23-1 preparation we have already covered these methods nucleophilic ring opening of epoxides by...
TRANSCRIPT
23-23-11
PreparationPreparation
We have already covered these methods• nucleophilic ring opening of epoxides by ammonia and
amines.• addition of nitrogen nucleophiles to aldehydes and
ketones to form imines• reduction of imines to amines
• reduction of amides to amines by LiAlH4
• reduction of nitriles to a 1° amine
• nitration of arenes followed by reduction of the NO2 group to a 1° amine
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PreparationPreparation
Alkylation of ammonia and amines by SN2 substitution.
• Unfortunately, such alkylations give mixtures of products through a series of proton transfer and nucleophilic substitution reactions.
CH3Br NH3
CH3NH3+Br
- (CH3)2NH2+Br
-(CH3)3NH+Br
-(CH3)4N+Br
-
+
+ + +
+ SN2
Methylammonium bromide
CH3Br NH3 CH3NH3+ Br-
polyalkylations
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Preparation via AzidesPreparation via Azides
Alkylation of azide ion.
-- + + -
Azide ion (a good nucleophile)
An alkyl azide
N NN NN NRN3-
RN3:: :
: : : :
Ph CH2ClK
+ N3
-
Ph CH2N31. LiAlH4
2. H2OPh CH2NH2
Benzyl chloride Benzyl azide Benzylamine
Overall Alkyl Halide Alkyl amine
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Example: Preparation via AzidesExample: Preparation via Azides
• Alkylation of azide ion.
Cyclohexene
trans-2-Amino-cyclohexanol
(racemic)
1,2-Epoxy-cyclohexane
trans-2-Azido-cyclohexanol
(racemic)
ArCO3H 1. K+ N3-
2. H2O
1. LiAlH42. H2ON3
OH
NH2
OH
O
Note retention of configuration, trans trans
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Reaction with HNOReaction with HNO22
Nitrous acid, a weak acid, is most commonly prepared by treating NaNO2 with aqueous H2SO4 or HCl.
In its reactions with amines, nitrous acid: • Participates in proton-transfer reactions.• A source of the nitrosyl cation, NO+, a weak
electrophile.
HNO2 H2O H3O+
NO2-
+ pKa = 3.37+
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Reaction with HNOReaction with HNO22
NO+ is formed in the following way.• Step 1: Protonation of HONO.
• Step 2: Loss of H2O.
• We study the reactions of HNO2 with 1°, 2°, and 3° aliphatic and aromatic amines.
H
H
N O+
OH
H+ OH N ONO OH
N O
+
+
+
+
The nitrosyl cation
(1) (2)
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Tertiary Amines with HNOTertiary Amines with HNO22
• 3° Aliphatic amines, whether water-soluble or water-insoluble, are protonated to form water-soluble salts.
• 3° Aromatic amines: NO+ is a weak electrophile and participates in Electrophilic Aromatic Substitution.
Me2N1. NaNO2, HCl, 0-5°C
2. NaOH, H2ON=OMe2N
N,N-Dimethyl-4-nitrosoanilineN,N-Dimethylaniline
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Secondary Amines with HNOSecondary Amines with HNO22
• 2° Aliphatic and aromatic amines react with NO+ to give N-nitrosamines.
N-H HNO2 N-N=O H2O
Piperidine N-Nitrosopiperidine
+ +
carcinogens
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Amines with HNOAmines with HNO22
Reaction of a 2° amine to give an N-nitrosamine.• Step 1: Reaction of the 2° amine (a nucleophile) with
the nitrosyl cation (an electrophile).• Step 2: Proton transfer.
N
H
N ON
H N=OH O
H
N
N=O
H O
H
H++
+ ++••
••
••••
••
• •
• •
• •••
• •• •
• •
• •
(1) (2)
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RNHRNH22 with HNO with HNO22
1° aliphatic amines give a mixture of unrearranged and rearranged substitution and elimination products, all of which are produced by way of a diazonium ion and its loss of N2 to give a carbocation.
Diazonium ion:Diazonium ion: An RN2+ or ArN2
+ ion
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1° RNH1° RNH22 with HNO with HNO22
Formation of a diazonium ion.Step 1: Reaction of a 1° amine with the nitrosyl cation.
Step 2: Protonation followed by loss of water.
:
:+
keto-enoltautomerism
A 1° aliphatic amine
An N-nitrosamine
R-NH2 N R-N-N=OO+ : :
:
H: :
:
A diazotic acid
R-N=N-O-H
: : ::
A diazotic acid
R-N=N-O-H
: : ::
+
A diazonium ion
++
A carbo- cation
O-HNR-N
H
N NR N N
H+
-H2O
R+
•• ••
••
••
••
••
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1° RNH1° RNH22 with HNO with HNO22
Aliphatic diazonium ions are unstable and lose N2 to give a carbocation which may:1. Lose a proton to give an alkene.
2. React with a nucleophile to give a substitution product.
3. Rearrange and then react by Steps 1 and/or 2.
(25%)
(5.2%)
(13.2%)
(25.9%) (10.6%)
0-5oC
NaNO2 , HClNH2OH
Cl
OH
+
+
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1° RNH1° RNH22 with HNO with HNO22
Tiffeneau-Demjanov reaction:Tiffeneau-Demjanov reaction: Treatment of a -aminoalcohol with HNO2 gives a ketone and N2..
