Lone Pair acting as Base.

Note the change in formal charges. As reactant oxygen had complete ownership of lone pair. In product it is shared. Oxygen more positive by 1.

Similarly, B has gained half of a bonding pair; more negative by 1.

An example: pi electrons as bases

Bronsted Lowry Acid

Bronsted Lowry Base

The carbocations are conjugate acids of the alkenes.

For the moment, just note that there are two possible carbocations formed.

Sigma bonding electrons as bases. Much more unusual!!

Super acid

A very, very electronegative F!!

A very positive S!! The OH becomes very acidic because that would put a negative charge adjacent to the S.

Trends for Relative Acid Strengths

Totally ionized in aqueous solution.

Aqueous Solution

Totally unionized in aqueous solution

Example

Ethanol, EtOH, is a weaker acid than phenol, PhOH.

It follows that ethoxide, EtO-, is a stronger base than phenolate, PhO-.

For reaction PhOH + EtO- PhO- + EtOH where does equilibrium lie?

pKa = 9.95

Stronger acid

H2O + PhOH H3O+ + PhO-

Ka = [H3O+][PhO-]/[PhOH] = 10-9.95

OH

phenol, PhOH

CH3CH2OH

ethanol, EtOH

Recall

H2O + EtOH H3O+ + EtO-

Ka = [H3O+][EtO-]/[EtOH] = 10-15.9

pKa = 15.9

Weaker acid

Stronger base

Weaker base.

Query: What makes for strong (or weak) acids?

K = 10-9.95 / 10-15.9 = 106.0

What affects acidity?1. Electronegativity of the atom holding the negative charge.

CH3OH CH3O - + H+

CH3NH2 CH3NH - + H+

CH3CH3 CH3CH2- + H+

Increasing electronegativity of atom bearing negative charge. Increasing stability of anion.

Increasing acidity.

Increasing basicity of anion.

2. Size of the atom bearing the negative charge in the anion.

CH3OH CH3O - + H+; pKa = 16

CH3SH CH3S - + H+; pKa = 7.0

Increasing size of atom holding negative charge. Increasing stability of anion.

Increasing acidity.

Increasing basicity of anion.

OO

What affects acidity? - 23. Resonance stabilization, usually of the anion.

OH

phenol, PhOH

OO

ethanol, EtOHCH3CH2OH CH3CH2O

- + H+

Increasing resonance stabilization. Increased anion stability.

Aci

dit

y

Increasing basicity of the anion.

No resonance structures!!

OH OH

etc.

Note that phenol itself enjoys resonance but charges are generated, costing energy, making the resonance less important. The more important resonance in the anion shifts the equilibrium to the right making phenol more acidic.

An example: competitive Bases & Resonance

• Two different bases or two sites in the same molecule may compete to be protonated (be the base).

O

O H

acetic acid

H+O

O H

HH+

O

O H

H

Acetic acid can be protonated at two sites.

Which conjugate acid is favored?

The more stable one! Which is that?

Recall resonance provides additional stability by moving pi or non-bonding electrons.

Pi bonding electrons converted to non-bonding.

O

O H

H

O

O H

H

Non-bonding electrons converted to pi bonding.

No valid resonance structures for this cation.

An example: competitive Bases & Resonance

H+O

O H

H

O

O H

H

O

O H

H

O

O H

acetic acid

All atoms obey octet rule!

All atoms obey octet rule!

The carbon is electron deficient – 6 electrons, not 8.

Lesser importance

Comments on the importance of the resonance structures.

What affects acidity? - 34. Inductive and Electrostatic Stabilization.

F3CCH2O - + H+

H3CCH2O - + H+H3CCH2OH

F3CCH2OH

Due to electronegativity of F small positive charges build up on C resulting in stabilization of the anion.

Increasing anion stability.Acidity.Increasing anion basicity.

Effect drops off with distance. EtOH pKa = 15.9

What affects acidity? - 45. Hybridization of the atom bearing the charge. H-A H+ + A:-.

sp3 sp2 sp

More s character, more stability, more “electronegative”, H-A more acidic, A:- less basic.

Incr

easi

ng

Aci

dit

y o

f H

A

Incr

easi

ng

B

asic

ity

of

A-

Note. The NH2-

is more basic than the RCC-

ion.

Know this order.

Example of hybridization Effect.

