Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
Soluzioni ai problemi proposti nel libro Capitolo 23
23.1 Le ammine sono classificate come 1o, 2o, o 3o secondo il numero di gruppi alchilici legati
all’atomo di azoto.
1o amine
2o amine
a. H N
2
1o
C6H5
H
N
N
H
NH2
b. CH3CH2O
2o amine
O
amine
O
NHCH3
O
2o amine
c.
N CH3
3o amine
23.2 L’atomo di N di un sale di ammonio quaternario è un centro stereogenico quando l’N è circondato
da quattro gruppi diversi. Tutti i centri stereogenici sono cerchiati.
CH3
OH
+
CH3
+
a. CH3 N CH2CH2 N CH2CH3
CH3
H
N
HO
CH3
OH
H
c.
b.
HO
H
N
O
N has 3 similar groups.
O
23.3
CH3
NH2
NH2
NH2
+ 3 more
resonance
structures
1.40 Å
1.47 Å
partial double
Because the lone pair on N can be delocalized
bond character
on the benzene ring, the C–N bond has partial double
bond character, making it shorter. Both the C and N atoms
must be sp2 hybridized (+ have a p orbital) for
delocalization to occur. The higher percent s-character
in both C and N shortens the bond as well.
The C–N bond is formed from two
sp3 hybridized atoms and the lone
pair is localized on N.
23.4
NHCH2CH3
a.
CH3CH2CH(NH2)CH3
2-butanamine
or
sec-butylamine
b.
(CH3CH2CH2CH2)2NH
dibutylamine
c.
N(CH3)2
N,N-dimethylcyclohexanamine
e.
N-ethyl-3-hexanamine
CH3
d.
f.
NHCH2CH2CH3
NH2
2-methyl-5-nonanamine
N-propyl-2-methylcyclopentanamine
23.5 Le ammine aromatiche sono definite come derivati dell’anilina.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
N-methylaniline
a.
b.
m-ethylaniline
NHCH3
3,5-diethylaniline
c.
NH2
CH3CH2
d. N,N-diethylaniline
N(CH2CH3)2
NH2
CH2CH3
CH2CH3
23.6 Un gruppo NH2 se è considerato un sostituente è detto gruppo amminico.
a. 2,4-dimethyl-3-hexanamine c. N-isopropyl-p-nitroaniline
e. N,N-dimethylethylamine g.
1-propylcyclohexylamine
NHCH(CH3)2
NH2
N
O2N
NH2
b. N-methylpentylamine
d. N-methylpiperidine
f. 2-aminocyclohexanone
h. p-butyl-N-ethylaniline
O
NHCH2CH3
NH3
NHCH3
N
23.7
a. CH3NHCH2CH2CH2CH3
e. (CH3CH2CH2)3N
NH2
1-octanamine
(octylamine)
c.
NH2
4,6-dimethyl-1-heptanamine
N
H
tripropylamine
N-methyl-1-butanamine
(N-methylbutylamine)
b.
i.
CH2CH3
2-ethylpyrrolidine
f. (C6H5)2NH
diphenylamine
N C(CH3)3
g.
j.
CH3CH2CH2CH(NH2)CH(CH3)2
2-methyl-3-hexanamine
k.
NH2
CH2CH3
3-ethyl-2-methylcyclohexanamine
N-tert-butyl-N-ethylaniline
CH3
d.
N
h. O
NH2
l.
N(CH2CH3)2
CH2CH2CH3
N-methyl-N-propylcyclohexanamine
23.8
4-aminocyclohexanone
N,N-diethylcycloheptanamine
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
a. cyclobutylamine
e. N-methylpyrrole
h. 3-methyl-2-hexanamine
NH2
NH2
N
b. N-isobutylcyclopentylamine
CH3
f. N-methylcyclopentylamine
N
i. 2-sec-butylpiperidine
NHCH3
H
c. tri-tert-butylamine
N
H
g. cis-2-aminocyclohexanol
N[C(CH3)3]3
j. (S)-2-heptanamine
NH2
d. N,N-diethylaniline
H NH2
OH
N(CH2CH3)2
23.9 [* indica un centro stereogenico]
a.
