wwu -- chemistry reactions of hydrogens: condensation reactions chapter 21

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Reactions of Hydrogens:Condensation Reactions

Chapter 21


Assignment for Chapter 21

DO: Sections 21.0 through 21.4 Sections 21.7 through 21.9 Sections 21.12 through 21.17 Section 21.22

SKIP: Sections 21.5 through 21.6 Sections 21.10 through 21.11 Sections 21.18 through 21.21


ALSO DO:

SummaryProblems


Problem Assignment

In-Text Problems: 21-1 through 21-9 21-12 through 21-17 21-25 through 21-45

End-of-Chapter Problems: 1 through 11 13 through 24


TautomerismCompounds whose structures differ markedly in the arrangement of atoms, but which exist in equilibrium, are called tautomers.Most often, tautomers are species that differ by the position of a hydrogen atom and which exist in equilibrium.The best-known example is keto-enol tautomerism.


Keto-Enol Tautomerism

C C

H

O

C C

O

H

keto enol

K



In general, the equilibrium favors the keto form very dramatically. If one calculates the relative energies of the keto and enol forms, one concludes that the formation of enol from keto should be endothermic by about 75 kJ/mole.


Keto-Enol TautomerismFrom this energy, one calculates an equilibrium constant for enolization:

6.3 x 10-14

Clearly, for most aldehydes and ketones, the ability to form an enol is an extremely minor property.



In the case of 1,3-dicarbonyl compounds, however, the equilibrium may shift to favor the enol form, since a stabilized, hydrogen-bonded structure is now possible.


Keto-Enol Tautomerism in 1,3-Dicarbonyl Compounds

O

CCH3 C

CCH3

O

H H

O

CCH3 C

CCH3

O

H

H

K

keto enol


Keto-Enol Tautomerism in 1,3-Dicarbonyl Compounds

The equilibrium lies substantially to the right.In simple ketones, such a hydrogen-bonded structure cannot form, and the percentage of enol found in an equilibrium mixture is very small.The following tables illustrate some typical enol percentages. Notice the difference between simple ketones and dicarbonyl compounds.


Some Representative Enol Percents

CH3 C CH2 C H

O O

CH3 C CH C H

OH O

CH3 C CH2 C CH3

O O

CH3 C CH C CH3

OH O

CH3 C CH2 C OC2H5

O O

CH3 C CH C OC2H5

OH O

% enol = 98

% enol = 80

% enol = 8


More Representative Enol Percents

O OH

O

CCH3 CH3 CH2 C CH3

OH

CH2 CH C CH3

O

CH2 CH C CH2

OH

% enol = 4.1 x 10

% enol = < 2 x 10

% enol = 2.5 x 10

-4

-4

-3


One last comment on this...

You may recognize some structural similarities between enols and enamines.Whenever an enol form can exist, it has the potential to be a nucleophile!


Acidity of -HydrogensReview material in Chapter 7, Section 7.7The acidity of a hydrogen attached to the -carbon of a carbonyl compound is much higher than the acidity of a typical C-H hydrogen.pKa values range from about 19 to 20 (compared with 48 to 50)


Acidity of -Hydrogens: The Reason

R C C

H

R

CH3

O

+

B

R C C

R

CH3

O

R C C

R

CH3

O

+ B H

an enolate ion


Acidity of -Hydrogens

Resonance stabilization of the enolate ion shifts the equilibrium to the right, thereby making the C-H bond more acidic.Once formed, the enolate ion is capable of reacting as a nucleophile. The -carbon of the enolate ion bears substantial negative charge.


Base-Promoted Halogenation of Ketones

CH2 C R

O

R

+

Br2

+

OH-

R CH C

Br

R

O

+

Br-

+

OHH


Base-Promoted Halogenation of Ketones

The experimental rate law is:

Rate = k[ketone][OH-]

Note that the rate law does not contain bromine!


Mechanism

1)

2)

R C CH2 H

O

+

O H

slowR C CH2

O

R C CH2

O

+O

H H

R C CH2

O

+ Br Brfast

R C CH2

O

Br

+

Br

Note that the first step is rate-determining


Example

C CH3

O

+

Br2

(one equivalent)

NaOH

C CH2

O

Br

+

Br-


But...

The halogenation is difficult to stop at the mono-substitution stage.Often, poly-halogenated products are formed in this reaction.


With an excess of bromine:

C CH3

O

+

Br2

(excess)

NaOHC C

O

Br

Br

Br


•There is also an acid-catalyzed halogenation reaction, which operates through the formation of the enol form of the ketone (recall that the enol is nucleophilic).

•Once formed, the enol displaces bromide ion from Br2, forming the brominated product.

•In the acid-catalyzed mechanism, mono-substitution is the predominant result.


Example

O

+ Br2

CH3COOH

O

Br

+ HBr


Alkylation of Enolate Ions

In the presence of a very strong base, the -hydrogen of an aldehyde or ketone can be replaced by an alkyl group.Once again, the strong base removes an -hydrogen to form an enolate ion.The enolate ion, acting as a nucleophile, participates in an SN2 substitution with an alkyl halide.


Alkylation of a Ketone

O

C CH3

O

C CH2

CH3 IO

C CH2 CH3

O

C CH2

strong base

THF

.. _

.. _


… and the “strong base” is:

CH3 C

CH3

O

CH3

K

potassium tert-butoxide

NaNH2 sodium amide

KH potassium hydride

NCHCH

CH3CH3

CH3CH3

Li

lithium diisopropylamide "LDA"

Best Choice


Mechanism

R C CH2 H

O

+

Base

1)

2)

R C CH2

O

+

Base-H

R C CH2

O

+ R X R C CH2 R

O

+

slow

SN2

X


Alkylation of Enolate Ions

Remember that enamines can also react with alkyl halides to give similar products.Review Chapter 16, Section 16.13.See also Chapter 21, Section 21.8.


Example

O

LDA

THF

CH3 CH2 Br

O

CH2 CH3


Here’s something different:

O

LDA

THF

Se Br

O

Se