Chemistry 51, Organic 1

Prerequisite: Chemistry 2

Grading Scheme

Recitation Grade: 100 pts average 75

Laboratory Grade 100 pts average 75

Lecture Exams 100 pts

Final 100 pts

Total 400 pts

Old exams and quizzes: academic.brooklyn.cuny.edu/chem/howell/jhowell.htm

Safety: goggles, pregnancy

Cheating: F

jhowell@brooklyn.cuny.edu

Goals of the course:

1. Structure of organic molecules, relation of structure to reactivity

2. Organic reaction patterns

3. Mechanism of reactions

4. Synthesis of organic compounds

5. Techniques of the trade

Scaling of Raw Grades for Recitation and Lab

or

1. Omit the student, B, that dropped the course.

Multiply these sums by 4.9342

Multiply these sums by 0.49342

2. Get the average to 75

In both cases get the same result…….

Note that student D was absent from two labs and will be low for that reason. We will deal with D later.

Now we have to spread the grades out so that there is a reasonable distribution about the desired average of 75. This is done by expanding or contracting the set of grades around the average so that the maximum turns out to be 95.

Now for the student D who is outside of the desirable limits (50 – 95). We move his raw grade up to 70 so that D does not skew the scaling process. D receives a scaled grade of about 51.

Atomic StructureElectrons occupy orbitals which are grouped into subshells and then shells.

Carbon atom: 1s22s22p2

Nitrogen: 1s22s22p3

Valence Shell Electrons: Used in chemical bond formation

C N

The pairing of electron spins

Electron Configuration of Atoms• Pauli Exclusion Principle:Pauli Exclusion Principle:

– No more than two electrons may be present in an orbital. If two electrons are present, their spins must be paired.

• Hund’s Rule:Hund’s Rule: – When orbitals of equal energy are available but there are

not enough electrons to fill all of them, one electron is added to each orbital before a second electron is added to any one of them; the spins of the electrons in degenerate orbitals should be aligned.

• Aufbau (“Build-Up”) Principle:Aufbau (“Build-Up”) Principle: – Orbitals fill in order of increasing energy from lowest

energy to highest energy.

Atomic Structure•1 shell consists of

•1s subshell holding the

•1s orbital

•2 shell consists of

•2s subshell holding the

•2s orbital and the

•2p subshell holding the

•2px,

•2py, and

•2pz orbitals.

s orbital

p orbital

Each orbital can hold two electrons.

An orbital can be •an atomic orbital (on an atom) or •a molecular orbital (on a molecule)

An orbital can be •an atomic orbital (on an atom) or •a molecular orbital (on a molecule)

Octet Rule, Lewis Structures

H atom can stabilize up to two electrons in the valence shell.

CF can stabilize up to 8 electrons in the valence shell.

Complete octet: two electrons around H; eight electrons for C F in the valence shell. Now we are providing maximum stabilization.

Electrons are stabilized by bond formation.

Electron Configuration of Atoms

Completing the Octet

Ionic Bonding: Electrons can be transferred to an atom to produce an anion and “complete the octet”.

Covalent Bonding: Electrons can be shared between atoms providing additional stabilization. The shared electrons typically “complete the octet” for each atom.

Number of Bonds

Neutral atoms Some positive ions Some negative ions

H: 1 more electron H+ 2 more H- 0 more

C: 4 more C2+ 6 more C- 3 more

N: 3 more N+ 4 more N- 2 more

O: 2 more O+ 3 more O- 1 more

F: 1 more F+ 2 more F- 0 more

Define: Unused stabilization of atom = (number of electrons that can be held in valence shell) – (number of electrons already present)

Bonds make use of the unused stabilizing capability of the atoms.

# Bonds in molecule= (Sum of unused stabilizing capability)/2

Bonds make use of the unused stabilizing capability of the atoms.

# Bonds in molecule= (Sum of unused stabilizing capability)/2

Formal Charge: the charge an atom in a Lewis bonding diagram.

Assign each valence shell electron to an atom:– Non-bonding electrons are assigned to the atom on

which they reside.– Bonding electrons are divided equally between the

atoms of the bond.

