Lecture4: 123.101

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Lecture 4 of "chemistry and living systems" 123.101

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Lecture4: 123.101

  1. 1. Unit One Parts 3 & 4:molecular bonding
  2. 2. Unit OneParts 3&4Bond strengthBond polarisation PagesResonance 34 & 46
  3. 3. Unit OneParts 3&4 ...today we continue to make our simple model more complex!Bond strengthBond polarisation PagesResonance 34 & 46
  4. 4. how strong arebonds? ...and we’re talking about covalent bonds...the important ones for organic chemists
  5. 5. bond the values aren’t strength important...only the concept / pattern C C C C C C > >C C C C C C 836 610 347kJ mol–1 kJ mol–1 kJ mol–1 Pg 40
  6. 6. bond strength C C C C C C > obviously, it takes >C C C C more energy to break C C an alkyne apart...breaking three bonds not one 836 610 347kJ mol–1 kJ mol–1 kJ mol–1 Pg 40
  7. 7. bond strength the differences are getting smaller...nearly twice as much energy needed to break 2 bonds but much less needed to break the third C C C C C C > >C C C C C C 836 610 347kJ mol–1 kJ mol–1 kJ mol–1 Pg 40
  8. 8. but...
  9. 9. bond C Cstrength C C a single σ bond is much stronger than a single π bond (head-to- head results in better overlap) C C C C Pg 40
  10. 10. bond C Cstrength C C ...this is the reason alkenes are functional groups but alkanes are not! C C C C Pg 40
  11. 11. are bond lengthand bond strength related? what about bond lengths?
  12. 12. bondstrength ←120→ shorter the bond the stronger it C C normally is... ←134pm→ C C ←154pm→ C C Pg 40
  13. 13. bond Pgstrength 40 C F ←138→ C F C Cl ←178pm→ C Cl ←193pm→ C Br C Br shorter the bond the stronger it normally is...better overlap of atoms / orbitals
  14. 14. how dowe explain? similar size and bond lengths but big difference in energy; why? ←134pm→ ←122pm→ C C C O 610 kJ mol–1 736 kJ mol–1
  15. 15. C O similar size so good orbital overlap...77 pm 73 pm
  16. 16. so far, our picture of bonds has said electrons are shared evenly between two atoms...bond polarisation
  17. 17. ...as always, we teach you a simple modelbond polarisation and then say “reality is more complex!” So lets take a step back...
  18. 18. two kinds ofbond...which one is it? is HCl covalent or ionic?
  19. 19. Polar covalent bondH Cl H+ Cl– electrons shared evenly in a covalent bond...or...H Cl H Cl Pg 34
  20. 20. Polar covalent bondH Cl H+ Cl– one electron lost from H and given to Cl (an ionic bond)H Cl H Cl Pg 34
  21. 21. Polar covalent bond ...or somewhere in the middle... H Cl δ+ δ–H Cl H Cl H Cl Pg 34
  22. 22. Polar covalent bond H Cl a covalent bond but with the electrons predominantly on one atom (ionic character) δ+ δ–H Cl H Cl H Cl Pg 34
  23. 23. H 2.1 Li Be B C N O F 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Na Mg Al Si P S Cl 0.9 1.2 1.5 1.8 2.1 2.5 3.0 K Ca Br 0.8 1.0 2.8 Rb Sr I 0.8 1.0 2.5 Bond Typerarely EN difference Examples electrons Calculation shared evenly in a covalent bond... ionic > 1.7 NaCl 3.0(Cl) - 0.9(Na) = 2.1 CH3O–H 3.5(O) - 2.1(H) = 1.4polar covalent 0.5 – 1.7 H–Cl 3.0(Cl) - 2.1(H) = 0.9 covalent 0 – 0.4 CH3–H 2.5(C) - 2.1(H) = 0.4 Pg H–H 2.1(H) - 2.1(H) = 0.0 35
