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S  0ݑ ``  X*  0 `   Z*  0 `   Z*z  bA޽h @ ?Parchment ̙33 Default Design 0 zr@ ( )   0xK P    P*    0z     R*  d  c $ ?    0DB  @  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  6r `P   P*    6 `   R*  H  0޽h ? ̙3380___PPT10. 0X(   $+ X X 0 P    X*  X 0`     Z*  X 6\ `P   X*  X 6૪ `   Z* H X 0޽h ? ̙3380___PPT10.1 0L0 P*(    0D ,AN INTRODUCTION TO THE CHEMISTRY OF ALCOHOLST!(2 82 f0f<fJ  C "A KTRE: . " <̙? x @ 0 0 ( <̙? x @L 0 0 ) 0Sx m OKNOCKHARDY PUBLISHING(2 f * 0X"`>B 2008 SPECIFICATIONS>G$G   B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 ` z(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0(   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. 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?     0){0P( OKNOCKHARDY PUBLISHING(2    0X gTHE CHEMISTRY OF ALCOHOLS2(2$f<fB  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   pq (    0HǕ  x CONTENTS Structure of alcohols Nomenclature Isomerism Physical properties Chemical properties of alcohols Identification using infra-red spectroscopy Industrial preparation and uses of ethanol Revision check listf 202 2dB  <D)P0Q  0Pp@ 0 0dB  @ <D)PXP   0X0@ 0 0  0{mC @  0 0  6~m( !@  0 0  6p]3@  0 0  6Km @  0 0  6> ]W @  0 0  6 m @  0 0  6 m @  0 0  0ٕX gTHE CHEMISTRY OF ALCOHOLS2(2$f<f  6 U @  0 0B  s *޽h ? ̙33y___PPT10Y+D=' = @B +H 0L0 WO(    0Hm  ^Before you start it would be helpful to& Recall the definition of a covalent bond Recall the difference types of physical bonding Be able to balance simple equations Be able to write out structures for simple organic molecules Understand the IUPAC nomenclature rules for simple organic compounds Recall the chemical properties of alkanes and alkenes H*(2C 2)BWdB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0\X gTHE CHEMISTRY OF ALCOHOLS2(2$f<fB  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 *"0 ( w dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0#q~ hCLASSIFICATION OF ALCOHOLS2(2ff  0@' 0 0 Aliphatic " general formula CnH2n+1OH - provided there are no rings " the OH replaces an H in a basic hydrocarbon skeleton I;PPn@ 3;  B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0    < (   dB   <D)P0Q   0Pp@ 0 0dB  @ <D)PXP   0X0@ 0 0   0t<q~ hCLASSIFICATION OF ALCOHOLS2(2ff   0C 0.  TAliphatic " general formula CnH2n+1OH - provided there are no rings " the OH replaces an H in a basic hydrocarbon skeleton Aromatic " in aromatic alcohols (or phenols) the OH is attached directly to the ring " an OH on a side chain of a ring behaves as a typical aliphatic alcohol The first two compounds are classified as aromatic alcohols (phenols) because the OH group is attached directly to the ring. I;PPn=@ 3;&# 3  P   C (ApenolsgH s B   s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0    t (  tdB t <D)P0Q t 0Pp@ 0 0dB t@ <D)PXP t 0X0@ 0 0 t 0D`q~ hCLASSIFICATION OF ALCOHOLS2(2ff  t 0b 0  4Aliphatic " general formula CnH2n+1OH - provided there are no rings " the OH replaces an H in a basic hydrocarbon skeleton Aromatic " in aromatic alcohols (or phenols) the OH is attached directly to the ring " an OH on a side chain of a ring behaves as a typical aliphatic alcohol The first two compounds are