17.2 how aldehydes and ketones react (part i) 1 ++ r = alkyl or aryl (c) y = alkyl, aryl or h (class...
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17.2 How Aldehydes and Ketones React (Part I)
1
C
O
C
O
C
O
YR
d+
R = alkyl or aryl (C)
Y = alkyl, aryl or H (class II) (No leaving group)
d-Electron rich (Lewis base, Nu)
Electron deficient (Lewis acid, E+)
Main Menu
Class I vs. Class II Carbonyl Compounds
2
Class II
Y = NR’2 (amide) = OR’ (ester, carboxylic acid) = OCOR’ (acid anhydride) = X (acyl halide)
C
O
YR
Class I
C
O
YR
Y = H (aldehyde) = R’’ (ketone)
H-H (pKa = 35)R-H (pKa = 50)
Hydride (H-) and carboanion are not
leaving groups
Relative Reactivity of Class I and Class II Carbonyl Compounds
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acyl halide
C
O
XRC
O
ORC
R'
O
C
O
OR'RC
O
NHR'R>> C
O
R'RC
O
HR> > > >
acid anhydrideketonealdehyde ester amide
Esters and amides are more stable than ketones and aldehydes due to their resonance stabilization.
H R’
Nucleophilic Addition (Class II)
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1. General mechanism in basic condition:
C
O
R'(H)R
+ Z+ H+
C
O
R
C
O
R'(H)
R Z- H+
C
OH
R'(H)
R Z
2. General mechanism in acidic condition:
C
O
R'(H)R
Z+ H+
C
O
R- H+
C
OH
R'(H)
R ZC
O
R'(H)R
H
Important pKa to Remember
5
Names AcidsH-Z
Approx. pKa
Conjugate Base, :Z
General Roles of :Z
Alkane (2°) 51 Base as Li+ saltNucleophile as Grignard reagent
Amine 38 Base and Nucleophile
Hydrogen 35Base in NaH, CaH2
Nucleophile in LiAlH4, NaBH4
Alcohol water 15-16 Often as a base but can be a
nucleophile
Ammonium 10-11 Weak base, but can be a nucleophile
Thiol 10-11 Nucleophile
Carboxylic Acid 4-5 Weak base, poor leaving group
Hydrochloric Acid -7 Leaving group, poor nucleophile
H3CCH
H3CH
H3CCH
H3C
HN
HH
HN
H
H H H
R O H R O
RNH
HH
RNH
H
SR H R S
HRCO2 RCO2
HClCl
Types of Nucleophile for Class II Carbonyl Groups
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1. Carbon as the nucleophilic atom
HC H+C +
pKa = 50
Basic condition
2. Hydrogen as the nucleophilic atom
carboanion
H hydride Mostly basic condition
3. Nitrogen as the nucleophilic atom
1° and 2° amines Mostly acidic condition
4. Oxygen as the nucleophilic atomAcidic condition
NH2
1° alcoholsOH
HCC H+CC +pKa = 25 Acetylide ion
Carbon as the Nucleophilic Atom: Grignard Reagents
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Carboanions are highly reactive.
HC B+ HBC +
pKa = 50
Hard to find a base to do the deprotonation.
carboanion
Formation of Grignard reagent
XC Mg+
X = Cl, Br or I
THF orEt2O
Mg2+X-CC MgX
THF: tetrahydrofuranEt2O: diethyl ether
The carbonanions can be stabilized.
O
Reactions of Grignard Reagents
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MgBrEt2O
MgBr MgBr
C
O
HH3C
C
O
HH3C
Mg+Br
C
OH
HH3CH3O+
Reactions of Grignard Reagents
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MgBr
Et2O
MgBr C
O
CH3H3C
2) H3O+
1)OH
MgBr
Et2O
MgBr
C
O
H
2) H3O+
1)
OH
3° alcohols
2° alcohols
Reactions of Grignard Reagents
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MgBr
Et2O
MgBr C
O
HH
2) H3O+
1)OH
MgBrEt2O
MgBrC OO
2) H3O+
1) CO2
OH
O
1° alcohols(one extra carbon)
Carboxylic acid
MgBr
Et2O
MgBr
2) H3O+
1) O OH
1° alcohols(two extra carbons)
Reactions of Grignard Reagents with Esters
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MgBr C
O
OCH3H3C
2) H3O+
1)
1 mol.
