Chapter 17Aldehydes and Ketones

Carbonyl Groups

C O C Opolarized

C O C O

(1) Aldehydes and Ketones

RC

O

H RC

O

R'

aldehydes ketones

H:– and R’:– are poor leaving groups

(2) Carboxylic Acid Derivatives

RC

O

ClRC

O

OH RC

O

OR' RC

O

NR'2

carboxylic acid substituent could be a leaving group (HO–, Cl–, R’O–, R’2N–); providesfor similar reactivity as aldehydes and ketones but sometimes different reactivity; wesaw a glimpse of this difference already

CH

O

COCH3

O

C H

OH

H

LiAlH4

C OCH3

O

H

CH

OLiAlH4

H2O

1) LiAlH4, Et2O2) H2O

1) LiAlH4, Et2O2) H2O

Nomenclature: prefix - parent - suffix

(1) Aldehydes:

need the longest chain that contains the (CHO) groupC

O

H

H3CC

O

Hbutanal

H3C C

O

H

CH3 CH3

1234

5 2,4-dimethylpentanal

CCH3H

O OH

CH3

12

34

56 5-hydroxy-4-methylhexanal

(2) Ketones: -one as suffix

H3CC

CH3

O

2-propanone (acetone)

H3CCH3

OH

Cl

O

4-chloro-5-hydroxy-3-hexanone1

234

56

There are a number of common carbonyl groups:

RC

O

H3CC

O

HC

O

C

O

Synthesis of Aldehydes

(1) From alcohols (oxidation)

RC

OH

HH

RC

O

H

COH

HH

PCCCH2Cl2

PCCCH2Cl2

(2) From alkenes (oxidation)

1) O3, CH2Cl22) Zn, HOAc

HOC

CH3

O

Ac

(3) From alkynes

C CR H

disiamylborane

R CH2

CH

OB H

1) 2) HO–, H2O2

(4) From esters via partial reduction

R

O

OR' R

O

H

1) DIBAH, toluene2) H3O+

(CH3)2CHCH2

Al

H

CH2CH(CH3)2

COCH 3

O

1) LiAlH4, Et2O2) H2O

C H

OH

H

LiAlH4

C OCH 3

O

H

CH

OLiAlH4

H2O

Synthesis of Ketones

(1) Oxidation of alcoholsO

O

Jones'reagent

PCCCH2Cl2

(2) Ozonolysis of alkenes (oxidation): can also use KMnO4 with acid

H3C CH3

CH2CH3

H3C CH3

OCH3

CH H

O+

1) O3, CH2Cl22) Zn, HOAc

(3) Hydration (addition of H2O) to terminal alkynes

C CR HR

O

CH3

H2O, H2SO4HgSO4

(4) Friedel-Crafts Acylation

R

O

Cl

AlCl3

R

O

(5) Organo-Copper (Cuprate) Coupling

R

O

Cl

1) (CH3)2CuLi, Et2O2) H2O

Oxidation of Aldehydes and Ketones

R

O

CH3

R

O

H R

O

OH

CrO3, H2SO4H2O, acetone

CrO3, H2SO4H2O, acetone

One can also use the Tollens reagent (Ag2O, NH4OH) to oxidize aldehydes, and it isa very gentle method for the selective oxidation of aldehydes

O

H

O

OH

Oxidation of aldehyde (RCHO) takes place via the hydrate

R

O

H R

O

OH

H2O CrO3H3O+

Nucleophilic Addition Reactions

C O

C is sp2

δ+ δ–

Nu

C ONu

C is sp3

R-C-R' ~ 120° R-C-R' ~ 109°

What about the nucleophile?

(1) Nucleophile can be

RC C N CHHO RO

(2) Nucleophile can be

H2O R-O-H R-NH 2 NH3

(3) Steric effects when nucleophile adds to carbonyl carbon

Which is more reactive: aldehydes or ketones?