classified as aromatic alcohols (phenols) because the OH group is attached directly to the ring. Structural differences " alcohols are classified according to the environment of the OH group " chemical behaviour, eg oxidation, often depends on the structural typeI;PPn=@ U@PP 3;&# 32V&@  1 P  t C (A123alcg o+  t 0܉O_ 7 PRIMARY 1 SECONDARY 2 TERTIARY 3x8 2 3  3 3P  t C (ApenolsgH s B t s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0    x (  xf x 0L0  "Alcohols are named according to standard IUPAC rules " select the longest chain of C atoms containing the O-H group; " remove the e and add ol after the basic name " number the chain starting from the end nearer the O-H group " the number is placed after the an and before the ol ... e.g butan-2-ol " as in alkanes, prefix with alkyl substituents " side chain positions are based on the number allocated to the O-H group e.g. CH3 - CH(CH3) - CH2 - CH2 - CH(OH) - CH3 is called 5-methylhexan-2-ol L6H@4    y3G     t           dB x <D)P0Q x 0Pp@ 0 0dB x@ <D)PXP x 0X0@ 0 0  x 0лq~ ]NAMING ALCOHOLS2(2ffB x s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0  hK(  hdB h <D)P0Q h 0Pp@ 0 0dB h@ <D)PXP h 0X0@ 0 0 h 0ģq~ n STRUCTURAL ISOMERISM IN ALCOHOLS2!(2 ff  h 0pʣ0P M}Different structures are possible due to... A Different positions for the OH group and B Branching of the carbon chain-.#x-'  ~ N  h C &Abutolge0 h 0֣` } X butan-1-ol  3   h 0@ۣ } X butan-2-ol  3   h 0ң @   a2-methylpropan-1-ol3   h 0 _  a2-methylpropan-2-ol3  B h s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 * "   (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ hBOILING POINTS OF ALCOHOLS2(2ffz   0\   PIncreases with molecular size due to increased van der Waals forces. Alcohols have higher boiling points than similar molecular mass alkanes This is due to the added presence of inter-molecular hydrogen bonding. More energy is required to separate the molecules. Mr bp / C propane C3H8 44 -42 just van der Waals forces ethanol C2H5OH 46 +78 van der Waals forces + hydrogen bonding GH@@8PxKPZ@Go 6 33333353333'333 3    L    -    +    " N   C &A alchb3X2  0!,dX2  0!OB  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 @ (( n dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0tq~ hBOILING POINTS OF ALCOHOLS2(2ff  0  Z Increases with molecular size due to increased van der Waals forces. Alcohols have higher boiling points than similar molecular mass alkanes This is due to the added presence of inter-molecular hydrogen bonding. More energy is required to separate the molecules. Mr bp / C propane C3H8 44 -42 just van der Waals forces ethanol C2H5OH 46 +78 van der Waals forces + hydrogen bonding Boiling point is higher for  straight chain isomers. bp / C butan-1-ol CH3CH2CH2CH2OH 118 butan-2-ol CH3CH2CH(OH)CH3 100 2-methylpropan-2-ol (CH3)3COH 83GH@@8PxKPZ@7~@G 5'  6333333333333333333 33    L    -    +    a  t N  C &A alchb3   0-N  2Greater branching = lower inter-molecular forces3@3 3 B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0  (`(  (dB ( <D)P0Q ( 0Pp@ 0 0dB (@ <D)PXP ( 0X0@ 0 0 ( 0lq~ hBOILING POINTS OF ALCOHOLS2(2ff ( 0@  Z Increases with molecular size due to increased van der Waals forces. Alcohols have higher boiling points than similar