RCCH + AgNO3 AgCCR (ppt)

acid base

terminal alkyne

non-terminal alkyne

RCCR + LiCH2CH2CH2CH3 No Reaction

RCCH + LiCH2CH2CH2CH3 HCH2CH2CH2CH3 + RCCLi

RCCR + AgNO3 NR

What affects acidity? - 5

6. Stabilization of ions by solvents (solvation).

H

O RO R + H

H

O

H

H

O

HH

OH

Solvation provides stabilization.

OH

ethanol

OH

propan-2-ol

OH

2-methylpropan-2-olCrowding inhibiting solvation

Solvation, stability of anion, acidity

pKa = 15.9 17 18

(CH3)3CO -, crowded

Comparison of alcohol acidities.

Example

Para nitrophenol is more acidic than phenol. Offer an explanation

OH

OH

N

O O

O

O

N

O O

+ H

+ H The lower lies further to the right.

Why? Could be due to destabilization of the unionized form, A, or stabilization of the ionized form, B.

A B

OH

N

O O

Examine the equilibrium for p-nitrophenol. How does the nitro group increase the acidity?

O

N

O O

+ H

Resonance structures A, B and C are comparable to those in the phenol itself and thus would not be expected to affect acidity. But note the + to – attraction here

OH

N

O O

OH

N

O O

OH

N

O O

OH

N

O O

A B C D

Structure D occurs only due to the nitro group. The stability it provides will slightly decrease acidity.

Examine both sides of equilibrium. What does the nitro group do?

First the unionized acid.

Note carefully that in these resonance structures charge is created: + on the O and – in the ring or on an oxygen. This decreases the importance of the resonance.

OH

N

O O

O

N

O O

+ H

Resonance structures A, B and C are comparable to those in the phenolate anion itself and thus would not be expected to affect acidity. But note the + to – attraction here

Structure D occurs only due to the nitro group. It increases acidity. The greater amount of significant resonance in the anion accounts for the nitro increasing the acidity.

Now look at the anion. What does the nitro group do? Remember we are interested to compare with the phenol phenolate equilibrium.

In these resonance structures charge is not created. Thus these structures are important and increase acidity. They account for the acidity of all phenols.

O

N

O O

O

N

O O

O

N

O O

O

N

O O

A B C D

3. (3 pts) Which is the stronger base and why?

HNvs

HN O

Sample Problem

H2N H2N O H2N O

Alkenes 1

ShapeAlkenes, 6 coplanar atoms.

All atoms in same plane except for

these hydrogens on sp3 carbon.

Arene shapesPlanar ring structure. 12 atoms coplanar.

2-phenyl propane

Phenyl group, C6H5,, Ph

Ph

Ph = C6H5

Pi bonds

pi orbitalsum

pi* orbitaldifference

Ene

rgy

Nomenclature

but-1-ene 3,4-dimethylhexa-1,5-diene

cis / trans

cis trans

Z / E generalization of cis / trans

Use R, S priorities to compare substituents on same carbon.

High priority on same side, Z. Opposite, E.

H Br

FCl

H F

BrCl

(E)-1-bromo-2-chloro-1-fluoroethene(Z)-1-bromo-2-chloro-1-fluoroethene

Cis / Trans in CycloalkenesFor small rings normally have cis double bonds.

trans cyclooctene

Terpenes and the isoprene Rule

• A terpene is composed of isoprene units joined head to tail (the isoprene rule).

This moleculehas additional cross links.

Note that location of functional groups such as OH or double bonds is not addressed.

Vitamin A

Four isprene units joined head to tail

One cross link (non-head to tail) linkage.

Fatty Acids

• Animal fats and vegetable oils are both triesters of glycerol, hence the name triglyceridetriglyceride.– Hydrolysis of a triglyceride in aqueous base followed by

acidification gives glycerol and three fatty acids.

– Fatty acids with no C=C double bonds are called saturated fatty acid.

– Those with one or more C=C double bonds are called unsaturated fatty acids.

CH2OCR

CH2OCR''

R'COCHO

O

1. NaOH, H2O

2. HCl, H2O

CH2OH

CH2OH

HOCH

RCOOH

R'COOH

R''COOH

+

Fatty acids

O

1,2,3-Propanetriol(glycerol)

A triglyceride(a triester of glycerol

Fatty Acids– The most common fatty acids have an even number of

carbons, and between 12 and 20 carbons in an unbranched chain.

– The C=C double bonds in almost all naturally occurring fatty acids have a cis configuration.