*
N
CH3
b.
CH2CH3
*
*N CH2CH2CH2CH3
CH3CH2CHCH
2CH2CH2
CH3
CH3
1 stereogenic center
2 stereoisomers
N
H
H CH3
CH3
N
CH3
CH2CH2CH2
H CH3
N
Cl
CH2CH2CH2
N
Cl
CH3 CH2CH3
C
CH3CH2
CH2CH2CH2CH3
CH2CH2CH2
H CH3
CH3 CH2CH3
C
CH3CH2
CH3
Cl
CH3 CH2CH3
C
CH3CH2
CH3
H CH3
H
CH3
2 stereogenic centers
4 stereoisomers
Cl
CH2CH2CH2CH3
CH3 CH2CH3
C
CH3CH2
CH2CH2CH2CH3
N
CH2CH2CH2
N
Cl
CH2CH2CH2CH3
23.10 Il pKa di molte ammine protonate è 10–11, il pKa dell’acido di partenza deve essere minore
di 10 perché l’equilibrio favorisca i prodotti. Le ammine pertanto vengono prontamente
protonate da acidi inorganici forti come HCl e H2SO4, ed anche da acidi carbossilici.
a. CH3CH2CH2CH2 NH2
+
HCl
pKa = –7
b. C6H5COOH
pKa = 4.2
+
(CH3)2NH
CH3CH2CH2CH2
NH3
+
Cl–
pKa ≈ 10
weaker acid
products favored
(CH3)2NH2
+
pKa = 10.7
weaker acid
products favored
C6H5COO–
c.
+ HO–
+ H 2O
N
H
N
H H
pKa ≈ 10
pKa = 15.7
weaker acid
reactants favored
23.11 Più debole è l’acido coniugato, maggiore è il suo pKa e più forte è la base. (I valori di pKa
sono dell'acido coniugato di una data ammina).
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
a. CH3NH2 (pKa = 10.7)
and
CH3CH2NH2 (pKa = 10.8)
weaker conjugate acid
stronger base
stronger conjugate acid
weaker base
b. (CH3CH2)3N (pKa = 11.0)
weaker conjugate acid
stronger base
and
(CH3)3N (pKa = 9.8)
stronger conjugate acid
weaker base
23.12 Le alchilammine 1o, 2o, e 3o sono più basiche di NH3 per l'effetto induttivo elettron
donatore dei gruppi R.
a. (CH3)2NH
and
NH3
b. CH3CH2NH2
1° alkylamine
stronger base
2° alkylamine
CH3 groups are electron donating.
stronger base
and
ClCH2CH2NH2
1° alkylamine
Cl is electron withdrawing.
weaker base
23.13 Le arilammine sono meno basiche delle alchilammine perché la coppia elettronica sull'N
è delocalizzato. Gruppi elettron donatori aumentano la densità elettronica del benzene,
rendendo le arilammine più basiche dell'anilina. Gruppi elettron attrattori sottraggono
densità elettronica dal benzene, rendendo le arilammine meno basiche dell'anilina.
NH2
NH2
NH2
NH2
a.
NH2
NH2
b.
O2N
CH3OOC
CH3O
electron
donating group
most basic
arylamine
intermediate
basicity
electron
withdrawing group
least basic
electron
withdrawing group
least basic
arylamine
intermediate
basicity
alkylamine
most basic
23.14 Le ammidi sono molto meno basiche delle ammine perché la coppia elettronica sull'N è
altamente delocalizzata.
CONH2
amide
least basic
NH2
arylamine
intermediate basicity
NH2
alkylamine
most basic
23.15
sp2 hybridized
more basic
CH3
a.
N
N
CH3
This N is also sp2 hybridized so
the electron pair can delocalize onto
the aromatic ring. Delocalization makes
this N less basic.
DMAP
4-(dimethylamino)pyridine
23.16
b.
H
N
sp3 hybridized N
CH3 stronger base
N
sp2 hybridized N
higher percent s-character
weaker base
nicotine
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
This electron pair is delocalized,
making it a weaker base.
