Upload
hoangdung
View
262
Download
5
Embed Size (px)
Citation preview
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+