  24. 24. H 2.1 Li Be B C N O F 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Na Mg Al Si P S Cl 0.9 1.2 1.5 1.8 2.1 2.5 3.0 K Ca Br 0.8 1.0 2.8 Rb Sr I 0.8 1.0 2.5 EN Bond Type difference Examples Calculation ionic > 1.7 NaCl 3.0(Cl) - 0.9(Na) = 2.1 CH3O–H 3.5(O) - 2.1(H) = 1.4polar covalent be0.5 – 1.7 ...electrons will closer to the more electronegative H–Cl 3.0(Cl) - 2.1(H) = 0.9 atom...given by the Pauli scale above (bigger number covalent 0 – 0.4 CH3–H 2.5(C) - 2.1(H) = 0.4 Pg more electronegative) H–H 2.1(H) - 2.1(H) = 0.0 35
  25. 25. H ...difference in 2.1 value indicates the nature of the Li Be B C N O F 1.0 1.5 2.0 2.5 3.0 3.5 4.0 bond... Na Mg Al Si P S Cl 0.9 1.2 1.5 1.8 2.1 2.5 3.0 K Ca Br 0.8 1.0 2.8 Rb Sr I 0.8 1.0 2.5 EN Bond Type difference Examples Calculation ionic > 1.7 NaCl 3.0(Cl) - 0.9(Na) = 2.1 CH3O–H 3.5(O) - 2.1(H) = 1.4polar covalent 0.5 – 1.7 H–Cl 3.0(Cl) - 2.1(H) = 0.9 covalent 0 – 0.4 CH3–H 2.5(C) - 2.1(H) = 0.4 Pg H–H 2.1(H) - 2.1(H) = 0.0 35
  26. 26. you do not need !to learn these values!
  27. 27. polarisationexplains carbonyl so, carbonyl stronger bond than alkenebond strength (& reactivity) because it has ionic bond character (electronic attraction between the two atoms) C C δ+δ+ δ– δ– CC OO
  28. 28. polarisationexplains carbonylbond strength (& reactivity) C C but carbonyl also more reactive because the δ+ charge attracts electrons δ+δ+ δ– δ– CC OO
  29. 29. electrons moveH
  30. 30. electrons moveH organic chemistry is all about the movement electrons
  31. 31. reactions are the movement of electrons H HH H H H
  32. 32. reactions are the movement of electrons reactions are the movement of H H electrons...H H H H
  33. 33. polarisation explains reactivity ...so if we can predict where the electrons start and where they finish (want to go)... I O H δ+ δ– δ+ δ– I O H
  34. 34. polarisation explains reactivity I O H ...then we can predict reactions I δ+ δ– δ– δ+ O H
  35. 35. polarisation δ– means moreexplains reactivity electrons or partial negative chargeof molecules δ+ δ– δ– H3C I δ+ O CH3 H δ+ O δ– Mg δ– δ+ δ+ Br CH3 H3C OMe δ–
  36. 36. polarisationexplains reactivity δ+ means lack of electrons or partialof molecules positive charge δ+ δ– δ– H3C I δ+ O CH3 H δ+ O δ– Mg δ– δ+ δ+ Br CH3 H3C OMe δ–
  37. 37. polarisationexplains reactivityof molecules δ+ δ– δ– H3C I δ+ O CH3 H δ+ (or slightly positive) part of a molecule will be δ+ attacked by... δ– O δ– Mg δ+ δ+ Br CH3 H3C OMe δ–
  38. 38. polarisationexplains reactivityof molecules ...the negative part of a Grignard reagent...in fact we can explain most chemical δ+ δ– reactions by these δ–/+ δ– H3C I charges δ+ O CH3 H δ+ O δ– Mg δ– δ+ δ+ Br CH3 H3C OMe δ–
  39. 39. so far, so good... i hope!
  40. 40. so lets apply what we know...