Br
2
H
HO
H
NH2
stronger base
sp3 hybridized N
25% s-character
CH3O
a.
b.
N
N
stronger base
sp hybridized N
33% s-character
N
N
stronger base
sp3 hybridized N
25% s-character
sp2 hybridized N
33% s-character
This compound is similar to
DMAP in Problem 25.27a.
N(CH3)2
c.
23.17
a. (CH3CH2)2NH
or
c. HCON(CH3)2 or (CH3)3N
N
amide
alkylamine
weaker base stronger base
sp3 hybridized N sp2 hybridized N
stronger base
weaker base
b.
C6H5NHCH3
or
arylamine
weaker base
d. (CH3CH2)2NH
C6H5CH2NH2
alkylamine
stronger base
or
(ClCH2CH2)2NH
2° alkylamine
Cl is electron withdrawing.
weaker base
2° alkylamine
stronger base
23.18
a.
NH2
NH3
NH2
arylamine intermediate
least basic
basicity
b.
NH2
c.
alkylamine
most basic
O2 N
d.
N
H
(C6H5)2NH
N
H
N
Cl
electron
withdrawing group
least basic
CH3
intermediate
basicity
C6H5NH2
diarylamine
least basic
delocalized
sp2 hybridized N sp3 hybridized N
electron pair on N intermediate
most basic
least basic
basicity
arylamine
intermediate
basicity
23.19
sp3 hybridized N
stronger base
sp3 hybridized N
stronger base
NH2
O
a.
C
NH2
(CH3CH2)2NCH2CH2O
delocalized
electron pair
b.
HO
CH3
OH
N
sp2 hybridized N
NH2
NH2
electron
donating group
most basic
NH2
alkylamine
most basic
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
23.20
Nc
O
C
a.
N
NHNH2
Nb < Na < Nc
Na
Nc
N
NH2
b.
Nb < Na < Nc
N
Nb
Nb H
Na
Order of basicity: Nb < Na < Nc
Nb – The electron pair on this N atom is delocalized
on the O atom; least basic.
Na – The electron pair on this N atom is not
delocalized, but is on an sp2 hybridized atom.
Nc – The electron pair on this N atom is on an sp3
hybridized N; most basic.
Order of basicity: Nb < Na < Nc
Nb – The electron pair on this N atom is delocalized
on the aromatic five-membered ring; least basic.
Na – The electron pair on this N atom is not
delocalized, but is on an sp2 hybridized atom.
Nc – The electron pair on this N atom is on an sp3
hybridized N; most basic.
23.21
The para isomer is the weaker base because the electron pair on its NH2 group can be delocalized onto the
NO2 group. In the meta isomer, no resonance structure places the electron pair on the NO2 group, and fewer
resonance structures can be drawn:
O2N
NH2
O2N
NH2
NH2
O2N
NH2
O2N
O2N
NH2
meta
NH2
O2N
NH2
O2N
NH2
O
para
N
O2N
O
NH2
O
N
O
23.22
NH2
NH2
O2N
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
N
This two-carbon bridge makes it
difficult for the lone pair on N to
delocalize on the aromatic ring.
N
A
B
pKa of the conjugate acid = 5.2 pKa of the conjugate acid = 7.29 Resonance structures that place a double bond
between the N atom and the benzene ring are
stronger conjugate acid
weaker conjugate acid
destabilized. Since the electron pair is more
weaker base
stronger base
localized on N, compound B is more basic.
The electron pair of this arylamine
is delocalized on the benzene ring,
decreasing its basicity.
N
N
B
Geometry makes having a
double bond here difficult.
23.23 La reazione SN2di un alogenuro alchilico con NH3 o un’ammina forma un ammina.
NH3
Cl
a.
NH2
b.
excess
CH2CH2NH2
CH3CH2Br
CH2CH2N(CH2CH3)3
Br–
excess
23.24
O
O
NH
KOH
N
O
A
O
C6H5CH2Cl
N
O
B
COO
–OH
H2O
H2N
COO
O
C
23.25 La sintesi di Gabriel converte un alogenuro alchilico in un’ammina 1o attraverso un processo a
due stadi: sostituzione nucleofila seguita da idrolisi.
