Unit One Parts 3 & 4:molecular bonding

Unit OneParts3&4

Bond strengthBond polarisationResonance Pages

34 & 46

Unit OneParts3&4

Bond strengthBond polarisationResonance Pages

34 & 46

...today we continue to make our simple

model more complex!

how strong are bonds?

...and we’re talking about covalent

bonds...the important ones for organic chemists

C C

C C

C C

C C

C C

C C> >

bond strength

836kJ mol–1

610kJ mol–1

347kJ mol–1 Pg

40

the values aren’t important...only the concept / pattern

C C

C C

C C

C C

C C

C C> >

bond strength

836kJ mol–1

610kJ mol–1

347kJ mol–1 Pg

40

obviously, it takes more energy to break

an alkyne apart...breaking

three bonds not one

C C

C C

C C

C C

C C

C C> >

bond strength

836kJ mol–1

610kJ mol–1

347kJ mol–1 Pg

40

the differences are getting smaller...nearly twice as

much energy needed to break 2 bonds but much less

needed to break the third

but...

C C C C

C C C C

bond strength

Pg40

a single σ bond is much stronger than a

single π bond (head-to-head results in better

overlap)

C C C C

C C C C

bond strength

Pg40

...this is the reason alkenes are functional groups but alkanes are

not!

what about bond lengths?

are bond length and bond strength

related?

C C

C C

C C

←120→

←134pm→

←154pm→

bond strength

Pg40

shorter the bond the stronger it normally is...

bond strength

C F

C Cl

C Br

←138→

←178pm→

←193pm→C Br

C F

C Cl

Pg40

shorter the bond the stronger it normally is...better overlap of

atoms / orbitals

how do we explain?

C C←134pm→

610 kJ mol–1C O←122pm→

736 kJ mol–1

similar size and bond lengths but big difference in

energy; why?

C O

77 pm 73 pm

similar size so good orbital

overlap...

bond polarisation

so far, our picture of bonds has said

electrons are shared evenly between two

atoms...

bond polarisation...as always, we teach you a simple model

and then say “reality is more complex!” So lets

take a step back...

is HCl covalent or ionic?

two kinds of bond...which one is

it?

H+ Cl–H Cl

H Cl H Cl

Polar covalent bond

Pg34

electrons shared evenly in a covalent

bond...or...

H+ Cl–H Cl

H Cl H Cl

Polar covalent bond

Pg34

one electron lost from H and given to Cl (an ionic bond)

H Cl

H Cl H ClH Clδ+ δ–

Polar covalent bond

Pg34

...or somewhere in the middle...

H Cl

H Cl H ClH Clδ+ δ–

Polar covalent bond

Pg34

a covalent bond but with the electrons

predominantly on one atom (ionic character)

H2.1Li1.0

Be1.5

B2.0

C2.5

N3.0

O3.5

F4.0

Na0.9

Mg1.2

Al1.5

Si1.8

P2.1

S2.5

Cl3.0

K0.8

Ca1.0

Br2.8

Rb0.8

Sr1.0

I2.5

Bond Type ENdifference Examples Calculation

ionic > 1.7 NaCl 3.0(Cl) - 0.9(Na) = 2.1

polar covalent 0.5 – 1.7 CH3O–HH–Cl

3.5(O) - 2.1(H) = 1.43.0(Cl) - 2.1(H) = 0.9

covalent 0 – 0.4 CH3–HH–H

2.5(C) - 2.1(H) = 0.42.1(H) - 2.1(H) = 0.0

Pg35

electrons rarely shared evenly in a covalent bond...

H2.1Li1.0

Be1.5

B2.0

C2.5

N3.0

O3.5

F4.0

Na0.9

Mg1.2

Al1.5

Si1.8

P2.1

S2.5

Cl3.0

K0.8

Ca1.0

Br2.8

Rb0.8

Sr1.0

I2.5

Bond Type ENdifference Examples Calculation

ionic > 1.7 NaCl 3.0(Cl) - 0.9(Na) = 2.1

polar covalent 0.5 – 1.7 CH3O–HH–Cl

3.5(O) - 2.1(H) = 1.43.0(Cl) - 2.1(H) = 0.9

covalent 0 – 0.4 CH3–HH–H

2.5(C) - 2.1(H) = 0.42.1(H) - 2.1(H) = 0.0

Pg35

...electrons will be closer to the more electronegative atom...given by the Pauli

scale above (bigger number more electronegative)

H2.1Li1.0

Be1.5

B2.0

C2.5

N3.0

O3.5

F4.0

Na0.9

Mg1.2

Al1.5

Si1.8

P2.1

S2.5

Cl3.0

K0.8

Ca1.0

Br2.8

Rb0.8

Sr1.0

I2.5

Bond Type ENdifference Examples Calculation

ionic > 1.7 NaCl 3.0(Cl) - 0.9(Na) = 2.1

polar covalent 0.5 – 1.7 CH3O–HH–Cl

3.5(O) - 2.1(H) = 1.43.0(Cl) - 2.1(H) = 0.9

covalent 0 – 0.4 CH3–HH–H

2.5(C) - 2.1(H) = 0.42.1(H) - 2.1(H) = 0.0

Pg35

...difference in value indicates the

nature of the bond...

you do not needto learn these values!!

polarisation explains carbonyl bond strength (& reactivity)

C C

C Oδ+ δ–C Oδ+ δ–

so, carbonylstronger bond than alkene because it has ionic bond

character (electronicattraction between the

two atoms)

polarisation explains carbonyl bond strength (& reactivity)

C C

C Oδ+ δ–C Oδ+ δ–

but carbonyl also more reactive because the δ+ charge attracts

electrons

Helectrons move

Helectrons move

organic chemistry is all about the

movement electrons

HHreactions are the movement of electrons

H H H H

reactions are the movement of electrons

H H H H

HHreactions are the

movement of electrons...

polarisation explains reactivity

I HO

HO

Iδ+ δ–

HO

δ+ δ–

...so if we can predict where the electrons start and where they finish (want to go)...

polarisation explains reactivity

IHO

HO

HO

δ+δ–δ+

δ–I...then we can

predict reactions

H3C I CH3OH

H3C

O

OMeCH3

MgBr

δ+δ+

δ+ δ+δ+ δ–

δ–

δ–

δ–δ–

polarisation explains reactivity of molecules

δ– means more electrons or partial

negative charge

H3C I CH3OH

H3C

O

OMeCH3

MgBr

δ+δ+

δ+ δ+δ+ δ–

δ–

δ–

δ–δ–

polarisation explains reactivity of molecules

δ+ means lack of electrons or partial

positive charge

H3C I CH3OH

H3C

O

OMeCH3

MgBr

δ+δ+

δ+ δ+δ+ δ–

δ–

δ–

δ–δ–

polarisation explains reactivity of molecules

δ+ (or slightly positive) part of a molecule will be

attacked by...

H3C I CH3OH

H3C

O

OMeCH3

MgBr

δ+δ+

δ+ δ+δ+ δ–

δ–

δ–

δ–δ–

polarisation explains reactivity of molecules

...the negative part of a Grignard reagent...in fact we can explain

most chemical reactions by these δ–/+

charges

so far, so good...

i hope!

so let's apply what we know...

draw nitromethaneCH3NO2

C + 3H + O2+N

Pg45

here are the constituent

atoms...

CH

HH N

O

OC + 3H + O2+N

Pg45

this structure obeys the octet

rule

≡ H3C NO

OCH

HH N

O

OC + 3H + O2+N

Pg45

doesn’t look quite right...need

to sort out formal charges

fc = 6-4-½(4)=0

≡ H3C NO

OCH

HH N

O

OC + 3H + O2+N

Pg45

top oxygen has no formal

charge

≡ H3C NO

OCH

HH N

O

OC + 3H + O2+N

fc = 5-0-½(8)=+1

Pg45

nitrogen has a +1 charge

≡ H3C NO

OCH

HH N

O

OC + 3H + O2+N

fc = 6-6-½(2)=-1

Pg45

bottom oxygen has –1

charge

H3C NO

O

≡ H3C NO

OCH

HH N

O

OC + 3H + O2+N

116 pm

130 pmPg45

so structure is this?? (one N=O bond and one

N–O bond)

so in theory it is all very easy...

but reality is a little more complex...

H3C NO

O122 pm

Pg46

turns out both N–O bonds

are identical

H3C NO

O122 pm

Pg46

...they are somewhere in between a N–O bond

and a N=O bond...a structure called a...

resonance h y b r i d

H3C NO

O122 pm

Pg46

H3C NO

OH3C N

O

O

resonance structures

H3C NO

O≡

Pg46

the two extremes are

resonance structures...

