General Class Information

http://stellar.mit.edu/S/course/5/fa08/5.12/index.html

• Instructors: B. Imperiali & S. O'Connor

• Lectures: Outline, Syllabus & Suggested Reading on Website

• Recitations: Start Second Week; See Handout for Policy on Changes

• Text: "Organic Chemistry" 7th Edition by McMurray

• Molecular Models: Highly Recommended! Allowed in Exams

• Problem Sets: Posted each wednesday (except exam weeks) Key posted following Monday Not graded, but doing them is required for good exam performance

• Exams: Four 50 minute Exams (Drop Lowest Score) & Final

• Grading: 65% In-Class Exams; 35% Final Exam

Study Skills

(How to Pass 5.12 with Flying Colors)

• Read before coming to lecture.

• Attend all lectures.

• Attend all recitations.

• Complete and understand all problem sets.

• Do suggested problems and practice exams without the key.

• Confused? Ask questions immediately.

Organic Chemistry: What is it?

1780: Organic compounds obtained from living sources (complex)

Vitalism: Belief that a "magic" force, present in plants and animals, is necessary for the synthesis of organic compounds

1789: Antoine Laurent Lavoisier observed that organic compounds are composed primarily of carbon and hydrogen

1828: Friedrich Wohler synthesized urea from inorganic compounds!!

Pb+NCO– NH4+OH– NH4

+NCO– C

O

H2N NH2

heat

leadcyanate

ammoniumhydroxide

ammoniumcyanate urea

inorganic organic

Modern Definition:

Organic chemistry is the chemistry of carbon compounds.

"The Age of Organic Chemistry"

• > 95% of All Known Compounds Composed of Carbon

• Crucial to Our Way of Life: Pharmaceuticals, Petroleum, Materials, Polymers (Clothing), OUR BODIES

STRUCTUREDetermining the Way in

Which Atoms Are Put

Together in Space to

Form Complex Molecules

SYNTHESISBuilding Complex

Molecules From Simple

Molecules Using

Chemical Reactions

MECHANISMUnderstanding the

Reactivity of Molecules:

How and Why Chemical

Reactions Take Place

METHODOLOGYDiscovering, Developing,

and Improving Chemical

Reactions

Why Carbon?

• Carbon forms a variety of strong covalent bonds to itself and other atoms.

INCREDIBLE STRUCTURAL DIVERSITY!

O

OH

H

H

HO

H

OH

OHHH

OH

N

N NH

N

NH2

NH

NH

O

O

H2N CH C

CH3

OH

OH2N CH C

CH2

OH

O

HO

H H

H

CH3OH

CH3 H

CH3OH

H H

O

Carbohydrates Amino Acids

DNA Bases Hormones

O

H

H2N

H

HO

H

OH

OHH

CH3

H

5.12 Organic Chemistry Fall 2008

Outline: Lectures 1 and 2

I. Chemical BondingA. HybridizationB. How to draw molecules and mechanismsC. Electronegativity and formal chargeD. Resonance

Reading: McMurray, 7th Edition Chapter 1 (review), 2.1-2.6

Problem Set 1 Posted September 3Key Posted Monday September 8

I. Chemical Bonding: Review from 5.111, 5.112, 3.091

Lewis Bonding Theory: Atoms transfer or share electrons to gain a full valence.(Carbon wants to form four bonds to gain a full octet.)

Molecular Orbital Theory: Atomic orbitals on different atoms combine to form stable molecular bonding orbitals.

(Carbon forms stable bonds to hydrogen, nitrogen, oxygen, sulfur, halides, etc.)

VSEPR Theory: Electron repulsion favors certain geometries (divalent = linear, trivalent = trigonal planar, tetravalent = tetrahedral).

(Carbon atoms can be linear (180°), trigonal planar (120°), or tetrahedral (109°).

I. Chemical Bonding in 5.12

Hybridization: atomic orbitals (s and p orbitals) combine or hybridize to form new atomic orbitals that can bond or overlap with one another more effectively

We are concerned with only s and p orbitals for majority of 5.12 course

A. Hybridization

2s 2px 2py 2pz 4 sp3 orbitals

2s 2px 2py 2pz 3 sp2 orbitals 2pz

• Trivalent carbon forms bonds using three sp2-hybridized orbitals.

• Tetravalent carbon forms bonds using four sp3-hybridized orbitals.

hybridization

hybridization

• Hybridization does not just apply to carbon. This theory can be applied to molecules containing any elements in the periodic table.

So, what do these hybridized orbitals look like?

• Hybrid orbitals are more directional (better overlap to form molecular orbitals).• Hybrid orbitals minimize electron-electron repulsion (think VSEPR).