NH2
a.
Br
CH3O
b. (CH3)2CHCH2CH2NH2
NH2
CH3O
Br
c.
(CH3)2CHCH2CH2Br
23.26 I nitrili sono ridotti ad ammine 1o LiAlH4. I gruppi nitro sono ridotti ad ammine 1o
per azione di una varietà di agenti riducenti. Le ammidi 1o, 2o, e 3o sono ridotte ad
ammine 1o, 2o, e 3o rispettivamente, con LiAlH4.
O
a.
CH3CHCH2NH2
CH3CHCH2NO2
CH3
CH3
CH3CHC N
CH3CHCNH2
CH3
CH3
O
b.
CH2NH2
C N
CH2NO2
C
NH2
c.
NH2
NO2
C
N
O
NH2
23.27 Le ammidi 1o, 2o, e 3o sono ridotte ad ammine 1o, 2o, e 3o rispettivamente, con LiAlH4.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
CONH2
O
CH2NH2
a.
NHCH3
c.
NHCH3
O
b.
N
N
23.28
General reaction:
(CH3)2CHNH2
[H]
R C N
RCH2NH2
isopropylamine
The amine needs 2 H's here.
The C bonded to the N must have 2 H's to be formed by reduction of a nitrile.
23.29 L’amminazione riduttiva è un metodo a due stadi che converte aldeidi e chetoni in ammine
1o, 2o, and 3o. L’amminazione riduttiva sostituisce un C=O con un legame C–H ed uno C–N
.
a.
NHCH3
CH3NH2
CHO
c.
NaBH3CN
O
b.
O
(CH3CH2)2NH
N(CH2CH3)2
NaBH3CN
NH2
NH3
NaBH3CN
23.30
a. C6H5CH2CH2CH2Br
b. C6H5CH2CH2Br
NH3
excess
C6H5CH2CH2CH2NH2
[1] LiAlH4
C6H5CH2CH2CN
c. C6H5CH2CH2CH2NO2
Pd/C
[1] LiAlH4
[2] H2O
NaCN
H2
d. C6H5CH2CH2CONH2
[2] H2O
C6H5CH2CH2CH2NH2
e. C6H5CH2CH2CHO
C6H5CH2CH2CH2NH2
NH3
C H CH CH CH NH
NaBH3CN 6 5 2 2 2 2
C6H5CH2CH2CH2NH2
23.31
a. (CH3CH2)2NH
NH2
b.
N(CH3)2
c.
O
CH3
C
N(CH3)2
NH2
NHCH2CH3
O
or
d.
H
N
H
N
O
O
CH3
N
C
H
O
23.32 In una amminazione riduttiva, un gruppo alchilico sull’N deriva dal composto carbonilico. Il resto
della molecola deriva dall’NH3 o dall’ammina.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
NH2
a.
H
+ NH3
O
b.
H
N
H
C6H5
NH2
C6H5
or
O
(CH3CH2CH2)2NH
or
H
H
N
H
C
O
CH3CH2CH2
CH2CH3
O
d.
C6H5
O
O
c. (CH3CH2CH2)2N(CH2)2CH(CH3)2
H2N
H
N
H
NH2
H
23.33
NH2
a. C6H5
O
b.
NaBH3CN
(CH3)2NH
O
C6H5
c. C6H5
b.
CH2NH2
NH3
[1] LiAlH4
CH2NH2
[2] H2O
CONH2
CH2NH2
[1] LiAlH4
c.
[2] H2O
CHO
d.
NH3
NaBH3CN
CH2NH2
d.
O
excess
CN
C6H5
NH2
N(CH3)2
23.34
Br
NH3
NaBH3CN
NH
NaBH3CN
a.
CHO
CH2NH2
NaBH3CN
NH
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
CH3
CH2Br
e.
CH2NH2
Br2
NH3
hν
excess
COOH
CONH2
[1] SOCl2
f.
[2] NH3
NH2
g.
[2] H2O
CN
[1] NaNO2/HCl
CH2NH2
[1] LiAlH4
[2] H2O
[2] CuCN
NO2
HNO3
h.