– The greater degree of unsaturation, the lower the melting point.

– Triglycerides rich in unsaturated fatty acids are generally liquid at room temperature and are called oilsoils.

– Triglycerides rich in saturated fatty acids are generally semisolids or solids at room temperature and are called fatsfats.

Fatty Acids

– the four most abundant fatty acidsCOOH

COOH

COOH

COOH

Stearic acid (18:0)(mp 70°C)

Oleic acid (18;1)(mp 16°C)

Linoleic acid (18:2)(mp-5°C)

Linolenic acid (18:3)(mp -11°C)

Alkene Reactions

Pi bonds

Plane of molecule

Reactivity above and below the molecular plane!

Addition ReactionsA - B A

B

Important characteristics of addition reactions

Orientation (Regioselectivity)

If the doubly bonded carbons are not equivalent which one get the A and which gets the B.

Stereochemistry: geometry of the addition.

Syn addition: Both A and B come in from the same side of the alkene. Both from the top or both from the bottom.

Anti Addition: A and B come in from opposite sides (anti addition).

No preference.

Reaction Mechanisms

Mechanism: a detailed, step-by-step description of how a reaction occurs.

A reaction may consist of many sequential steps. Each step involves a transformation of the structure.

For the step C + A-B C-A + B

ReactantsProducts

Transition State

Energy of Activation. Energy barrier.

Three areas to be aware of.

Energy Changes in a Reaction

• Enthalpy changes, H0, for a reaction arises from changes in bonding in the molecule.– If weaker bonds are broken and stronger ones

formed then H0 is negative and exothermic.– If stronger bonds are broken and weaker ones

formed then H0 is positive and endothermic.

Gibbs Free Energy

Gibbs Free Energy controls the position of equilibrium for a reaction. It takes into account enthalpy, H, and entropy, S, changes.

An increase in H during a reaction favors reactants. A decrease favors products.

An increase in entropy (eg., more molecules being formed) during a reaction favors products. A decrease favors reactants.

G0: if positive equilibrium favors reactants (endergonic), if negative favors products (exergonic).

G0 = H0 – TS0

Multi-Step ReactionsStep 1: A + B Intermediate

Step 2: Intermediate C + D

Step 1: endergonic, high energy of activation. Slow process

Step 2: exergonic, small energy of activation. Fast Process.

Step 1 is the “slow step”, the rate determining step.

Characteristics of two step Reaction 1. The Intermediate has

some stability. It resides in a valley.

2. The concentration of an intermediate is usually quite low. The Energies of Activation for reaction of the Intermediate are low.

3. There is a transition state for each step. A transition state is not a stable structure.

4. The reaction coordinate can be traversed in either direction: A+B C+D or C+D A+B.

Hammond PostulateThe transition state for a step is close to the high energy end of the curve.

For an endothermic step the transition state resembles the product of the step more than the reactants.

For an exothermic step the transition state resembles the reactants more than the products.

Reaction coordinate.

Example

•Endothermic

•Transition state resembles the (higher energy) products.

CH3 - H + Br CH3 + H - Br H = 109 kJ

[H3C H Br]

Almost broken.

Almost formed.

Almost formed radical. Only a small

amount of radical character remains.

Electrophilic Additions

– Hydrohalogenation using HCl, HBr, HI

– Hydration using H2O in the presence of H2SO4

– Halogenation using Cl2, Br2

– Halohydrination using HOCl, HOBr

– Oxymercuration using Hg(OAc)2, H2O followed by reduction

Electrophilic Addition

We now address regioselectivity….

Regioselectivity (Orientation)

The incoming hydrogen attaches to the carbon with the greater number of hydrogens. This is regioselectivity. It is called Markovnikov orientation.

Mechanism

Step 2

Step 1

Now examine Step 1 Closely

Rate Determining Step. The rate at which the carbocation is formed controls the rate of the overall reaction. The energy of activation for this process is critical.

Electron rich, pi system.

Showed this reaction earlier as an acid/base reaction. Alkene was the base.

New term: the alkene is a nucleophile, wanting to react with a positive species.

Acidic molecule, easily ionized.

We had portrayed the HBr earlier as a Bronsted-Lowry acid.

New term: the HBr is an electrophile, wanting to react with an electron rich molecule (nucleophile).

The carbocation intermediate is very reactive. It does not obey the octet rule

(electron deficient) and is usually present only in low concentration.