  41. 41. draw nitromethane CH3NO2
  42. 42. here are the constituentC + 3H + N + 2 O atoms... Pg 45
  43. 43. H O this structure obeys the octetC + 3H + N + 2 O H C N rule H O Pg 45
  44. 44. H O OC + 3H + N + 2 O H C N O ≡HC 3 N H O doesn’t look quite right...need to sort out formal charges Pg 45
  45. 45. H O OC + 3H + N + 2 O H C N O ≡HC 3 N H O fc = 6-4-½(4)=0 top oxygen has no formal charge Pg 45
  46. 46. H O OC + 3H + N + 2 O H C N O ≡HC 3 N H O fc = 5-0-½(8)=+1 nitrogen has a +1 charge Pg 45
  47. 47. H O OC + 3H + N + 2 O H C N O ≡HC3 N H O fc = 6-6-½(2)=-1 bottom oxygen has –1 charge Pg 45
  48. 48. H O OC + 3H + N + 2 O H C N O ≡HC 3 N H O 116 pm O H3C N so structure is this?? (one N=O bond and one 130 pm N–O bond) O Pg 45
  49. 49. so in theory it is allvery easy...
  50. 50. but reality is a little more complex...
  51. 51. Pg O 46H3C N 122 pm O turns out both N–O bonds are identical
  52. 52. Pg O 46H3C N 122 pm O ...they are somewhere in between a N–O bond and a N=O bond...a structure called a...
  53. 53. Pg O 46 H3C N 122 pm Oresonancehybrid
  54. 54. the two Pg extremes are resonance 46 structures... O O OH3C N O H3C N ≡ H3C N O Oresonancestructures
  55. 55. reality is a Pg resonance hybrid 46 O O OH3C N O H3C N ≡ H3C N O Oresonancestructures
  56. 56. we can convert Pg the extremes by pushing electrons 46 (not atoms) O O OH3C N O H3C N ≡ H3C N O Oresonancestructures
  57. 57. Pg 46 O O OH3C N O H3C N ≡ H3C N O Oresonancestructures lets try and explain the relationship between resonance structures and resonance hybrids...
  58. 58. imagine you took one man...lets call him Peter...as one of your resonance structuresresonancestructures
  59. 59. ...and one spider as the other resonanceresonance structure...and now you combine them...structures
  60. 60. the resulting cross is resonance no longer either a man or a spider... structures
  61. 61. ...and it certainlyisn’t something that resonanceflicks back and forthbetween the two...no instead you have a structures hybrid or...
  62. 62. resonance© Marvel Comics hybrid
  63. 63. H H H H C C C C H C H H C H H Hresonance structuresDO NOT EXISTbut are useful
  64. 64. H H H H C C C C H C H H C H H Hresonance structuresDO NOT EXIST &eas ierbut are useful t od raw
  65. 65. H H H H C C C C H C H H C H H Hresonance structuresDO NOT EXIST &eas ier they are Lewis structures so obeybut are useful raw octet rule so we can od draw them... t
  66. 66. resonance hybridE X I SimpossibleLewis structure TS H H C C H C H H
  67. 67. resonance hybridE X I SimpossibleLewis structure TS H H C C ...the Lewis structure H C H no longer obeys octet rule (how many electrons on central H C?)
  68. 68. resonance hybridE X I SimpossibleLewis structure TS H H ...the circle you draw in C Cthe centre of benzene is aresonance hybrid but the H C Hdouble bonds I draw make its chemistry easier to predict... H
  69. 69. only electrons move only electrons move between resonance structures (and in reactions)
  70. 70. all atoms are stationary ...the atoms remain stationary
  71. 71. this is used to represent the movement of two electrons...curlyarrow
  72. 72. it is possibly the most importantcurly ‘scribble’ an organic chemist ever learns...arrow
  73. 73. with this you can bin most text books and just predict reactions instead of learning them...curlyarrow
  74. 74. words cannot describe howcurly wonderful I think this little doodle is!arrow
  75. 75. how do we draw resonance structures? so now we know what a resonance structureis...we need to be able to spot them and draw them...
