Transcript 23 Amines
Amines
Chapter 23
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Structure & Classification
Amines are classified as:
•
1 °, 2°, or , 3° amines: Amines in which there are 1, 2, or 3 alkyl or aryl groups.
CH 3 - NH 2 Methylamine (a 1° amine) CH 3 - NH CH 3 Dimethylamine (a 2° amine) CH 3 CH 3 - N : CH 3 Trimethylamine (a 3° amine)
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Structure & Classification
Amines are further divided into aliphatic, aromatic, and heterocyclic amines:
•
Aliphatic amine: An amine in which nitrogen is bonded only to alkyl groups.
•
Aromatic amine: An amine in which nitrogen is bonded to one or more aryl groups.
N H 2 CH 3 N -H CH 3 CH 2 - N- CH 3 Aniline (a 1° aromatic amine) N-Methylaniline (a 2° aromatic amine) Benzyldimethylamine (a 3° aliphatic amine)
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Structure & Classification
•
Heterocyclic amine: An amine in which nitrogen is one of the atoms of a ring .
N N H H Pyrrolidine Piperid ine (h eterocyclic alip hatic amines) N N H Pyrrole Pyridine (h eterocyclic aromatic amin es)
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Nomenclature
Aliphatic amines: replace the suffix -e parent alkane by -amine .
of the NH 2 NH 2 2-Propanamine (S)-1-Phen yl ethanamine H 2 N NH 2 1,6-Hexanediamine
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Nomenclature
The IUPAC system retains the name aniline.
N H 2 N H 2 N H 2 N H 2 Aniline N O 2 4-Nitroaniline (
p-
Nitroaniline) CH 3 4-Methylaniline (
p-
Toluidine) OCH 3 3-Methoxyaniline (
m-
Anisidine)
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Nomenclature
Among the various functional groups, -NH 2 one of the lowest in order of precedence.
is OH H 2 N 2-A min oeth anol OH NH 2 (S )-2-Amin o-3-methyl 1-bu tanol H 2 N COOH 4-A minoben zoic acid Amine vs alcohol Amine vs acid
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Nomenclature
Common names for most aliphatic amines are derived by listing the alkyl groups bonded to nitrogen in one word ending with the suffix amine .
H CH 3 NH 2 NH 2 N Et 3 N Meth ylamine
tert -
Butylamine D icyclopen tylamine Triethylamin e
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Nomenclature
When four groups are bonded to nitrogen, the compound is named as a salt of the corresponding amine.
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Chirality of Amines
• •
Consider the unshared pair of electrons on nitrogen as a fourth group, then the arrangement of groups around N is approximately tetrahedral.
An amine with three different groups bonded to N is chiral and exists as a pair of enantiomers and, in principle, can be resolved.
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Chirality of Amines
•
In practice, however, they cannot be resolved because they undergo inversion, which converts one enantiomer to the other.
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Chirality of Amines
•
Pyramidal inversion is not possible with quaternary ammonium ions, and their salts can be resolved.
Cl N Et Me
S
Enan tiomer Cl N Et Me
R
Enan tiomer
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Physical Properties
Amines are polar compounds, and both 1 ° and 2° amines form intermolecular hydrogen bonds.
•
N-H- - -N hydrogen bonds are weaker than O-H- - -O hydrogen bonds because the difference in electronegativity between N and H (3.0 - 2.1 =0.9) is less than that between O and H (3.5 - 2.1 = 1.4).
Using bp as an indication of H bonding MW (g/mol) bp (°C) CH 3 CH 3 30.1
-88.6
CH 3 NH 2 31.1
-6.3
CH 3 OH 32.0
65.0
Increasing strength
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Basicity
All amines are weak bases, and aqueous solutions of amines are basic.
H CH 3 - N H + Methylamine H- O-H H + CH 3 - N- H O-H H Methylammonium hydroxide
•
It is common to discuss their basicity by reference to the acid ionization constant of the conjugate acid.
K
a = CH 3 NH 3 + + H 2 O [ CH 3 NH 2 ] [H 3 O + ] [CH 3 NH 3 + ] = CH 3 NH 2 2.29 x 10 -1 1 + H 3 O + p
K
a = 10.64
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Basicity
•
Using values of p
K
a , we can compare the acidities of amine conjugate acids with other acids.