Carbocations

Electron deficient.

Does not obey octet rule.

Lewis acid, can receive electrons.

Electrophile.

sp2 hybridized.

p orbital is empty and can receive electrons.

Flat, planar. Can react on either side of the plane.

Very reactive and present only in very low concentration.

Step 2 of the Mechanism

Br

Br

Mirror objects

:Br-

:Br-

Regioselectivity (Orientation)

H - Br

H

+ Br

H

+ Br

HBr

2-Bromo-propane

HBr

1-Bromo-propane

Secondary carbocation

Primary carbocation

Secondary carbocation more more stable and more easily formed.

Or

Carbocation Stabilities

Order of increasing stability:

Methyl < Primary < Secondary < Tertiary

Order of increasing ease of formation:

Methyl < Primary < Secondary < Tertiary

Increasing Ease of Formation

Factors Affecting Carbocation Stability - Inductive

1. Inductive Effect. Electron redistribution due to differences in electronegativities of substituents.

• Electron releasing, alkyl groups, -CH3, stabilize the carbocation making it easier to form.

• Electron withdrawing groups, such as -CF3, destabilize the carbocation making it harder to form.

HF

F

F H

-

+

-

-

Factors Affecting Carbocation Stability - Hyperconjugation

2. Hyperconjugation. Unlike normal resonance or conjugation hyperconjugation involves bonds.

H

H

H H

HHH

H

H H

ethyl carbocation

Hyperconjugation spreads the positive charge onto the adjacent alkyl group

Hyperconjugation Continued

Drifting of electrons from the filled C-H bond into the empty p orbital of the carbocation. Result resembles a pi bond.

Another description of the effect.

Factors Affecting Carbocation Stability - Resonance

allylic carbocation

Utilizing an adjacent pi system.

H HH H

benzylic carbocation

H H H H

Positive charge delocalized through resonance.

Another very important example.

Positive charge delocalized into the benzene ring. Increased stability of carbocation.

Note: the allylic carbocation can react at either end!

The benzylic carbocation will react only at the benzylic position even though delocalization occurs!

Another Factor Affecting Carbocation Stability – Resonance

Utilizing an adjacent lone pair.

CH2

O

HCH2

O

H

Look carefully. This is the conjugate acid of formaldehyde, CH2=O.

Production of Chiral Centers. Goal is to see all the possibilities.

The H will attach here.Regioselectivity Analysis:

the positive charge will go here and be stabilized by resonance

with the phenyl group.

Ph Me

EtMe

Me is methyl groupEt is ethyl groupPh is phenyl group

Ph

MeEt

Me

H

Ph MeEt

MeH

mirror plane

Enantiomeric carbocations.

Br-

Ph MeEtMe

HBr

Br-

Ph

MeEt

Me H

Br

H+

H+

Br-

Ph MeEtMe

HBr

Ph

MeEt

MeH

Br

Br-

What has been made?Two pairs of enantiomers.

React alkene with HBr.

Note that the ends of the double bond are different.

Production of Chiral Centers - 2

Ph MeEtMe

HBr Ph

MeEt

Me H

Br

Ph MeEtMe

HBrPh

MeEt

MeH

Br

Racemic Mixture 1 Racemic Mixture 2

The product mixture consists of four stereoisomers, two pairs of enantiomers

The product is optically inactive.

Distillation of the product mixture yields two fractions (different boiling points). Each fraction is optically inactive.

Rule: optically inactive reactants yield optically inactive products (either achiral or racemic).

diastereomers

Acid Catalyzed Hydration of Alkenes

What is the orientation??? Markovnikov

Mechanism

Step 1

Step 2

Step 3

Note the electronic structure of the oxonium ion.

Carbocation Rearrangements

Expected product is not the major product; rearrangement of carbon skeleton occurred.

The methyl group moved. Rearranged.

Also, in the hydration reaction.

The H moved.

Mechanism including the “1,2 shift”

Step 1, formation of carbocation

Step 2, the 1,2 shift of the methyl group with its pair of electrons.

Step 3, the nucleophile reacts with the carbocation

Reason for Shift: Converting a less stable carbocation (20) to a more stable carbocation (30).

Addition of Br2 and Cl2

Stereochemistry

Anti Addition (halogens enter on opposite sides); Stereoselective

Syn addition (on same side) does not occur for this reaction.

Mechanism, Step 1Step 1, formation of cyclic bromonium ion.