H3C NO

OH3C N

O

O

resonance structures

H3C NO

O≡

Pg46

reality is a resonance

hybrid

H3C NO

OH3C N

O

O

resonance structures

H3C NO

O≡

Pg46we can convert

the extremes by pushing electrons

(not atoms)

H3C NO

OH3C N

O

O

resonance structures

H3C NO

O≡

Pg46

lets try and explainthe relationship between resonance structures and

resonance hybrids...

resonance structures

imagine you took one man...lets call him

Peter...as one of your resonance structures

resonance structures

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

resonance structures

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

resonance structures

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

hybrid or...

resonance hybrid

© Marvel Comics

HC

CC

H

H

H

H

resonance structuresDO NOT EXISTbut are useful

HC

CC

H

H

H

H

HC

CC

H

H

H

H

resonance structuresDO NOT EXISTbut are useful

HC

CC

H

H

H

H

& easier to draw

HC

CC

H

H

H

H

resonance structuresDO NOT EXISTbut are useful

HC

CC

H

H

H

H

& easier to draw

they are Lewis structures so obey

octet rule so we can draw them...

resonance hybridE X I S T SLewis structure impossible

HC

CC

H

H

H

H

resonance hybridE X I S T SLewis structure impossible

HC

CC

H

H

H

H

...the Lewis structure no longer obeys

octet rule (how many electrons on central

C?)

resonance hybridE X I S T SLewis structure impossible

HC

CC

H

H

H

H

...the circle you draw in the centre of benzene is a resonance hybrid but the double bonds I draw make

its chemistry easier to predict...

only electrons move

only electrons move between resonance structures (and in

reactions)

all atoms are stationary...the atoms remain

stationary

curly arrow

this is used to represent the

movement of two electrons...

curly arrow

it is possibly the most important

‘scribble’ an organic chemist

ever learns...

curly arrow

with this you can bin most text books and just predict reactions instead of learning

them...

curly arrow

words cannot describe how

wonderful I think this little doodle

is!

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...

1L e w i s structure

H3C CO

O

Pg46

...first part is relatively easy (or at

least covered in earlier material!)

2..pushable electrons‘ .’

N , Clone pairs

C Cπ bonds

Pg48

now we need to identify which electrons can be moved or pushed (for

some reason we always talk about pushing

‘curly arrows’

N , Clone pairs

C Cπ bonds

Pg50

the source of electrons

2..pushable electrons‘ .’

2..pushable electrons‘ .’

N , Clone pairs

C Cπ bonds

Pg48

lone pairs of electrons are often

‘pushable’

2..pushable electrons‘ .’

N , Clone pairs

C Cπ bonds

Pg48

as are double (or triple) bonds...

3receptors

positive charges C

electronegative atoms C O

atoms with ‘pushable’ electrons

C

Pg48

the next step is to find a target for the

electrons...somewhere they want to go...

3receptors

positive charges C

electronegative atoms C O

atoms with ‘pushable’ electrons

C

Pg50

where electrons are ‘happy’

3receptors

positive charges C

electronegative atoms C O

atoms with ‘pushable’ electrons

C

Pg48

all of the above will happily accept

electrons so are good receptors

3receptors

positive charges C

electronegative atoms C O

atoms with ‘pushable’ electrons

C

Pg48

this is only a receptor because it can also loss

electrons (remember we do not want more than eight 8 electrons around an atom)

3receptors

positive charges C

electronegative atoms C O

atoms with ‘pushable’ electrons

C

Pg48

NOTE: the donor and acceptor must be

one bond apart (no more no less)

4H3C C

O

OH3C C

O

OH3C C

O

O

resonance f o r m s Pg48

finally, move the electrons and form a

new, valid Lewis structure

4H3C C

O

OH3C C

O

OH3C C

O

O

resonance f o r m s Pg48

here are three resonance structures

for the molecule in steps one (the

ethanoate anion)

4H3C C

O

OH3C C

O

OH3C C

O

O

resonance f o r m s Pg48

remember the resonance hybrid will

be somewhere in between all these as shown on the next

slide...

H3C CO

Oor H3C C

O

O

–1/2

–1/2

delocalisation

C–O 130 pm

all bond lengths are the same...showing that the compound

never has a single C–O bond or a double C=O

bond

delocalisation

C–O 130 pmthe electrons are said

to be delocalised over the three atoms

(O–C–O)

H3C CO

Oor H3C C

O

O

–1/2

–1/2

delocalisation

C–O 130 pmelectrons are happy

when they are delocalised as they are

spread over a larger area...so are further

apart

H3C CO

Oor H3C C

O

O

–1/2

–1/2

H3CC

CCH3

N

H

Ph

H3CC

CCH3

N

H

Ph

examples...

Xwhy is this wrong?

...because it has 10 electrons in valence shell of C, which is

never allowed!

H3CC

CCH3

N

H

Ph

examples...the correct way involvespushing the lone pair of the

nitrogen anion down one bond to make a double C=N bond and then

pushing the electrons off thecarbon (so that it doesn’t have

10 valence electrons)and...

H3CC

CCH3

N

H

Ph

examples...

H3CC

CCH3

N

H

Ph

...moving them to the carbon at the end of the

double bond (we can’t move them two bonds) and

forming this new anion

≡H3C

CC

CH3

N

H

Ph δ–

δ–H3C

CC

CH3

N

H

Ph

examples...