CH2NH2
OH
HNO2
O
H2O N2
+ + +
A-aminoalcohol Cycloheptanone
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Mechanism of Tiffeneau-DemjanovMechanism of Tiffeneau-Demjanov
• Reaction with NO+ gives a diazonium ion.
• Concerted loss of N2 and rearrangement followed by proton transfer gives the ketone.
:OH
CH2NH2HNO2
O-H
CH2 N N+
(A diazonium ion)
-N2
O
+ CH2
OH
CH2
O H+ proton transfer to H2O
A resonance-stabilized cation Cycloheptanone
:
:
: : :
:
:
:
Similar to pinacol rearrangement
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Pinacol Rearrangement: an example of Pinacol Rearrangement: an example of stabilization of a carbocation by an adjacent stabilization of a carbocation by an adjacent lone pair.lone pair.
Overall:
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MechanismMechanism
Reversible protonation.
Elimination of water to yield tertiary carbocation.
1,2 rearrangement to yield resonance stabilized cation.
Deprotonation.
This is a protonated
ketone!
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1° 1° ArArNHNH22 with HNO with HNO22
The -N2+ group of an arenediazonium salt can be
replaced in a regioselective manner by these groups.
Ar-NH2HNO2
Ar-N2+ (-N2)
HCl, CuCl
H2O
HBF4
HBr, CuBr
KCN, CuCN
KI
H3PO2
Ar-I
Ar-F
Ar-H
Ar-Cl
Ar-Br
Ar-CN
Ar-OHSchiemannreaction
Sandmeyerreaction0-5°C
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1° ArNH1° ArNH22 with HNO with HNO22
A 1° aromatic amine converted to a phenol.
2-Bromo-4-methylaniline
2-Bromo-4-methylphenol
1. HNO2
2. H2O, heat
NH2
Br
CH3
OHBr
CH3
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1° ArNH1° ArNH22 with HNO with HNO22
Problem:Problem: What reagents and experimental conditions will bring about this conversion?
(1) (2) (3) (4)
CH3 CH3
NO2
COOH
NO2
COOH
NH2
COOH
OH
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1° ArNH1° ArNH22 with HNO with HNO22
Problem:Problem: Show how to bring about each conversion.
NH2
CH3
ClCH3
CCH3
N
NH2
CH3
ClCl
CH2NH2
CH3
CH3
ClCl
(5)
(6) (7)
(8)
(9)
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Hofmann EliminationHofmann Elimination
Hofmann elimination:Hofmann elimination: Thermal decomposition of a quaternary ammonium hydroxide to give an alkene.• Step 1: Formation of a 4° ammonium hydroxide.
(Cyclohexylmethyl)trimethyl-ammonium hydroxide
Silveroxide
(Cyclohexylmethyl)trimethyl- ammonium iodide
+
+ H2 OAg2O
AgI
CH2-N-CH3
CH3
CH3
I-
+
+CH2-N-CH3
CH3
CH3
OH-
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Hofmann EliminationHofmann Elimination
• Step 2: Thermal decomposition of the 4° ammonium hydroxide.
(Cyclohexylmethyl)trimethyl- ammonium hydroxide
TrimethylamineMethylene-cyclohexane
++CH2 (CH3 )3N H2 O
160°+CH2-N-CH3
CH3
CH3
OH-
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Hofmann EliminationHofmann Elimination
Hofmann elimination is regioselective - the major product is the least substituted alkene.
Hofmann’s rule:Hofmann’s rule: Any -elimination that occurs preferentially to give the least substituted alkene as the major product is said to follow Hofmann’s rule.
CH3
N(CH3)3 OH- CH2 (CH3)3N H2O++
heat+
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Hofmann EliminationHofmann Elimination
• The regioselectivity of Hofmann elimination is determined largely by steric factors, namely the bulk of the -NR3
+ group.
• Hydroxide ion preferentially approaches and removes the least hindered hydrogen and, thus, gives the least substituted alkene.
• Bulky bases such as (CH3)3CO-K+ give largely Hofmann elimination with haloalkanes.
+
E2 reaction (concertedelimination)
C C
H
N(CH3)3H
H H
CH
HC
H
CH3 CH2
HO-
N(CH3)3
HOH
CH3 CH2
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Cope EliminationCope Elimination
Cope elimination:Cope elimination: Thermal decomposition of an amine oxide.Step 1: Oxidation of a 3° amine gives an amine oxide.
Step 2: If the amine oxide has at least one -hydrogen, it undergoes thermal decomposition to give an alkene.
CH2 N-CH3
CH3
H2O2
O
CH3
CH2 N-CH3 H2O++
+
-
An amine oxide
O
CH3
CH2 N-CH3
H 100-150°CCH2 (CH3)2NOH+
N,N-Dimethyl-hydroxylamine
Methylene-cyclohexane
+
-
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Cope EliminationCope Elimination
• Cope elimination shows syn stereoselectivity but little or no regioselectivity.
• Mechanism: a cyclic flow of electrons in a six-membered transition state.
:O
heat+
-
Transition state
an alkene
N,N-dimethyl-hydroxylamine
C C
H NCH3
CH3N
CH3
CH3
OH
C C
: