Cap. 4 - Ateneonline

Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
Soluzioni ai problemi proposti nel libro Capitolo 4
4.1 La formula generale per un alcano aciclico è CnH2n + 2.
a. C12H26
2n + 2 = # H's
2(12) + 2 = 26
yes
b. C8H16
c. C30H64
2n + 2 = # H's
2(8) + 2 = 18
no
2n + 2 = # H's
2(30) + 2 = 62
no
4.2 Il butano ha 4 Carboni in fila. L’isobutano ha 3 Carboni in fila ed 1 carbonio in catena laterale.
H H
H H H
a. H C C C
b. (CH3)3CH
H
CH3
H H CH3
CH3
C
H
H C C H
c.
H
H
H C C H
e.
H H
CH3
4 C's in a row.
butane
d.
3 C's in a row.
isobutane
4 C's in a row.
butane
4 C's in a row.
butane
3 C's in a row.
isobutane
4.3 L’isopentano ha 4 carboni in fila ed 1 carbonio in catena laterale.
H
H
a. CH3CH2 C CH3
CH3
b.
H C CH3
CH3
C CH3
H
c. CH3CH2CH(CH3)2
redraw
H
isopentane
isopentane
d.
H
H C
CH3
C H
e.
C H CH3
H
isopentane
5 C's in a row.
pentane
isopentane
4.4 Per classificare un atomo di carbonio come 1°, 2°, 3° o 4° determinare a quanti atomi di
carbonio è legato (1° C = legato ad un altro C, 2° C = legato a due altri C, 3° C = legato a tre altri
C, 4° C = legato a quattro altri C). Ridisegnare se necessario per osservare ogni carbonio
chiaramente.
1°C
1°C's
1°C
a. CH3CH2CH2CH3
b. (CH3)3CH
c.
d.
4°C
CH3
2°C's
1°C's
CH3
C H
CH3 3°C
1°C
4°C's
All other C's are 1°C's.
All other C's are 2°C's.
3°C
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
4.5 Per classificare un atomo di idrogeno come 1°, 2°, or 3°, determinare se è legato ad un carbonio
1°, 2° o 3° C (un H 1° è legato ad un C 1°, un H 2° è legato ad un C 2°, un H 3° è legato ad un C
3°). Ridisegnare se necessario.
1° H
1° H
a. CH3CH2CH3
2° H's
2° H's
CH3
CH2
H
CH3
C
C CH3
CH2
CH3
CH
redraw
3° H
1° H's
CH3
3° H
redraw
c.
b. CH3CH2CH(CH3)C(CH3)3
CH2
C
CH
CH3
CH
3° H's
CH2
2° H's
CH3 CH3
CH3
CH3
All other H's are 1° H's.
2° H's
1° H's
4.6 Gli isomeri costituzionali differiscono nel modo in cui gli atomi sono legati tra di loro. Per
disegnare tutti gli isomeri costituzionali:
[1] Disegnare tutti i carboni della catena principale.
[2] Togliere un C ed usarlo come sostituente. (Non aggiungerlo al carbonio terminale: questo
rigenera la catena precedente.)
[3] Togliere due C ed usarli come sostituenti, etc.
Five constitutional isomers of molecular formula C6H14
[1] long chain
[2] with one C as a substituent
CH3CH2CH2CH2CH2CH3
CH3CH2CH2
[3] using 2 C's as substituents
CH3
CH3
CH3
C CH3 CH3CH2
C CH2CH3
H
H
CH3CH2
C CH3
CH3
4.7
Molecular formula C8H18 with one CH3 substituent.
CH3CH2CH2CH2CH2
CH3
CH3
C CH3 CH3CH2CH2CH2
C CH2CH3
H
H
4.8 Usare le regole delle risposte 4.4 e 4.5.
CH3
CH3CH2CH2
C CH2CH2CH3
H
CH3
H
H
C
C CH3
CH3 CH3
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
1°
4°
3°
CH3 CH3
2°
a.
1)
2°
2)
4°
b. 1) CH3
CH2
CH2
CH
CH3
2)
CH3
3°
1°
C
C
CH2
CH3
CH2
CH
All CH3's have 1° H's.
All CH2's have 2° H's.
All CH's have 3° H's.
2°
1°
CH3
CH2
CH2
CH
CH3
CH3
4.9
One possibility:
H H H H
c. H C C C H
CH3
a. CH3 C CH3
b.
CH3
CH3
d.
HH
CH3
CH
CH3
4.10 Usare i passaggi della risposta 4.6 per disegnare gli isomeri costituzionali.
Five constitutional isomers of molecular formula C5H10 having one ring
[1]
[2]
[3]
CH3
4.11
4.12
CH3
CH3
CH3
CH2CH3
CH3
I cicloalcani hanno formula molecolare CnH2n. Per un cicloalcano con 288 carboni, ci
sarebbero 2(288) = 576 H. Formula molecolare = C288H576.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
a. Five consitutional isomers of molecular formula C4H8:
CH2
CH2
CH3CH CHCH3
CHCH2CH3
CH3
CH3
C
CH3
b. Nine constitutional isomers of molecular formula C7H16.