CH 3 NH 2 + CH 3 COOH p
K
a 4.76
(s tronger acid) CH 3 NH 3 + p
K
a 10.64
(w eak er acid ) + CH 3 COO p
K
eq = -5.88
K
e q = 7.6 x 10 5
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Basicity-Aliphatic Amines
Aliphatic Amines
•
note that pK a + pK b = 14 Amine Ammonia Primary Amines methylamine ethylamine cyclohexylamine Secondary Amines dimethylamine diethylamine Tertiary Amines trimethylamine triethylamine Structure N H CH CH C 6 3 3 3 H N H CH 1 1 2 2 NH 2 NH 2 ( CH 3 ( CH 3 ) 2 N H CH 2 ) 2 NH ( CH 3 ( CH 3 ) 3 N CH 2 ) 3 N p
K
a 9.26
10.64
10.81
10.66
10.73
10.98
9.81
10.75
p
K
4.74
3.36
3.19
3.34
3.27
3.02
4.19
3.25
b Stronger bases
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Basicity-Aromatic Amines
Amine Aromatic A min es An iline Structure NH 2 p
K
a of Conju gate A cid 4.63
4-Methylanilin e CH 3 NH 2 5.08
Weaker bases 4-Ch loroaniline Cl NH 2 4.15
4-N itroanilin e O 2 N Heterocyclic Aromatic A min es Pyrid ine N N Imidazole N H NH 2 1.0
5.25
Intermediate 6.95
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Basicity-Aromatic Amines
•
Aromatic amines are considerably weaker bases than aliphatic amines.
NH 2 + H 2 O Cycloh exylamine NH 3 + OH Cyclohexylammonium hydroxid e p
K
a = 10.66
NH 2 + H 2 O A niline NH 3 + OH A niliniu m h yd roxide p
K
a = 4.63
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Basicity-Aromatic Amines
Aromatic amines are weaker bases than aliphatic amines because of two factors:
•
Resonance stabilization of the free base, which is lost on protonation.
H H H H H N H H + N + N + N H .
H .
H .
H .
.
H .
H . .
N un hybridized 2
p
orb ital of N H H nitrogen is
sp
2 h yb rid ized
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Basicity-Aromatic Amines
•
The greater electron-withdrawing inductive effect of the
sp
2 -hybridized carbon of an aromatic amine compared with that of the
sp
3 -hybridized carbon of an aliphatic amine.
And note the effect of substituents
Electron-releasing groups, such as alkyl groups, increase the basicity of aromatic amines.
Electron-withdrawing groups, such as halogens, the nitro group, and a carbonyl group decrease the basicity of aromatic amines by a combination of resonance and inductive effects.
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Example: Basicity-Aromatic Amines
3-nitroaniline is a stronger base than 4-Nitroaniline.
O 2 N NH 2 3-N itroanilin e p
K
a 2.47
O 2 N NH 2 4-N itroaniline p
K
a 1.0
de localiz ation of th e n itrog en lon e p air on to th e oxyg en atom s of th e n itro grou p O O + N NH 2 O + N O Cannot do this kind of resonance in 3 nitroaniline NH 2 +
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Basicity-Aromatic Amines
Heterocyclic aromatic amines are weaker bases than heterocyclic aliphatic amines.
N H Piperidine p
K
a 10.75
N Pyridin e p
K
a 5.25
N N H Imid azole pK a 6.95
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Basicity-Aromatic Amines
•
In pyridine, the unshared pair of electrons on N is not part of the aromatic sextet.
H .
H .
.
.
.
H .
N :
an
sp
2 h yb rid orbital; th e electron p air in this orbital is n ot a p art of the aromatic s extet
H H
nitrogen is
sp
2 h yb rid ized
•
Pyridine is a weaker base than heterocyclic aliphatic amines because the free electron pair on N lies in an
sp
2 hybrid orbital (33%
s
character) and is held more tightly to the nucleus than the free electron pair on N in an
sp
3 hybrid orbital (25%
s
character).
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Basicity-Aromatic Amines
Imidazole Which N lone pair is protonated? The one which is not part of the aromatic system.
This electron pair is not a part of the aromatic sextet This electron pair is a part of the aromatic sextet N N H Imidazole + H 2 O Aromaticity is maintained when imidazole is protonated H N + + N H Imidazolium ion OH -
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Basicity-Guanidine
Guanidine is the strongest base among neutral organic compounds.
NH NH 2 + H 2 N C NH 2 Guan idine + H 2 O H 2 N C NH 2 + OH Guanid inium ion p
K
a = 13.6
•
Its basicity is due to the delocalization of the positive charge over the three nitrogen atoms.
H 2 N + N H 2 C N H 2 + H 2 N N H 2 C N H 2 Three equivalent contributing structures H 2 N N H 2 C + N H 2
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Reaction with Acids
All amines, whether soluble or insoluble in water, react quantitatively with strong acids to form water-soluble salts.