Step 2

Detailed Stereochemistry, addition of Br2

H3C CH3

C3H7 C2H5

Br

Br

(S) (R)H3C CH3

C3H7 C2H5

Br

Br

Br

(S)

(S)

H3C

CH3C3H7

C2H5

Br

Br

(R)

(R)

H3C

CH3

C3H7

C2H5

Br

Br

(R) (S)

H3C CH3

C3H7 C2H5

Br

Br

Br

Br(R)

(R)

H3C

CH3

C3H7

C2H5

Br

Br

(S)

(S)

H3C

CH3

C3H7

C2H5

Br

Br

enantiomers

Alternatively, the bromine could have come in from the bottom!

enantiomers

S,S

S,S

R,R

R,R

Only two compounds (R,R and S,S) formed in equal amounts. Racemic mixture.

Bromide ion attacked the carbon on the right.

But can also attack the left-side carbon.

Number of products formed.

(S)

(S)

H3C

CH3

C3H7

C2H5

Br

Br

enantiomers

enantiomers

S,S

S,S

R,R

R,R

We have formed only two products even though there are two chiral carbons present. We know that there is a total of four stereoisomers. Half of them are eliminated because the addition is anti. Syn (both on same side) addition does not occur. (R)

(R)

H3C

CH3

C3H7

C2H5

Br

Br

(R)

(R)

H3C

CH3

C3H7

C2H5

Br

Br

(S)

(S)

H3C

CH3C3H7

C2H5

Br

Br

Attack of the Bromide Ion

(S) (R)H3C CH3

C3H7 C2H5

Br

Br

(S)

(S)

H3C

CH3C3H7

C2H5

Br

Br

Starts as R Becomes S

The carbon was originally R with the Br on the top-side. It became S when the Br was removed and a Br attached to the bottom.

In order to preserve a tetrahedral carbon these two substituents must move upwards. Inversion.

Progress of Attack

Things to watch for:

•Approach of the red Br anion from the bottom.

•Breaking of the C-Br bond.

•Inversion of the C on the left; Retention of the C on the right.

R1R2

R3R4

Br2

anti addition

R1R2

R3R4

Br

Br

+ enantiomer

Using Fischer Projections

Not a valid Fischer projection since top vertical bond is coming forward.

Convert to Fischer by doing 180 deg rotation of top carbon.

+ enantiomer

Br

R1 R2

Br

R4 R3

=

There are many variations on the addition of X2 to

an alkene. Each one involves anti addition.

Br -

+ enantiomer

Br

R1 R2

Br

R4 R3

R2 R4

R1 R3

Br

I -

+ enantiomer

I

R1 R2

Br

R4 R3

+ enantiomer

Br

R1 R2

I

R4 R3

+

The iodide can attach to either of the two carbons.

I -I -

Instead of iodide ion as nucleophile can use alcohols to yield ethers, water to yield alcohols, or amines.

R1R2

R3R4

Br2

Regioselectivity

If Br2 is added to propene there is no regioselectivity issue.

Br2

Br

Br

If Br2 is added in the presence of excess alternative nucleophile, such as CH3OH, regioselectivity may become important.

Br - BrOCH3

BrCH3O-H

Br

OCH3and/or

+ H + + Br - + H + + Br -

Regioselectivity - 2Consider, again, the cyclic bromonium ion and the resonance structures.

R

BrWeaker bond

More positive charge

Stronger bond

Expect the nucleophile to attack here. Remember inversion occurs.

Regioselectivity, Bromonium Ion

– Bridged bromonium ion from propene.

Et

H Me

Me

Me

Cl2/H2O

H

Example

Regioselectivity, addition of Cl and OH

Cl, from the electrophile Cl2, goes here

OH, the nucleophile, goes here

Stereochemistry: anti addition

Note: non-reacting fragment unchangedEt

H Me

Me

Me

Et

H Me

Me

Me

+

Cl

HH

OH

Cl

OH

Put in Fisher Projections. Be sure you can do this!!

Et

H Me

Me OH

Me

H Cl

Et

H Me

HO Me

Me

Cl H

+

Bromination of a substituted cyclohexene

Consider the following bromination.

C(CH3)3

Br2

Expect to form two bromonium ions, one on top and the other on bottom.

C(CH3)3

Br+

C(CH3)3

Br+

+

Expect the rings can be opened by attack on either carbon atom as before.

But NO, only one stereoisomer is formed. WHY?