molecular mass alkanes This is due to the added presence of inter-molecular hydrogen bonding. More energy is required to separate the molecules. Mr bp / C propane C3H8 44 -42 just van der Waals forces ethanol C2H5OH 46 +78 van der Waals forces + hydrogen bonding Boiling point is higher for  straight chain isomers. bp / C butan-1-ol CH3CH2CH2CH2OH 118 butan-2-ol CH3CH2CH(OH)CH3 100 2-methylpropan-2-ol (CH3)3COH 83GH@@8PxKPZ@7~@Go 6 33333353333'333 6333333333333333333 33    L    -    +    a  t   ( 0hxN  2Greater branching = lower inter-molecular forces3@3 3 ~  ( 0S" ? X0 P  ( C (Abut1olg P  ( C (Atbutolgf 0P ( C (Abut2olguB ( s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   4 (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  04q~ lSOLVENT PROPERTIES OF ALCOHOLS2(2ff   0ċ  Solubility Low molecular mass alcohols are miscible with water Due to hydrogen bonding between the two molecules Heavier alcohols are less miscibleX@ =*&  N   C &Aalchb4=> N   C &A alchb2C   0k   qSolvent properties Alcohols are themselves very good solvents They dissolve a large number of organic moleculesBj@_ r X2  0!z X2  0! X2  0!l4lX2  0!W  0 dU  <Show the relevant lone pair(s) when drawing hydrogen bonding=(2= &    B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0  ( ̙33 dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0 q~ mCHEMICAL PROPERTIES OF ALCOHOLS2 (2ff(   0 H FThe OXYGEN ATOM HAS TWO LONE PAIRS; this makes alcohols... BASES Lewis bases are lone pair donors Bronsted-Lowry bases are proton acceptors The alcohol uses one of its lone pairs to form a co-ordinate bond NUCLEOPHILES Alcohols can use the lone pair to attack electron deficient centres8;Hl@PP@3C 3C&{   P   C (Aolbaseg*s' B  s *޽h ? ̙33y___PPT10Y+D=' = @B +!  0L0 0 ( ` ( U dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0Tq~ p"ELIMINATION OF WATER (DEHYDRATION)2#(2"ff4   0,  Reagent/catalyst conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4) Conditions reflux at 180C Product alkene Equation e.g. C2H5OH(l)   > CH2 = CH2(g) + H2O(l) Mechanism Step 1 protonation of the alcohol using a lone pair on oxygen Step 2 loss of a water molecule to generate a carbocation Step 3 loss of a proton (H+) to give the alkene Alternative Method Pass vapour over a heated alumina (aluminium oxide) catalystQ@Px@ M@      33333333 333  Fty  b   [   +  \ R  C *A dehydr3gB  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0  p ( T' dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0 q~ p"ELIMINATION OF WATER (DEHYDRATION)2#(2"ff   0 ! pMECHANISM Step 1 protonation of the alcohol using a lone pair on oxygen Step 2 loss of a water molecule to generate a carbocation Step 3 loss of a proton (H+) to give the alkene Note 1 There must be an H on a carbon atom adjacent the carbon with the OH Note 2 Alcohols with the OH in the middle of a chain can have two ways of losing water. In Step 3 of the mechanism, a proton can be lost from either side of the carbocation. This gives a mixture of alkenes from unsymmetrical alcohols... Px@Px@@  3C3t   [   +     B L   C $A alc1g5 R   C *A dehydr3gzB  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0  R( w dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0*q~ cOXIDATION OF