Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
Soluzioni ai problemi proposti nel libro Capitolo 20
20.1 Il numero di legami C–N determina la classificazione come ammidi 1o, 2o, o 3o.
2° amide
O
H
N
N
2° amide
N
OCH3 O
O
3° amide
S
H
N
N
N
OCH3 O
3° amide
20.2 All’aumentare della basicità di Z, aumenta la stabilità di RCOZ, a causa dell’aumento della
stabilizzazione che deriva dalla risonanza.
R
O
O
C
C
R + Br
Br
O
R
C
+
Br
The basicity of Z determines how much
this structure contributes to the hybrid.
Br– is less basic than –OH, so RCOBr
is less stable than RCOOH.
20.3
O
CH3
Cl
CH3
NH2
CH3
H
C
O
Cl
CH3
O
O
C
C
NH2
H
C
Cl
NH2
This resonance structure contributes little to the hybrid
since Cl– is a weak base. Thus, the C–Cl
bond has little double bond character, making it similar in
length to the C–Cl bond in CH3Cl.
This resonance structure contributes more to the hybrid
since –NH2 is more basic. Thus, the C–N bond
in HCONH2 has more double bond character, making
it shorter than the C–N bond in CH3NH2.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
20.4
a. (CH3CH2)2CH COCl
O
d.
redraw
H
O
C
OCH2CH3
alkyl group = ethyl
Cl 2-ethylbutanoyl chloride
acyl group =
formate
2-ethyl
b.
ethyl formate
O
C6H5COOCH3
e. CH3CH2
redraw
C
O
O
C
benzoic propanoic anhydride
O
OCH3
methyl benzoate
acyl group =
propanoic
alkyl group = methyl
acyl group =
benzoate
c. CH3CH2CON(CH3)CH2CH3
redraw
f.
O
acyl group =
benzoic
CN
3-ethylhexanenitrile
6 carbon chain =
hexanenitrile
N
N-ethyl-N-methyl
N-ethyl-N-methylpropanamide
acyl group =
propanamide
20.5
a. 5-methylheptanoyl chloride
d. N-isobutyl-N-methylbutanamide g. sec-butyl 2-methylhexanoate
O
O
O
N
Cl
b. isopropyl propanoate
O
h. N-ethylhexanamide
e. 3-methylpentanenitrile
O
CN
c. acetic formic anhydride
O
f. o-cyanobenzoic acid
O
O
H
O
O
CH3
OH
CN
O
N
H
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
20.6
O
Cl
a.
g.
2,2-dimethylpropanoyl chloride
N
O
N-cyclohexylbenzamide
H
O
b.
h.
cyclohexyl pentanoate
O
m-chlorobenzonitrile
Cl
CN
c.
O
O
i.
Cl 3-phenylpropanoyl chloride
isobutyl 2,2-dimethylpropanoate
O
d.
2-ethylhexanenitrile
O
j.
C
CN
O
Cl
O
e.
cis-2-bromocyclohexanecarbonyl chloride
Br
O
O
k.
N
cyclohexanecarboxylic anhydride
N,N-diethylcyclohexanecarboxamide
O
O
f.
cyclopentyl
cyclohexanecarboxylate
l.
phenyl phenylacetate
O
O
20.7
a. propanoic anhydride
O
e. isopropyl formate
H
b. α-chlorobutyryl chloride
O
O
O
O
i. benzoic propanoic anhydride
O
O
O
f. N-cyclopentylpentanamide
O
j. 3-methylhexanoyl chloride
NH
O
Cl
O
Cl
c. cyclohexyl propanoate
k. octyl butanoate
O
d. cyclohexanecarboxamide
h. vinyl acetate
O
O
O
NH2
O
CN
O
C
Cl
g. 4-methylheptanenitrile
O
l. N,N-dibenzylformamide
CH3
O
H
20.8 Classificare i composti usando le regole della risposta 20.13.
N
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
O
O
a.
OCH2CH2CH3
NH2
−NH
2
Cl
−OCH CH CH
2
2
3
stongest base
least reactive
−Cl
weakest base
most reactive
intermediate
O
b.