C(CH3)3

Br-Br

Br

C(CH3)3

Br

Br

+

Addition to substituted cyclohexene

HH

Br2

The tert butyl group locks the conformation as shown.

Br

Br

H

H

H H

+

The cyclic bromonium ion can form on either the top or bottom of the ring.

How can the bromide ion come in? Review earlier slide showing that the bromide ion attacks directly on the side opposite to the ring.

Progress of Attack

Things to watch for:

•Approach of the red Br anion from the bottom.

•Breaking of the C-Br bond.

•Inversion of the C on the left; Retention of the C on the right.

Notice that the two bromines are maintained anti to each other!!!

Addition to substituted cyclohexene

Br2

Br

Br

+

ObserveRing is locked as shown. No ring flipping.

Attack as shown in red by incoming Br ion will put both Br into equatorial positions, not anti.

Br

Br

This stereoisomer is not observed. The bromines have not been kept anti to each other but have become gauche as displacement proceeds.

Br-

Br-

Be sure to allow for the inversion motion at the carbon attacked by the bromide ion.

Addition to substituted cyclohexene

Br2

Br

Br

+

Attack as shown in green by the incoming Br will result in both Br being axial and anti to each other

Br

Br

This is the observed diastereomer. We have kept the bromines anti to each other.

Br-

Br-

Oxymercuration-Reduction

Regioselective: Markovnikov Orientation

Occurs without 1,2 rearrangement, contrast the following

3,3-dimethylbut-1-ene

H2O

H2SO4

OH

formed viarearrangement

1 Hg(OAc)2

2. NaBH4OH

No rearrangement

Alkene Alcohol

Mechanism1

2

3

4

Hydroboration-Oxidation

Alkene Alcohol

Anti-Markovnikov orientation

Syn addition

1. BH3

2. H2O2

HHO

HHO

Borane, a digression

Isoelectronic with a carbocation

B B

H

HH

H

HH

MechanismSyn stereochemistry, anti-Markovnikov orientation now established.

Two reasons why anti-Markovnikov:

1. Less crowded transition state for B to approach the terminal carbon.

2. A small positive charge is placed on the more highly substituted carbon.

Just call the circled group R. Eventually have BR3.

Next…

Cont’d

Oxidation and Reduction Reactions

We think in terms of Half Reactions

• Write reactants and products of each half reaction.

Cr2O7 2- + CH3CH2OH Cr 3+ + CH3CO2H

Cr2O7 2- 2 Cr 3+

Balance oxygen by adding water

+ 7 H2O

In acid balance H by adding H +

14 H+ +

Balance charge by adding electrons

6 e - +

Inorganic half reaction…

If reaction is in base: first balance as above for acid and then add OH- to both sides to neutralize H +. Cancel extra H2O.

Will be oxidized.

Will be reduced.

Cont’d

Now the organic half reaction…

Balance oxygen by adding water

In acid balance H by adding H +

Balance charge by adding electrons

CH3CH2OH CH3CO2HH2O + + 4 H+ + 4 e-

Combine half reactions so as to cancel electrons…

CH3CH2OH CH3CO2HH2O + + 4 H+ + 4 e-

Cr2O7 2- 2 Cr 3+ + 7 H2O14 H+ +6 e - +

3 x ( )

16 H+ + 2 Cr2O7 2- + 3 CH3CH2OH 4 Cr 3+ + 3 CH3CO2H + 11 H2O

2 x ( )

Formation of glycols with Syn Addition

Osmium tetroxide

Syn addition

KMnO4cold, dilute, slightly alkaline

also KMnO4

Anti glycols

PhCO3H, a peracid

O

H+

O

H

H2O

HO

OH

Using a peracid, RCO3H, to form an epoxide which is opened by aq. acid.

Peracid: for example, perbenzoic acid

O O

OH

The protonated epoxide is analagous to the cyclic bromonium ion.

epoxide

An example

chiral, optically active

(S)-3-methylcyclohex-1-ene

PhCO3HO + O

aq. acid

OH

OH

OH

OHOH

OH

OH

OH

Are these unique?

Diastereomers, separable (in theory) by distillation, each optically active

OzonolysisR3

R4

R1

R2

1. O3

2. (CH3)2SR4

R3

O

R1

R2

O+

Reaction can be used to break larger molecule down into smaller parts for easy identification.