2s 2px 2py 2pz 2 sp orbitals 2py 2pz

• Divalent carbon forms bonds using two sp-hybridized orbitals.

hybridization

1. sp Hybridization (Linear)

s p 2 sp

X

• 180° bond angle minimizes electron repulsion (VSEPR)

• enhanced e– density

in bonding regions

• two sp-orbitals

But, we only used one s- and one p-orbital!There are two more p-orbitals.

2s 2px 2py 2pz 2 sp orbitals 2py 2pz

hybridization

X

pY

pZ

sp

sp

Complete Orbital Picture of an sp Hybridized Atom

What do these orbitals look like in a molecule?

CH2 CH C H

C CH H

• For simplicity, draw lines connecting p-orbitals to represent !-bonds.

• 2 sp-orbitals• 2 p-orbitals

e.g. Acetylene

"

!y

"""

!z

2s 2px 2py 2pz 2 sp orbitals 2py 2pz

hybridization

2. sp2 Hybridization (Trigonal Planar)

X

120°

s 2 p 3 sp2

• enhanced electron density in bonding

regions

• three sp2-orbitals

X sp2

pY

sp2

Complete Orbital Picture

• For simplicity, leave out small back lobes.• Draw the molecular orbital picture for

ethylene (C2H4).

2s 2px 2py 2pz 3 sp2 orbitals 2pz

hybridization

2s 2px 2py 2pz 4 sp3 orbitals

hybridization

3. sp3 Hybridization (Tetrahedral)

109°

s 3 p 4 sp3

X

C

H

H H

H

e.g. Methane (CH4)

N

H H

H

e.g. Ammonia (NH3)

• No p orbitalsleft over!

!

! !!

! !!

Hybridized Orbitals: All σ-bonds (single bonds) and lone pairsUnhybridized p-Orbitals: All π-bonds (double/triple bonds)sp3: 0 π-bond (single bond)sp2: 1 π-bond (double bond 1 σ-bond + 1 π-bond)sp: 2 π-bonds (triple bond 1 σ-bond + 2 π-bonds)

4. Assigning Hybridization to Atoms in a Molecule(You need to be able to do this!)

Count the hybrid atomic orbitals.# of hybrid orbitals = # of !-bonds + # of lone pairs

# hybrid orbitals hybridization geometry

4 sp3 tetrahedral

3 sp2 trigonal planar

2 sp linear

approx. bond angles

109°

120°

180°

e.g. What is the hybridization of each non-hydrogen atom in acetonitrile? Draw the bonding orbitals (leaving out the small back lobes).

H3C C Na b

4. Assigning Hybridization to Atoms in a Molecule(You need to be able to do this!)

Count the hybrid atomic orbitals.# of hybrid orbitals = # of !-bonds + # of lone pairs

# hybrid orbitals hybridization geometry

4 sp3 tetrahedral

3 sp2 trigonal planar

2 sp linear

approx. bond angles

109°

120°

180°

e.g. What is the hybridization of each non-hydrogen atom in acetonitrile? Draw the bonding orbitals (leaving out the small back lobes).

H3C C Na b

Ca: sp3 (4 !-bonds) C

H

H

H

!

!!

!

4. Assigning Hybridization to Atoms in a Molecule(You need to be able to do this!)

Count the hybrid atomic orbitals.# of hybrid orbitals = # of !-bonds + # of lone pairs

# hybrid orbitals hybridization geometry

4 sp3 tetrahedral

3 sp2 trigonal planar

2 sp linear

approx. bond angles

109°

120°

180°

e.g. What is the hybridization of each non-hydrogen atom in acetonitrile? Draw the bonding orbitals (leaving out the small back lobes).

H3C C Na b

Ca: sp3 (4 !-bonds)

Cb: sp (2 !-bonds)

CC

H

H

H

!

!!

! !

4. Assigning Hybridization to Atoms in a Molecule(You need to be able to do this!)

Count the hybrid atomic orbitals.# of hybrid orbitals = # of !-bonds + # of lone pairs

# hybrid orbitals hybridization geometry

4 sp3 tetrahedral

3 sp2 trigonal planar

2 sp linear

approx. bond angles

109°

120°

180°

e.g. What is the hybridization of each non-hydrogen atom in acetonitrile? Draw the bonding orbitals (leaving out the small back lobes).

H3C C Na b

Ca: sp3 (4 !-bonds)

Cb: sp (2 !-bonds)

N: sp (1 !-bond, 1 lone pair)

NCC

H

H

H

!

!!

! !