Formal charge = (# valence shell electrons in neutral atom) - (# nonbonding electrons)

- ½ (# bonded electrons)

Recipe to calculate the Formal Charges.

Formal Charge and Bonding Patterns. If the octet rule obeyed…

Formal

chargeC N O

+1

0

-1

C

C

C

N

N

N

O

O

O

What is “wrong” with the yellow box? Give examples of common molecules.

Lewis DiagramsTypical Problem: Given a compound of molecular formula CH3CHCH2 draw a Lewis bonding structure.

How many bonds in the molucule?(3 * 4 + 6 * 1) / 2 = 9 bonds

Draw a bonding structure making use of single bonds to hold the molecule together.

C

C

C

H

H H

H

H

H

How many bonds left to draw? 9 – 8 = 1 bond left

Put remaining bond(s) any place where the octet rule is not violated.

C

C

C

H

H H

H

H

H

C H

Another Example

Draw Lewis structure for CH2N2 with formal charges.

# bonds = ½ (4 + 2 *1 + 2*3) = ½ x 12Let’s characterize the valence electrons# electrons present = 4 + 2*1 + 2*5 = 16# bonding electrons = 12# non-bonding electrons = 16 – 12 = 4 =

2 lone pairs

Unused stabilizing capacities.

CH2N2 (continued)

6 bonds, 2 lone pairs

Skeleton uses 4 bonds: C N N

H

H

C N N

H

H

C N N

H

H

More important due to negative charge being on electronegative N

OR

Which of these two structures is more important? Which provides the better description?

Curved Arrows to Reposition Electrons

Curved arrows are the standard way of showing electron reorganizations, repositioning of lone pairs and bonds.

C N N

H

H

C N N

H

H

Transforms the lone pair into a bonding pair. Still counts towards terminal N but also towards middle

N. Too many electrons around middle N.

Bonding pair must retreat from the middle N and become a lone pair on the C.

Some principals of using curved arrows.

•Reactions: Restructure the molecular system, creating bonds between unbonded atoms.

•Resonance: Moving pi bonds and lone pairs but leaving single bond framework intact.

An overly simplified presentation involves simply the motion shown below and its reverse.

A lone pair on B is converted to bonding electrons.

What happens at B: The octet count at B is not changed but it does become more positive by 1. The actual charge depends on the initial charge.

What happens at A: Two more electrons are added to the octet count of A. We cannot exceed the octet count of 8.

•If A originally had an octet count of only 6 then we have simply completed the octet.

•If the octet of A was originally complete at 8 then we must withdraw some other pi electrons from A to make room for the newly formed bond. A becomes more negative by 1.

Now for its reverse

A B A B

The reverse motion, conversion of bond into a lone pair, is shown below.

Here we convert a pi bond to a lone pair.

What happens at B: The pi electrons move away from B and no longer count towards either the octet of B or its formal charge. B becomes more positive by 1 and no longer satisfies the octet rule (electron deficient).

What happens at A: The octet count is not changed at A but A does become more negative by 1.

A BA B

Example: resonance in amides

N H

O

H

H

N H

O

H

H

Formamide, NH2CHO, could be written with two electron diagrams as shown below.

Each structure obeys the octet rule (provides maximum stabilization by bonding). The structure on the right is less stable because of the charges that are present.

The problem is how to smoothly interconvert one to the other.

Proceeding with our elementary steps

N H

O

H

H

N H

O

H

H But this structure violates the octet rule on the high side. 10

ELECTRONS AT THE CARBON!! FORBIDDEN!!

We have to make room for the incoming electrons by retreating

the pi electrons of the CO double bond onto the O.

N H

O

H

H

N H

O

H

H

These two steps would be combined into one (always) and avoid that repugnant structure which violates the octet rule on the high side.

N H

O

H

H

N H

O

H

H

As the lone pair on the N moves in towards the central C the pi electrons in the C=O double bond retreat onto the O.

Maximum number of bonds permitted using valence shell orbitals:

4 H 4 more

1 O 2 more

1 C 4 more

1 N+ 4 more

HO-CH-NH2+

14/2 = 7 bonds

N is positive to allow for overall positive charge. Could have

made any atom positive.

