lecture from monday ch14c, 4/9/12 topics covered: -functional groups -chemdraw, chem3d (computer...

Lecture from Monday

Ch14C, 4/9/12

Topics covered:-Functional groups-ChemDraw, Chem3D (computer modeling)-Hybridization picture of H2C=C=CH2 (allene)-Drawing resonance structures (Klein handout)-Assessing “good” vs. “bad” resonance structures-Introduction to Conjugation (Part 1 of 2)

Practice Problems for Hybridization

CH3 O1. Identify the functional groups present in the molecule

2. Identify the hybridization of every atom (solutions available in Klein, Problem 4.12)

3. Build a model in 3D to visualize what the molecule actually looks like

C C C

H

H H

H1. Draw out the hybridized orbitals for this molecule.

2. Are the 4 hydrogens all in the same plane? This problem illustrates the power of being able to predict 3-D structure by understanding the principles of hybridized orbitals

RHN

Intro to Functional GroupsFurther reading:- A basic intro in Klein, Sec. 5.1 (p. 84-85)- A detailed list of functional groups in Appendix A of the Thinkbook (p. 326)

R OH

OCarboxylic acid

R O

O

R' Ester

R H

O

R and R' = a simple carbon substituent (i.e. methyl, cyclohexyl, benzene, etc.). In examples below, R and R' can be the same or different substituents

Aldehyde

R R'

O Ketone

R OH Alcohol

R NH2 Amine (1o)

Alkene

Alkyne

Benzene ring

R X Alkyl halide

R O R' Ether

R-SH Thiol

R S R' Sulfide

R S S R' DisulfideR NH2

O

R NH

O

R'or Amide

R H

NR'

Imine

R' Amine (2o)

R N R'

R''

Amine (3o)

R Cl

OAcid chloride

R C N Nitrile

R NO

O

Nitro

R O

O

R'

O

Acid anhydride

OHPhenol

Practice Problems for Hybridization

CH3 O alkyne

ketone

alkene

benzene ringAcetylsalicylic acid (Aspirin)- used to treat headaches

O

OH

O

O CH3

carboxylic acid

ester

O

O

CH3

NH2

HN

H3C O

O

H3C

H3C

Oseltamivir (Tamiflu)- used to treat flu (e.g. H5N1)

ester

amine

amide

ether alkene

Penicillin G- antibiotic produced by the Penicillium fungus

N

HN

O

S Me

Me

OHO

O

Ph

amide

carboxylic acid

amide

sulfide

Made using ChemDraw and Chem3DPrograms available on computers in UCLA Science Learning Center (Young 4335 and others)Intro tutorial to how to use ChemDraw: http://www.chem.umass.edu/~samal/267/chemdraw.pdf

CH3 O

http://www.chem.umass.edu/~samal/267/chemdraw.pdf

Vollhardt, Figure 14-4 (p. 618)

•So, the hydrogens of allene are NOT in the same plane, but are instead in two planes. •This is not something we might have been able to predict just from looking at the Lewis structure

IUPAC name: propanedieneCommon name: allene H

C

H

C C

H

H

C O C O

Which Resonance Contributor Represents Reality?

The carbon monoxide case:

Neither contributor fully represents CO

Resonance hybrid: A weighted average or blend of resonance contributors; the most accurate representation of the electronic structure of a molecule.

Peach Plum

“True” structure:

Nectarine

Your friend asks you to describe what a nectarine is because he’s never seen or eaten one… (Klein, Sec 2.1)

C O C OX

Which Resonance Contributor Represents Reality?

Once upon a study break...

Real creatureFantasy creaturesNeither fully represents reality A unicorn-dragon hybrid?

The carbon monoxide case:

Neither contributor fully represents CO

Resonance hybrid: A weighted average or blend of resonance contributors; the most accurate representation of the electronic structure of a molecule.

X

Drawing the Resonance Hybrid

Example: Draw the resonance hybrid for acetate ion, CH3CO2-.

2. Draw the features that are the same for all contributors• Sigma and pi bonds, lone pairs, and formal charges

3. Add features that are not the same for all contributors

CH3 O

O

CH3 O

O

CH3 O

O

CH3 O

O

1. Draw contributors

•Partial (shared) pi bonds shown as ----•Partial (shared) charges shown as + or - CH3 O

O

Resonance hybrid

Do All Contributors Have Equal Importance?Is a rhinoceros more unicorn or more dragon?

contributor “stability” = resemblance to reality = contribution to hybrid

Therefore we need contributor preference (“stability”) rules:

As the number and/or magnitude of rules violations ,

= importance of individual contributor

= contribution to resonance hybrid

Resonance Contributor Preference Rules

Rule #1: The most important contributor has the maximum number of atoms with full valence shells.