OH HO NH 2 HO OH NH 3 + Cl + HCl H 2 O HO HO (
R
)-N orepinep hrine (on ly s ligh tly s olu ble in w ater) (
R
)-N orepinep hrine h yd roch loride (a w ater-s olu ble salt)
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Reaction with acids
Separation and purification of an amine and a neutral compound.
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Preparation
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 LiAlH 4 reduction of nitriles to a 1 ° amine nitration of arenes followed by reduction of the NO 2 group to a 1 ° amine
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Preparation
Alkylation of ammonia and amines by S N 2 substitution.
CH 3 Br + N H 3 S N 2 CH 3 NH 3 + Br Methylammonium bromide
•
Unfortunately , such alkylations give mixtures of products through a series of proton transfer and nucleophilic substitution reactions.
CH 3 Br + NH 3 CH 3 NH 3 + Br + (CH 3 ) 2 NH 2 + Br + (CH 3 ) 3 NH + Br + (CH 3 ) 4 N + Br polyalkylations
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Preparation via Azides
Alkylation of azide ion.
N 3 : N + N Azide ion (a good nucleophile) N : RN 3 R N An alkyl azide + N N : Ph CH 2 Cl Benzyl ch loride K + N 3 Ph CH 2 N 3 Ben zyl azid e 1 . LiAlH 4 2 . H 2 O Ph CH 2 NH 2 Benzylamin e Overall Alkyl Halide
Alkyl amine
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Example: Preparation via Azides
•
Alkylation of azide ion.
Cyclohexene A rCO 3 H O 1 . K + N 3 2 . H 2 O 1,2-Epoxy cyclohexane OH N 3
trans-
2-Azido cyclohexanol (racemic) OH 1 . LiA lH 4 2 . H 2 O N H 2
trans-
2-Amino cyclohexanol (racemic) Note retention of configuration,
trans
trans
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Reaction with HNO
2
Nitrous acid, a weak acid, is most commonly prepared by treating NaNO 2 or HCl.
with aqueous H 2 SO 4 HNO 2 + H 2 O H 3 O + + NO 2 p
K
a = 3.37
In its reactions with amines, nitrous acid:
• •
Participates in proton-transfer reactions.
A source of the nitrosyl cation, NO + , a weak electrophile.
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Reaction with HNO
2
NO + is formed in the following way.
• •
Step 1: Protonation of HONO.
Step 2: Loss of H 2 O.
H + + H O N O (1) H + O N O (2) H H O + + N H O N + O The nitrosyl cation
•
We study the reactions of HNO 2 aliphatic and aromatic amines.
with 1 °, 2°, and 3°
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Tertiary Amines with HNO
2
• •
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.
Me 2 N 1 . NaNO 2 , HCl, 2 . NaOH, H 2 O
N ,N
-D imethylaniline 0-5°C Me 2 N N=O
N,N
-D imethyl-4-nitrosoaniline
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Secondary Amines with HNO
2
•
2 ° Aliphatic and aromatic amines react with NO + to give
N
-nitrosamines.
carcinogens N-H + HNO 2 Piperidin e Mechanism: • • N H
N
-N itrosopiperidin e + • • + N O • • (1) • • H O • • H H + N • • • • N=O • • (2) N-N=O + H 2 O • • • • N • • N=O • • + • • H O + H H
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RNH
2
with HNO
2
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 N give a carbocation.
2 to
Diazonium ion: An RN 2 + or ArN 2 + ion
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1 ° RNH
2
with HNO
2
Formation of a diazonium ion.
Step 1: Reaction of a 1 ° amine with the nitrosyl cation.
R-NH 2 + A 1° aliphatic amine : + N O : H R-N-N=O : An N-nitrosamine keto-enol tautomerism R-N=N-O-H A diazotic acid Step 2: Protonation followed by loss of water.
R-N=N-O-H A diazotic acid H + •• R-N •• N R + N N A diazonium ion H •• + O-H - H 2 O R + A carbo cation + N N
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1 ° RNH
2
with HNO
2
(Aliphatic)
Aliphatic diazonium ions are unstable and lose N 2 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.
Cl (5.2%) OH N H 2 N aN O 2 , H Cl 0-5 o C (13.2%)
+
(25%) OH (25.9%)
+
(10.6%)
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1 ° RNH
2
with HNO
2
Tiffeneau-Demjanov reaction: aminoalcohol with HNO 2
OH
CH 2 NH 2 + HNO 2 Treatment of a
gives a ketone and N 2 .