ALCOHOLS2(2ff   0|1  <All alcohols can be oxidised depending on the conditions Oxidation is used to differentiate between primary, secondary and tertiary alcohols The usual reagent is acidified potassium dichromate(VI) Primary Easily oxidised to aldehydes and then to carboxylic acids. Secondary Easily oxidised to ketones Tertiary Not oxidised under normal conditions. They do break down with very vigorous oxidationD9H\@933 3 3$32@   =  d P   C (A123alcg% g +   0G 7 PRIMARY 1 SECONDARY 2 TERTIARY 3x8 2 3  3 3B  s *޽h ? ̙33y___PPT10Y+D=' = @B +8  0L0 G ? P(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0tVq~ kOXIDATION OF PRIMARY ALCOHOLS2(2ff  0]0> JPrimary alcohols are easily oxidised to aldehydes e.g. CH3CH2OH(l) + [O]   > CH3CHO(l) + H2O(l) ethanol ethanal it is essential to distil off the aldehyde before it gets oxidised to the acid CH3CHO(l) + [O]   > CH3COOH(l) ethanal ethanoic acid 2Hw@Rn@2333333333+P333336 (   p  %  p     B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0   (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0dq~ kOXIDATION OF PRIMARY ALCOHOLS2(2ffL   00 Primary alcohols are easily oxidised to aldehydes e.g. CH3CH2OH(l) + [O]   > CH3CHO(l) + H2O(l) ethanol ethanal it is essential to distil off the aldehyde before it gets oxidised to the acid CH3CHO(l) + [O]   > CH3COOH(l) ethanal ethanoic acid Practical details the alcohol is dripped into a warm solution of acidified K2Cr2O7 aldehydes have low boiling points - no hydrogen bonding - they distil off immediately if it didn t distil off it would be oxidised to the equivalent carboxylic acid to oxidise an alcohol straight to the acid, reflux the mixture compound formula intermolecular bonding boiling point ETHANOL C2H5OH HYDROGEN BONDING 78C ETHANAL CH3CHO DIPOLE-DIPOLE 23C ETHANOIC ACID CH3COOH HYDROGEN BONDING 118C2Hw@Rn@@-@@Px2333333333+P333336 <   !P/   3;3    (   p  %  p    _    B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   0 ' ( ;;bbb dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0χq~ kOXIDATION OF PRIMARY ALCOHOLS2(2ff0   0և14 2Controlling the products e.g. CH3CH2OH(l) + [O]   > CH3CHO(l) + H2O(l) then CH3CHO(l) + [O]   > CH3COOH(l)nK@5P333333333 33333  T   C ,A distil1bgOiD    0 x T OAldehyde has a lower boiling point so distils off before being oxidised further(PHO H    0طH 8 #OXIDATION TO ALDEHYDES DISTILLATION:$H  $    0x y8 $OXIDATION TO CARBOXYLIC ACIDS REFLUX:%H %   0 T FAldehyde condenses back into the mixture and gets oxidised to the acidGHG ? P  C (Adist1cg=R B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   ,(  ,dB , <D)P0Q , 0Pp@ 0 0dB ,@ <D)PXP , 0X0@ 0 0 , 0q~ mOXIDATION OF SECONDARY ALCOHOLS2 (2ffz , 0`1P Secondary alcohols are easily oxidised to ketones e.g. CH3CHOHCH3(l) + [O]   > CH3COCH3(l) + H2O(l) propan-2-ol propanone The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment with a powerful oxidising agent they can be further oxidised to a mixture of acids with fewer carbon atoms than the original alcohol.2HV@233333333 333*(   f@*  v    B , s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 &  @  (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  00q~ mOXIDATION OF SECONDARY ALCOHOLS2 (2ffh   0\61P' Secondary alcohols are