4. Assigning Hybridization to Atoms in a Molecule(You need to be able to do this!)

Count the hybrid atomic orbitals.# of hybrid orbitals = # of !-bonds + # of lone pairs

# hybrid orbitals hybridization geometry

4 sp3 tetrahedral

3 sp2 trigonal planar

2 sp linear

approx. bond angles

109°

120°

180°

e.g. What is the hybridization of each non-hydrogen atom in acetonitrile? Draw the bonding orbitals (leaving out the small back lobes).

H3C C Na b

Ca: sp3 (4 !-bonds)

Cb: sp (2 !-bonds)

N: sp (1 !-bond, 1 lone pair)

NCC

H

H

H

!

!!

! !

"y

"z

B. Short-Hand for Chemists: Easy Communication

1. Line-Angle Formulas: Drawing Complex Molecules Quickly

C

CCC

CC

CC

CC C

CC

C

C

C

C

O

CH3 H

CH3

OH

H

HH

HH H

H

H HH

H

H

HH

HH H

H

H

H

Testosterone(not so easy)

O

OH

H3C H

H3C

H H

Vincristine

CC

C

H H

Propane(pretty easy)

much easier!

H

HH

H

HH

Just imagine...

N

OH

CO2MeNH

NCO2Me

OH

OAc

N

H

MeO

CHO

H

Rules for Drawing Line-Angle Formulas

• Bonds are represented by lines (one line = two shared electrons)

• Do not draw carbon or hydrogen atoms, except at termini (for aesthetics)

• Assume carbon atoms are at ends of lines and where they meet

• Assume enough C–H bonds to give each neutral carbon atom four bonds (an octet)

• Draw heteroatoms and attached hydrogen atoms (N,O,S,P,F,Cl,Br,I, etc.)

e.g. isopropanol: CH3CH(OH)CH3

CC

C

H O

H HH H

HH

H

OH

H3C CH3

OHor

e.g. cyclohexanone

C

CC

C

CC

O OH

H

H

HH HH

HH

H

2. Using Dashes and Wedges: Molecules Are Not Flat!

• Tetra-Substituted Carbon Is Tetrahedral (more on this later).

H

CH H

H

CH4109.5°

H

H HH

lines

wedge

dash

methane

lines: in the plane of the paperdashes: going back into the paper (away from you)wedges: coming out of the paper (toward you)

e.g. Propane

HH

HH

HH

HH

OH

OH

OH

OH

e.g. Isomers of 1,2-Cyclohexanediol

2. Using Dashes and Wedges: Molecules Are Not Flat!

• Tetra-Substituted Carbon Is Tetrahedral (more on this later).

H

CH H

H

CH4109.5°

H

H HH

lines

wedge

dash

methane

lines: in the plane of the paperdashes: going back into the paper (away from you)wedges: coming out of the paper (toward you)

e.g. Propane

HH

HH

HH

HH

OH

OH

OH

OH

e.g. Isomers of 1,2-Cyclohexanediol

H

H

H

H

Lewis/Kekulé Structures: Represent atoms sharing electrons to form bonds

Line-Angle Structures: Simplify the drawing of complex molecular structures

Dashes and Wedges: Allow chemists to draw molecules in 3-D

BUT! These simplified structures do not always represent the electronic nature or reactivity of organic molecules!

Electrons are not stationary objects,so it sometimes helps to think about electrons "in motion" . . .

BUT HOW DO WE REPRESENT ELECTRONS IN MOTION?

Representing Molecules

3. Curved Arrow Formalism (Arrow Pushing)

• Chemists use arrows to represent the motion of electrons within and between molecules.

fishook arrow:1 electron moving

double arrow:2 electrons moving

1. The tail starts at the electrons that are moving (lone pair or bond).2. The head shows where the electrons end up (lone pair or bond).

H2N H3C–Cl H2N–CH3 Cl

(l.p. to bond)

(bond to l.p.)

e.g. electron motion in a substitution reaction (much more detail later)

• Use what you know about Lewis bonding to predict the product of the following reaction. Remember to indicate formal charge. Use curved arrows to show the mechanism (movement of electrons).

H3N BH3?

NH H

H

8 electrons(1 lone pair)

B

H

H H6 electrons

(wants 2 more)

• Nitrogen atom is nucleophilic (Lewis basic).

nucleophile: electron-rich atom, often negatively charged, with a free lone pair to donate to another atom

• Boron atom is electrophilic (Lewis acidic).

electrophile: electron-poor atom with a low-lying vacant or easily vacated orbital; wants to accept electrons from a nucleophile

H3N BH3 H3N–BH3

Sample Problem: Using What You Know

C. Electronegativity and Formal Charge

• covalent bonds have a degree of ionic character to them“polar covalent bonds”bonding electrons are attracted more strongly to one atom than the othermore electronegative atoms attract bonding electrons more strongly(see table 2.2 page 36)

C

O

H HH

H!"