Rule #1 is more influential than all the other preference rules.

Rules #2-6 have no particular order of preference.

Open valence shell on carbonLess important contributor

Example: H2C OHH2C OH

In some cases it may not be possible for all atoms to have full valence shells.

For practice: see Klein Sec 2.8


Rule #2: The most significant contributor has the maximum number of covalent bonds.

Three covalent bondsLess important contributor

Example: C

H

H

O C

H

H

O


Rule #3: The most significant contributor has the least number of formal charges.

No formal chargesMore important contributor

Two formal chargesLess important contributor

Example: C

H

H

O C

H

H

O


Rule #4: If a contributor must have formal charge(s), the most important contributors has these charges on the atom(s) that can best accommodate them.

Carbon EN = 2.5Less important contributor

Oxygen EN = 3.5More important contributor

•Negative formal charges best on atoms of high electronegativity O- better than C-

•Positive formal charges best on atoms of low electronegativity C+ better than O+

•Minimize formal charge magnitude +1 better than +2

H3C CH2

O

H3C CH2

O

H3C CH2

O


Rule #5: Resonance interaction (i.e., pi bond) is strongest between atoms in the same row of the periodic table.•Usually CNOF

•Usually outweighs electronegativity considerations (rule #4)

F, C both 2nd rowMore important contributorEven though EN F > EN Cl

C 2nd row; Cl 3rd rowLess important contributor

Example:F C Cl

H

F C Cl

H


Rule #6: Other factors (such as aromaticity) that we will encounter later.

Violations to the resonance contributor preference rules exist, but are uncommon.

Resonance Structure Examples

N O N O

Reasonable Resonance Structure

Unreasonable Resonance Structure

N O

H

- positive charge on oxygen, but oxygen doesn't have even have octet

- has formal charges, but at least octet rule is satisfied and charges are distributed on the appropriate atom (based on EN)

OH

OH

H

OH

H

H

H

- one carbon doesn't have full octet- two carbons directly adjacent both have negative charge

- all atoms have complete octet

Nitro Compounds

Napoleon's Buttons, Chap 5: Nitro Compounds

4 KNO3 (s) + 7 C(s) + S (s) 3 CO2 (g) + 3 CO (g) + 2 N2 (g) + K2CO3 (s) + K2S (s)- potassium nitrate- saltpeter- "Chinese snow"

carbon (e.g. wood charcoal)

sulfur carbon dioxide

carbon monoxide

nitrogen potassium carbonate

potassium sulfide

shockwave of gas (6000 meters per sec for TNT)

CH3

NO2

OHO

NH2

Chemical Formula: C7H7NO2

p-nitro toluene(explosive)

p-amino benzoic acid (PABA)(not explosive)- sunscreens

CH3

NO2

NO2O2N

trinitrotoluene (TNT)- patented by Alfred Nobel (Nobel Prize)

O

O

O

NO2

NO2

NO2

Nitroglycerin

body chemistry

N=O

nitric oxide- dilates blood vessels- treatment of heart disease angina pectoris

2 NH4NO3 (s) 2 N2 (g) + O2 (g) + 4 H2O (g)

ammonium nitrate (fertilizer)

O

O

O

O

NO2

NO2

NO2

O2N

pentaerythritol tetranitrate (PETN)- plastic explosive

N

R

O

ONitro functional group

What does this molecule have to do with resonance?

Conjugated Molecules - Part 1 Lecture Supplement page 28

What is Conjugation?

For any molecule best structure = lowest energy

Lowest energy from minimizing electron repulsion and maximizing p orbital overlap•Maximum p orbital overlap allows...

Square planar or tetrahedral? Staggered or eclipsed?

MethaneH

CH H

HH

H

CH

H

What are these?}ResonanceConjugationAromaticity

•Familiar examples: Which structure is lowest energy?