O
O
O
F3C
O
O
ester
least reactive
anhydride
intermediate
O
O
O
O
CF3
anhydride with electron
withdrawing F's
most reactive
O
O
c.
OH
−
SH
Cl
−
OH strongest base
least reactive
−
SH
intermediate
Cl weakest base
most reactive
20.9
Better leaving groups make acyl compounds more reactive. A has an electron withdrawing NO2
group, which stabilizes the negative charge of the leaving group, whereas B has an electron
donating OCH3 group, which destabilizes the leaving group.
O
CH3
C
O
O
CH3
NO2
A
C
O
OCH3
B
an electron withdrawing substituent an electron donating substituent
O
O
O
N
O
N
O
O
O
OCH3
OCH3
O
one possible
resonance structure
leaving group from A
one possible
resonance structure
leaving group from B
Adjacent negative charges destabilize
the leaving group.
Delocalizing the negative charge on the
NO2 stabilizes the leaving group
making A more reactive than B.
20.10
resonance structures for the leaving group
O
O
C N
N
R
imidazolide
Nu
R C N
Nu
O
N
R
C
Nu
N
N
N
N
N
N
N
N
N
The leaving group is both resonance stabilized and aromatic
(6 electrons), making it a much better leaving group than
exists in a regular amide.
N
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
20.11
Reaction as an acid:
O
CH3
O
B
C
CH3
NH2
O
C
NH
CH3
C
CH3CH2
B
NH2
NH
CH3CH2
NH
no resonance stabilization
of the conjugate base
These two resonance structures
make the conjugate base more stable,
and therefore CH3CONH2 a stonger acid.
Reaction as a base:
CH3
O
O
C
C
CH3
NH2
CH3CH2
This electron pair is
localized on N.
NH2
NH2
This electron pair is delocalized by resonance,
making it less available for electron donation.
Thus CH3CONH2 is a much weaker base.
20.12 Per disegnare i prodotti di queste reazioni di sostituzione nucleofila acilica, individuare il
nucleofilo ed il gruppo uscente. Poi sostituire il gruppo uscente con il nucleofilo e disegnare un
prodotto neutro.
nucleophile
O
a.
CH3
C
O
CH3OH
Cl
nucleophile
CH3
C
O
+ HCl
OCH3
b. CH
3
leaving group
C
O
NH3
OCH2CH3
CH3
C
NH2
+ HOCH2CH3
leaving group
20.13 Migliore è il gruppo uscente, più reattivo risulta il derivato dell'acido carbossilico. La base più
debole è il miglior gruppo uscente.
O
O
C
C
a.
−NH
NH2
stongest base
least reactive
b.
−
20.14
C
OCH3
−OCH
3
2
CH3CH2
O
−Cl
weakest base
most reactive
intermediate
O
O
O
C
C
C
NHCH3
NHCH3 stongest base
least reactive
CH3CH2
−
OH
OH
intermediate
Cl
CH3CH2
−
O
O
C
CH2CH3
OOCCH2CH3 weakest base
most reactive
I composti acilici più reattivi possono essere convertiti nei composti acilici meno reattivi.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
a.
CH3COCl
c.
CH3COOH
more reactive
YES
b. CH3CONHCH3
less reactive
NO
CH3COCl
more reactive
CH3COOCH3
NO
less reactive
CH3COOCH3
less reactive
more reactive
20.15
CH3
O
O
O
O
C
C
C
C
O
CH3
Cl3C
O
CCl3
acetic anhydride
trichloroacetic anhydride
The Cl atoms are electron withdrawing,
which makes the conjugate base (the
leaving group, CCl3COO–) weaker
and more stable.
20.16
O
a.
O
H2O
O
OH
+
N
–
H Cl
pyridine
Cl
O
b.
CH3COO−
+ NH4+Cl–
NH2
NH3
c.
excess
O
O
O
+ Cl–
CH3
(CH3)2NH
N(CH3)2
d.
+ (CH3)2NH2 Cl–
excess
20.17 Il meccanismo ha 3 stadi: [1] attacco nucleofilo dell'O; [2] trasferimento di un protone; [3]
eliminazione del gruppo uscente Cl− per formare il prodotto.
O
O
Cl
CH3OH
O
Cl
OCH3
O CH3
O CH3
pyridine
O
Cl
H
H
N
N
20.18
O
O
O
O
H2O
a.
O
OH
NH3
HO
c.
pyridine
O
b.
CH3OH
O
O
HO
NH2
NH4
O
excess
O
CH3
O
O
(CH3)2NH
d.
excess
O
N(CH3)2
O
(CH3)2NH2
Cl
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
20.19
La reazione di un acido carbossilico con cloruro di tionile lo converte in un cloruro
dell'acido.
O
a. CH3CH2
O
SOCl2
C
OH
CH3CH2
C
O
C
b.
Cl
O
[1] SOCl2
OH
O
[2] (CH3CH2)2NH (excess)
C
C
Cl
N(CH2CH3)2
20.20
COOH + CH3CH2OH
a.
H2SO4
C
OCH2CH3
+ H2O
O
O
COOH
b.
+
OH
C
H2SO4
+ H2O
O
O
COOH
c.
+
C
O
+ CH3OH
O
H2SO4
d. HO
Na+
O
NaOCH3
OH
+ H2O
O
20.21
O
O
C
CH318OH
OH
C
18
OCH3
+ H2O
20.22
O
HO
C
OH
O
H
H−OSO3H
C
HO
O
HO OH
H
O
–
+ HSO4
HO OH
H
O
–OSO H
3
O
O H
O
+ H2SO4
H−OSO3H
H2O +
O
HSO4–
HO OH2
O
+ HSO4–
20.23
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
O
a. O
CH3NH2
+
O
O
CH3NH2
O− NH3CH3
OH
b.
c.
DCC
O
O
CH3NH2
NHCH3
Δ
OH
NHCH3
OH
+ H2O
20.24
CH3CH2CH2CH2COCl
NH3
H 2O
a.
CH3CH2CH2CH2COOH
pyridine
b.
excess
CH3CH2CH2CH2CONH2
(CH3CH2)2NH
CH3CH2OH
e.
CH3CH2CH2CH2COOCH2CH3
pyridine
O
CH3COO−
c.
d.
CH3CH2CH2CH2
O
O
excess
CH3CH2CH2CH2CON(CH2CH3)2
C6H5NH2
CH3
f.
excess
CH3CH2CH2CH2CONHC6H5
20.25
O
O
O
a.
SOCl2
d.
no reaction
H2O
2
no reaction
O
O
b.
NaCl
(CH3CH2)2NH
OH
e.
excess
N(CH2CH3)2
O
O
CH3OH
c.
O H2N(CH2CH3)2
OCH3
O
O
f.
OH
CH3CH2NH2
excess
NHCH2CH3
O
O
20.26
H3NCH2CH3
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
O
OH
O
NaHCO3
a.
+ H2CO3
g.
O Na
O
NaOH
b.
+ H 2O
h.
c.
O
O
OH
[1] NaOH
i.
O
SOCl2
OCH3
CH3OH
–
O Na
O
CH3OH
H2SO4
[2] CH3COCl
O
Cl
d.
NaCl
no reaction
j.
O
CH3NH2
DCC
NH3
e.
O
O NH4
k.
NH2
l.
O
NH3
f.
Δ
NHCH3
[1] SOCl2
(1 equiv)
O
O
O
[2] CH3CH2CH2NH2
NHCH2CH2CH3
O
[1] SOCl2
[2] (CH3)2CHOH
OCH(CH3)2
20.27
O
a.
SOCl2
b.
H3O+
O
no reaction
d.
NH3
O
O
O
H2O/–OH
c.
HO
NH2
OH
HO
O
HO
e. CH3CH2NH2
NHCH2CH3
O
20.28
NH2
a.
H3O+
CH2COOH
b.
H2O/–OH
CH2COO
O
20.29
a.
H3O+
COOH
d.
[1] CH3CH2Li
[2] H2O
CN
b.
H2O/–OH
COO
[2] H2O
C
[2] H2O
O
[1] LiAlH4
CH2NH2
f.
O
H
e. [1] DIBAL-H
CH3
c. [1] CH3MgBr
O
[2] H2O
HO
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
20.30
O
COOH
C
SOCl2
a.
pyridine
OH
g.
[2] H2O/–OH
h.C6H5CH2COOH
+
O
[1] CH3CH2CH2MgBr
C
C6H5
[2] H2O
d. (CH3)2CHCOOH +
N
C6H5
N
H
(excess)
c. C6H5CN
(CH3)2CH
C
NHCOCH3
O
COOH
O
(CH3CO)2O +
NH2
NHCOCH3
(excess)
NH2
O
+ CH3COO– H3N
O
O
f.
HOOC
OCH(CH3)CH2CH3
+
O
C6H5CH2CH2CH2COOH
Δ
j.
O
–OH
H3O+
O
k.
H2O
+
H3O
OH
C6H5CH2CH2COOCH2CH3
l.
OH
H2O
–OH
C6H5CH2CH2COO–
+
HOCH2CH3
O
O
20.31
R
OH
OH
OH
O H
O H
C
C
C
C
C
OR'
R
R
OR'
OR'
R
OH
product of step [1]
R
O H
R
OH
C
OH
product of step [5]
20.32
O
C
H−OSO3H
O
CH3
O
[2] CH3CH2CH2CH2NH2
[3] LiAlH4
C6H5CH2CH2NHCH2CH2CH2CH3
[4] H2O
CH2CH2CH3
CH3
e.
CH3CH2CH2CH2
[1] SOCl2
C6H5CH2CH2CH2CN
i.
O
H2SO4
CH3CH2CHOH
CH3CH2CH2CH2Br
OH
O
b. C6H5COCl
O
[1] NaCN
Cl
OH
OH
H−OSO3H
C
O
CH3
C
OH
+ HSO4–
OH
O
CH3
C
OH H
O
CH3
C
OH
O
CH3
OH
H
+ HSO4–
H2O
+ HSO4–
O
C
O H
O
H
+ H2SO4
C
OH
HSO4–
CH3OH
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
20.33 In queste condizioni l'estere è idrolizzato per formare un acido carbossilico e un alcol.
O
a.
CH3
C
O
[1] −OH
OCH2CH3
[2] H3O+
COOCH(CH3)2
b.
O
C
CH3
OH
OCH3 [2] H3O+
OH
+ HOCH3
COOH
[1] −OH
+ HOCH(CH3)2
[2] H3O+
OCH3
O
[1] −OH
c.
+ HOCH2CH3
OCH3
20.34
O
a.
CH3CH2
–OH/H O
2
C 18
OCH3
18
O
O
CH3CH2
+
C
H18
b.
OCH3
O–
CH3CH2
C
18
O
–
OH/H2O
OCH3
CH3CH2
C
O
CH3CH2
O
+ HOCH3
20.35
CH2OCO(CH2)15CH3
CH2 OH
Na+ −OOC(CH2)15CH3
CH
Na+ −OOC(CH2)15CH3
hydrolysis
CHOCO(CH2)15CH3
CH2OCO(CH2)7CH=CH(CH2)7CH3
OH
Na+ −OOC(CH2)7CH=CH(CH2)7CH3
CH2 OH
cis
glycerol
cis
soap
20.36
O
O
NH
H2O/
OH
OH
O
O
N H
H
O
O
NH
NH2
20.37
NaCN
a. CH3CH2CH2 Br
CH3CH2CH2
CN
H2O/–OH
c.
CN
CN
COOH
H2O/H+
b.
COO
CN
COOH
20.38
a.
CH3
O
OH
NH
C
C
C
NH2
O
b.
20.39
C
CH3
c.
NH
OH
NHCH3
C
NCH3
CH3CH2
OH
NH2
CH3CH2
C
O
C 18
O
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
a. CH3CH2 Br
[1] NaCN
[2] LiAlH4
[3] H2O
CH3CH2
b. CH3CH2CH2 CN
CH2NH2
O
[1] DiBAL–H
CH3CH2CH2
C
[2] H2O
H
20.40
Br NaCN
CN
H3O+
COOH SOCl2
A
COCl [1] (CH3)2CuLi
C
[2] H2O
O
D
B
[1] LiAlH4
CH3
[2] H2O
CH3
E
OH/H+
PCC
CH2NH2
CO2CH3 [1] DIBAL–H
CHO [1] CH3Li
[2] H2O
C
OH
[2] H2O
G
F
H
[1] LiAlH4
(CH3CO)2O
[2] H2O
CH2OTs NaCN
CH2OH
CH2NHCOCH3
CH2CN
TsCl/pyridine
I
K
J
L
[1] CH3MgBr
[2] H2O
O
M
20.41
O
H D
C CH3
a.
CH3COCl
OH
c.
O
pyridine
CH3
CH3
b.
Br
H D
20.42
CH3
NaCN
C
H D
CH3CH2OH
H+
COOH
CH3
O
CN
D H
d.
CH3
C
CH3
+
Cl C H
6 5
C
H
NH2
(2 equiv)
C
CH3
C6H5
C
COOCH2CH3
CH3
C
H +
H
C6H5
NH
NH3
+
C
Cl–
CH3
O
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
O
a.
Cl
Cl
Cl
CH3NH
excess
O
C
c.
NHCH3
Cl
O
O
b.
O
CH3NH2
C
Cl
O
O
H2O
excess
Cl
HO
C
+ HCl
OH
Reaction with H2O vapor in the lungs
results in the formation of HCl, increasing
the acidity of the lungs.
HOCH2CH2OH
C
C
O
All H's on these C's
are identical.
20.43
O
H 2N
H
N
H
H
COOH
O
H 2O
OCH3
H 2N
H
O
H
OH
H2N
OH
+ CH3OH
O
COOH
phenylalanine
aspartame
20.44
O
C
a.
Cl
O
C Cl
C Cl
NH2NH
NH2NH2
b.
O
NHNH2 + Cl
+ HN
N
HO
C
NH2NH
H
O
O
O
O
O
O
O
O
HO
OH
O
OH
O
c.
H–OSO3H
OH
OH
OH
OH
OH
HO
HO
OH
O H
OH H–OSO3H
HO
OH2
O
O
–
+ HSO4–
HSO4
+ H2SO4
HSO4–
O
H2SO4
+
O
O H
O
+ H2O
20.45
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
H+
O
CH3
C
OH
OH
C
CH3
OH
OH
CH3
OH
CH3
C OH
H
OH
C OH2
CH3
18
O H
O H
18
CH3
18
18
+ H2O
OH
H+
C OH
18
O H
O H
A
+ H3O+
+ H2O
+ H2O
Two possibilities for A:
CH3
OH
O H
C
C
CH3
18
O H
C
O
H2O
18
CH3
OH
C
18
+ H3O+
OH
A
CH3
OH
OH
C
C
CH3
18
O H
OH
18
O H
CH3
H2O
C
18
+ H3O+
O
A
20.46
O
C
OH
H OSO3H
C
OCH3
OCH3
OH
C OCH3
C OCH3
OCH2CH2CH2CH3
OCH2CH2CH2CH3
H OSO3H
H
CH3CH2CH2CH2OH
+ HSO4–
OH
HSO4
HSO4
O
C
O H
OCH2CH2CH2CH3
+ H2SO4
C
C
OCH2CH2CH2CH3
OCH2CH2CH2CH3
+ CH3OH
20.47
OH H
OCH3
+ HSO4–
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
O
R
O
C
–
O R' +
OH
H2O
SN2
This bond is not broken.
H CH3
O
CH3
C
O R CH2CH3
X
C
mechanism
CH3
C
+ R'OH
O
H CH3
O
accepted
C
R
HO
O
H CH3
O
C
CH2CH3
(R)-2-butanol
CH3
C
O
X
C
CH2CH3
According to the accepted
mechanism, the stereochemistry
around the stereogenic center is
retained in the product.
This bond is cleaved.
H CH3
O
CH3
C
C
O
–
OH
CH3
H2O
SN2 alternative
CH2CH3
X
CH3 H
O
C
O
CH3CH2
C
OH
(S)-2-butanol
This SN2 mechanism would form
the product of inversion leading to
(S)-2-butanol. Since (R)-2-butanol
is the only product formed, the SN2
mechanism does not occur during
ester hydrolysis.
20.48
Reaction in base:
O
C
CH3
O
C
O
NH2
C
O
NH
H
A
O
CH3
or
O
C
Reaction in acid:
O
N
H
B
C
H A
CH3
OH
N
H
A
Intermediated X can be re-converted
to starting material B or go on to form
A by a stepwise process.
O
H
A
C
CH3
NH
C
O
C
CH3
CH3
poorer leaving group
O
O
C
OH
NH
CH3
O
O
NH3
20.49
C
NH
CH3
Once this product forms it cannot
revert to B since RNH3+ no longer
has a lone pair so it is not a
nucleophile.
N
H
CH3
B
X
O
C
O
This product is favored since the
negatively charged O atom is a weaker
base and therefore better leaving group
than the negatively charged N atom.
CH3
N
H
O
OH
C
N
H
Once this intermediate forms,
there are two possibilities.
B
OH
O
O
N
H
OH
O
CH3
NH2
CH3
OH
A
O
CH3
C
CH3
O H
NH2
O
NH2
A
C
OH
H A
C
O
CH3
H A
This NH2 is more basic than the OH
group in B, so it is protonated in acid.
A
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
O OCH2CH3
NH2 C OCH2CH3
O
NH2
OH
+
CH3CH2O
C
O
OCH2CH3
O
H
+ HB+
B
X
diethyl carbonate
(any base)
O
O
HN
O OCH2CH3
NH2 C OCH2CH3
O
+ CH3CH2O–
HN
B
OCH2CH3
H
O
H N
O
O
NH2 C OCH2CH3
OCH2CH3
O
O
+ HB+
Y
+ CH3CH2O–
20.50
sp3 C
sp2 C
O
RCH2
Cl
less electrophilic C
more crowded C since it is
surrounded by four atoms
R
C
O
Cl
R
more electrophilic C
due to electron withdrawing O
more reactive
C
Cl
This resonance structure illustrates
how the electronegative O atom
withdraws more electron density
from C.
The sp2 hybridized C of RCOCl is much less crowded,
and this makes nucleophilic attack easier as well.
20.51 L'esterificazione di Fischer è il trattamento di un acido carbossilico con un alcol in presenza
di un catalizzatore acido per formare un estere.
a. (CH3)3CCO2CH2CH3
(CH3)3CCOOH
+ HOCH2CH3
b.
O
OH
O
20.52
OH
O
c.
O
O
HO
d.
O
OH
O
O
HO
HO
O
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
NH3
Br excess
a.
NH2
H3O+
NaCN
Br
H
N
OH
NH2
CN
O
DCC
O
NaOH
b.
Br
OH
H2SO4
OH
OH
O
O
[from (a)] O
[1]
c.
CN
[from (a)]
MgBr
Mg
Br
[2] H2O
MgBr
O
O
H2SO4
CrO3
d.
OH
[from (b)]
OH
H2SO4/H2O
O
20.53
–
a.
Br
CN
H2
CN
O/H+
COOH
CN
b.
[from (a)]
SOCl2
c.
COOH
COCl
[from (b)]
d.
[from (b)]
e.
COOH
HOCH2CH3
CO2CH2CH3
H2SO4
[1] CH3MgBr
[from (a)]
CN
C
[2] H2O
CH3
O
[1] DIBAL–H
f.
CN
[from (a)]
g.
