Carbenes, :CH2

1.

2.

carbene

Preparation of simple carbenes

Mechanism of the elimination.

Reactions of Carbenes, :CH2 (not for synthesis)

Addition to double bond.

Insertion into C-H bond

Formation of ylide (later)

liquid

Simmons Smith Reaction (for synthesis, addition to alkenes to yield cyclopropanes)

CH2I2 + Zn(Cu) ICH2ZnI

Carbenoid, properties similar to carbenes.

Electronic Structure

Electrons paired, singlet

Triplet and Singlet Methylene

CH2N2

singlet carbene

stereospecificaddition

pi electrons

triplet carbene

CH2

diradical

+

non-stereospecific

Dominant form in solution

Gas phase

Rotation can occur around this

bond.

Aldehydes and Ketones

Chapter 16

Structure

Aldehydes

R H

O

Ketone

R R

O

sp2

Carbonyl group

pentanalO

O

2-pentanone

Examples of Naming

O

pentan-2-oneO

OH

2-(1-hydroxyethyl)pentanal

O

O

3-oxopentanal

OO

(E)-3-(but-1-enyl)hexane-2,4-dione

H

CH3

CO2H

CHO

CH3

H

2,3-dimethyl-4-oxobutanoic acid

CHO

benzaldehyde

Resonance

O O

O

result

O O

Extension of resonance

Boiling pointsFor compounds of comparable molecular weight…

Alkanes, ethers < aldehydes, ketones < alcohols < carboxylic acids

Water SolubilityKetones and Aldehydes, like ethers, can function as hydrogen bond acceptors and smaller compounds have significant water solubility.

Hydrogen BondingDipole-dipoleDispersion Forces

Recall Preparation from AlcoholsOH

CH2OH

HO

Na2Cr2O7

acid, 35 deg.

OH

CO2H

O

Be sure you can balance this kind of reaction.

RCH2OH RCH=OPCC

Use PCC to limit oxidation of primary alcohol to the aldehyde. Secondary are oxidized to ketone.

Can also be done using KMnO4 in base with heat or bleach in acid solution (HOCl).

PCCSecondary

R2CHOH R2C=O

Primary alcohol

Preparations, con’d

• Reaction of acid chloride and Gilman

R Cl

O

R' 2CuLi

R R'

O

R OH

O

SOCl2

R'X

1. Li

2. CuX

RCH2OH

Na2Cr2O7

But where do we get these??

Note that we have two possible disconnects available

RR'

O RHO

O

+ R'X

OHR'

O

+ RX

HOCH2R

R'OH

HOCH2R'

ROH

Example: Prepare 2-butanone from ethyl alcohol

O

CH3CO2H + CH3CH2X

via Gilman andacid chloride

CH3CH2OH

Requirement to start with ethanol suggests a disconnect into two carbon fragments.

Done!

Aldehydes from carboxylic acids

R Cl

O

R OH

O SOCl2 LiAlH(OC(CH3)3)3

R H

O

And from alcohols, as before:

RCH2OH

PCC

RCHO

Reduction

Oxidation

A Common Sequence

RCH2OH

PCC

RCHO

R'OHPX3

RX

MgR'MgX

R R'

OH

Na2Cr2O7

R R'

O

Observe these parts at this moment.

Reactions

Addition of a nucleophile: Nucleophilic Addition

O

Nu: -

O

Nu

O

Nu

tetrahedral intermediate

OH

Nu

OH

Nu

Good nucleophile, usually basic

+

+

Attack of nucleophile occurs on both sides of carbonyl group.

Produces both configurations.

Overall: H – Nu was added to carbonyl group double bond.

Notice that the CO bond order was reduced from 2 to 1. The addition reduced the bond order. We will use this idea later.

Reaction can also be done in acid environment.

Nucleophiles not expected to be as strong (why?) but the oxygen may become protonated making the carbonyl a better electrophile (why?).

OH

Nu:

OH

Nu

OH

Nu

O acid

+

Very electronegative, protonated oxygen. Pulls the pi electrons into itself

strongly.

Problem: If there is too much acid present the nucleophile may become protonated, deactivating it

Addition of Grignard (Trumpets Please)

Recall the formation of a Grignard and its addition to an oxirane

R-XMg

etherR-Mg-X

O

OMgX

R

R-XMg

etherR-Mg-X

O

OMgX

R

OMgX

R

+

OH

R

OH

R

mild acid

Carbonyls may be added to in same way…

If a new chiral center is created both configurations will be produced.