EthaneH

C C

HH

H

HH

C C

HH

H

H H

H

What is Conjugation?Case #1: Relative stability of C4H6 isomers

• Restriction: Hof comparisons only valid among isomers

4 C (graphite) + 3 H2 (g) Hof = 29.9 kcal mol-1

4 C (graphite) + 3 H2 (g) Hof = 47.7 kcal mol-1

More stable isomer

Isomers

•Isomers: Same molecular formula, different structure

• Importance: Lower Hof = more stable isomer

•Heat of formation (enthalpy of formation; Hof): Hypothetical enthalpy change when a

substance is synthesized from elements in their standard states

29.9

30.7

32.0 C C CH3H3C

37.5

91.8 1,2-butadiene (an allene)

s-cis-1,3-butadiene

45.8

C C CH2CH3H

s-trans-1,3-butadiene

47.7

64.6

67.7

H2C C C

CH3

H

1,3-dienes

E n t h a l p y o f f o r m a t i o n ( H o f ) , k c a l m o l - 1 ( c a l c u l a t e d )

0.0 4 C (graphite) + 3 H2 (g)


Why this order?•Ring strain?

No ring strain

•Position of pi bonds?

Two pi bonds

One pi bondRing strain

Two pi bonds

No ring strain

•Number of pi bonds?

Case #1: Relative stability of C4H6 isomers

Alternating pi-sigma-pi bonds


Case #2: Catalytic Hydrogenation of 1,3-Dienes versus 1,4-Dienes

Thermodynamics

•Lose: H-H sigma bond, C-C pi bond

•Gain: 2 x C-H sigma bond

•Sigma bonds usually stronger than pi bonds

•Therefore catalytic hydrogenation is exothermic (H < 0)

H = -30 kcal mol-1Example: CC

H

CH3CH2

H

H

H2

PtCCH3CH2

H

C

H

H

H

H

Catalytic hydrogenation: Addition of H2 to a pi bond with a catalyst

What is Conjugation?Case #2: Catalytic hydrogenation of 1,3-dienes versus 1,4-dienes

H = -65.1 kcal mol-1H = -56.5 kcal mol-1

1,3-butadiene more stable than 2-butyne

E n t h a l p y ( H ; k c a l m o l - 1 )

C CH3C CH3

Same molecule = same enthalpy (H)

+ 2 H2

+ 2 H2

•Observation: H (1,3-butadiene butane) < H (2-butyne butane) by 8.6 kcal/mol

•Conclusion: Lowest H (for catalytic hydrogenation) belongs to most stable isomer

How can we use catalytic hydrogenation to probe C4H6 isomer stability?

•Fact: 1,3-butadiene more stable than 2-butyne

What is Conjugation?Case #2: Catalytic hydrogenation of 1,3-dienes versus 1,4-dienes

•Use H (cat H2) to compare pi-sigma-pi (1,3-diene) versus pi-sigma-sigma-pi (1,4-diene)

H = -30 kcal mol-1

Predict: H = 2 x (-30) = -60 kcal mol-1Observe: H = -60 kcal mol-1Conclusion: No special stability for 1,4-diene

1,4-Pentadiene (a 1,4-diene)

1-Pentene (alkene energy benchmark)

1,3-Butadiene (a 1,3-diene) Predict: H = -60 kcal mol -1 if stability = 1,4-dieneObserve: H = -56.5 kcal mol-1 3.5 kcal mol-1 less than expectedConclusion: 1,3-diene more stable than 1,4-diene.

2 H2

Pt1

23

4

2 H2

Pt12

34

5

General observation: 1,3-dienes more stable than similar 1,4-dienes

H2

Pt

The experiment

Lecture from Wed, 4/11/12

-Conformations of 1,3-butadiene (s-cis vs. s-trans)-Discussion of how dihedral angle relates to extent of p-orbital overlap-Resonance picture of 1,3-butadiene correlates well to physical properties-Conjugated molecules and color

C

CH

HCH

C

H

H

H

C

CC

H

HCH

H

H

H

1,3-Butadiene: A Closer Look

What is origin of special 1,3-diene stability?

Major conformation? Torsional strain: s-trans < s-cis

Stability: s-trans > s-cis

Planarity: Nearly always planar

s-cis*5%

s-trans95%

C

CC

H

HCH

H

H

H

2.21 Å

C

CH

HCH

C

H

H

H

2.50 ÅK eq > 1

* s-cis = two pi bonds, separated by sigma bond, in cis arrangement

29.5

30

30.5

31

31.5

32

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180

E n e r g y ( k c a l m o l - 1 )

C=C-C=C Dihedral angle (degrees)

1,3-Butadiene: A Closer LookRotation around Csp2 - Csp2 bond:

Pi bondsperpendicular

Barri

er to

rota

tion

Highest energy point

Conclusion: More than just torsional strain is at work

Angle between planes formed by three atoms each

Energy difference between most and

least stable conformations

0 o 90 o

180 o

H

H

H

H

C

CH

HCH

C

H

H

H

1,3-Butadiene: Resonance Model

What is the origin of this extra stability, planarity, and barrier to rotation?