O + H 2 O + N 2 A
-aminoalcohol Cycloheptan on e
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Mechanism of Tiffeneau-Demjanov
• •
Reaction with NO + gives a diazonium ion.
Concerted loss of N 2 and rearrangement followed by proton transfer gives the ketone. : OH CH 2 NH 2 HNO 2 : O-H CH 2 + N N : -N 2 (A diazonium ion) : O H + CH 2 + O H CH 2 proton transfer to H 2 O O A resonance-stabilized cation Cycloheptanone Similar to pinacol rearrangement
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Pinacol Rearrangement: an example of stabilization of a carbocation by an adjacent lone pair.
Overall:
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Mechanism
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 ° Primary Amines with HNO 2 (Aromatic)
The -N 2 + group of an arenediazonium salt can be replaced in a regioselective manner by these groups. H 2 O Ar-OH HBF 4 Ar-F Sch ieman n reaction HCl, CuCl Ar-Cl Ar-NH 2 HNO 2 0-5°C Ar-N 2 + (-N 2 ) HBr, CuBr Ar-Br San dmeyer reaction KCN, CuCN Ar-CN KI Ar-I H 3 PO 2 Ar-H
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1 ° ArNH
2
with HNO
2
A 1 ° aromatic amine converted to a phenol.
NH 2 Br CH 3 2-Bromo-4 methylanilin e 1 . HNO 2 2 . H 2 O, heat OH CH 3 Br 2-Bromo-4 meth ylp henol
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1 ° ArNH
2
with HNO
2 Problem: What reagents and experimental conditions will bring about this conversion?
CH 3 CH 3 COOH COOH COOH (1) (2) (3) (4) N O 2 N O 2 N H 2 OH
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1 ° ArNH
2
with HNO
2
Problem: Show how to bring about each conversion.
CH 3 Cl (5) CH 3 NH 2 (6) CH 3 C N (7) CH 3 CH 2 NH 2 (8) Cl CH 3 NH 2 (9) Cl Cl CH 3 Cl
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Hofmann Elimination
Hofmann elimination: Thermal decomposition of a quaternary ammonium hydroxide to give an alkene.
•
Step 1: Formation of a 4 ° ammonium hydroxide.
+ CH 3 I CH 2 -N - CH 3 + A g 2 O H 2 O CH 3 (Cyclohexylmethyl)trimethyl ammonium iodide Silver oxide + CH 3 OH CH 2 -N - CH 3 CH 3 (Cyclohexylmethyl)trimethyl ammonium hydroxide + A gI
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Hofmann Elimination
•
Step 2: Thermal decomposition of the 4 ° ammonium hydroxide.
CH 3 + CH 2 -N - CH 3 OH 160° CH 3 (Cyclohexylmethyl)trimethyl ammonium hydroxide CH 2 + ( CH 3 ) 3 N + H 2 O Methylene cyclohexane Trimethylamine
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Hofmann Elimination
Hofmann elimination is regioselective - the major product is the least substituted alkene.
CH 3 + N(CH 3 ) 3 OH heat CH 2 + (CH 3 ) 3 N + H 2 O
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.
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Hofmann Elimination
HO H C H CH 3 CH 2 H H C + N ( CH 3 ) 3 E2 reaction (concerted elimination) HO H H CH 3 CH 2 C C H H N ( CH 3 ) 3
• • •
The regioselectivity of Hofmann elimination is determined largely by steric factors, namely the bulk of the -NR 3 + group.
Hydroxide ion preferentially approaches and removes the least hindered hydrogen and, thus, gives the least substituted alkene.
Bulky bases such as (CH 3 ) 3 CO K + give largely Hofmann elimination with haloalkanes .
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Cope Elimination
Cope elimination: amine oxide.
Thermal decomposition of an Step 1: Oxidation of a 3 ° amine gives an amine oxide.
O CH 2 N-CH 3 + H 2 O 2 CH 2 + N-CH 3 + H 2 O CH 3 CH 3 An amin e oxid e Step 2: If the amine oxide has at least one
-hydrogen, it undergoes thermal decomposition to give an alkene.
H O CH 2 + N-CH 3 100-150°C CH 2 + (CH 3 ) 2 NOH CH 3 Meth ylen e cyclohexane
N,N
-D imethyl hydroxylamine
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Cope Elimination
• •
Cope elimination shows syn stereoselectivity but little or no regioselectivity.
Mechanism: a cyclic flow of electrons in a six membered transition state.
H C :O C N + CH 3 CH 3 Transition s tate heat C H O C N CH 3 CH 3 an alken e
N,N
-d imeth yl hydroxylamin e
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