Aldehydes (RCHO):

(i) nucleophile can easily approach carbonyl carbon

(ii) addition product is less sterically hindered

(iii) transition state is less crowded: low ∆G≠

(iv) greatest polarization of the C=O bond with C being most δ+ since H cannotstabilize the positive charge on carbon very well

Ketones (RC(O)R’):

(i) R groups can stabilize the partial positive charge on the carbonyl carbon

Let us consider a simple nucleophile such as H2O

hydrate(gem diol)

R

O

R'

H2O

R

HO

R'

OH

geminal = "same carbon"

an equilibrium: position of equilibrium depends on R and R’ (on stability of aldehydeand ketone vs the hydrate product)

R R’ Keq

H H

CH3 H

CH3 CH3

ClCH2 H

Cl3C H

ClCH2 ClCH2

CF3 CF3

(1) addition of water is more favorable for aldehydes than for ketones

(2) electronegative groups attached to carbonyl carbon make addition more favorable

C

O

C

O

C

O

δ+

δ–

Why is this useful? The more a compound favors addition at equilibrium, the morerapidly it will react in addition reactions -- the transition state has similar preferencesas the addition intermediate (Hammond postulate again...)

∆G°

transitionstate

E n

e

r

g

y

Reaction Coordinate

∆G°rxn

∆G

The equilibrium can be established with either base or acid catalysis:

Base: HO

C

OHO

C

O

OH

H OHC

OH

OH+ HO

Acid: H3O

δ+

δ–

C

OH

C

O H OH2 H2OC

OH

OH

H

H2OC

OH

OH

C

OH

(a) If nucleophile is strong enough, there is

(b) If nucleophile is not strong enough, need toby coordination to a Bronsted acid (H+) or a Lewis acid

HCN addition

(1) HCN is a weak acid pKa ~ 9.1 (not much CN– at equilibrium)

(2)

RC

O

H

H CNC

OH

CNRHδ+

δ–

catalyzed by base (in order to make NC–)or by direct addition of NC–

CH

O

+ HCN–CN

1) LiAlH4, THF2) H2O

OH

CH2NH2

H3O+

heat (∆)

CO2H

OH

+ NCN C

Grignard addition

C

OC

O

RC

OH

Rδ+

δ–

R Mg X H OHEt2O

so not an equilibrium

magnesium coordinates carbonyl oxygen (Mg is Lewis acidic) and makes carbonylcarbon even more electrophilic -- magnesium helps to “activate” the carbonyl carbon

Hydride Addition: are sources of H:– and both reagents canreduce aldehydes and ketones to alcohols

CH

O

1) NaBH4, EtOH

2) H3O+

not reversible so not an equilibrium

Amines: nucleophiles with attached hydrogens (good nucleophiles)

R'C

O

CH

R'C

N

CH

RR NH2

R NH

R

secondary aminesprimary amines

R NH2

R'C

N

C

RR

R'C

O

CH R2NH

O NCH3

+ H2O+ CH3NH2

(1) Primary Amines (product is an imine): pH of the reaction is very important (bestat about pH = 4.5)

C

O

δ+

δ–

H3O+

C

OH2

NHRC

NR

H2O

imine

RNH2

C

NHR

protonationto yield a

good leaving group

C

O

NH2R

O NCH3

NOH

N

HN

O

NH2

N

HN

NO2

NO2

all are R-NH2 for formation of C

NR

CH3NH2

NH2OH

NH

O

NH2H2N

NH

H2N NO2

O2N

(a) all are equilibria

O

+ CH3NH2

NCH3

+ H2O

(b) equilibrium favored to reactants for imine so need to

(c) oximes, semicarbazones, and hydrazones are (reaction can even be done in H2O!)

(2) Secondary amines R2NH: enamines

C

O

CH

R2NH C

OH

NR2CH

H3O+

C

OH2

NR2CH

C

N

CH

RR

C

N

C

RR

C

O

NHR2CH

H2O

--

-- there is no proton to lose on N when you start with a secondary amine (R2NH)

O

(CH3)2NH

O

CH3

O

(CH3)2NH

NCH3H3C

NH3CN

CH3

H

Wolff-Kishner Reaction

RC

O

R' RC

H

R'

H

H2N-NH2

RC

N

R'

NH2

RC

N

R'

NH

RC

N

R'

NHH2O

RC

N

R'

N

H

H

H2O

HOHO

RC

R'H

+ N2+ H2O

O

1) H2NNH2, KOH

2) H3O+

Acetal Formation

RC

O

R'C

OR"

OR"RR'