CH2NH2
[1] LiAlH4
NH2
H2
then as in (g).
Pd/C
H2SO4
23.35 Le ammine attaccano i gruppi carbonilici per formare prodotti di addizione nucleofila o di
sostituzione.
CH3CH2CH2NH2
O
a.
O
NCH2CH2CH3
O
O
O
b. CH3 C O C CH3
COCl
c.
CH3CH2CH2NH2
CH3
CH3CH2CH2NH2
C
O
NHCH2CH2CH3
CH3
CONHCH2CH2CH3
C
(CH3CH2)2NH
N(CH2CH3)2
O
O
C
O
(CH3CH2)2NH
CH3
CH3
C
N(CH2CH3)2
CON(CH2CH3)2
COCl
(CH3CH2)2NH
23.36
N2+ Cl–
NH2
NaNO2
a.
HCl
CH3
b. CH3CH2 N CH3
H
NaNO2
c.
NaNO2
HCl
CH3CH2
HCl
N
H
CH3
N
NO
NaNO2
N CH3
d.
HCl
NO
–
N2 Cl
NH2
23.37
+
N N
Y
+
N N
H
H
N N
H
23.38
Y
+
N N
Y
H
+
Y
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Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
OH
NH2
N N
a.
b.
NH2
c.
HO
OH
N N
HO
HO
N N
OH
23.39 Per determinare quali reagenti sono necessari per sintetizzare un particolare azo composto, dividere
sempre la molecola in due componenti: uno ha un anello benzenico con lo ione diazonio, e
l'altro ha un anello benzenico con un gruppo fortemente elettron donatore.
O 2N
a. H2N
Cl
b. HO
N N
N N
CH3
O2N
H2N
Cl
HO
N2
CH3
N2
23.40
H
N
N(CH3)2
f. CH3I (excess), followed by Ag2O and Δ
N-ethylaniline
H H
N
a. HCl
+ CH2=CH2
O
Cl–
H H
N
g. CH3CH2COCl
N
CH3COO–
b. CH3COOH
O
h. The product in (g), then HNO3/H2SO4
O
N
c. (CH3)2C=O
N
N
O 2N
CH3
NO2
i. The product in (g), then [1] LiAlH4; [2] H2O
N
N
d. CH2O/NaBH3CN
e. CH3I (excess)
CH3 CH3
N
I–
j. The product in (h), then H2/Pd
O
O
N
NH2 H2N
23.41
N
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Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
CH3
NH2 p-methylaniline
AlCl3
f. CH3COCl/AlCl3
a. HCl
CH3
NH2
NH3 Cl–
O
CH3
b. CH3COCl
C CH3
NH
CH3
O
c. (CH3CO)2O
C CH3
NH
CH3
g. CH3COOH
CH3
NH3 CH3COO–
h. NaNO2/HCl
CH3
N2+Cl–
O
i. step (b), then CH3COCl/AlCl3
d. excess CH3I
e. (CH3)2C=O
N(CH3)3 I–
CH3
CH3
N C(CH3)2
b. CH3CH2CH2CH2NH2
c. CH3CH2CH2CH2NH2
d. CH3CH2CH2CH2NH2
e. CH3CH2CH2CH2NH2
f. CH3CH2CH2CH2NH2
23.43
ClCOC6H5
j. CH3CHO/NaBH3CN
CH3CH2CH2CH2NHCOC6H5
O=C(CH2CH3)2
CH3CH2CH2CH2N=C(CH2CH3)2
[1] CH3I (excess)
[2] Ag2O
[3] Δ
C6H5CH2Br
CH3CH2Br
CH3I
excess
C CH3
NH
C O
CH3
23.42
a. CH3CH2CH2CH2NH2
CH3
CH3CH2CH=CH2
CH3CH2CH2CH2NHCH2C6H5
CH3CH2CH2CH2NHCH2CH3
[CH3CH2CH2CH2N(CH3)3]+I–
CH3
NHCH2CH3
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Janice Gorzynski Smith
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a. CH3(CH2)6NH2
CH3
[1] CH3I (excess)
[2] Ag2O
[3] Δ
[2] Ag2O
NH2
[3] Δ
[2] Ag2O
[3] Δ
CH3
CH3
CH3
CH3
CH2
major product
CH3
major product
[1] CH3I (excess)
N
H
[1] CH3I (excess)
(E + Z)
e.
c.