!+

C

Li

H HH

!+

!"

oxygen EN = 3.5carbon EN = 2.5

lithium EN = 1.0carbon EN = 2.5

dipole moment

formal charge = # valence electrons - # bonding electrons/2 - nonbonding electrons

H3N BH3 H3N–BH3

Formal Charge:N: 5 – 4 – 0 = +1B: 3 – 4 – 0 = –1

Calculating Formal Charge

Formal Charge:N: 5 – 3 – 2 = 0B: 3 – 3 – 0 = 0

calculating formal charge for C, O, and N should become second nature in this course

Common Formal Charges

See table 2.2, page 42

C C C N N O O

valence e 4 4 4 5 5 6 6

# bonds 3 3 3 4 2 3 1

# nonbonding e 1 0 2 0 4 2 6

formal charge 0 1 -1 1 -1 1 -1

C. Resonance: Electronic Motion Within a Molecule

• The reactivity of a molecule is not always explained by one Lewis structure.

• Molecules can be thought of as hybrids or weighted averages of two or more Lewis structures, each with a different placement of electrons. • These structures, called resonance structures, are not real or detectable, but they are a useful conceptual tool for understanding the reactivity of molecules.

• Chemical reactions generally involve the movement of electrons between two or more molecules, but the concept of electron motion is also important

within a molecule.

e.g. How can you predict where a nucleophile (such as H2N–) will react

with formaldehyde (CH2O)?

O

H H

O

H H

O

H H!+

!–

• The minor resonance structure suggests that the carbon atom is electron-deficient (electrophilic).

O

H H!+

H2NH H

H2N O

major minor

O

H HH2N ?

• Use resonance to better understand the electronic nature of formaldehyde . . .

The nucleophile will react with the electrophilic carbon atom.

1. Only electrons move! Nuclei and the sigma- (single bond-) framework are

unchanged (Resonance occurs in the pi-system: conjugated lone pairs and pi-bonds).

2. Every resonance structure must be a valid Lewis structure.

3. Keep track of lone pairs and formal charges.

4. Use arrow-pushing formalism to interconvert and identify possible resonance

structures.

5. Always use double-headed arrow ( ) in between resonance structures.

6. Lower energy resonance structures contribute most to the overall structure of the

molecule.

1. Rules for Drawing Resonance Structures

How do you predict the relative energies of resonance structures?

i) Filled Octets: Second row elements (C, N, O , F) want an octet (filled valence

shell of electrons). Because C is the least electronegative, structures in which C has

6 electrons, 3 bonds and a positive charge are possible (not possible with N, O, F).

ii) Minimum # of Formal Charges & Maximum # of Bonds

iii) Negative Charges on Most Electronegative Atoms: O > N > C

iv) Minimize Charge Separation: Try to keep the formal charges close together.

2. Guidelines for Predicting Energies of Resonance Structures

(In Order of Importance)

H2CN

CH3

CH3

H2CN

CH3

CH3

H2CN

CH3

CH3

majorfollows guidelines

minorviolates i

minorviolates i

H3CO

CN

H3CO

CN

H3CO

CN

H3CO

CN

H3CO

CN

A B C

D E

A: follows all guidelines

B: violates ii: 2 formal charges

C: violates ii & iii: 2 formal charges; negative charge on C

D: violates i & ii: 6 electrons on C; 2 formal charges

E: violates i, ii & iv: 6 electrons on C; 2 formal charges;

more charge separation than D

Relative Energy: A << B < C << D < E

Relative Importance: A (major) > B > C > D > E

minor

Delocalization of Charge = Stabilization

OH O–H+

OH–H+ O

phenolpKa 10

cyclohexanolpKa 17

• In general, the more resonance structures there are, the greater the stabilization.

• Equivalent resonance structures provide more stabilization than inequivalent ones.

• Localized charge on oxygen not very stable

• Not a very stable conjugate base, so cyclohexanol is not very acidic

• Negative charge delocalized by resonance

• Stabilized conjugate base, so phenol is more acidic than cyclohexanol

3. Delocalization of Charge

Delocalization of Charge = Stabilization

O O O O

O

O

• Delocalization (resonance) stabilizes the negative charge in the conjugate base of phenol (phenoxide).

• The localized charge in the conjugate base of cyclohexanol is much less stable (no resonance!).