H

O

C

N

H

H

H

Set-up single bond framework.

6 bonds used in single bond framework.

1 left to assign. It will be a pi bond, part of a double bond. The additional bond is used to provide maximum stabilization; completing the octet.

Where does it go? Only two possibilities….

H

O

C

N

H

H

H

H

O

C

N

H

H

H

How are they interconverted?

So what is the third structure?

It won’t have the double bond. Will not obey the octet rule. Can collapse the pi bond into a lone pair.

H

O

C

N

H

H

H

H

O

C

N

H

H

H

Could we have collapsed in other direction?

H

O

C

N

H

H

H

H

O

C

N

H

H

H

2

No lone pair on this N. High energy

structure; not very important.

Our three structures…..

HO

CN

H

H

H

HO

CN

H

H

H

HO

CN

H

H

H

Does not obey octet rule. Least stable, least important.

Most stable with positive charge on the electropositive atom, N.

Intermediate stable with positive charge on more electronegative oxygen.

H

O

C

N

H

H

H

H

O

C

N

H

H

H

Examine one of the interconversions shown earlier.

Looking at it in reverse.

Higher energy structure Lower energy, one more bond.

This is a common and significant resonance motion.

A supply of electrons moving towards a place where needed and providing a new bond in the process.

Exceptions to Octet Rule

But the S violates the octet rule by having more than 8 electrons around the S. Valence shell is 3s and 3p. S now would have to use additional atomic orbitals

This means S would have to use the 3d shell orbitals. Luckily, they are low in energy and can be utilized. Octet is expanded due to use of 3d orbitals.

Example: sulfate ion.Working as usual (use S2- and four O): # bonds = 1/2 (0 + 4*2) = 4

Sulfur (2+) is very electronegative pulling the high energy lone pairsfrom oxygen into itself. A better representation would be

S

O

OO

O

-1

-1

S

O

OO

O

-1

-1

-1

-1

+2Result:

Donation from O to S.

S O

S

o

oo

o

-1

-1

-1

-1

+2

Sulfur (2+) is very electronegative pulling the high energy lone pairsfrom oxygen into itself. A better representation would be

Empty 3d Filled 2p

Observe that this is another example of donation of available electrons (lone pair) into a neighboring acceptor orbital.

Survey of Classes of Molecules

1. Hydrocarbons, C & HA Alkanes: only single bonds

1. Acyclic (no rings) Alkanes: CnH2n+2

suffix: -ane

2. Cyclic alkanes: example cyclohexane

CnH2n+2-2(# rings) Presence of a ring eliminates two hydrogens

butane 2-methylpropane

CH3

CH3

- 2 H CH2

H2C

2. Alkenes: double bonds, pi bondsa. Suffix: -ene

i acyclic

ii cycloalkenes

A pi bond eliminates two hydrogens

C4H10 (butane) C4H8 (butene) + 2 H

4-methylpent-1-ene2-methylpent-1-ene

cyclohexene

general formula: CnH2n+2-2(# rings) -2(#pi bonds)

Alkynes: C to C triple bond = 2 pi bonds + sigmaCH3CH2CCH

but-1-yne

hepta-1,3-dien-6-yne

Oxygen Containing Molecules

Incorporation of an oxygen into a molecule does not change the relationship between the number of carbons and number of hydrogens.

CnH2n+2-2(#pi bonds) – 2(# rings)Oa

H C

H

H

HO H

Watch how this makes sense.

CnH2n+2

Hydrogen Deficiency

2n + 2: Number of hydrogens for an acyclic compound with n carbons and no pi bonds.

If a compound has a different number of hydrogens it must be due to the presence of pi bonds or rings.

# pi bonds

+

# rings

= 1/2 ( (2n + 2) - (actual number of hydrogens present) )

- 2(#pi bonds) – 2(# rings)

hydrogen deficiency.