  76. 76. ...first part is relatively easy (or at least covered in earlier material!) O H3C C1 OLewis Pgstructure 46
  77. 77. lone pairs N ,C π bonds C C2 now we need to identify which electrons can be moved or pushed (forelectrons .’..pushable Pg‘ some reason we always talk about pushing 48 ‘curly arrows’
  78. 78. lone pairs N ,C π bonds C C of electronsthe source2electrons .’ ..pushable ‘ Pg 50
  79. 79. lone pairs of electrons are often ‘pushable’ lone pairs N ,C π bonds C C2electrons .’..pushable‘ Pg 48
  80. 80. as are double (or triple) bonds... lone pairs N ,C π bonds C C2electrons .’..pushable‘ Pg 48
  81. 81. positive charges C electronegative C O atoms atoms with C‘pushable’ electrons3 the next step is to find a target for the electrons...somewhere they want to go... Pg receptors 48
  82. 82. positive charges C electronegative C O atoms atoms with C ‘pushable’ electrons3 ‘happy’ arewhere electrons Pg receptors 50
  83. 83. positive charges C electronegative C O atoms atoms with C‘pushable’ electrons3 all of the above will happily accept electrons so are good receptors Pg receptors 48
  84. 84. positive charges C this is only a receptor because it can also loss electronegativeelectrons (remember we do C O atomsnot want more than eight 8electrons around an atom) atoms with C‘pushable’ electrons3 receptors Pg 48
  85. 85. positive charges C electronegative C O atoms atoms with C‘pushable’ electrons3 NOTE: the donor and acceptor must be one bond apart (no more no less) Pg receptors 48
  86. 86. O O OH3C C H3C C H3C C O O O4 finally, move the electrons and form a new, valid Lewis resonancestructure Pg forms 48
  87. 87. O O OH3C C H3C C H3C C O O O4 here are three resonance structures for the molecule in resonance steps one (the ethanoate anion) Pg forms 48
  88. 88. O O OH3C C H3C C H3C C O O O4 remember the resonance hybrid will be somewhere in between all these as resonance shown on the next Pg forms slide... 48
  89. 89. delocalisation all bond lengths are the same...showing that the compound O never has a single C–O bond or a double C=O O –1/2 bondH3C C or H3C C O O –1/2C–O 130 pm
  90. 90. delocalisation O O –1/2H3C C or H3C C O O –1/2C–O 130 pm the electrons are said to be delocalised over the three atoms (O–C–O)
  91. 91. delocalisation O O –1/2H3C C or H3C C O O –1/2C–O 130 pm electrons are happy when they are delocalised as they are spread over a larger area...so are further apart
  92. 92. examples...Ph Ph X N N C CH3 C CH3H3C C H3C C H H why is this wrong? ...because it has 10 electrons in valence shell of C, which is never allowed!
  93. 93. examples... the correct way involves pushing the lone pair of the nitrogen anion down one bond to make a double C=N bond and then pushing the electrons off the carbon (so that it doesn’t havePh 10 valence electrons) N and... C CH3H3C C H
  94. 94. examples...Ph Ph N N C CH3 C CH3H3C C H3C C H H ...moving them to the carbon at the end of the double bond (we can’t move them two bonds) and forming this new anion
  95. 95. examples...Ph Ph Ph δ– N N NH3C C C CH3 H3C C C CH3 ≡ H3C C δ– CH3 C H H H the resonance hybrid shares (delocalises) the electrons over two bonds or three atoms...