easily oxidised to ketones e.g. CH3CHOHCH3(l) + [O]   > CH3COCH3(l) + H2O(l) propan-2-ol propanone The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment with a powerful oxidising agent they can be further oxidised to a mixture of acids with fewer carbon atoms than the original alcohol.2HV@233333333 333*(   @*  v    ^B  6DjJ    0OK E  ZOXIDATION OF TERTIARY ALCOHOLS (2f  0>X 0O<  3Tertiary alcohols are resistant to normal oxidation,4H3 4 B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 40 O(  0dB 0 <D)P0Q 0 0Pp@ 0 0dB 0@ <D)PXP 0 0X0@ 0 0 0 0h[q~ cOXIDATION OF ALCOHOLS2(2ff$ 0 0`g1P~ Why 1 and 2 alcohols are easily oxidised and 3 alcohols are not For oxidation to take place easily you must have two hydrogen atoms on adjacent C and O atoms.CH`@!&!!!"! & !! !_H  # q B 0 s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   08k (  8dB 8 <D)P0Q 8 0Pp@ 0 0dB 8@ <D)PXP 8 0X0@ 0 0 8 0pwq~ cOXIDATION OF ALCOHOLS2(2ff$ 8 0x1P~ Why 1 and 2 alcohols are easily oxidised and 3 alcohols are not For oxidation to take place easily you must have two hydrogen atoms on adjacent C and O atoms.CH`@!&!!!"! & !! !_H  # q DF    8 N ( H  8     8 0(( H N` H H R C O + [O] R C O + H2O H H a 2 ffZB  8 s *DԔ  ZB  8 s *DԔ  ZB  8 s *DԔ X ZB 8 s *DԔ X ZB 8 s *DԔ h ZB 8 s *DԔ   ZB 8 s *DԔ   ZB 8 s *DԔ h ZB 8 s *DԔ  `B 8 0D> P P  .8 6 b1 2f   B 8 s *޽h ? ̙33y___PPT10Y+D=' = @B +. 0L0 =5#%@(  @dB @ <D)P0Q @ 0Pp@ 0 0dB @@ <D)PXP @ 0X0@ 0 0 @ 0ܫq~ cOXIDATION OF ALCOHOLS2(2ff$ @ 0䲉1P~ Why 1 and 2 alcohols are easily oxidised and 3 alcohols are not For oxidation to take place easily you must have two hydrogen atoms on adjacent C and O atoms.CH`@!&!!!"! & !! !_H  # q DF    @ N ( H  @     @ 0T( H N` H H R C O + [O] R C O + H2O H H a 2 ffZB  @ s *DԔ  ZB  @ s *DԔ  ZB  @ s *DԔ X ZB @ s *DԔ X ZB @ s *DԔ h ZB @ s *DԔ   ZB @ s *DԔ   ZB @ s *DԔ h ZB @ s *DԔ  `B @ 0D> P P F    @ V N ( H @    @ 0( H ` H H R C O + [O] R C O + H2O R R a 2 ff ffZB @ s *DԔ  ZB @ s *DԔ  ZB @ s *DԔ X ZB @ s *DԔ X ZB @ s *DԔ h ZB @ s *DԔ   ZB @ s *DԔ   ZB @ s *DԔ h ZB  @ s *DԔ  `B !@ 0D> P P  #@ 6| b1 2    $@ 6 0  b2 2f   B @ s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 $0<U(  <dB < <D)P0Q < 0Pp@ 0 0dB <@ <D)PXP < 0X0@ 0 0 < 0lq~ cOXIDATION OF ALCOHOLS2(2ff$ < 0\1P~ Why 1 and 2 alcohols are easily oxidised and 3 alcohols are not For oxidation to take place easily you must have two hydrogen atoms on adjacent C and O atoms.CH`@!&!!!"! & !! !_H  # q DF    < N ( H  <     < 0 ( H N` H H R C O + [O] R C O + H2O H H a 2 ffZB  < s *DԔ  ZB  < s *DԔ  ZB  < s *DԔ X ZB < s *DԔ X ZB < s *DԔ h ZB < s *DԔ   ZB < s *DԔ   ZB < s *DԔ h ZB < s *DԔ  `B < 0D> P P F    < V N ( H <    < 0( H ` H H R C O + [O] R C O + H2O R R a 2 ff ffZB < s *DԔ  ZB < s *DԔ  ZB < s *DԔ X ZB < s *DԔ X ZB < s *DԔ h ZB < s *DԔ   ZB < s *DԔ   ZB < s *DԔ h ZB  < s *DԔ  `B !< 0D> P P  ,< 0 8W  &>This is possible in 1 and 2 alcohols but not in 3 alcohols.|?H !! ! H      .< 60 b1 2    /< 65 0  b2 2   B < s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 004Y(  4dB 4 <D)P0Q 4 0Pp@ 0 0dB 4@ <D)PXP 4 0X0@ 0 0 4 0>q~ cOXIDATION OF ALCOHOLS2(2ff$ 4 0T1P~ Why 1 and 2 alcohols are easily oxidised and 3 alcohols are not For oxidation to take place easily you must have two hydrogen atoms on adjacent C and O atoms.CH`@!