H
CH3
CH3CH2CH2CH2CH2CH2CH3
H
C CH2CH2CH2CH3
CH3CH2
CH3
CH3
CH3
CH3
CH3
H
H
C CH2CH2CH3 CH3CH2
C CH2CH3 CH3
C
C CH2CH3
CH3
CH3
CH3 CH3
H
CH3CH2
C CH2CH3
CH2CH3
CH3
H
CH3
C
C CH3
C CH2CH2CH3
CH3
H
H
C
CH2 C CH3
CH3
CH3
CH3 CH3
c. Twelve constitutional isomers of molecular formula C6H12 containing one ring.
4.13 Seguire questi passaggi per attribuire il nome ad un alcano:
[1] Attribuire il nome alla catena principale attraverso l’individuazione della catena carboniosa
più lunga.
[2] Numerare la catena in modo che il primo sostituente abbia il numero più basso.
Successivamente nominare e numerare tutti i sostituenti, assegnando a sostituenti simili un
prefisso (di, tri, etc.).
[3] Combinare tutte le parti, ordinando sostituenti in ordine alfabetico, ignorando tutti i prefissi
eccetto iso.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
4-tert-butyl
[1]
CH3
a. CH3CH2CH2
[2]
CH3
C CH3
C CH2CH2CH2CH3
CH3
[3] 4-tert-butyl-4-methyloctane
CH3
CH3 C CH3
CH3CH2CH2 C CH2CH2CH2CH3
1 2 3 4 CH3 6 7 8
8 carbons = octane
4-methyl
[1]
b.
H
H C CH3
H C CH2 CHCH3
CH3
CH3
[2]
5 H 6
[3] 2,4-dimethylhexane
H C CH3
H C CH2 CHCH3
4 CH3 2 CH31
6 carbons = hexane
4-methyl 2-methyl
[1]
CH3
c. CH3CH2CH2
6-isopropyl
H
C CH3
C CH2 CH2
H
H
CH3 C CH3
CH3CH2CH2 C CH2 CH2
9 8 7 6H 5
4
[2]
CH2CH3
C CH3
H
9 carbons = nonane
4.14
2
1
[3] 6-isopropyl-3-methylnonane
CH2CH3
C CH3
H
3
3-methyl
Usare i passaggi della risposta 4.10 per assegnare il nome ad ogni alcano.
a. CH3CH2CH(CH3)CH2CH3
[1]
[2]
redraw
CH3CH2
CH CH2CH3
1
2
CH3CH2
CH3
3
4
[3] 3-methylpentane
5
CH CH2CH3
CH3
3-methyl
5 carbons = pentane
b. (CH3)3CCH2CH(CH2CH3)2
redraw
[1]
[2] 2
CH3
1
CH3
CH3 C CH2 CH CH2CH3
CH3
[3] 4-ethyl-2,2-dimethylhexane
4
5
6
CH3 C CH2 CH CH2CH3
CH2CH3
CH3
6 carbons = hexane
CH2CH3
4-ethyl
2,2-dimethyl
c. CH3(CH2)3CH(CH2CH2CH3)CH(CH3)2
redraw
[1]
4-isopropyl [3] 4-isopropyloctane
[2]
CH3
CH3CH2CH2CH2 CH CH CH3
CH3
8 7 6 5 4
CH3CH2CH2CH2 CH CH CH3
CH2CH2CH3
8 carbons = octane
2,2,4,4-tetramethyl
d. [1]
5 carbons = pentane
[2]
CH2CH2CH3
3
2
1
[3] 2,2,4,4-tetramethylpentane
1 2 3 4 5
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
2-methyl
[2]
[1]
1
[3] 3-ethyl-2,5-dimethylheptane
34 5 6 7
e.
2
or
3-ethyl 5-methyl
longest chain = 7 carbons = heptane
Number so there are more substituents.
Pick the upper option.
2-methyl
f.
[2] 1
[1]
3
5
3-ethyl
6
[3] 5-sec-butyl-3-ethyl-2,7-dimethyldecane
5-sec-butyl
2
10 carbons = decane
8 9 10
7-methyl
4.15 Per disegnare una struttura dal nome:
[1] Trovare la radice corrispondente e disegnare il numero di atomi di carbonio. Usare il suffisso
per identificare il gruppo funzionale. (-ano = alcano)
[2] Numerare arbitrariamente gli atomi di C nella catena. Aggiungere i sostituenti ai carboni
appropriati.
[3] Ridisegnare con gli H in modo che i carboni abbiano 4 legami.
a. 3-methylhexane
[1] 6 carbon alkane
[2]
CH3
C
C C C C C
C
methyl on C3
C C C C C
[3]
CH3
CH3CH2
CH CH2CH2CH3
b. 3,3-dimethylpentane
[1]
5 carbon alkane
[2]
methyl groups on C3
[3]
CH3
CH3
C C C C C
c. 3,5,5-trimethyloctane
[1]
8 carbon alkane
C
C
C C C
[2]
methyl groups on C3 and C5
CH3
CH2CH3
[3]
CH3
C C C C C C C C
C
CH3
CH3
CH3
C C C C C C C C
CH3CH2
CH3
CH3CH2
CH3
CH CH2 C
CH2CH2CH3
CH3
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
d. 3-ethyl-4-methylhexane
[1]
[2]
6 carbon alkane
[3]
ethyl group on C3
CH2CH3
CH3CH2
CH2CH3
C
C
C C C C C
C C C C C
CH3
CH CH CH2CH3
CH3
methyl group on C4
e. 3-ethyl-5-isobutylnonane
[1]
[2]
C C C C C C C C C
[3]
