Summary Sheet 2: Enols and Enolates · PDF fileSummary Sheet 2: Enols and Enolates version...

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Page 1: Summary Sheet 2: Enols and Enolates · PDF fileSummary Sheet 2: Enols and Enolates version 1.0, ... (will react with strong acid) ... amide R O N 2 carboxylic acid R O OH acid chlo

Summary Sheet 2: Enols and Enolates

version 1.0, 4/13/10copyright James Ashenhurst, 2010. suggestions/questions/comments? [email protected]

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Structural Features of the Carbonyl Group:•Carbon, oxygen: sp2 hybridized•O–C–C bond angle ~120 °•C=O bond strongly polarized toward oxygen. •Carbonyl carbon is partially positive therefore electrophilic!•Lone pairs render oxygen weaklynucleophilic (will react with strong acid)

The carbonyl is an electron withdrawing π system with low-lying π* orbitals. It stabilizes adjacent negative charge.

H3C CH3

CH3O

OMe

EthanepKa = ~50

Methyl propionatepKa = ~25 ~1025 more acidic just by

replacing H with a carbonyl!Why the huge difference in acidity? The lone pair is stabilized by donation into the carbonyl π system.

H2C CH3M

Conjugate base

CH3O

OMe

MConjugate base (enolate)

CH3O OMe

MCH3

O OMe

M

Answer: The more electron-poor the carbonyl, the greater will be its ability to stabilize negative charge. Conversely, the greater the donating ability of a substitutent on the carbonyl, the less it will be able to stabilize negative charge.

Enolate (Carbanion resonance form)

Enolate (oxy-anion resonance form)

Question: How do you explain the relative acidity of thefollowing series?

H3C

O

CF3 H3C

O

CH3 H3C

O

OR H3C

O

NR2> > >

most acidic least acidic

The aromatic electrophilic substitution chart is a good proxyfor the ability of a functional group to donate to a carbonyl:

Substitution of the α-carbon by a second carbonyl derivative makes the α-proton even more acidic:

O

Me

O

Me

O

OMe

O

Me

O

OMe

O

MeO

pKa = 9 pKa = 11 pKa = 13

NR2 , O > OH > OR, NHAc > CH3, R > Cl, Br, F, I > C(O)ORCF3, etc.

CH2H H

H H

LiNEt2

NEt2

M

highly unstable

CH3

NEt2does not form

H

H H

LiNEt2

O

O H

Et2N

O

O t-BuLi

Et2N

O

O t-Bu

H

H

X

X

forms readily!

t-Bu

Likewise, the presence of a carbonyl group activates alkenes towardnucleophilic attack:

enolate: stabilized!

O

CF3

O

Me

O

OR

O

NR2

The reactivity of the alkene toward nucleophilic attack is directly related to the stability of the enolate that forms -

> > >

increasing reactivity toward nucleophilic attackincreasing stability of enolate

Can predict the course of the reaction by pKa!

Enolate = deprotonated enol

Carbonyl compound

H3C

O

OEt

O

OEtH

H

ester enolateO-bound form

O

OEtH

HN

i-Pri-Pr

Li (LDA)

Li Li

C-bound form(or other strong base)

Important: the Enolate is a NUCLEOPHILE

Two key examples:

O

OEt

OR H

M O

OEt

OH

R

O

OEt

OR OR

M O

OEt

OH

R

note: though an ester enolateis shown here, the reaction ofany enolate with an aldehydeis generally called an "Aldol".

Aldol reaction

Claisen Condensation

Key Reaction: Enolate Formation

O

OR

O

RO>

Amphiphilic =nucleophilic at both O and C; here we focus on the reactions at C.

Name Structure

aldehydeR

O

H

ketoneR

O

R'

esterR

O

OR

amideR

O

NR2

carboxylicacid R

O

OH

acidchloride R

O

Cl

anhydrideR

O

O

nitrile R–CN

lactone(cyclicester)

O

O( )n

lactam(cyclic amide)

O

NH(R)( )n

*

*

NH2 = 1° amideNHR = 2 ° amideNR1R2 = 3 ° amide

O

R

1)evans.harvard.edu/pdf/evans_pKa_table.pdf2) www.chem.wisc.edu/areas/reich/pkatable/

Key Concept: Tautomerism

O

CH2

R

OH

RH

H

keto form enol form

Tautomerism: a form of isomerism wherea keto converts to an enol through the movement of a proton and shifting of bonding electrons

H

For acetone (R=CH3) the keto:enol ratio is ~6600:1 at 23 °C. Main reason is the difference in bond strengths between the two species.The enol tautomer is most significant for ketones and aldehydes. (You may also encounter it with acid chlorides in the mechanism of the Hell-Vollhard-Zolinsky reaction). Esters and amides are less acidic and exist almost exclusively as the keto form (e.g. >106 : 1 keto: enol for ethyl acetate)Acetone in D2O will slowly incorporate deuterium at the α−carbon. The enol form is responsible for this behavior. The rate of keto/enol tautomerism is greatly increased by acid (see below right)

Example pKA* ofH

H3C

O

H17

H3C

O

Me20

H3C

O

OMe

H2C O

O

H2C N

OMe

O

NMe2H2C

H3C

O

OH

H3C

O

Cl

H3C

O

O

O

Me

Me-CN

For a comprehensive list of pKa values see:

25

Note that some of of these values differ slightly from textbook to textbook and instructor to instructor. However, there is universal agreement that for a given structure, pKas increase in the order aldehyde < ketone < ester < amide

If you have a vastlyl different set of values Iwould appreciate it if you brought it to my attention.

The goal here is clarity and consistency. I would greatly appreciate feedback on any errors, omissions, or suggestions for improvements. Thanks!

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Note 2

Note 2

Note 1

Note 1: The carboxylic acid is deprotonated first. Subsequent deprotonation of the α-carbon would form a dianion, which has a high activation barrier due to charge repulsion. Itcan be done, but it requires a very strong base.Note 2: It is difficult to measure the pKa of these species due to their reactivity.

Effects on acidity of alkyl groups Effect on reactivity of alkenes:

The rate of keto/enol interconversion isgreatly enhanced by acid:

Five factors that influence the relative proportion of keto/enol:

O

RR

H H

H X O

RR

H

H

X:

Acid makes carbonyl more electrophilic, increasing acidity of α-protons, facilitating formation of enol: this increases K1

O

RR

HHKeto Enol

Enol is nucleophilic at α-carbon, acid isexcellent electrophile: this increases K2

K1

K2

H H X

Net result: Addition of acid speeds proton exchange between the keto and enol forms.

*source: P. Y. Bruice, "Organic Chemistry"

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1. AromaticityO OH

ExclusiveNot observed

2. Hydrogen bonding stabilizes the enol form.

O O

MeMe

O O

MeMe

HH-bond

24 : 76 (at 23 °C)3. Strongly hydrogen bonding solvents can disrupt this, however. The above equilibrium is 81:19 using water as solvent.

O

Me

4. . Conjugation is stabilizing.

OH

MePreferred enol form

5. As with alkenes, increasing substitution increasesthermodynamic stability (assuming equal steric factors)