The document provides information about molecular bonding, including bond strength, bond polarization, and resonance. It discusses how bond strength decreases from triple to double to single bonds. It also explains that polar covalent bonds have electrons predominantly located on one atom due to differences in electronegativity. Resonance structures are presented as alternative Lewis structures that represent different locations of electrons in a molecule, with the actual structure being a resonance hybrid that is an average of the contributing structures. Nitromethane is used as an example to illustrate these concepts.

1. Unit One Parts 3 & 4:
molecular bonding

2. Unit One
Parts 3&4
Bond strength
Bond polarisation Pages
Resonance 34 & 46

3. Unit One
Parts 3&4
...today we continue
to make our simple
model more
complex!
Bond strength
Bond polarisation Pages
Resonance 34 & 46

4. how strong are
bonds?
...and we’re talking
about covalent
bonds...the
important ones for
organic chemists

5. bond the values aren’t
strength important...only the
concept / pattern
C C C C C C
> >
C C C C C C
836 610 347
kJ mol–1 kJ mol–1 kJ mol–1
Pg
40

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
836 610 347
kJ mol–1 kJ mol–1 kJ mol–1
Pg
40

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
836 610 347
kJ mol–1 kJ mol–1 kJ mol–1
Pg
40

8. but...

9. bond C C
strength
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. bond C C
strength
C C
...this is the reason
alkenes are functional
groups but alkanes are
not!
C C C C
Pg
40

11. are bond length
and bond strength
related?
what about bond
lengths?

12. bond
strength
←120→ shorter the bond
the stronger it
C C normally is...
←134pm→
C C
←154pm→
C C
Pg
40

13. bond Pg
strength 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. how do
we 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. C O
similar size so
good orbital
overlap...
77 pm 73 pm

16. so far, our picture of
bonds has said
electrons are shared
evenly between two
atoms...
bond polarisation

17. ...as always, we teach
you a simple model
bond polarisation and then say “reality is
more complex!” So lets
take a step back...

18. two kinds of
bond...which one is
it?
is HCl covalent or
ionic?

19. Polar covalent bond
H Cl H+ Cl–
electrons shared
evenly in a covalent
bond...or...
H Cl H Cl
Pg
34

20. Polar covalent bond
H Cl H+ Cl–
one electron lost
from H and given to
Cl (an ionic bond)
H Cl H Cl
Pg
34

21. Polar covalent bond
...or somewhere in
the middle...
H Cl
δ+ δ–
H Cl H Cl H Cl
Pg
34

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. 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) = 1.4
polar 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. 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) = 1.4
polar 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. 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) = 1.4
polar 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. you do not need
!
to learn these values!

27. polarisation
explains carbonyl
so, carbonyl
stronger bond than alkene
bond strength (& reactivity)
because it has ionic bond
character (electronic
attraction between the
two atoms)
C C
δ+δ+ δ–
δ–
CC OO

28. polarisation
explains carbonyl
bond strength (& reactivity)
C C but carbonyl also
more reactive because
the δ+ charge attracts
electrons
δ+δ+ δ–
δ–
CC OO

29. electrons move
H

30. electrons move
H organic chemistry is
all about the
movement electrons

31. reactions are the movement of electrons
H H
H H H H

32. reactions are the movement of electrons
reactions are the
movement of
H H
electrons...
H H H H

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. polarisation explains reactivity
I O
H
...then we can
predict reactions
I δ+
δ–
δ–
δ+ O
H

35. polarisation δ– means more
explains reactivity
electrons or partial
negative charge
of molecules
δ+ δ– δ–
H3C I δ+ O CH3
H
δ+ O δ–
Mg δ– δ+ δ+
Br CH3
H3C OMe
δ–

36. polarisation
explains reactivity δ+ means lack of
electrons or partial
of 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. polarisation
explains reactivity
of 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. polarisation
explains reactivity
of 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. so far, so good...
i hope!

40. so let's apply what
we know...

41. draw nitromethane
CH3NO2

42. here are the
constituent
C + 3H + N + 2 O atoms...
Pg
45

43. H O this structure
obeys the octet
C + 3H + N + 2 O H C N rule
H O
Pg
45

44. H O O
C + 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. H O O
C + 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. H O O
C + 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. H O O
C + 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. H O O
C + 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. so in theory it is all
very easy...

50. but reality is a little
more complex...