CN
CH2NH2
[2] H2O
h.
CH3COOH
CH2NH2
[from (g)]
20.54
CHO
[2] H2O
[1] LiAlH4
[from (a)]
O
OH
DCC
CH2NHCOCH3
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
CN
CH3
+ NaCN
a. CH3Cl
H3O+
CH3
COOH
[1] CO2
CH3 MgCl
Cl + Mg
CH3
Br
CH3
COOH
This method can't be used because an SN2 reaction
can't be done on an sp2 hybridized C.
+ NaCN
b.
[2] H3O+
sp2
MgBr
Br
+ Mg
COOH
[1] CO2
[2] H3O+
This method can't be used because an SN2 reaction
can't be done on a 3° C.
c. (CH3)3CCl + NaCN
[1] CO2
(CH3)3C MgCl
(CH3)3C Cl + Mg
d. HOCH2CH2CH2CH2Br
HOCH2CH2CH2CH2
+ NaCN
H3O+
CN
HOCH2CH2CH2CH2
COOH
This method can't be used because you can't make
a Grignard reagent with an acidic OH group.
Br + Mg
HOCH2CH2CH2CH2
(CH3)3C COOH
[2] H3O+
20.55
O
CH3Cl
CH3Cl
AlCl3
AlCl3
CH3
COOH
KMnO4
CH3
HOCH2CH2CH2CH3
C OCH CH CH CH
2
2
2
3
H2SO4
C OCH2CH2CH2CH3
COOH
O
dibutyl phthalate
20.56
CH3CH2Cl
a.
AlCl3
OH
COH
KMnO4
salicylamide
NO2
H2/Pd/C
HO
HO
(+ ortho isomer)
H
N
c.
C
CH3
O
HO
acetaminophen
[from (b)]
20.57
H
N
NaH
O
OH
OH
NH2
H
N
CH3COCl
H2SO4
OH
CNH2
NH3
Cl
OH
(+ para isomer)
b.
SOCl2
2 equiv
OH
HNO3
O
O
O
C
CH3
acetaminophen
H
N
CH3CH2Br
O
CH3CH2O
CH3
O
HO
(More nucleophilic
NH2 reacts first.)
C
C
CH3
O
p-acetophenetidin
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
CH3Cl
a.
HCl
CH3OH
Br2
CH3
AlCl3
CH2Br
hν
NaCN
H2O
H+
CH2CN
CH2COOH
HOCH2CH3
H2SO4
CH3Cl
ZnCl2
CH2COOCH2CH3
ethyl phenylacetate
CH3
CH3Cl
b.
NH2
NO2
[from (a)]
CH3
HNO3
H2SO4
AlCl3
NH2
CH3
H2
NH2
COOH
COOCH3
HOCH3
KMnO4
H2SO4
Pd
methyl anthranilate
(+ para isomer)
[from (a)]
O
CH3
CH3Cl
AlCl3
c.
CH3CH2OH
COOH
KMnO4
OH CH3COOH
H2SO4
[1] LiAlH4
[2] H2O
O
benzyl acetate
CrO3
H2SO4/H2O
CH3COOH
20.58
[1] LiAlH4
Cl
Cl
–CN
CN
NC
excess
NH2
H2N
[2] H2O
O
H2O
OH
HO
H2SO4
O
20.59
O
a. CH3 C OH
H2SO4
b. CH313CH2OH
CrO3
H2SO4/H2O
c. CH3CH2Br
O
CH313CH2OH
H218O
d. CH313CH2OH
13
OCH2CH3
O
O
13
CH3
C
CH3CH2OH
OH
18OH
+ base (H18O–)
PBr3
C
CH3
CH3CH2
CH313CH2Br
–
H2SO4
CH3
OCH2CH3
O
CH3COCl
CH3
C
18
OCH2CH3
18
18
OH
13C
CH313CH218OH
CrO3
18O
O
13 C 18
H2SO4/H218O CH3
OH
CH3CH2OH
H2SO4
13 C
CH3
+
OCH2CH3
H218O