isobutyl group on C5
9 carbon alkane
CH3
CH3
CH2 CH CH3
CH2 CH CH3
C C C C C C C C C
CH3CH2
CH CH2 CH CH2CH2CH2CH3
CH2CH3
CH2CH3
ethyl group on C3
4.16 Usare i passaggi della risposta 4.10 per assegnare il nome ad ogni alcano.
[1]
[3] hexane
[2]
H H H H H H
H C C C C C C H
no substituents , skip [2]
H H H H H H
6 carbons = hexane
[1]
2-methyl
[2]
H H H CH3 H
H C C C C
C H
H H H H
H
H C C C C
5
[1]
H H H
[2] H H CH3 H H
H C C C
C C H
H H
5
H C C C
C
H
H H CH3 H
H H
2,2-dimethyl
[2]
H H CH3 H
[3] 3-methylpentane
C C H
H H H
5 carbons = pentane
[1]
1
H
3-methyl
H H CH3 H H
H C C C
[3] 2-methylpentane
C H
H H H H
5 carbons = pentane
1
H H H CH3 H
1
H H CH3 H
H C C C
C
[3] 2,2-dimethylbutane
H
4
H H CH3 H
[2]
H H
H
H
H C C
C
C H
4 carbons = butane
[1]
H H
H
H
H C C
C
C H
H CH3 CH3 H
4
1
[3] 2,3-dimethylbutane
H CH3 CH3 H
4 carbons = butane
2,3-dimethyl
4.17
Seguire questi passaggi per assegnare il nome ad un cicloalcano:
[1] Assegnare il nome alla radice del cicloalcano contando i carboni dell’anello ed aggiungendo
ciclo-.
[2] Numerazione:
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
[2a] Numerare l’anello iniziando da un sostituente ed assegnando al secondo sostituente il
numero più basso.
[2b] Assegnare il numero più basso al sostituente che precede in ordine alfabetico.
[2c] Denominare e numerare tutti i sostituenti, assegnando ai sostituenti uguali un prefisso
(di, tri, ecc.).
[3] Combinare tutte le parti, disponendo in ordine alfabetico tutti i sostituenti, ed ignorando tutti I
prefissi ad eccezione di iso.
(Ricordare: se una catena carboniosa ha più carboni dell’anello, la catena diventa la radice e
l’anello un sostituente.)
1
2
CH3
[2] 3 C C C CH
3
[1]
a.
C
4
6 carbons in ring =
cyclohexane
C
1,1-dimethyl
[3] 1,1-dimethylcyclohexane
C 6
5
Number so the
substituents are at C1.
1,2,3-trimethyl
[1]
[2]
3
4C
b.
CH3
[3] 1,2,3-trimethylcyclopentane
C 2
C CH3
5C C
CH3
5 carbons in ring =
cyclopentane
1
Number so the first substituent
is at C1, second at C2.
1
2
3 C
C
[2] C
[1]
c.
C
CH3 4 C
5
6 carbons in ring =
cyclohexane
4-methyl
C
[3] 1-butyl-4-methylcyclohexane
6
1-butyl
Number so the earliest alphabetical
substituent is at C1, butyl before methyl.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
1
[1]
[2]
C
5C
d.
4C
C
3
6 carbons in ring =
cyclohexane
1-sec-butyl
6
[3] 1-sec-butyl-2-isopropylcyclohexane
C
C
2
2-isopropyl
Number so the earliest alphabetical
substituent is at C1, butyl before isopropyl.
[1]
[2]
1
3
5
C
C
C
e.
C
C
[3] 1-cyclopropylpentane
4
2
1-cyclopropyl
longest chain =
5 carbons =
pentane
Number so the
cyclopropyl is at C1.
4.18 Per disegnare le strutture, usare i passaggi della risposta 4.15.
a. 1,2-dimethylcyclobutane
[1] 4 carbon cycloalkane
[2]
methyl groups
on C1 and C2
C C
C C
CH3 1
C C 4
[3] CH3
C C 3
CH3 2
CH3
b. 1,1,2-trimethylcyclopropane
[1] 3 carbon cycloalkane
c. 4-ethyl-1,2-dimethylcyclohexane
[1] 6 carbon cycloalkane
[2]
C
C
C
[3]
CH3
CH3
[1] 5 carbon cycloalkane
C
C
C C
CH3
3
CH3CH2 4 C 2 CH3
C
C
[3] CH CH
3
2
C
C
ethyl 5 C 1 CH3
on C4
6
2 CH3's
C
d. 1-sec-butyl-3-isopropylcyclopentane
C
3 CH3's
3
C C
C
2C
C C CH3
1 CH
3
C
C
CH3
[2]
[2]
CH3
CH
CH3
isopropyl
[3]
C3
4C
C 2
C
C
5
1
CH CH3
sec-butyl
CH3 CH2
CH3
CH3
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
e. 1,1,2,3,4-pentamethylcycloheptane
[1] 7 carbon cycloalkane
[2]
5 CH3's
CH3
CH3 3 C 4
C
C5
C
C
C
C
C
C C
CH3
2C
CH3
1
C6
C C7
CH3
CH3
[3]