Common Reactions of Grignards

R-Mg-X

acid,weak acid

R-H

H H

O

RCH2OH

primary alcohol

R' H

O

RCHR'

OH

secondary alcohol

O

OH

R

primary alcohol

R' R''

O

RR'R''C-OHtertiary alcohol

R'CO2Etan ester

R2R'COHa tertiary alcohol

Examine reaction with ester further.

CO2

RCO2Hcarboxylic acid

Both of these reactions extend carbon chain & keep -OH functionality at end of chain. Can extend further.

ROHRXRH(D)

ROHRXRCO2H

ROH + R’CH2OHRX + R’CHORCH(OH)R’

ROH + R’R”CHOH

RX + R’R”CO RR’C (OH)R”

ROH + R’CH2OHRX + R’CO2HR2C(OH)R’

Grignard Reacting with an Ester.Look for two kinds of reactions.

R' OEt

OR-Mg-X

R' OEt

O

RR' R

O

R-Mg-X

R' R

OMgX

R

R' R

OH

R

But where does an ester come from?

R'COCl

EtOH

Acid chloride

R'CO2H

SOCl2 Perhaps this carboxylic acid comes from the oxidation of a primary alcohol or reaction of a Grignard with CO2.

Substitution

Addition

Any alcohol will do here.

Synthetic Planning… Use of epoxides and carbonyls offer different disconnect sites.

R-Mg-XO OH

R

Pattern

HO-C-R

New bond. Disconnect

site

Want this to be the nucleophile

(Grignard).

O1. RMgX HO

R

2. H2O, HClepoxide

New bond. Disconnect site.

Pattern

HO-C-C-R

Nucleophile

carbonyl

Patterns to recognize: carbonyl vs oxirane

We can create the following fragments of target molecules by using an organometallic (carbon nucleophile)

O RMgX dil. acid OH

R

reaction with a carbonyl

RMgX dil. acid OH

reaction with a oxirane

O

R

The difference is the extra CH2 when using an oxirane.

Synthetic Planning…

R1

C

R3R2

OHGive synthetic routes to

R1 R2

O

+ R3-Mg-X

R1 R3

O

+ R2-Mg-X

R2 R3

O

+ R1-Mg-X

If none of the Rs are H then these three synthetic routes may be available.

Three different disconnects possible

Example: Synthesize from ethanol

O

OH

CH3CH2X

CH3CH2OH

MgX2

CH3CO2EtOH

CH3CO2H CH3CH2OHCH3CH2OHCH3CH2MgX

O

Preliminary Analysis

•Hmmm, even number of carbons, at least that is good; ethanol is a two carbon molecule.

•Now the problem is to divide it up into smaller fragments.

•Ether linkage is easily constructed. Williamson.

•Two butyl groups attached to the central 2 carbon fragment. Grignard + ester.

Done

Bisulfite Addition

O

SOHO

O

O

SHO O

O

OH

SO O

O

Addition product.

Practical importance: liquid carbonyl compounds can be difficult to purify. The bisulfite addition products will be crystalline and may be recrystallized.

Addition of Organolithium Compounds to Carbonyls

Generally the reactions are the same as for Grignards but the lithium compounds are more reactive (and more difficult to handle).

Br

Li

Li

O

O

OH

mild acid

bromocyclohexane

Decreased reactivity of electrophile due to steric hindrance to attack. So we used the alkyl lithium instead of a Grignard.

Nucleophiles derived from terminal alkynes

For example, once formed, the new alkynyl alcohol can be hydrated in two ways, Markovnikov and anti Markovnikov.

Carefully observe the structure of the products, the relationship of the OH and the carbonyl.

Note that the regioselectivity used here is only effective if this alkyne is terminal. Otherwise get a mixture.

Can do all the reactions of an alkyne and an alcohol. But remember that we have two acidic groups: the more acidic OH and the less acidic terminal alkyne. We discussed this problem earlier.

Addition of hydrogen cyanide

Think of what the mechanism should be….

Followed by protonation of the alkoxide ion (perhaps by unionized HCN).

pH issue. Slightly basic media so that HCN has partially ionized to cyanide ion, the actual nucleophile.

basic

Follow-up reactions on the cyanohydrins…

R CH2R

O NaCN

RCH2R

OH

N

dil acid

acid catalysisdehydration

R

RHC

H2O

N

H2

catalyst RCH2R

OH

NH2

Hydrolysis. Acid, heatorbase, heat

RCH2R

OH

CO2H

We saw this hydrogenation before.