NH2
CH3
[1] CH3I (excess)
b.
d.
CH3(CH2)4CH=CH2
β
N
H
CH2=CH2
major product
+ (CH3)2CHN(CH3)2
[2] Ag2O
[3] Δ
β
[1] CH3I (excess)
β
[2] Ag2O
[3] Δ
+
N(CH3)2
major product
(CH3)2N
CH2=CHCH3 + CH3CH2N(CH3)2
+
(E + Z)
(CH3)2N
(E + Z)
23.44
N(CH3)2
A: HN(CH3)2
mild acid
O
B: NH2CH2CH2CH3
NCH2CH2CH3
mild acid
OH
D: H2SO4
C: NaBH4/CH3OH
E: mCPBA
NH2
H: [1] CH3I (excess)
G: NH3
[2] Ag2O/Δ
NaBH3CN
O
O
Br
I: Br2
CH3COOH
23.45
OH
O
NHCH2CH2CH2CH3
J
NH2CH2CH2CH2CH3
F: NH2CH3
NHCH3
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
a.
NH3
CH2CH2Cl
f.
CH2CH2NH2
C6H5CH2CH2NH2 + (C6H5CO)2O
C6H5CH2CH2NHCOC6H5
excess
O
b.
–OH
Cl
N
H2O
CO2
g.
CO2
h.
NO2
Sn
Br
NH2
[2] H2O
CN
CONHCH2CH3
NaBH3CN
N CH2C6H5
O
+
N
N
H
[1] LiAlH4
d.
e.
NH + C6H5CHO
i.
HCl
N N O
HCl
NH2
O
c.Br
NaNO2
NH
j. CH3CH2CH2 N CH(CH3)2
CH2NH2
[1] LiAlH4
[1] CH3I (excess)
[2] Ag2O
[3] Δ
H
CH2NHCH2CH3
CH3CH=CH2
[2] H2O
(CH3)2NCH(CH3)2
CH3CH2CH2N(CH3)2
23.46
N2+ Cl–
Cl
a. H2O
HO
H
Cl
d. CuBr
Cl
b. H3PO2
Br
A
Cl
h. C6H5NH2
N N
NH2
N N
OH
Cl
NC
Cl
F
Cl
e. CuCN
i. C6H5OH
Cl
Cl
c. CuCl
Cl
f. HBF4
I
Cl
j. KI
I
Cl
g. NaI
23.47
Under the acidic conditions of the reaction, aniline is first protonated to form an ammonium salt that has a
positive charge on the atom bonded to the benzene ring. The –NH3+ is now an electron withdrawing meta
director, so significant amounts of meta substitution occurs.
NH2
H OSO3H
NH3
+ HSO4–
This group is now a meta director.
23.48
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
N N
aryl diazonium salt
N N
N N
N N
N N
R N N
alkyl diazonium salt
The N2+ group on an aromatic ring is stabilized by
resonance, whereas the alkyl diazonium salt is not.
23.49
Br
a. Br
H H Na+
N
Br
OH
H
N
Br
N
H CH2CH3
Br
Br
+ CH3CH2NH2
+ H2O
Na+
OH
N
+ H2O + NaBr
CH2CH3
H
O
O
H
N
b.
NH2
O
H
N
proton
transfer
H
OH2
proton
source
N
H3B–H
H
H
N
+ H2O
H
N
+ BH3
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
23.50
NaNO2
HCl/H2O
NH2
NH2
Cl
H
+ HCl
+
N O
N N O
N N O
H
H
N N O H
H
+
H Cl
+
H Cl
+
OH
+
N N O
OH
+
–
N N OH2 + Cl
N N O H
H
Cl
+
O H
OH
Cl–
+ HCl
H
+
H
+
1,2-H shift
B
A
O H
H
H OH
Cl–
B
H Cl–
+ HCl
N N
+
+
+N N
A H OH
OH
+ HCl
+ H2O