1 mol. OH
H3CC
O
O
CH3+
0.5 mol.0.5 mol.
MgBr C
O
OCH3H3C
2) H3O+
1)
2 mol.
1 mol. OH
1 mol.
Reactions of Grignard Reagents with Esters
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C
O
OH3C
CH3
MgBr C
O
H3C
OCH3
Why two equivalents of Grignard reagent are needed?
C
O
H3C
O CH3
A ketone (more reactive than ester)
MgBr
C
O
H3CC
OH
H3C H3O+
Carbon as the Nucleophilic Atom: Acetylide Ions
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HC H+C +
pKa = 50 carboanion
HCC H+CC +pKa = 25 Acetylide ion
Why the pKa of acetylide is much lower?
2Px
2Pz
2Py
2S
The radius of 2S orbital is smaller than the radius of 2P orbitals.
Order for the radius of hybridized orbitals: SP3 > SP2 > SPOrder for the electronegativity of hybridized orbitals: SP3 < SP2 < SPOrder for the acidity of H’s of hybridized orbitals: SP3 < SP2 < SP
HCCpKa = 40
Reactions of Carbonyl Groups with Acetylide Ions
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C
O
CH3H3C
2) H2O
1) OH
HCC + Na+NH2- NH3CC +
pKa = 25 Acetylide ion pKa = 38
CC
Carbon as the Nucleophilic Atom: Cyanide
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Hydrogen cyanide is weakly acidic.
HCN B+ HBCN +
pKa = 9.1Cyanide is highly poisonous.
cyanide
Addition of cyanide to aldehydes or ketones:
OC
R'(H)
R
CN-+HCl
C
HO
R
R'(H) CNStable in acidic condition but unstable in basic condition.
H+, H2Oheat
H2, Pt/C
C
HO
R
R'(H) CH2NH2C
HO
R
R'(H) CO2H
a-hydroxy carboxylic acid
Hydrogen as the Nucleophilic Atom: Hydride Reagents
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Reagents that can provide hydrides as nucleophiles:
Al
H
H
H H+Al-
H
H
H H
Li+
Theoretically, one molecule of LiAlH4 or NaBH4 can provide four hydrides.
Lithium aluminum hydride
LiAlH4NaBH4
NaH CaH2
B
H
H
H H+B-
H
H
H H
Na+
Sodium boroydride
Reagents that can provide hydrides as bases:
Diisobutylaluminum hydride (DIBAL)
Al H
Reactions of Aldehydes and Ketones with Hydride Reagents
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General Reactions:
Examples:
OC
R'(H)
R 1) LiAlH4 or NaBH4
2) H2OC
HO
R
R'(H) H
O1) LiAlH42) H2O
OH
H
O1) NaBH42) H2O
OH
General Mechanism for the Reduction of Aldehydes and Ketones
Using Hydride Reagents
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OC
R'(H)
R
C
H
R
R'(H) OAlH3
Al-
H
H
H Hd-
d-
The three H’s can still act as hydrides.