CH3
CH3
CH3
CH3
4.19 Per assegnare il nome ai cicloalcani, usare i passaggi della risposta 4.17.
[1]
5 carbons in ring =
cyclopentane
[1]
[2]
C C
[3] methylcyclobutane
C C
CH3
CH3
4 carbons in ring =
cyclobutane
[1]
methyl
[2]
CH3
CH3
CH3
3 carbons in ring =
cyclopropane
[1]
[2]
CH3
3 carbons in ring =
cyclopropane
4.20
CH3
C
C C
[3] ethylcyclopropane
CH2CH3
ethyl
3 carbons in ring =
cyclopropane
CH3
C C
1,2-dimethyl
CH2CH3
[1]
[3] 1,2-dimethylcyclopropane
C
[2]
CH3
[3] 1,1-dimethylcyclopropane
CH3
1,1-dimethyl
Usare I passaggi delle risposte 4.13 e 4.17 per dare il nome agli alcani.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
a. CH3CH2CHCH2CHCH2CH2CH3
[1]
CH3
CH2CH3
1
2
3 4 5 6
CH3
5-ethyl
CH3
7-methyl
7
b. CH3CH2CCH2CH2CHCHCH2CH2CH3
[2] 1
[3] 5-ethyl-3-methyloctane
CH2CH3
3-methyl
[1]
8
[2] CH3CH2CHCH2CHCH2CH2CH3
8 carbons = octane
CH2CH3
7
CH3
2 3CH2CH3
[3] 3,3,6-triethyl-7-methyldecane
CH3CH2CCH2CH2CHCHCH2CH2CH3
CH2CH3 CH2CH3
CH2CH3 CH2CH3
10 carbons = decane
3,3,6-triethyl
c. CH3CH2CH2C(CH3)2C(CH3)2CH2CH3
[1]
[2]
redraw
CH3CH2CH2
CH3 CH3
CH3CH2CH2
C
[3] 3,3,4,4-tetramethylheptane
CH3 CH3 3
C
C CH2CH3
2
4 CH3 CH3
C CH2CH3
CH3 CH3
1
3,3,4,4-tetramethyl
7 carbons = heptane
d. CH3CH2C(CH2CH3)2CH(CH3)CH(CH2CH2CH3)2
3,3-diethyl
redraw
[1]
[2]
CH3CH2 H
CH3CH2
C
CH3CH2 H
H
CH3CH2
C CH2CH2CH3
C
1
CH3CH2 CH3 CH2CH2CH3
C
5
4
C CH2CH2CH3
C
CH3CH2 CH3 CH2CH2CH3
4-methyl
8 carbons = octane
e. (CH3CH2)3CCH(CH3)CH2CH2CH3
[3] 3,3-diethyl-4-methyl-5-propyloctane
H
5-propyl
3,3-diethyl
4
[1]
redraw
[2]
CH3CH2
CH3CH2 H
CH3CH2
C
1
C CH2CH2CH3
[3] 3,3-diethyl-4-methylheptane
CH3CH2 H
C
C CH2CH2CH3
CH3CH2 CH3
6
7
4-methyl
CH3CH2 CH3
7 carbons = heptane
f. CH3CH2CH(CH3)CH(CH3)CH(CH2CH2CH3)(CH2)3CH3
3 4 5
redraw
H H H
[1]
[2]
CH3CH2
H
H
H
C
C
C CH2CH2CH2CH3
CH3 CH3 CH2CH2CH3
9 carbons = nonane
CH3CH2
1 2
C
C
[3] 3,4-dimethyl-5-propylnonane
C CH2CH2CH2CH3
CH3 CH3 CH2CH2CH3
3,4-dimethyl
5-propyl
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
g. (CH3CH2CH2)4C
4,4-dipropyl
redraw
[1]
[2]
CH2CH2CH3
CH3CH2CH2
4
C CH2CH2CH3
CH2CH2CH3
CH3CH2CH2
C CH2CH2CH3
3
1 2
CH2CH2CH3
[3] 4,4-dipropylheptane
CH2CH2CH3
7 carbons = heptane
h.
1
2
[1]
[2]
5
3
7
6-isopropyl
3-methyl
10 carbons =decane
i.
[3] 6-isopropyl-3-methyldecane
6
[2]
[1]
10 carbons = decane
8
6
4-isopropyl
4
[3] 8-ethyl-4-isopropyl-2,6-dimethyldecane
1
8-ethyl
2,6-dimethyl
4-isopropyl
j.
4
[2]
[1]
[3] 4-isopropyloctane
1
8 carbons = octane
2,2,5-trimethyl
k.
2,2,5-trimethylheptane
1 2
1
2
CH(CH2CH3)2
l.
=
3
3-cyclobutylpentane
4
5
3-cyclobutyl
5
m. 4
3
1
1-sec-butyl
2
1-sec-butyl-2-isopropylcyclopentane
2-isopropyl
5
n.