H3CC

CCH3

N

H

Ph

the resonance hybrid shares (delocalises) the

electrons over two bonds or three atoms...

examples...

H3CC

CCH3

NPh

H HH3C

CC

CH3

NPh

H HX

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)

example...

H3C CO

O Hethanoic acid

H3C CO

CH3propanone

H3CH2C

O Hethanol

124pm129pm 146pm

122pm

the bonds in ethanoic acid are not what we would

predict compared to other simple molecules...(C=O

longer & C–O shorter)...why?

example...

H3C CO

OH3C C

O

O H H

δ–

δ+≡H3C C

O

O H

H3C CO

O Hethanoic acid

H3C CO

CH3propanone

H3CH2C

O Hethanol

124pm129pm 146pm

122pm

...reason is that lone pair of electrons are pushable and

the C=O is a good receptor...

example...

H3C CO

OH3C C

O

O H H

δ–

δ+≡H3C C

O

O H

H3C CO

O Hethanoic acid

H3C CO

CH3propanone

H3CH2C

O Hethanol

124pm129pm 146pm

122pm...which allows a new resonance structure that

can contribute to the resonance hybrid and gives

the C–OH bond double bond character so causes it

to shrink...

example...

H3C CO

O H

H3C CO

O H5 bonds!

H3C CO

O H2 molecules

H3C CO

O H2 molecules

X we cannot start from the lone pair on the carbonyl

example...

H3C CO

O H

H3C CO

O H5 bonds!

H3C CO

O H2 molecules

H3C CO

O H2 molecules

Xcarbon can never have 5

bonds (or 10 valence electrons)

example...

H3C CO

O H

H3C CO

O H5 bonds!

H3C CO

O H2 molecules

H3C CO

O H2 molecules

Xas soon as we split the

molecule in two we have performed a reaction and not looking at resonance

anymore.

NH

O

HONH2

CO2H HN NH2

NH

NH

O

HONH2

CO2H HN NH2

NH

example...

kytotorphinpain regulation

CC

CCC

CH

H

HH

H

HCC

CCC

CH

H

HH

H

HCC

CCC

CH

H

HH

H

H≡

example...

why is phenol acidic?

or why is phenol a separate functional group

and not an alcohol?

why is phenol acidic?

for a group to be acidic it must be able to give away

H+ (a proton)

CC

CCC

CO

H

HH

H

H

delocalisation......if phenol losses H+ then we are left

with O–...is this stable (will it readily form)?

CC

CCC

CO

H

HH

H

HCC

CCC

CO

H

HH

H

H

delocalisation......we can move the lone pair to form C=O as we

can push the electrons of C=C...we have spread the electrons over three atoms

so they are happy...

CC

CCC

CO

H

HH

H

HCC

CCC

CO

H

HH

H

HCC

CCC

CO

H

HH

H

HCC

CCC

CO

H

HH

H

H

delocalisation...

turns out we can form many other resonance structures so the

electrons are delocalised over 7 atoms...they are really jolly. So

anion stable so loss of H+ easy so phenol is acidic

CC

CCC

CO

H

HH

H

H

δ–

δ–δ–

δ–

H3C CH3

H3C CH3CH3

CH3 CH3

CH3 CH3

H3C

H3C CH3

H3C CH3CH3

CH3 CH3

CH3 CH3

H3C

conjugationany double bonds separated by a one

single bond can delocalise

H3C CH3

H3C CH3CH3

CH3 CH3

CH3 CH3

H3C

H3C CH3

H3C CH3CH3

CH3 CH3

CH3 CH3

H3C

conjugationsuch double bonds

are said to be conjugated

H3C CH3

H3C CH3CH3

CH3 CH3

CH3 CH3

H3C

H3C CH3

H3C CH3CH3

CH3 CH3

CH3 CH3

H3C

conjugationconjugation

leads to coloured compounds...this is carotene from (you guessed it)

carrots

NN

OSO2

HO

SO2O

Na

NN

OSO2

HO

SO2O

Na

NN

OSO2

HO

SO2O

Na

NN

OSO2

HO

SO2O

Na

Cl food green 4 E142

NN

OSO2

HO

SO2O

Na

NN

OSO2

HO

SO2O

Na

NN

OSO2

HO

SO2O

Na

NN

OSO2

HO

SO2O

Na

Cl food green 4 E142

hopefully, you can see that if we have alternating double and single bonds

we can form multiple resonance structures or delocalise the electrons

what have....we learnt?

•e l e c t r o n s where they are

•b o n d s& their strength

•resonance Image created by Cary Sandvig of SGI

©the bbp@flickr

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