C

CH

HCH

C

H

H

H

C

CH

HCH

C

H

H

H

Resonance hybrid

C

CH

HCH

C

H

H

H

explains barrier to rotation

explains planarity

Resonance hybrid observations •Partial C2-C3 pi bond

•C2-C3 pz orbital overlap

Resonance is often a strong influence on molecular structure, so start there

1,3-Butadiene: Resonance Model

The resonance model looks useful, but simplistic. Is it accurate?

C2-C3 barrier to rotation (kcal mol-1)

C2-C3 bond length (Å)

1,3-ButadieneModel molecules

1.54 1.33

4.5 ~60

Conclusion:

1.48

7.5

Resonance model is accurate despite its simplicity

The resonance model predicts...

Molecule is more stableMolecule is less stable

•Pi electrons have longer wavelength

•No significant resonance •Has some resonance

1,3-Butadiene: Resonance ModelHow does resonance explain why a 1,3-diene is more stable than a 1,4-diene?

1,3-diene1,4-diene

CC C C

C

H

H H

H

HH

H H

CC

H H

C

H

CCH

H H

H

He- e-

•Pi electrons have shorter wavelength

CC C C

C

H

H H

H

HH

H H

CC

H H

C

H

CCH

H H

H

H

•Pi electrons confined between two carbons •Pi electrons roam over four carbons

E = hc/ so wavelength = energy

Discussion of Handout:“p orbital overlap: Resonance, Conjugation, and

Aromaticity”

Conjugated Molecules - Part 2Lecture Supplement page 37

chlorophyll a

CH3

CH3

CH3CH3

CH3CH3

CH3

CH3

H3C

H3C

CH3 CH3

CH3 CH3 CH3

H3C CH3

H3C CH3

CH3

lycopene

-carotene

NN

NN

CH3

CH3

CH3H3C

H3CO

O OR

Mg2+

H3C

CH3

CH3CH3CH3CH3

O

O

R =

Part 1 Summary

•...more resonance•...more electron delocalization•...lower electron energy

Consequences of p orbital overlap•Atoms with p orbitals must be planar•Partial pi bond(s)•Barrier to rotation

more stable thanExample:

Adjacent, overlapping p orbitals allows for...

Greater stability

H NH2

O

Extra Stability Limited to 1,3-Dienes?

Amide resonance contributors:

Pi electron delocalizationprovides increased stability

H NH2

O

Amide resonance hybrid:

H NH2

O

•1,3-diene has four adjacent p orbitals

•Three adjacent p orbitals is enough to provide extra stability

Example: An amide

Extra Stability Limited to 1,3-Dienes?Another special stabilization example: An amide

•Predict: Four attachments = sp3

•sp3 lacks p orbital needed for resonance•Therefore sp2 to accommodate resonance•Therefore sp2 to increase stability

I thought hybridization is controlled only by the number of attachments?!•Energy causes geometry; geometry causes hybridization

Influenced by electron repulsion, resonance, etc.

Resonance hybridH NH2

O

Nitrogen hybridization?

H NH2

O

C

N

O

H

H

H

C C C CH

H

H

HH

H

Extra Stability Limited to 1,3-Dienes?More adjacent p orbitals = larger electron “playground”

An amide has three adjacent, parallel p orbitals:

Compare with 1,3-butadiene: Four adjacent, parallel p orbitals:

Build your own model

Build your own model

C

N

O

H

H

H

p orbital overlap forms pi bonds

C C C CH

H

H

HH

H

In general: Adjacent, parallel p orbitals improve molecular stability

Sigma bonds

p orbital overlap gives delocalized pi bonds

Conjugation: Special stability provided by electron delocalization in three

or more adjacent, parallel, overlapping p orbitals.

Conjugation: A DefinitionFinally!

Decrease in electron energy

Not limited to pi-sigma-pi(four carbon pz orbitals)

Consequences of Conjugation

Conjugation influences widespread

•Chemical reactivity: Is molecule reactive or inert?

•Molecular structure: Is a molecule (or portion of molecule) flat? Is bond rotation hindered?

•Physical properties: Color

•Etc.