Let’s see what we can do with the mechanism of the hydrolysis of the nitrile group to a carboxylic acid.

The action is at the nitrile group, CN --> CO2H.

But how does a nitrile group behave? What could be happening?

R

N

aq. acid, heat

R

CO2H

+ NH3

Chemically speaking: the nitrogen of the nitrile is basic (lone pair) and can be protonated. This makes it a better electrophile (cf. carbonyl). Multiple bond can undergo addition (cf. carbonyl) reducing bond order.

Goal: Break the C to N bonding and create C-O bonds.

Considerations: neither the electrophile (RCN) nor the nucleophile (water) is very reactive. Since we are in acid protonate the CN group to make it a better electrophile. Then attack it with the water nucleophile to add water. This results in reduction of C-N bond order and creation of C to O bonds .

Overall

We are breaking the CN bond; bond order goes from 3 to 0. Probably stepwise.

R

N

R

NH

acid

Again, we are in acid environment. Let’s protonate something…. Protonate the multiple bonded N atom to make better electrophile and attack with the nucleophile, water.

R

NH

HOH

R

NHH2O

R

NHHO

+ H+

What have done so far? Reduced the CN bond order from 3 to 2 and added one O to the C. Moving in the right direction!

Want to reduce the CN bond order to zero and introduce more O on the C. Keep going! To induce the water to attack again (adds another O) need to increase the reactivity of the electrophile. Protonate again!! On the O.

R

NH2O

+ H+

Note the bonding pattern here. We have seen it before.

HO NH similar toHO CH2

which tauermerizes in acid or base, keto-enol

HO CH2O

R

NH2HO

OH2

R

NH2HO

OH2

reposition the H+

R

NH3HO

O

H

R

+ NH3

O

HO

Now want to get rid of the NH2. We have all the O’s we need.

We know what we have to do. Have to get the N protonated to make it a good leaving group.

Initial equilibrium with acid

Done.

R

NH2O

R

NH2O

acid

H

HOH

Wittig Reaction

H I PPh3Bu-Li R R

O

R Ror NaH

Example, synthesize CHCH3

or combine them the other way…

Substitution Elimination

Wittig Reaction Mechanism

Ph3P: +

H I

Ph3P

H Acidic hydrogen

Phosphonium ylidePh3P

H

strong base, BuLiPh3P

Nucleophilic center

Ph3P+

R R

O

Ph3P

R

R

O

betaine

Ph3P

R

R

O

oxaphosphetane

Ph3P

R

R

O

oxaphosphetane

Ph3PO

RR

Nucleophilic substitution

Friedel Crafts Acylation

R Cl

O

AlCl3

R O

And then all the reactions of ketones…

Formation of Hydrates, carbonyls and water.

Carbonyl side of equilibrium is usually favored.

Hemiacetals and Acetals, carbonyls and alcohols

(Unstable in Acid; Stable in base)

(Unstable in Acid; Unstable in base)

Addition reaction.

Substitution reaction

Formation of Hemiacetals, catalyzed by either acid or base. Let’s do it in Base first.

hemiacetal

Use Base to set-up good necleophile.

An addition of the alcohol to the carbonyl has taken place. Same mechanism as discussed earlier.

Poor nucleophile Good nucleophile

But first let’s take stock.

We have an addition reaction.

Just mixing a carbonyl and an alcohol do not cause a reaction.

One of them must be made a better reactant.

Carbonyl can be made into a better electrophile by protonating in acid.

Alcohol can become a better nucleophile in base by ionization.

Alternatively, hemiacetal formation in Acid

Protonation of carbonyl (making the oxygen more electronegative)

Attack of the (poor) nucleophile on (good) electrophile.

Deprotonation

Overall, we have added the alcohol to the carbonyl.

Hemiacetal to Acetal, Acid Only

Protonate the hemiacetal, setting up leaving group.

Departure of leaving group.

Attack of nucleophile

DeprotonationSubstitution reaction, cf SN1.

Equilibria

Generally, the hemiacetals and acetals are only a minor component of an equilibrium mixture. In order to favor formation of acetals the carbonyl compound and alcohol is reacted with acid in the absence of water. Dry HCl) The acetals or hemiacetals maybe converted back to the carbonyl compound by treatment with water and acid.

An exception is when a cyclic hemiacetal can be formed (5 or 6 membered rings).

Hemiacetal of D-Glucose

Try following the stereochemistry here for yourself

The hemiacetal can form with two different configurations at the carbon of the carbonyl group. The carbon is called the anomeric carbon and the two configurations are called the two anomers. The two anomers are interconverted via the open chain form.