&!!!"! & !! !_H  # q DF    4 N ( H  4     4 0@X( H N` H H R C O + [O] R C O + H2O H H a 2 ffZB  4 s *DԔ  ZB  4 s *DԔ  ZB  4 s *DԔ X ZB 4 s *DԔ X ZB 4 s *DԔ h ZB 4 s *DԔ   ZB 4 s *DԔ   ZB 4 s *DԔ h ZB 4 s *DԔ  `B 4 0D> P P F    4 V N ( H 4    4 0m( H ` H H R C O + [O] R C O + H2O R R a 2 ff ffZB 4 s *DԔ  ZB 4 s *DԔ  ZB 4 s *DԔ X ZB 4 s *DԔ X ZB 4 s *DԔ h ZB 4 s *DԔ   ZB 4 s *DԔ   ZB 4 s *DԔ h ZB  4 s *DԔ  `B !4 0D> P P F  O "4  ` #4 0| O > R H R C O + [O] R ? 2f f fZB $4 s *DԔ0ZB %4 s *DԔhZB &4 s *DԔ( (_ZB '4 s *DԔp p_ZB (4 s *DԔ((o`B )4 0D>  ZB *4 s *D> 8P ZB +4B s *D>x 9H  ,4 0$ 8W  8>This is possible in 1 and 2 alcohols but not in 3 alcohols.?H !!! ! H     8  04 -4 0t b1 2    .4 0@ 0  b2 2    /4 0  b3 2f   B 4 s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   P (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0 q~ hESTERIFICATION OF ALCOHOLS2(2ff   0,0 !Reagent(s) carboxylic acid + strong acid catalyst (e.g conc. H2SO4 ) Conditions reflux Product ester Equation e.g. CH3CH2OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l) ethanol ethanoic acid ethyl ethanoate Notes Concentrated H2SO4 is a dehydrating agent - it removes water causing the equilibrium to move to the right and increases the yield >h@Q@9@@ ( 33333  3333 33333333 333:  4@      F ?    H8N    fB   6D3olB   <BD3o{>T  # L)?fB  6D3olB  <BD3o{B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 $`(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0ٌq~ hESTERIFICATION OF ALCOHOLS2(2ff  00 r Reagent(s) carboxylic acid + strong acid catalyst (e.g conc. H2SO4 ) Conditions reflux Product ester Equation e.g. CH3CH2OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l) ethanol ethanoic acid ethyl ethanoate Notes Concentrated H2SO4 is a dehydrating agent - it removes water causing the equilibrium to move to the right and increases the yield Uses of esters Esters are fairly unreactive but that doesn t make them useless Used as flavourings Naming esters Named from the alcohol and carboxylic acid which made them... CH3OH + CH3COOH CH3COOCH3 + H2O from ethanoic acid CH3COOCH3 from methanol METHYL ETHANOATEJh@Q@9@@@L@9@9#x@5  TsV 3 333 3333   33 3 3 3 3  3 3 t          L F ?   H8N    fB   6DolB   <BDo{>T   # L)?fB   6DolB  <BDo{F ?     8N   fB  6DolB  <BDo{>T  # L)?fB  6DolB  <BDo{B  s *޽h ? ̙33y___PPT10Y+D=' = @B +` 0L0 og` ( 0l4 dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0 5q~ iOTHER REACTIONS OF ALCOHOLS2(2ff   0,< t OXYGEN Alcohols make useful fuels C2H5OH(l) + 3O2(g)    > 2CO2(g) + 3H2O(l) Advantages have high enthalpies of combustion do not contain sulphur so there is less pollution can be obtained from renewable resources :#@@#33333333 333   B  s *޽h ? ̙33y___PPT10Y+D=' = @B +X 0L0 g _ p  (   dB   <D)P0Q   0Pp@ 0 0dB  @ <D)PXP   0X0@ 0 0   0Pq~ iOTHER REACTIONS OF ALCOHOLS2(2ff   0W+ l nOXYGEN Alcohols make useful fuels C2H5OH(l) + 3O2(g)    > 2CO2(g) + 3H2O(l) Advantages have high enthalpies of combustion do not contain sulphur so there is less pollution can be obtained from renewable resources SODIUM Conditions room temperature Product sodium alkoxide and hydrogen Equation 2CH3CH2OH(l) + 2Na(s)   > 2CH3CH2O Na + + H2(g) sodium ethoxide