Repeat 3 times
C
H
R
R'(H) OAl
4
H2O C
H
R
R'(H) OH Al(OH)4-+
Comparison of LiAlH4, DIBAL and NaBH4
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Relative Reactivity
>LiAlH4 DIBAL NaBH4>
Amide Ester Carboxylic acid Ketone Aldehyde
LiAlH4 yes yes yes yes yes
DIBAL no yes ? yes yes
NaBH4 no no no yes yes
NaBH3CN>Stable in weak acidUnstable in weak acid
Reduction of Ester with LiAlH4
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General reaction
O
COR'R
1) LiAlH4
2) H2O CH2 OHR R'OH+
Al-
H
H
H H
MechanismO
COR'R
O
COR'R
H
AlH3O
C OR'R H
AlH2HC
H
R
H OAlH2
OR'
H2O C
H
R
H OH R'OH+ Al(OH)4-+
Reduction cannot stop at the stage of aldehyde
Reduction of Carboxylic Acids with LiAlH4
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General reaction
O
COHR
1) LiAlH4
2) H2O CH2 OHR
Al-
H
H
H H
MechanismO
COR
H
O
COR
AlH2
H
O
CR H
AlH3H
AlH2O
C
O
R
H OAlH2
H2O C
H
R
H OH + Al(OH)4-
C
H
R
H OAlH3
Reduction cannot stop at the stage of aldehyde
Reduction of Amides with LiAlH4
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General reactionO
CNHR'R
1) LiAlH4
2) H2O CH2 NHR'R
H3Al H
MechanismO
CNR
H
R' O
CNR
AlH2
H
R'
N
CR H
AlH3H
AlH2O
R'
C
N
R
H OAlH2
R'
H2O C
H
R
H NH
+ Al(OH)4-
R'
C
H
R
H NAlH3
R'
Reduction of Ester with DIBAL
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General reaction
O
COR'R
1) DIBAL, -78°C2) H2O, -78°C
CHR
O
O
COR'R
1) DIBAL, -78° - 0°C2) H2O, 0°C
CH2 OHR R'OH+
Reduction can stop at the stage of aldehyde
Control of temperature is important for the reduction to stop at the stage of aldehyde.
Examples
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O 1) NaBH42) H2O
OH
H
O 1) NaBH42) H2O
OH
O
1) LiAlH42) H2O
OH
O
O 1) LiAlH42) H2O
OH+ OH
Examples
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O 1) DIBAL2) H2O
OH
Cl
O 1) NaBH42) H2O
OH
1) LiAlH42) H2OOH
O
NH
O 1) LiAlH42) H2O
NH
OH+ CH3OH
Examples
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1) LiAlH42) H2OOCH3
O
OH+ CH3OH
1) DIBAL, -78oC2) H2O, -78oCOCH3
O
H+ CH3OH
O
1)NaBH42) H2OOCH3
O
No reaction
1)NaBH42) H2ONHCH3
O
No reaction
Selective Reduction
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In most of the cases, hydride reducing reagents cannot reduce C=C.
1) LiAlH42) H2OOCH3
OO
OH
OH
1) NaBH42) H2OOCH3
OO
OCH3
OH O
H2 (1 atm)Pd/COCH3
OO
OCH3
O O
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Learning Check1. What could be the reagent needed for this transformation?
2. What could be reagent needed for this transformation?
O
OH
1) reagent2) HCl, H2O
O
OH
OH
(a) LiAlH4 (b) H2 (1 atm), Pt (c) NaBH4 (d) mCPBA (e) none of the above
O
MgBr1) reagent2) HCl, H2O
H
O(a) (b) (c) (d) CO2 (e) none of the above
OH
OH
H
OH
H
OCH3
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Learning Check3. What could be the product for the following reaction?
4. What could be the product for the following reaction?
Br 1) Mg, Et2O2) D2O
(a) (b) (c) (d) (e) none of the above
Product
CH2D
OD
OH
D
O
CH3
1) NaCN2) H+, H2O, heat
(a) (b) (c) (d)
(e) none of the above
Product
CH3
OH
CO2H
CN
O
CH3
OH
CN
CH3
OH
CH2NH2
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Learning Check
6. What could be the product for the following reaction?
5. What could be the reagent needed for the following reaction?O
NHCH3
reagent
(a) LiAlH4 then H+, H2O (b) NaBH4 then H+, H2O (c) H2 (1 atm), Pd (d) NaBH3CN then H+, H2O
(e) none of the above
NHCH3
O
OH
1) NaBH42) H+, removal of water
H
O
OH
(a) (c)
product
(e) none of the above
O
H
(b) OOHO O (d)
OH
OH
31
Learning Check
O
OCH3
1) CH3CH2MgBr(2 equivalents)2) HCl, H2O
OH(a) (b) (c) (d)
(e) none of the above
product ?
O
OH
O
CH2CH3
OH
CH2CH3
7. What could be the product for the following reaction?
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Learning CheckMain Menu
O
OCH3
1) DIBAL, -78oC2) HCl, H2O, -78oC
O
H
(a) (b) (c) (d)
(e) none of the above
product ?
O
OH
O
OCH3
OH
8. What could be the product for the following reaction?