6
1
4
3
1-isobutyl-3-isopropylcyclohexane
2
1-isobutyl
3-isopropyl
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
4.21
2,2-dimethyl
CH3
3,3-dimethyl
CH3
4,4-dimethyl
CH3
CH3
C CH2CH2CH2CH2CH3
CH3CH2
C CH2CH2CH2CH3
CH3CH2CH2
C CH2CH2CH3
1
CH3
1
CH3
1
CH3
2
3
3,3-dimethylheptane
2,2-dimethylheptane
H
H
H
H
CH2CH2 C CH2CH3
CH3 C CH2 C CH2CH2CH3
CH3
C
1
1
CH3
2
CH3
4
CH3
2
H
H
CH3
C CH2CH2CH2 C CH3
1
CH3
6
CH3
2,6-dimethyl
2,6-dimethylheptane
H
1
C
1
CH3 CH3
C CH2CH2CH3
3
H
C CH2CH2CH2CH3
CH3CH3
2,3-dimethyl
2,3-dimethylheptane
H
H
CH3CH2
3
H
CH3 C
CH3
2,5-dimethylheptane
2,4-dimethylheptane
4.22
2
2,5-dimethyl
2,4-dimethyl
2
5
4
4,4-dimethylheptane
H
CH3CH2
C CH2 C CH2CH3
1
CH3
4
3,4-dimethyl
3,4-dimethylheptane
Usare i passaggi della risposta 4.15 per disegnare le strutture.
3
5
CH3
3,5-dimethyl
3,5-dimethylheptane
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
a. 3-ethyl-2-methylhexane
[1]
6 C chain
C C
[2]
C C C C C C
b. sec-butylcyclopentane
5 C ring
[1]
C C C C
H
CH3 CH2CH3
methyl
on C2
[3] CH3 C
H
C CH2CH2CH3
CH3 CH2CH3
ethyl on C3
[2]
isopropyl on C4
c. 4-isopropyl-2,4,5-trimethylheptane
[1]
7 C chain
[2]
CH3
CH
C C C C
CH3
C C C C C C C
CH3
[3]
C C C
CH3 CH3
methyls on C2, C4, and C5
d. cyclobutylcycloheptane
[1] 7 C cycloalkane
[2]
CH3
CH3
CH3
CH
CH CH2 C CHCH2CH3
CH3
CH3 CH3
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
e. 3-ethyl-1,1-dimethylcyclohexane
[1]
6 C cycloalkane
[2]
[3]
CH3CH2
C
C
C
C
C
ethyl on C3 C
C
2 ethyl groups
8 C cycloalkane
C
C
C
C
C
C C
g. 6-isopropyl-2,3-dimethylnonane
[1]
9 C alkane
C
[2]
C
C
C C
C
C
C
C
C
C
C
C
C
8 C alkane
C
C
C
C
C
C
C
C
C
C
CH3
C
CH3
[2] CH3 CH3
C
C
C
C
C
[1]
C
C
Disegnare i composti.
[3]
ethyl on C1
[2]
CH2CH3
or
methyl on C3
C(CH3)3
[2]
CH3CH2
4.23
C
CH3 CH3
C
C
C
C
CH2CH3
j. trans-1-tert-butyl-4-ethylcyclohexane
6 C ring
C
C
methyl
CH3
[1]
[3]
CH3
CH3
i. cis-1-ethyl-3-methylcyclopentane
5 C ring
CH
5 methyl groups
h. 2,2,6,6,7-pentamethyloctane
[1]
isopropyl
methyl
CH3
C
[3]
C C
C
C
2 methyl
groups on
C1
CH3
C
C C
C
C
CH3
C
C C
[2]
C
C
C
f. 4-butyl-1,1-diethylcyclooctane
[1]
C
CH3
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
a. 2,2-dimethyl-4-ethylheptane
CH2CH3
CH2 C CH2CH2CH3
CH3
CH3 C
alphabetized incorrectly
ethyl before methyl
CH3
e. 1-ethyl-2,6-dimethylcycloheptane
CH2CH3
2
CH3
Numbered incorrectly.
Renumber so methyls
are at C1 and C4.
H
1
4
CH3
4-ethyl-2,2-dimethylheptane
2-ethyl-1,4-dimethylcycloheptane
b. 5-ethyl-2-methylhexane
f. 5,5,6-trimethyloctane
Longest chain was not
chosen = heptane
3
Numbered incorrectly.
Renumber so methyls
are at C3 and C4.
4
2,5-dimethylheptane
3,4,4-trimethyloctane
H
c. 2-methyl-2-isopropylheptane
g. 3-butyl-2,2-dimethylhexane
CH3C CH3
longest chain was not
chosen = octane
CH3
C CH2CH2CH2CH2CH3
CH3
2,3,3-trimethyloctane
d. 1,5-dimethylcyclohexane
CH3
1
Numbered incorrectly.
Renumber so methyls
are at C1 and C3.
1
4-tert-butyloctane
h. 1,3-dimethylbutane
longest chain not
chosen = pentane
3
CH3
H
CH3 CCH2
CH3
1,3-dimethylcyclohexane
4.24
4
longest chain not
chosen = octane
CH2
CH3
2-methylpentane
Confrontare i pesi molecolari per determinare i punti di ebollizione relativi.
gasoline: C5H12 - C12H26
lowest molecular weight:
lowest boiling point
kerosene: C12H26 - C16H34
middle molecular weight:
intermediate boiling point
diesel fuel: C15H32 - C18H38
highest molecular weight:
highest boiling point
4.25 Confrontare i numeri dei carboni e l’area superficiale per determinare I punti di
ebollizione relativi. Regole:
[1] Aumento del numero dei carboni = aumento del punto di ebollizione.