The alcoholThe carbonyl

Stabilities of the Anomers…

More stable form, with the OH of the anomeric carbon is equatorial

Less stable form.

Here note the alternating up-down relationships.

Here see the cis relationship of these two OH groups, one must be axial.

Acetals as Protecting Groups

Target molecule

Form this bond by reacting a nucleophile with an electrophile. Choose Nucleophile and Electrophile centers.

E

N

Br-Mg

The nucleophile could take the form of an organolithium or a Grignard reagent. The electrophile would be a carbonyl. Do you see the problem with

the approach??

Grignard would react with this

carbonyl.

Synthetic Problem, do a retrosynthetic analysis

Use Protecting Group for the carbonyl… Acetals are stable (unreactive) in neutral

and basic solutions.Create acetal as protecting group.

Now create Grignard and then react Grignard with the aldehyde to create desired bond.

Remove protecting group.

Same overall steps as when we used silyl ethers: protect, react, deprotect.

protect

react

deprotect

Tetrahydropyranyl ethers (acetals) as protecting groups for alcohols.

Recall that the key step in forming the acetal was creating the carbocation as shown…

Recall that we can create carbocations in several ways:

1. As shown above by a group leaving.

2. By addition of H+ to a C=C double bond as shown next.This cation can now react with an alcohol to yield an acetal. The alcohol becomes part of an acetal and is protected.

This resonance stabilized carbocation then reacts with an alcohol molecule to yield the acetal.

An acid

There are other ways to create carbocations……

Sample Problem

Provide a mechanism for the following conversion

O

HO OHHCl/H2O

O O

OH

First examination:

have acid present and will probably protonate

Forming an acetal. Keep those mechanistic steps in mind.

Ok, what to protonate? Several oxygens and the double bond. Protonation of an alcohol can set-up a better leaving group. Protonation of a carbonyl can create a better electrophile.

We do not have a carbonyl but can get a similar species as before.

O H+ O

H

O

The protonation of the C=CStrongly electrophilic center, now can do addition to the C=O

Now do addition, join the molecules

O

HO OH

O O

HO

O O

OH

Product

Now must open 5 membered ring here. Need to set-up leaving group.

H+

O O

HO

O O

HO

H

Leaving group leaves….

O O

HO

HO O

HO

H

O O

HO

H

Followed by new ring closure.

O O

HO

HO O

HO O

OH OH

Done. Wow!

Sulfur Analogs

Consider formation of acetal O

acetaldehydeethanal

OH OH

dry HCl

OO

Sulfur AnalogO

acetaldehydeethanal

SH SH

dry HCl

SS

dithiane

SS

H

Bu Li

SS

+ BuH

The aldehyde hydrogen has been made acidic

Why acidic?

Sulfur, like phosphorus, has 3d orbitals capable of accepting electrons: violating octet rule.

SS SS

empty sulfur 3d orbital

f illed carbon p orbital

Recall early steps from the Wittig reaction discussed earlier

Ph3P: +

H I

Ph3P

H

Ph3P

H

strong base, BuLiPh3P

This hydrogen is acidic.

Why acidic? The P is positive and can accept charge from the negative carbon into the 3d orbitals

PH

Some Synthetic Applications

Umpolung – reversed polarity

What we have done in these synthetic schemes is to reverse the polarity of the carbonyl group; change it from an electrophile into a nucleophile.

O CNO

CN

electrophile

OSS

O

OH

O

nucleophilic

Can you think of two other examples of Umpolung we have seen?

Nitrogen Nucleophiles

Mechanism of Schiff Base formation

Attack of nucleophile on the carbonyl

Followed by transfer of proton from weak acid to strong base.

Protonation of –OH to establish leaving group.

Leaving group departs, double bond forms.

Hydrazine derivatives

Note which nitrogen is nucleophilic

H2N

NH

NH2

O

Nucleophilic nitrogen

H2N

NH

NH2

O

Favored by resonance

Less steric hinderance

Reductive Amination

Pattern:

R2C=O + H2N-R’ R2CH-NH-R’

EnaminesRecall primary amines react with carbonyl compounds to give Schiff bases (imines), RN=CR2.

Primary amine

Secondary AmineSee if you can write the mechanism for the reaction.

But secondary amines react to give enamines

Acidity of Hydrogens hydrogens are weakly acidic

Weaker acid than alcohols but stronger than terminal alkynes.

Learn this table….