Singly bonded Oxygen

Alcohols, ROH, -ol

CH3CH2OH ethanol

OH

OH

OH

12

5

hex-5-ene-1,2-diol

Two OH groups: a diol

Useful Classiification of Alcohols (and other things): Primary,

Secondary, Tertiary

1o, 2o, 3o

Alcohols and many other groups may be classified as

Primary: one carbon directly bonded to the C bearing the –OH. CH3CH2CH2OH

Secondary: two carbons directly bonded to the C bearing the –OH. (CH3CH2)2CHOH

Tertiary: three carbons directly bonded to the C bearing the –OH. (CH3CH2)3COH

Ethers: ROR, isomeric with alcohols

H3C

O

CH3

methoxymethane

dimethyl ether

H3C

O

C2H5

methoxyethanemethyl ethyl ether

Methoxy group (alkoxy)

Doubly Bonded Oxygen

Carbonyl GroupO

This is a pi bond, resulting in the elimination of two hydrogens.

Aldehydes, RCHO, -al

H3C H

O

acetaldehydeethanal

O

2-methylbutanal

12

3

4

2-methylbut-3-enal

Start numbering the chain at

CHO

DoubleBonds part of Parent Chain

A more complex exampleStill more complex, now we put a double bond in the parent chain.

Both carbonyls must be included in parent chain –dione.

Ketones, R(CO)R, -one

O

pentan-2-one

More complex example:

O O

Double bond included as part of parent chain

3-butyl-5-methylhex-5-ene-2,4-dione

Butyl Side chain.

Simple example

1

Both carboxylic groups must be in parent chain

Ethyl Group

Methyl group

Vinyl side chain

Another more complex problem

CO2H CO2H

Carboxylic Acids, RCO2H, -oic acid

CO2H

pentanoic acid

CO2-

+ H+Ka = 10-5

pentanoate carboxylate anion

2-ethyl-3-methyl-4-vinylpentanedioic acid

Start numbering here

Attached to or part of the Parent Chain are:

• Functional Groups: -OH, carbonyl (aldehydes, ketones), carboxylic acids

• Substituents: alkyl groups, halogens, alkoxy groups

Some Simple Substituents

• Alkyl, -R: An alkane minus one hydrogen (providing the point of attachment to the parent chain).

-CH2CH3: ethyl (-Et);

-CH2CH2CH3: propyl;

-CH(CH3)2: isopropyl• Alkoxy, -OR. –OCH2CH3: ethoxy (-OEt)• Halo, -X. –Cl: chloro

More complex substituents

Systematically named.

CH2Cl

OCH3

12

34

6-(2-(chloromethyl)-4-methoxy-1-methylbutyl)

6

Indicates position of attachment on parent chain

Substituents on the butyl substituent!!

Parent chain

Substituent

Note that the atom directly attached to the parent chain is atom 1 in the substituent.

Functional Groups

Non-Functional Groups (alkyl, alkoxy, halide) will always appear as substituent and as a prefix.

Functional Groups (pi bonds, alcohol, aldehyde, ketone, carboxylic acid) will determine the suffix of parent chain. If more than one functional group then we must prioritize. Highest priority determines suffix. The rest ( alcohol, aldehyde, ketone, carboxylic acid) appear a prefix. But note unsaturation (pi bonding) in the parent chain is always specified as a suffix.

Functional Group Priorities

Group Suffix if highest Prefix if not highest

Triple bond -yne NA. Always as suffix

Double Bond -ene NA. Always as suffix

Alcohol -ol hydroxy

Ketone -one oxo

Aldehyde -al oxo

Carboxylic Acid -oic Acid

OH

O O

alcohol hydroxy prefix

ketone oxo prefix

aldehyde highest -al suffix

4-hydroxy-3-oxopentanal

1

Highest priority at bottom

HO

CHO

HO2C

CO2H

Highest Priority Functional Group: two carboxylic acids. Pentanedioic acid

Double bond in parent chain: Pent-2-enedioic acid

Ethyl side chain bearing a hydroxy at position 1: 3-(1-hydroxyethyl)

Ethyl side chain bearing an oxo at position 2: 4-(2-oxoethyl)

Butyl side chain with double bond at position 2: 2-(but-2-enyl)

2-(but-2-enyl)-3-(1-hydroxyethyl)-4-(2-oxoethyl)pent-2-enedioic acid

One more example. 1

2

34

5

Polar Bonds and Electronegativity

Electronegativity increases towards the upper right.

Electronegativity differences results in polar bonds.

Example:

Li Be B C N O F

Na Mg Al Si P S Cl

H

H Cl

Polar Bonds can lead to Polar MoleculesCombine the bond dipoles to get the overall dipole by means of vector addition.Bond dipoles are put tail to head to provide an overall result.

Resultant Vector

O

H H+ =

O C O

resulting molecular dipole = 0

+ = zero

Some Examples

Combine these three vectors

Predicting Molecular dipoles

Clearly, molecular geometry is very important in determining molecular dipoles.

Problem: Rank the following for size of molecular dipole. Bond angles are 120o.

F

H

F

H

F

H

H

F

F

F

H

HA B C

A > C > B(0)

Molecular geometry (VSEPR)

The number of groups of electrons around an atom determines the shape.

Each lone pair, single, double, or triple bond counts as a group.

# groups 2 3 4 5

hybridization sp sp2 sp3 dsp3

Geometry of the groups

Linear Trigonal planar

Tetrahedral Trigonal bipyramid

Example 1

• CH4

• Four bonds = Four groups

• Tetrahedral

C H

H

H

H

Example 2

H

O

H

Twobonds

Twolone pairs

Four groups of electrons.

Tetrahedral geometry for groups.

Bent Molecule

OH

H 104o

Nature of Chemical Bonds

So far, we have the octet rule which tells us how many bonds we can make. But how do we understand the nature of the bonds?

Three models for bonding: ionic, valence bond, molecular orbitals.

Ionic Bonding

Requires very different electronegativities to make the complete transfer of electrons worthwhile.

Not discussed further although it occurs in salts of organic acids. For example sodium acetate.

Quantum or Wave Mechanics• Albert Einstein: E = h (energy is quantized)

– light has particle properties.• Erwin Schrödinger: wave equation

– wave function, wave function, : A solution to a set of equations that depicts the energy of an electron in an atom.

– each wave function is associated with a unique set of quantum numbers.

– each wave function represents a region of three-dimensional space and is called an orbitalorbital.

– 22 is the probability of finding an electron at a given point in space.

Quantum or Wave Mechanics

• Characteristics of a wave associated with a moving particle. Wavelength is designated by the symbol .

Quantum or Wave Mechanics

• When we describe orbital interactions, we are referring to interactions of waves. Waves interact – constructively or – destructively.

• When two waves overlap, if they are of the same sign then they combine constructively, build-up. Opposite sign overlap combines destructively, meaning they cancel.

Shapes of Atomic s and p Orbitals

– All s orbitals have the shape of a sphere with the center of the sphere at the nucleus.

– Figure 1.8 (a) Calculated and (b) cartoon representations showing an arbitrary boundary surface containing about 95% of the electron density.

Shapes of Atomic s and p Orbitals

– Three-dimensional representations of the 2px, 2py, and 2pz atomic orbitals. Nodal planes are shaded.

Shapes of Atomic s and p Orbitals

2px, 2py, and 2pz atomic orbitals.

Molecular Orbital Theory

• MO theory begins with the hypothesis that – electrons in atoms exist in atomic orbitals

and – electrons in molecules exist in molecular

orbitals.

Molecular Orbital Theory• Rules:

– Combination of n atomic orbitals (mathematically adding and subtracting wave functions) gives n MOs (new wave functions).

– MOs are arranged in order of increasing energy.

– MO filling is governed by the same rules as for atomic orbitals:

• Aufbau principle: fill beginning with lowest energy orbital

• Pauli exclusion principle: no more than 2e- in a MO

• Hund’s rule: when two or more MOs of equivalent energy (degenerate) are available, add 1e- to each before filling any one of them with 2e-.

Molecular Orbital Theory

• MOs derived from combination by (a) addition and (b) subtraction of two 1s atomic orbitals.

Covalent Bonding

• Bonding molecular orbital: A MO in which electrons have a lower energy than they would have in isolated atomic orbitals.