51. Pg
O 46
H3C N
122 pm
O
turns out both
N–O bonds
are identical

52. Pg
O 46
H3C N
122 pm
O
...they are somewhere
in between a N–O bond
and a N=O bond...a
structure called a...

53. Pg
O 46
H3C N
122 pm
O
resonance
hybrid

54. the two Pg
extremes are
resonance 46
structures...
O O O
H3C N
O
H3C N ≡ H3C N
O O
resonance
structures

55. reality is a Pg
resonance
hybrid 46
O O O
H3C N
O
H3C N ≡ H3C N
O O
resonance
structures

56. we can convert
Pg
the extremes by
pushing electrons
46
(not atoms)
O O O
H3C N
O
H3C N ≡ H3C N
O O
resonance
structures

57. Pg
46
O O O
H3C N
O
H3C N ≡ H3C N
O O
resonance
structures lets try and explain
the relationship between
resonance structures and
resonance hybrids...

58. imagine you took one
man...lets call him
Peter...as one of your
resonance structures
resonance
structures

59. ...and one spider as
the other resonance
resonance
structure...and now
you combine them...
structures

60. the resulting cross is
resonance
no longer either a
man or a spider...
structures

61. ...and it certainly
isn’t something that
resonance
flicks back and forth
between the two...no
instead you have a
structures
hybrid or...

62. resonance
© Marvel Comics
hybrid

63. H H H H
C C C C
H C H H C H
H H
resonance structures
DO NOT EXIST
but are useful

64. H H H H
C C C C
H C H H C H
H H
resonance structures
DO NOT EXIST
&eas ier
but are useful t od raw

65. H H H H
C C C C
H C H H C H
H H
resonance structures
DO NOT EXIST
&eas ier
they are Lewis
structures so obey
but are useful raw
octet rule so we can
od
draw them...
t

66. resonance hybrid
E X I Simpossible
Lewis structure
TS
H H
C C
H C H
H

67. resonance hybrid
E X I Simpossible
Lewis 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. resonance hybrid
E X I Simpossible
Lewis structure
TS
H H
...the circle you draw in C C
the centre of benzene is a
resonance hybrid but the
H C H
double bonds I draw make
its chemistry easier to
predict...
H

69. only electrons
move
only electrons move
between resonance
structures (and in
reactions)

70. all atoms are
stationary
...the atoms remain
stationary

71. this is used to
represent the
movement of two
electrons...
curly
arrow

72. it is possibly the
most important
curly
‘scribble’ an
organic chemist
ever learns...
arrow

73. with this you can bin
most text books and
just predict reactions
instead of learning
them...
curly
arrow

74. words cannot
describe how
curly
wonderful I think
this little doodle
is!
arrow

75. how do we draw
resonance structures?
so now we know what a
resonance structure
is...we need to be able to
spot them and draw
them...

76. ...first part is
relatively easy (or at
least covered in
earlier material!)
O
H3C C
1
O
Lewis Pg
structure 46

77. lone pairs N ,C
π bonds C C
2
now we need to identify
which electrons can be
moved or pushed (for
electrons .’
..pushable Pg
‘
some reason we always
talk about pushing
48
‘curly arrows’

78. lone pairs N ,C
π bonds C C
of electrons
the source
2electrons .’
..pushable
‘
Pg
50

79. lone pairs of
electrons are often
‘pushable’
lone pairs N ,C
π bonds C C
2
electrons .’
..pushable
‘
Pg
48

80. as are double (or
triple) bonds...
lone pairs N ,C
π bonds C C
2
electrons .’
..pushable
‘
Pg
48

81. positive charges C
electronegative C O
atoms
atoms with C
‘pushable’ electrons
3
the next step is to find
a target for the
electrons...somewhere
they want to go...
Pg
receptors 48

82. positive charges C
electronegative C O
atoms
atoms with C
‘pushable’ electrons
3 ‘happy’
are
where electrons
Pg
receptors 50

83. positive charges C
electronegative C O
atoms
atoms with C
‘pushable’ electrons
3
all of the above will
happily accept
electrons so are good
receptors
Pg
receptors 48

84. positive charges C
this is only a receptor
because it can also loss
electronegative
electrons (remember we do
C O
atoms
not want more than eight 8
electrons around an atom)
atoms with C
‘pushable’ electrons
3 receptors
Pg
48