[2] Aumento dell’area superficiale = aumento del punto di ebollizione (la ramificazione diminuisce
l’area superficiale).
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
CH3(CH2)6CH3
CH3(CH2)5CH3
CH3CH2CH2CH2CH(CH3)2 (CH3)3CCH(CH3)2
7 C's
linear
8 C's
linear
largest number of C's
no branching
highest bp
7 C's
one branch
7 C's
three branches
increasing branching
decreasing surface area
decreasing bp
increasing boiling point: (CH3)3CCH(CH3)2 < CH3CH2CH2CH2CH(CH3)2 < CH3(CH2)5CH3 < CH3(CH2)6CH3
4.26
Usare le regole della risposta 4.25.
a.
CH3CH2CH3, CH3CH2CH2CH3, CH3CH2CH2CH2CH3
4 C's
3C's
5 C's
lowest boiling point
highest boiling point
b. (CH3)2CHCH(CH3)2,
CH3CH2CH2CH(CH3)2,
most branching
lowest boiling point
CH3(CH2)4CH3
least branching
highest boiling point
4.27
CH3(CH2)6CH3
no branching = higher surface area
higher boiling point
4.28
(CH3)3C(CH3)3
branching = lower surface area
lower boiling point
more spherical, better packing =
higher melting point
Per disegnare una proiezione di Newman, visualizzare i carboni come uno davanti ed uno dietro
rispettivamente. Il legame C−C non viene disegnato. C’è una sola conformazione sfalsata ed
una eclissata.
Br
rotation here
H H
H C C Br
C in front
H H
H
H
C behind
H
H
60o
H
H
H Br
H
H
H
1
staggered
2
eclipsed
4.29 Le conformazioni sfalsate sono più stabili delle conformazioni eclissate.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
H
H
H CH3 HH
H
H
H
HH H
H H HCH3
H
H
H
H
CH3
H
H CH3
eclipsed
energy maximum
H H
H C C CH3
H H
Energy
rotation here
H
H
H
H
H
0o
60o
CH3
H
H
H H
H
H
H
CH3 CH
3
120o
H
H
180o
240o
H
300o
Dihedral angle
360o = 0o
staggered
energy minimum
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
4.30
1 kcal/mol
H
H
H CH3
To calculate H,CH3 destabilization:
3.5 kcal/mol (total) −
2 kcal/mol for 2 H,H eclipsing interactions
= 1.5 kcal/mol for one H,CH3 eclipsing interaction
H
H,H eclipsing
1 kcal/mol of destabilization
H
1 kcal/mol
4.31 Per determinare l’energia dei conformeri ricordare due fattori:
[1] I conformeri eclissati sono più stabili di quelli eclissati.
[2] Minimizzare le interazioni steriche mettendo I gruppi più ingombranti lontano uno dall’altro.
La conformazione a più alta energia è la conformazione eclissata in cui i due gruppi più
ingombranti sono eclissati. La conformazione a minore energia è la configurazione sfalsata
in cui i due gruppi più ingombranti sono anti.
CH3
H
H
H
CH3
CH3
H CH3
CH3
CH3
H
H
most stable
least stable
1
2
2
2
2
Energy
2
1
0
o
60o
1
120
o
180
1
o
240
o
300 o
360
o
dihedral angle
4.32
Per determinare i conformeri più stabili e quelli meno stabili, usare le regole della risposta 4.31.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
rotation here
H
CH3
H
CH3
CH3
60o
C CH2CH3
CH3
H
CH3
CH3
H
H
CH3
H
CH3
H
H
1
staggered
most stable
H
2
eclipsed
3
staggered
most stable
60o
CH3
CH3
CH3
CH3
CH3
60o
H
H
CH3
CH3
60o
CH3
H
H
H
6
eclipsed
least stable
H
CH3
60o
H
CH3
CH3
CH3
60o
H
H
H
5
staggered
4
eclipsed
least stable
4.33 Per determinare i conformeri più stabili e quelli meno stabili, usare le regole della risposta 4.31.
Cl
1,2-dichloroethane
H
H
ClCH2 CH2 Cl
60o
H
H
H
H
H
H Cl
Cl
H
Cl
1
staggered, anti
2
eclipsed
rotation here
60o
H
H
Cl
H
Cl
3
staggered, gauche
60o
60o
H
Cl
H
HH
H
o
60
H
Cl
6
eclipsed
Cl
H
H
Cl
5
staggered, gauche
60o
HH
HH
Cl
Cl
4
eclipsed
Fondamenti di chimica organica
Janice Gorzynski Smith
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highest energy
Cl groups eclipsed
least stable
4
2
Eclipsed forms are
higher in energy.
Energy
6
3
180o
Staggered forms
are lower in energy.
5
1
most stable
1
most stable
Cl groups anti
120o
60o
0o
60o
120o
180o
Dihedral angle between 2 Cl's
4.34 Aggiungere l’aumento di energia per ogni interazione eclissata per determinare la
destabilizzazione.
