Hemiacetals and Acetals, carbonyls and alcohols

(Unstable in Acid; Stable in base)

(Unstable in Acid; Unstable in base)

Addition reaction.

Substitution reaction

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