0 (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ eINVESTIGATING MOLECULES2(2ff   00  PChemical 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 relative numbers of hydrogen atoms present in a molecule Confirmation By comparison of IR or NMR spectra and mass spectrometry $Q@? f;fe   Q ^B ` 6DjJ9+B  s *޽h ? ̙33~  0L0   @  (    0tS JREVISION CHECK (2f  0W ZWhat should you be able to do? 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