H
H
CH3
H
H CH3
H CH3
b.
a.
CH3
H
H
1 H,H interaction =
2 H,CH3 interactions
(2 x 1.5 kcal/mol) =
1 kcal/mol
3 kcal/mol
CH3
3 H,CH3 interactions
(3 x 1.5 kcal/mol) =
4.5 kcal/mol
Total destabilization
Total destabilization = 4 kcal/mol
4.35
CH3
a.
H
H
CH3
H
CH3
H
and
H
H
1 gauche CH3,CH3
= 0.9 kcal/mol
of destabilization
CH3
H
CH3
CH3
higher energy
2 gauche CH3,CH3
0.9 kcal/mol x 2 = 1.8 kcal/mol
of destabilization
Energy difference =
1.8 kcal/mol – 0.9 kcal/mol =
4.36
b.
H
H
and
CH3
CH3
H CH3
H
CH3
CH3
2 gauche CH3,CH3
0.9 kcal/mol x 2 =
1.8 kcal/mol
of destabilization
CH3
H
higher energy
3 eclipsed H,CH3
1.5 kcal/mol x 3 = 4.5 kcal/mol
of destabilization
Energy difference =
0.9 kcal/mol
4.5 kcal/mol – 1.8 kcal/mol =
2.7 kcal/mol
Usare le regole della risposta 4.31 per determinare i conformeri più stabili e meno stabili.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
a. CH3 CH2CH2CH2CH3
b. CH3CH2CH2 CH2CH2CH3
H CH CH CH
2
2
3
H
CH2CH2CH3
H
H
H
H
H
HH
H
staggered
most stable
H
CH2CH3
H
eclipsed
least stable
CH3CH2
CH2CH3
CH2CH3
H
H
staggered
ethyl groups anti
most stable
All staggered conformers are equal in energy.
All eclipsed conformers are equal in energy.
H
HH
H
eclipsed
ethyl groups eclipsed
least stable
4.37
CH3CH2
H
H
60°
(1)
CH3CH2
H
CH3
H
CH2CH2CH3
H
CH2CH3
H
H
H
60°
H
H
CH3
H
1
60°
60°
H
H
H
H
H
60°
H
CH3CH2
CH3
H
6
60°
H
CH3
5
Energy
1
most stable
120o
Staggered forms
are lower in energy.
5
3
180o
Eclipsed forms are
higher in energy.
6
1
most stable
60o
0o
60o
120o
180o
Dihedral angle between two alkyl groups
H
H
H
H
least stable
4
2
CH2CH3
CH3
3
2
CH3CH2
H
CH3
CH2CH3
4
Fondamenti di chimica organica
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Copyright © 2009 – The McGraw-Hill Companies srl
most stable
CH3CH2
H
H
(2) CH3CH2
CHCH2CH3
60°
H
H
H
60°
CH3
CH3
H
CH3
H
CH2CH3
H
H
CH2CH3
CH3
CH3
CH3
1
CH3
3
2
60°
60°
H
CH3
H
CH3CH2
H
60°
H
CH3CH2
CH3
H
H
H
H
CH3
CH3
60°
H
CH3
CH2CH3
CH3
5
6
least stable
4
least stable
4
Energy
2
Staggered forms
are lower in energy.
5
3
1
most stable
180o
Eclipsed forms are
higher in energy.
6
1
most stable
120o
60o
0o
60o
120o
180o
Dihedral angle
(between CH3 CH2 in back and CH 3 in front)
4.38 Due tipi di tensione:
• La tensione torsionale è dovuta a gruppi eclissati su atomi di carbonio adiacenti.
• La tensione sterica è dovuta a sovrapposizione delle nuvole elettroniche di gruppi voluminosi
(es: interazioni gauche).
H
CH3
H
CH3
H
a.
b.
CH3
H
CH3
two sites
three bulky methyl groups close =
steric strain
H
H
H CH3
HH
c.
H
CH3
eclisped conformation=
torsional strain
CH2CH3
CH3CH2
two bulky ethyl groups close =
steric strain
eclipsed conformation =
torsional strain
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
4.39 La barriera di energia rotazionale è uguale alla differenza di energia tra la più alta energia
eclissata e la più bassa energia sfalsata della molecola.
a. CH3 CH(CH3)2
CH3
H
H
b. CH3 C(CH3)3
CH3
H
H CH3
H
H
CH3
CH3
H
H
H
most stable
CH3
CH3
H
least stable
Destabilization energy =
CH3
H
H CH3
H
CH3
H
most stable
least stable
Destabilization energy =
2 H,CH3 eclipsing interactions
2(1.5 kcal/mol) =
1 H,H eclipsing interaction =
3 kcal/mol
1 kcal/mol
Total destabilization = 4 kcal/mol
4 kcal/mol = rotation barrier
3 H,CH3 eclipsing interactions
3(1.5 kcal/mol) =
4.5 kcal/mol
Total destabilization = 4.5 kcal/mol
4.5 kcal/mol = rotation barrier
4.40
Cl
H
H
H
H
H
most stable
H
H
H Cl
H
H
least stable
2 H,H eclipsing interactions = 2(1 kcal/mol) = 2 kcal/mol
Since the barrier to rotation is 3.7 kcal/mol,
the difference between this value and the
destabilization due to H,H eclipsing is the
destabilization due to H,Cl eclipsing.