Notes alcohols are organic chemistry s equivalent of water water reacts with sodium to produce hydrogen and so do alcohols the reaction is slower with alcohols than with water. Alkoxides are white, ionic crystalline solids e.g. CH3CH2O Na+#@@@@#33333333 333  !3333 333333 333"6333333Z9  q     : B   s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0     (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0Ԏq~ eBROMINATION OF ALCOHOLS2(2ff   00g LnReagent(s) conc. hydrobromic acid HBr(aq) or sodium (or potassium) bromide and concentrated sulphuric acid Conditions reflux Product haloalkane Equation C2H5OH(l) + conc. HBr(aq)    > C2H5Br(l) + H2O(l) Mechanism The mechanism starts off similar to that involving dehydration (protonation of the alcohol and loss of water) but the carbocation (carbonium ion) is attacked by a nucleophilic bromide ion in step 3.@R@@O@  B  3333)3333 333      a   !  v   +          Q   0̹ ( Step 1 protonation of the alcohol using a lone pair on oxygen Step 2 loss of a water molecule to generate a carbocation (carbonium ion) Step 3 a bromide ion behaves as a nucleophile and attacks the carbocationnPn8DD   Z      (       R   C *A dehydr2g( m B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 (   (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0ʎq~ dINFRA-RED SPECTROSCOPY2(2ff   0ю0t  2Chemical bonds vibrate at different frequencies. When infra red (IR) radiation is passed through a liquid sample of an organic molecule, some frequencies are absorbed. These correspond to the frequencies of the vibrating bonds. Most spectra are very complex due to the large number of bonds present and each molecule produces a unique spectrum. However the presence of certain absorptions can be used to identify functional groups. BOND COMPOUND ABSORBANCE RANGE O-H alcohols broad 3200 cm-1 to 3600 cm-1 O-H carboxylic acids medium to broad 2500 cm-1 to 3500 cm-1 C=O ketones, aldehydes strong and sharp 1600 cm-1 to 1750 cm-1 esters and acids@P<PUP@ZPxo@73 :  > < @     O B  s *޽h ? ̙33y___PPT10Y+D=' = @B +u  0L0  |   (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ dINFRA-RED SPECTROSCOPY2(2ff   0 0|  zIDENTIFYING ALCOHOLS USING INFRA RED SPECTROSCOPY Differentiation Compound O-H C=O ALCOHOL YES NO ALDEHYDE / KETONE NO YES CARBOXYLIC ACID YES YES ESTER NO YES ALCOHOL ALDEHYDE CARBOXYLIC ACID PROPAN-1-OL PROPANAL PROPANOIC ACID O-H absorption C=O absorption O-H + C=O absorption 3H"PnXPx@23 3F3T   P   S (A 1r1g 0P   S (A 1r2g `P   S (A 1r3g B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 (  ( w dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0*q~ p"INDUSTRIAL PREPARATION OF ALCOHOLS2#(2"ff   01 &^FERMENTATION Reagent(s) GLUCOSE - produced by the hydrolysis of starch Conditions yeast warm, but no higher than 37C Equation C6H12O6   > 2 C2H5OH + 2 CO2  H@  0 (333333333333  B  s *޽h ? ̙33y___PPT10Y+D=' = @B +A  0L0 PHD(  DdB D <D)P0Q D 0Pp@ 0 0dB D@ <D)PXP D 0X0@ 0 0 D 0|Pq~ p"INDUSTRIAL PREPARATION OF ALCOHOLS2#(2"ff D 0 XM  NFERMENTATION Reagent(s) GLUCOSE - produced by the hydrolysis of starch Conditions yeast warm, but no higher than 37C Equation C6H12O6   > 2 C2H5OH + 2 CO2 Advantages LOW ENERGY PROCESS USES RENEWABLE RESOURCES - PLANT MATERIAL