2 R"–OH

+

acid

(1) equilibrium so:

(a) lots of R”-OH and also remove H2O then favor equilibrium to the(b) lots of H2O then favor equilibrium to the

(2) protection of ketone

O

OCH 3

O ? O

OH

for ester reduction: LiAlH4 is needed but ketone would be reduced at the same timeso need to protect (mask, hide) the ketone group

O

OCH3

O

OCH3

OOCH3CH3O

OH

OCH3CH3O

1) LiAlH4, Et2O 2) H2O

O

OH

CH3OHH+

H2OH+

CH3

O

2 CH3OH, H+

Equilibrium: ∆G° = ∆H° – T ∆S° =

Acetal formation has an unfavorable entropy (3 moles <=> 2 moles), so use diol

CH3

O

O

Br Br

OO

CH3

OOH3O+

O

CH3

+ H2OH+HO OH

1) Mg, Et2O2) CH3Br

H+HO OH

How?

(a) hemiacetal formation: acidic or basic conditions are fine

R

O

R'

H3O+

R

O

R'

H OH

OCH3RR'

R

O

R'

CH3OHOH

OCH3RR'

H

CH3OH

CH3OO

OCH3RR'

CH3OHOH

OCH3RR'

(b) hemiacetal conversion to acetal works in

OH

OCH3RR'

H3O+ OH

OCH3RR'

H

R

O

R'

CH3

R

O

R'

CH3

OCH3

OCH3RR'

acetal

CH3OH

OCH3

OCH3RR'

H

CH3OH

Thioacetals

R R'

O

R R'

H HRaney NiH2, EtOH

H+HS SH

same mechanism as for acetal formation

CH3

O

SHHS

H+CH3

SS

Raney NiH2, EtOH

CH3

H H

Wittig Reaction

triphenylphosphineoxide

ylid

C

O+ O P

PhPh

Ph

R

R

O PPhPh

Ph+C

O O PPh3

RR

betaine

R2C PPhPh

Ph

Ph =

Ph3P CH3–Br

O PPh3

RR

oxaphosphatane

R2C PPhPh

Ph

Ph3P–CH3NaHEt2O Ph3P–CH2

acidic hydrogens

O

Ph3P–CH2THF

1)

2) H2O

CH2

+ Ph3P=O

+ Ph3P=O

CH2

2) H2O

1) Ph3P–CH2THF

O

Cannizzaro Reaction:

H

O

HO H

O OH H

O

OH

O

+

O

H H

O

O

OH

H H

+

(a) H:– as a leaving group

(b) reaction is driven by the formation of stable carboxylate anion (irreversible)

Similar action as NADPH in biology:

N

R

O

H2N

H HO

not aromatic some aromatic character

N

R

O

H2N

H

R

H

R'

O

1,2-Addition to Carbonyl Group:

C

ONu+

O

NuH2O

OH

Nu1

2

addition of Nu and H to C1 and O2

1,4-Addition to αααα,ββββ-Unsaturated Enone:

O

12

3

4 OH

HNu

O

HNu

Nu

O

HNu

H2O

(a) final step is a

(b) overall reaction is addition of Nu-H to C1 and O4 of enone

(c) need the carbonyl group (C=O) to have 1,4-addition to the C=C double bond

O O

Nu

no reaction

1) Nu2) H2O

O O O

1) Nu2) H2O

What kinds of nucleophiles work?

(1) Amines

O

O

H3CEt2NHEtOH

O

H3C NEt2H H

CH3NH2EtOH

O

NHCH3

(2) HCN addition

O O

CN

1) Et2Al-CN, toluene

2) H3O+

Nagata reaction

(3) Organo-copper (Cuprate) Reactions

O

1) R2CuLi

2) H3O+

How to make R2CuLi?

O

1) (CH3)2CuLi

2) H3O+

O

CH3

O

H3C

R-X 2 Lipentane

R-Li + Li-X

2 R-Li R2CuLi + LiI

1)

CuLi )

)

2

2) H3O+

CuIEt2O

O HO CH3

1) CH3Li, Et2O2) H2O

HO CH3

O

CH3

1) CH3MgBr, Et2O

2) H3O+

1) (CH3)2CuLi, Et2O

2) H3O+