  96. 96. examples... we cannot move this double bond as there is no electron acceptor (and we can’t have 5 bonds or 10 valence electrons on C) Ph Ph X N N C CH3 C CH3 H3C C H3C C H H H H
  97. 97. example... the bonds in ethanoic acid are not what we would predict compared to other simple molecules...(C=O longer & C–O shorter)...why?124pm 122pm O H2 O 129pm H3C C 146pm H3C C H3C C O H O H CH3 ethanoic acid ethanol propanone
  98. 98. example... ...reason is that lone pair of O Oelectrons are pushable and O δ–H3C C H3C C O H ≡ the C=O is a good receptor... H3C C δ+ O H O H124pm 122pm O H2 O 129pm H3C C 146pm H3C C H3C C O H O H CH3 ethanoic acid ethanol propanone
  99. 99. example... O O O δ–H3C C H3C C O H ≡ H3C C δ+ O H O H124pm ...which allows a new 122pm O H2 resonance structure that O 129pm 146pm can contribute to the H3C C resonance hybrid and gives C H3C H3C C O H O the C–OH bond double H CH3 bond character so causes it ethanoic acid ethanol to shrink... propanone
  100. 100. example... O H3C C 5 bonds! O H X we cannot start from the lone pair on the carbonyl O OH3C C H3C C 2 molecules O H O H O H3C C 2 molecules O H
  101. 101. example... carbon can never have 5 bonds (or 10 valence electrons) O H3C C 5 bonds! O H X O OH3C C H3C C 2 molecules O H O H O H3C C 2 molecules O H
  102. 102. example... as soon as we split the O molecule in two we have performed a reaction and H3C C 5 bonds! not looking at resonance O H X anymore. O OH3C C H3C C 2 molecules O H O H O H3C C 2 molecules O H
  103. 103. example... O CO2H H N NH2 N H NH2 NH HO O CO2H H N NH2 N H NH2 NH HO kytotorphin pain regulation
  104. 104. example... H H HH C H H C H H C H C C C C C C C C ≡ C C C CH C H H C H H C H H H H
  105. 105. why is phenolacidic? or why is phenol a separate functional group and not an alcohol?
  106. 106. why is phenolacidic? for a group to be acidic it must be able to give away H+ (a proton)
  107. 107. delocalisation... ...if phenol losses H+ then we are left O with O–...is this stable (will itH C H readily form)? C C C CH C H H
  108. 108. delocalisation... ...we can move the lone pair to form C=O as we O O can push the electrons of H C H C=C...we have spread theH C H electrons over three atoms C C C C so they are happy... C C C CH C H H C H H H
  109. 109. delocalisation... O O O OH C H H C H H C H H C H C C C C C C C C C C C C C C C CH C H H C H H C H H C H H H H H turns out we can form many other Oδ– resonance structures so the electrons are delocalised over 7 H δ– C δ– H atoms...they are really jolly. So C C anion stable so loss of H+ easy so C C phenol is acidic H C H δ– H
  110. 110. H3CH3C CH3 CH3 CH3 CH3 CH3 H3C CH3 CH3 H3CH3C CH3 CH3 CH3 CH3 CH3 H3C CH3 CH3 any double bonds separated by a one single bond canconjugation delocalise
  111. 111. H3CH3C CH3 CH3 CH3 CH3 CH3 H3C CH3 CH3 H3CH3C CH3 CH3 CH3 CH3 CH3 H3C CH3 CH3 such double bonds are said to be conjugatedconjugation
  112. 112. H3CH3C CH3 CH3 CH3 CH3 CH3 H3C CH3 CH3 H3CH3C CH3 CH3 CH3 CH3 H3C CH CH3 conjugation 3 CH3 leads to coloured compounds...this is carotene fromconjugation (you guessed it) carrots
  113. 113. N N N N Na Na HO HO OSO2 SO2O OSO2 SO2ON N N N Na Na HO HO OSO2 SO2O OSO2 SO2O Cl food green 4 E142
  114. 114. N N N N Na Na HO HO OSO2 SO2O OSO2 SO2ON N N N hopefully, you can see that if we have alternating double and single bonds Na Na HO can form multiple resonance we HO structures or delocalise the electrons OSO2 SO2O OSO2 SO2O Cl food green 4 E142
  115. 115. what have ....we learnt? •e l e c t r o n s where they are •b o n d s & their strengthImage created by Cary Sandvig of SGI • re s o n a n c e
  116. 116. readpart 5 ©the bbp@flickr

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