SIMPLE EQUIPMENT Disadvantages SLOW PRODUCES IMPURE ETHANOL BATCH PROCESS H>@  x T 0 K B D s *޽h ? ̙33y___PPT10Y+D=' = @B +y 0L0  (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0pTq~ p"INDUSTRIAL PREPARATION OF ALCOHOLS2#(2"ff   0s HYDRATION OF ETHENE Reagent(s) ETHENE - from cracking of fractions from distilled crude oil Conditions catalyst - phosphoric acid high temperature and pressure Equation C2H4 + H2O   > C2H5OH H@ > ?3333333333333  B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   $ (  $dB $ <D)P0Q $ 0Pp@ 0 0dB $@ <D)PXP $ 0X0@ 0 0 $ 0q~ p"INDUSTRIAL PREPARATION OF ALCOHOLS2#(2"ffZ $ 0K (HYDRATION OF ETHENE Reagent(s) ETHENE - from cracking of fractions from distilled crude oil Conditions catalyst - phosphoric acid high temperature and pressure Equation C2H4 + H2O   > C2H5OH Advantages FAST PURE ETHANOL PRODUCED CONTINUOUS PROCESS Disadvantages HIGH ENERGY PROCESS EXPENSIVE PLANT REQUIRED USES NON-RENEWABLE FOSSIL FUELS TO MAKE ETHENE Uses of ethanol ALCOHOLIC DRINKS SOLVENT - industrial alcohol / methylated spirits FUEL - petrol substitute in countries with limited oil reservesH@   3b&   K B $ s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 $ (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0 q~ ^USES OF ALCOHOLS2(2ff   0  4|ETHANOL DRINKS SOLVENT industrial alcohol / methylated spirits (methanol is added) FUEL used as a petrol substitute in countries with limited oil reserves METHANOL PETROL ADDITIVE improves combustion properties of unleaded petrol SOLVENT RAW MATERIAL used as a feedstock for important industrial processes FUEL Health warning Methanol is highly toxicn@@ @@33C3K373 3=3 &1   B B  s *޽h ? ̙33y___PPT10Y+D=' = @B +T 0L0 c[ ( w dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0tq?@~ p"LABORATORY PREPARATION OF ALCOHOLS2#(2"ff   0ߐ afrom haloalkanes - reflux with aqueous sodium or potassium hydroxide from aldehydes - reduction with sodium tetrahydridoborate(III) - NaBH4 from alkenes - acid catalysed hydration using concentrated sulphuric acid Details of the reactions may be found in other sections. @9H@ < ; y Z   <      B  s *޽h ? ̙33y___PPT10Y+D=' = @B +n  0L0 } u   (    00 JREVISION CHECK (2f  08 ZWhat should you be able to do? (2   0X0  Recall and explain the physical properties of alcohols Recall the different structural types of alcohols Recall the Lewis base properties of alcohols Recall and explain the chemical reactions of alcohols Write balanced equations representing any reactions in the section Understand how oxidation is affected by structure Recall how conditions and apparatus influence the products of oxidation Explain how infrared spectroscopy can be used to differentiate between functional groups 2fffff%ff,ff'ffffff#ff?f f)ffBffRf   0   ~,CAN YOU DO ALL OF THESE? YES NO6- 2!    0 H @P 0 0   0f h @U 0 0dB   <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0H  0޽h ? ̙33y___PPT10Y+D=' = @B + 0L0  ;(  9  0X!  YYou need to go over the relevant topic(s) again Click on the button to return to the menudG(282/$f f) f  s * x  P @S 0 0dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP   0X0@ 0 0H  0޽h ? ̙33y___PPT10Y+D=' = @B + 0L0  D(    01xK (WELL DONE! 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