• Sigma () bonding molecular orbital: A MO in which electron density is concentrated between two nuclei along the axis joining them and is cylindrically symmetrical.

Covalent Bonding• A MO energy diagram for H2. (a) Ground

state and (b) lowest excited state.

Covalent Bonding

• Antibonding MO: A MO in which electrons have a higher energy than they would in isolated atomic orbitals.

VB: sp3 Hybridization of Atomic Orbitals

– The number of hybrid orbitals formed is equal to the number of atomic orbitals combined.

– Elements of the 2nd period form three types of hybrid orbitals, designated sp3, sp2, and sp.

– The mathematical combination of one 2s atomic orbital and three 2p atomic orbitals forms four equivalent sp3 hybrid orbitals.

sp3 Hybridization, with electron population for

carbon to form four single bonds

2s

2p

sp3

En

erg

y

VB: sp3 Hybridization of Atomic Orbitals

• sp3 Hybrid orbitals. (a) Computed and (b) cartoon three-dimensional representations. (c) Four balloons of similar size and shape tied together, will assume a tetrahedral geometry.

VB: sp2 Hybridization of Atomic Orbitals

• The mathematical combination of one 2s atomic orbital wave function and two 2p atomic orbital wave functions forms three equivalent sp2 hybrid orbitals.

sp2 Hybridization, with electron

population for carbon to form double

bonds

2s

2p

sp2En

erg

y 2p

VB: sp2 Hybridization of Atomic Orbitals

• Hybrid orbitals and a single 2p orbital on an sp2 hybridized atom.

VB: sp Hybridization of Atomic Orbitals

• The mathematical combination of the 2s atomic orbital and one 2p atomic orbital gives two equivalent sp hybrid orbitals.

sp Hybridization, with electron

population for carbon to form

triple bonds

2s

2p

spEn

erg

y 2p

• sp Hybrid orbitals and two 2p orbitals on an sp hybridized atom.

VB: sp Hybridization of Atomic Orbitals

Combining VB & MO Theories

• VB theory views bonding as arising from electron pairs localized between adjacent atoms. These pairs create bonds.

• Further, organic chemists commonly use atomic orbitals involved in three hybridization states of atoms (sp3, sp2, and 2p) to create orbitals to match the experimentally observed geometries.

• How do we make orbitals that contain electrons that reside between adjacent atoms? For this, we turn back to MO theory.

Combining VB & MO Theories

• To create orbitals that are localized between adjacent atoms, we add and subtract the atomic and hybrid orbitals on the adjacent atoms, which are aligned to overlap with each other.

• Consider methane, CH4. The sp3 hybrid orbitals of carbon each point to a 1s orbital of hydrogen and, therefore, we add and subtract these atomic orbitals to create molecular orbitals.

• As with H2, one resulting MO is lower in energy than the separated atomic orbitals, and is called a bonding orbital. The other is higher in energy and is antibonding.

Combining VB & MO Theories• Molecular orbital mixing diagram for creation of a C-C

bond.

Combining VB & MO Theories• A double bond uses sp2 hybridization.

• Consider ethylene, C2H4. Carbon (and other second-period elements) use a combination of sp2 hybrid orbitals and the unhybridized 2p orbital to form double bonds.

• Now the atomic and hybrid orbitals before mixing into MOs.

Combining VB & MO Theories• MO mixing diagram for the creation of C-C bond.

Present in double and triple bonds.

Combining VB & MO Theories

• A carbon-carbon triple bond consists of one bond formed by overlap of sp hybrid orbitals and two bonds formed by the overlap of parallel 2p atomic orbitals.

Kinds of Hybridization

spn hybridization obtained by mixing the 2s atomic orbital with n different 2p atomic orbitals to yield (n+1) hybrids.

geometry VSEPR

groups

Orbitals Where

found

#

Pi bonds

sp linear 2 2

sp2 trigonal planar

3 1

sp3 tetrahedral 4 0

C

C

C

C

Hybrids are in black, colored orbitals are p orbitals not used in hybridization.

Example of how hybridization determines geometry

CH2=C=CH2

sp2 sp sp2

C C C

H

HH

H

Assign hybridization

Match up p orbitals for pi bonds