Keto-Enol Tautomerism(Note: we saw tautomerism before in the hydration of alkynes.)

Fundamental process

CH3

O

CH2

HO

keto form enol formusually small component

acid or basecatalysis

Mechanism in base:

CH3

O :OH-

CH2

O

Negative carbon, a carbanion, basic, nucleophilic carbon.

CH2

O

Additional resonance form, stabilizing anion, reducing basicity and nucleophilicity.

CH2

HOH-O-H

Protonation to yield enol form.

Details…

Base strengthAlkoxides will not cause appreciable ionization of simple carbonyl compounds to enolate.

Strong bases (KH or NaNH2) will cause complete ionization to enolate.

Double activation (1,3 dicarbonyl compounds) will be much more acidic.

O O

H HFor some 1,3 dicarbonyl compounds the enol form may be more stable than the keto form.

More details…

nucleophilicity CH2

O

Nucleophilic carbon

Some examples:O O O Obase

R-X O O

R

CH3

O :OH-

CH2

O

Br-Br

CH2Br

O

Some reactions related to acidity of hydrogens

Racemization

Exchange

Oxidation: Aldehyde CarboxylicRecall from the discussion of alcohols.

Milder oxidizing reagents can also be used

RCHOAg(NH3)2

+

RCO2- + Ag

Tollens Reagent test for aldehydes

“Drastic Oxidation” of Ketones

O

dichromate, etc

at high temperature

CO2H

HO2C

CO2H

HO2C

Obtain four different products in this case.

Reductions: two electron

O

OH

HNaBH4

Or LiAlH4

O

OH

H

H2/Pt

Reductions: Four Electron

OH H

Zn(Hg), HCl

acid

OH H

base

H2N-NH2

KOH, heat

Clemmenson

Wolf-Kishner

Mechanism of Wolf-Kishner, C=O CH2

O H2N-NH2N

N

H

H

Recall reaction of primary amine and carbonyl to give Schiff base. Here is the formation of the Schiff base. We expect this to happen.

N

N

H

HN

N H

N

N H

-OH

These hydrogens are weakly acidic,

just as the hydrogens to a

carbonyl are acidic.Weakly acidic hydrogen removed. Resonance occurs. Same as keto/enol tautomerism.

N

N HH-O-H N

N H

H

Protonation (like forming the enol)

N

N H

H

-OH

NN

H

N

N

H

Perform an elimination reaction to form N2.

H-O-HH

H

O

C

C H

O

C

C H

Here is the resonance for the anion from the keto-enol systemO

C

C

H

Haloform Reaction, overall

CH3O

X2

CX3O

NaOH

NaOH

CO2-

+ HCX3

The last step which produces the haloform, HCX3 only occurs if there is an methyl group, a methyl directly attached to the carbonyl.

If done with iodine then the formation of iodoform, HCI3, a bright yellow precipitate, is a test for an methyl group (iodoform test).

CH3

O

methyl

Steps of Haloform ReactionThe first reaction:

CH3O

X2

CX3O

NaOH

CH3O

X2

CH2XO

NaOH

All three H’s replaced byX. This must happen stepwise, like this:

Pause for a sec: We have had three mechanistic discussions of how elemental halogen, X2, reacts with a hydrocarbon to yield a new C-X bond. Do you recall them?

Radical Reaction: R. + X-X R-X + X. (initiation required)

Addition to double bond: C=C + X-X + Br- (alkene acts as nucleophile, ions)Br

Nucleophilic enolate anion: CH3

O :OH-

CH2

O

Br-Br

CH2Br

O

Mechanism of Haloform Reaction-1

C H3

O

R

:OH-

C H2

O

R

Br-Br

CH2Br

O

R

Br-

Repeat twice again to yield C Br3

O

R

Where are we? The halogens have been introduced. First reaction completed.

Using the last of the three possibilities

One H has been replaced by halogen.

But now we need a substitution reaction. We have to replace the CBr3 group with OH.

Mechanism of Haloform - 2

R

CH3O X2

R

CX3O

NaOH

NaOH

R

+ HCX3

OO-

R

CX3O

-OH

R

CX3O

OH

Attack of hydroxide nucleophile. Formation of tetrahedral intermediate. Anticipate the attack…

R

CX3O

OH

Reform the carbonyl double bond. CX3

- is ejected. The halogens stabilize the negative carbon.

R+ -:CX3

OOH

R+ HCX3

OO-

Neutralization.

This is a substitution step; OH- replaces the CX3 and then ionizes to become the carboxylate anion.