85. positive charges C
electronegative C O
atoms
atoms with C
‘pushable’ electrons
3
NOTE: the donor and
acceptor must be
one bond apart (no
more no less)
Pg
receptors 48

86. O O O
H3C C H3C C H3C C
O O O
4
finally, move the
electrons and form a
new, valid Lewis
resonancestructure
Pg
forms 48

87. O O O
H3C C H3C C H3C C
O O O
4
here are three
resonance structures
for the molecule in
resonance
steps one (the
ethanoate anion) Pg
forms 48

88. O O O
H3C C H3C C H3C C
O O O
4
remember the
resonance hybrid will
be somewhere in
between all these as
resonance
shown on the next
Pg
forms
slide...
48

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
bond
H3C C or H3C C
O O –1/2
C–O 130 pm

90. delocalisation
O O –1/2
H3C C or H3C C
O O –1/2
C–O 130 pm
the electrons are said
to be delocalised
over the three atoms
(O–C–O)

91. delocalisation
O O –1/2
H3C C or H3C C
O O –1/2
C–O 130 pm
electrons are happy
when they are
delocalised as they are
spread over a larger
area...so are further
apart

92. examples...
Ph Ph
X
N N
C CH3 C CH3
H3C C H3C C
H H
why is this wrong?
...because it has 10
electrons in valence
shell of C, which is
never allowed!

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 have
Ph 10 valence electrons)
N and...
C CH3
H3C C
H

94. examples...
Ph Ph
N N
C CH3 C CH3
H3C 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. examples...
Ph Ph Ph δ–
N N N
H3C
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. 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. 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. 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 H
124pm 122pm
O H2 O
129pm H3C C 146pm H3C C
H3C C
O H O H CH3
ethanoic acid ethanol propanone

99. example...
O O O δ–
H3C C H3C C
O H
≡ H3C C
δ+
O H O H
124pm ...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. example...
O
H3C C 5 bonds!
O H
X
we cannot start from the
lone pair on the carbonyl
O O
H3C C H3C C 2 molecules
O H O H
O
H3C C 2 molecules
O H

101. example... carbon can never have 5
bonds (or 10 valence
electrons)
O
H3C C 5 bonds!
O H
X
O O
H3C C H3C C 2 molecules
O H O H
O
H3C C 2 molecules
O H

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 O
H3C C H3C C 2 molecules
O H O H
O
H3C C 2 molecules
O H

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. example...
H H H
H C H H C H H C H
C
C
C
C
C
C
C
C
≡ C
C
C
C
H C H H C H H C H
H H H

105. why is phenol
acidic? or why is phenol a
separate functional group
and not an alcohol?

106. why is phenol
acidic? for a group to be acidic it
must be able to give away
H+ (a proton)

107. delocalisation...
...if phenol losses
H+ then we are left
O with O–...is this
stable (will it
H C H readily form)?
C C
C C
H C H
H

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 the
H C H electrons over three atoms
C C C C
so they are happy...
C C C C
H C H H C H
H H

109. delocalisation...
O O O O
H 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 C
H 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. H3C
H3C CH3 CH3 CH3
CH3 CH3 H3C CH3
CH3
H3C
H3C CH3 CH3 CH3
CH3 CH3 H3C CH3
CH3 any double bonds
separated by a one
single bond can
conjugation
delocalise

111. H3C
H3C CH3 CH3 CH3
CH3 CH3 H3C CH3
CH3
H3C
H3C CH3 CH3 CH3
CH3 CH3 H3C CH3
CH3 such double bonds
are said to be
conjugated
conjugation

112. H3C
H3C CH3 CH3 CH3
CH3 CH3 H3C CH3
CH3
H3C
H3C CH3 CH3 CH3
CH3 H3C CH
CH3 conjugation 3
CH3 leads to coloured
compounds...this
is carotene from
conjugation
(you guessed it)
carrots

113. N N N N
Na Na
HO HO
OSO2 SO2O OSO2 SO2O
N N N N
Na Na
HO HO
OSO2 SO2O OSO2 SO2O
Cl food green 4 E142

114. N N N N
Na Na
HO HO
OSO2 SO2O OSO2 SO2O
N 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. what have
....we learnt?
•e l e c t r o n s
where they are
•b o n d s
& their strength
Image created by Cary Sandvig of SGI
• re s o n a n c e

116. read
part 5
©the bbp@flickr