O

CH3

CH3

CH3

O

CH3

HO

CH3

CH3

H

O

CH3

HO

CH3

CH3

H

O

CH3

HO

CH3

CH3


•Influences distribution of products in chemical reaction

•Reaction products: stability = amount produced

•Example: Determine major product of this reaction:

- H2O

Ten conjugated p orbitals Six conjugated p orbitalsMore stable Less stable

Produced in greatest amount

O

CH3

CH3

CH3

or

Which product isomer is more stable?

Consequence #1: More extensive conjugation = greater stability

N

O

H R


•Resistance to conformational change (barrier to rotation)

More stable Less stable

Barrier to rotation and planarity critical to protein function.

N

O

H

RN

O

H R

Resonance hybrid

Barrier torotation

Planar

Less torsional strain More torsional strain

Consequence #2: Partial pi bond character

•Example: Amide linkage between two amino acids in a protein

•Causes planarity of atoms conjugated p orbitals


Consequence #3: Highly conjugated molecules may be colored

Examples:

Chlorophyll Lycopene -Carotene

Origin of color: •Some portion of visible (white) light spectrum is absorbed

•Brain perceives remaining light as color

So how does molecular structure control energy of photons absorbed?

molecular orbitals as descriptors of conjugated systems

Recall: Here is how we described the atomic orbitals of one carbon atom

2p

2s

sp3

hybridizationsp3

4 atomic orbitals in 4 hybrid atomic orbitals out

Here: The number of molecular orbitals is equal to that of the component p orbitals (which is the number of carbon atoms with a p orbital)

example: 1,3-butadiene

4 p atomic orbitals in 4 molecular orbitals out

four 2p atomic orbitals

2p

1

2

3

4

HOMO

LUMO the lower this gap, the longer the wavelength absorbed

Consequences of ConjugationHow does molecular structure control energy of photons absorbed?

•Molecule absorbs photon (h)

E

E l e c t r o n e n e r g y

Lowest UnoccupiedMolecular Orbital

(LUMO)

Highest OccupiedMolecular Orbital

(HOMO)

Absorb photon (E = h)

Ground state Excited state

•Electron is excited to higher energy molecular orbital

•Energy of photon absorbed must equal HOMO/LUMO orbital energy difference (E)

Controlled by electron energies


number of conjugated p orbitals E•WhenE low enough, photons of visible light absorbed•Unabsorbed portion of visible light spectrum perceived as color

P h o t o n e n e r g y

Photon absorbed Observed color

Ultraviolet (UV)

Visible light

YellowOrange

RedVioletIndigoBlue

Green

Colorless

VioletIndigoBlue

GreenYellowOrange

Red


Example Molecules

Four conjugated p orbitalsE = ultraviolet

Perceived color = colorless

1,3-Butadiene

Six conjugated p orbitalsE = ultraviolet


1,3,5-Hexatriene

Eight conjugated p orbitalsE = ultraviolet


Styrene

YellowOrange

RedVioletIndigoBlue

Green

VioletIndigoBlue

GreenYellowOrange

Red


Example Molecules

Twelve conjugated p orbitalsPhotons of h = E are indigo

Perceived color = orange

Ten conjugated p orbitalsPhotons of h = E are violet

Perceived color = yellow

Retinal

O

Retinol (vitamin A)

OH

YellowOrange

RedVioletIndigoBlue

Green

VioletIndigoBlue

GreenYellowOrange

Red


Example Molecules

22 conjugated p orbitalsPhotons of h = E are blue

Perceived color = red

H3C CH3

CH3 CH3 CH3

CH3 CH3 CH3

CH3H3C

LycopeneNot conjugated

Not conjugated

YellowOrange

RedVioletIndigoBlue

Green

VioletIndigoBlue

GreenYellowOrange

Red

Friday’s topic: Aromaticity (pre-read!!!)

1,3-Butadiene: A Closer LookRotation around the C2-C3 bond•The s-cis conformation is planar

Indicates movie to play

•Filename: s-cis change perspective.mov

1,3-Butadiene: A Closer LookRotation around the C2-C3 bond•Perpendicular conformation has lowest torsional strain

•Filename: s-cis to perpendicular.mov

1,3-Butadiene: A Closer LookRotation around the C2-C3 bond•The s-trans conformation is planar

•Filename: perpendicular to s-trans_prof.mov

lecture from monday ch14c, 4/9/12 topics covered: -functional groups -chemdraw, chem3d (computer...

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xwhich resonance contributor

contributor stability

significant contributor

resonance hybridexample

important contributors

number of formal charges

important contributoroxygen

formal charges3