Here’s how:

Cannizaro Reaction

Overall:2 RCHO

conc. KOH

heatRCO2

- + RCH2OH

Restriction: no hydrogens in the aldehydes.

H3C

O

CHO

H

CHO

hydrogens No hydrogens

Why the restriction? The hydrogens are acidic leading to ionization.

Mechanism

What can happen? Reactants are the aldehyde and concentrated hydroxide.

Hydroxide ion can act both as

Base, but remember we have no acidic hydrogens (no hydrogens).

Nucleophile, attacking carbonyl group.

R

O

H

HO-:

R

O

H

OH

R

OH

R

O

OH

+

R

OH

H

R

O

O

+R

OHH

H

Attack of nucleophilic HO-

Re-establish C=O and eject H- which is immediately received by second RCHO

Acid-base

Experimental Evidence

2 RCDO RD2OH + RCO2-

KOH, H2O

These are the hydrogens introduced by the reaction. They originate in the aldeyde and do not come from the aqueous hydroxide solution.

Kinetic vs Thermodynamic Contol of a Reaction

Examine Addition of HBr to 1,3 butadiene

HBrH

Br

+

Br

H

1,2 product 1,4 product

Mechanism of reaction.

H-Br

HH

Allylic resonance

1,2 product 1,4 product

Br Br

H

BrBr

H

But which is the dominant product?

HBrH

Br

+

Br

H

1,2 product 1,4 product

Nature of the product mixture depends on the temperature.

Product mixture at -80 deg 80% 20%Product mixture at + 40 deg 20% 80%

Goal of discussion: how can temperature control the product mixture?

Thermodynamic Control: Most stable product dominates

Kinetic Control: Product formed fastest dominates

When two or more products may be formed in a reaction A X or A B

Thermodynamic control assumes the establishing of equilibrium conditions and the most stable product dominates.

Kinetic Control assumes that equilibrium is not established. Once product is made it no longer changes.

Equilibrium is more rapidly established at high temperature. Thermodynamic control should prevail at high temperature where equilibrium is established.

Kinetic Control may prevail at low temperature where reverse reactions are very slow.

HBrH

Br

+

Br

H

1,2 product 1,4 product

Nature of the product mixture depends on the temperature.

Product mixture at -80 deg 80% 20%Product mixture at + 40 deg 20% 80%

Thermodynamic ControlMore stable product

Kinetic Control

Product formed most quickly, lowest Ea

Formation of the allylic carbocation.

Can react to yield 1,2 product or 1,4 product.

Most of the carbocation reacts to give the 1,2 product because of the smaller Ea leading to the 1,2 product. This is true at all temperatures.

At low temperatures the reverse reactions do not occur and the product mixture is determined by the rates of forward reactions. No equilibrium.

Most of the carbocation reacts to give the 1,2 product because of the smaller Ea leading to the 1,2 product. This is true at all temperatures.

At higher temperatures the reverse reactions occur leading from the 1,2 or 1,4 product to the carbocation. Note that the 1,2 product is more easily converted back to the carbocation than is the 1,4. Now the 1,4 product is dominant.

Diels Alder Reaction/Symmetry Controlled Reactions

Quick Review of formation of chemical bond.

HO- + H+ H - O - H

Electron donor

Electron acceptor

Note the overlap of the hybrid (donor) and the s orbital which allows bond formation.

HO- + H+ H - O H

For this arrangement there is no overlap. No donation of electrons; no bond formation.

Diels Alder Reaction of butadiene and ethylene to yield cyclohexene.

We will analyze in terms of the pi electrons of the two systems interacting. The pi electrons from the highest occupied pi orbital of one molecule will donate into an lowest energy pi empty of the other. Works in both directions: A donates into B, B donates into A.

new bonds

A B

A

B

HOMOdonor

HOMOdonor

LUMOacceptor

LUMOacceptor

B HOMO donates into A LUMO

A HOMO donates into B LUMO Note the

overlap leading to bond formation

Note the overlap leading to bond formation

Try it in another reaction: ethylene + ethylene cyclobutane

new bonds

A B

A B

LUMO

HOMO

LUMO

HOMO

Equal bonding and antibonding interaction, no overlap, no bond formation, no reaction

Br

Br

excess sodium methoxide

Reaction Problem

HO

OEt

using only compoundshaving two carbons as the source of all carbons in the target molecule

Synthesis problem

Give the mechanism for the following reaction. Show all important resonance structures. Use curved arrow notation.

aq. acid

heatOEt

OHO

+ EtOH

Mechanism Problem