Ozonolysis Example

For example, suppose an unknown compound had the formula C8H12 and upon ozonolysis yielded only 3-oxobutanal. What is the structure of the unknown?

The hydrogen deficiency is 18-12 = 6. 6/2 = 3 pi bonds or rings.

The original compound has 8 carbons and the ozonolysis product has only 4

Conclude: Unknown two 3-oxobutanal.

Unknown

C8H12

ozonolysysO

O

O

O

Simply remove the new oxygens and join to make double bonds.But there is a second possibility.

O

O

Another Example

2. An unknown compound (derived from the gall bladder of the gila monster) has the formula C10H14 . When subjected to ozonolysis the following compound is isolated

O

O O

O

Suggest a reasonable structure for the unknown.

Hydrogen Deficiency = 8. Four pi bonds/rings.

Unknown has no oxygens. Ozonolysis product has four. Each double bond produces two carbonyl groups. Expect unknown to have 2 pi bonds and two rings.

To construct unknown cross out the oxygens and then connect. But there are many ways the connections can be made.

a

bc

d

a-b & c-d

a b

c

da-c & b-d

ac

d

b

a-d & b-c

ad

c

b

Look for a structure that obeys the isoprene rule.

Mechanism

OO

O OO

O OO

O OO

O

Consider the resonance structures of ozone.

These two, charged at each end, are the useful ones to think about.

Electrophile capability.

Nucleophile capability.

Mechanism - 2

OO

O OO

O OO

O OO

O

Mechanism - 3

Mechanism - 4

Hydrogenation

No regioselectivity

Syn addition

Heats of Hydrogenation

Consider the cis vs trans heats of hydrogenation in more detail…

Heats of Hydrogenation - 2The trans alkene has a lower heat of hydrogenation.

Conclusion:

Trans alkenes with lower heats of hydrogenation are more stable than cis.

We saw same kind of reasoning when we talked about heats of combustion of isomeric alkanes to give CO2 and H2O

Heats of Hydrogenation

Incr

easi

ng s

ubst

itutio

n

Red

uced

hea

t of

Hyd

roge

natio

n

By same reasoning higher degree of substitution provide lower heat of hydrogenation and are, therefore, more stable.

Acid Catalyzed Polymerization

Principle: Reactive pi electrons (Lewis base) can react with Lewis acid. Recall

Which now reacts with a Lewis base, such as halide ion to complete addition of HX yielding 2-halopropane

Variation: there are other Lewis bases available. THE ALKENE.

+ HH

The new carbocation now reacts with a Lewis base such as halide ion to yield halide ion to yield 2-halo-4-methyl pentane (dimerization) but could react with another propene to yield higher polymers.

the carbocation is an acid!

+

Examples of Synthetic Planning

Give a synthesis of 2-hexanol from any alkene.OH

Planning:

Alkene is a hydrocarbon, thus we have to introduce the OH group

How is OH group introduced (into an alkene): hydration

What are hydration reactions and what are their characteristics:

•Mercuration/Reduction: Markovnikov

•Hydroboration/Oxidation: Anti-Markovnikov and syn addition

What alkene to use? Must involve C2 in double bond.

Which reaction to use with which alkene?

Markovnikov rule can be applied here. CH vs CH2.

Want Markovnikov!

Use Mercuration/Reduction!!!

Markovnkov Rule cannot be used here. Both are CH.

Do not have control over regioselectivity.

Do not use this alkene.

For yourself : how would you make 1 hexanol, and 3-hexanol?

Another synthetic example…

How would you prepare meso 2,3 dibromobutane from an alkene?

Analysis:

Alkene must be 2-butene. But wait that could be either cis or trans!

We want meso. Have to worry about stereochemistry

Know bromine addition to an alkene is anti addition (cyclic bromonium ion)

trans

Br2

BrBr

H

Br

Br

rotate lower unit

Br H

Br H

meso

This worked! How about starting with the cis?

cis

Br2

H Br

Br H

racemic mixture

+ enantiomer

This did not work, gave us the wrong stereochemistry!

Addition Reaction General Rule…

Characterize Reactant as cis or trans, C or T

Characterize Reaction as syn or anti, S or A

Characterize Product as meso or racemic mixture, M or R

Relationship

C RA

cis

Br2H Br

Br H

racemic mixture

+ enantiomer

Characteristics can be changed in pairs and C A R will remain true.

Want meso instead?? Have to use trans. Two changed!!

AT M

trans

Br H

Br H

meso

Br2