3.7 kcal/mol - 2 kcal/mol = 1.7 kcal/mol
destabilization due to H,Cl eclipsing
4.41 Il conformero gauche può formare un legame idrogeno intramolecolare, che lo rende il
conformero più stabile.
4.42 Due punti:
• I legami assiali sono direzionati in alto o in basso, mentre quelli equatoriali verso l’esterno.
• Un carbonio in alto ha un legame assiale in alto, ed un carbonio in basso ha un legame assiale in
basso.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
equatorial
axial up CH3
Br
H
H
HO
H
equatorial
H
Cl H
H
OH
CH3
axial up
Br
HO
H
axial down
Cl
OH
Up carbons are dark circles.
Down carbons are clear circles.
equatorial
4.43 Disegnare un secondo conformero a sedia per inversione dell’anello.
• I carboni in alto diventano carboni in basso, e i legami assiali diventano legami
equatoriali.
• I legami assiali diventano equatoriali, ma i legami in alto rimangono in alto; cioè un
legame assiale in alto diventa un legame equatoriale in alto.
• Il conformero con i gruppi più ingombranti in posizione equatoriale è più stabile ed è
presente in maggior concentrazione all’equilibrio.
axial
H eq
Draw in the H
Br
a.
Draw second conformer.
Br
H
Up carbons switch
and label the C
to down carbons.
more stable as up or down. Axial bond is up =
Br is equatorial.
up carbon
Br
axial
eq
Axial bond is down =
down carbon
axial
eq
Cl
b.
Cl
Draw in the H
Draw second conformer.
Cl
Up carbons switch
and label the C
to down carbons. axial Haxial bond is down =
as up or down. eq
down carbon
Axial bond is up =
up carbon
more stable
Cl is equatorial.
H
Axial bond is up =
up carbon
eq
H
c.
CH2CH3
more stable
CH2CH3 is equatorial.
4.44
Draw second conformer.
Draw in the H
CH2CH3
and label the C
as up or down.
eq
Up carbons switch
to down carbons.
H
CH2CH3
Axial bond is down =
down carbon
Fondamenti di chimica organica
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Copyright © 2009 – The McGraw-Hill Companies srl
axial
H
OH axial
[a]
(1)
[c] HO
[d]
down HO
HO
eq
eq HO
H
H
eq
[a]
(2)
[b]
H
axial
H eq
CH3eq
H
H
OH
OH
down
up
CH
eq Br 3
H eq
ax
ax
[c]
HO
ax
Br
[d]
CH3
H
[b]
OH eq
eq
Br
CH3 up
H
axial
HO
[c]
both up =
cis
H
OH ax
H
ax
ax
H
H
axial
[a]
OH
Br
H eq
CH3 eq
H eq
eq HO
eq
up
Br
H ax
OH
H
one up, one down =
trans
axial
(3)
ax
H up
OH
[b]
HO
[d]
ax
ax
H
OH
OH eq
eq HO
eq H
H eq
H
one up, one down =
trans
H
ax
OH
ax
4.45 Un isomero cis ha due gruppi dalla stessa parte dell’anello. I due gruppi possono essere
disegnati entrambi su sia entrambi giù. Viene rappresentata solo una possibilità. Un isomero
trans ha un gruppo su una parte dell’anello ed uno dalla parte opposta. Ciascun gruppo può essere
disegnato da ciascun lato. Viene rappresentata solo una possibilità.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
(1)
[a]
cis
(2)
(3)
[a]
[a]
trans
cis
[b] cis isomer
trans
cis
[b] cis isomer
trans
[b] cis isomer
ax
ax
ax
ax
ax
ax
eq
eq
eq
both groups equatorial
more stable
[c] trans isomer
ax
eq
eq
eq
larger group equatorial
more stable
[c] trans isomer
larger group equatorial
more stable
[c] trans isomer
ax
ax
eq
eq
eq
eq
ax
larger group equatorial
more stable
[d]
ax
both groups equatorial
more stable
[d]
The cis isomer is more
stable than the trans
since one conformer has
both groups equatorial.
eq
eq
ax
both groups equatorial
more stable
[d]
The trans isomer is more
stable than the cis
since one conformer has
both groups equatorial.
The trans isomer is more
stable than the cis
since one conformer has
both groups equatorial.
4.46 Confrontare gli isomeri disegnandoli nella conformazione a sedia. I sostituenti equatoriali sono
più stabili. Confrontare le definizioni del problema 4.53.
Fondamenti di chimica organica
Janice Gorzynski Smith
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(a) 1,2-diethylcyclohexane
cis
H
CH2CH3 ax
H
cis
CH2CH3 eq
H
(CH3)2CH
H
CH3CH2
CH2CH3
eq
ax
eq
H
H
H
both groups equatorial
most stable of all conformers
cis isomer
eq
CH2CH3
H
trans
H
H
(CH3)2CH
CH2CH3
(CH3)2CH
both groups equatorial
most stable of all conformers
trans isomer
CH2CH3
H
CH2CH3
eq
H
H
eq
eq
H
ax
The cis isomer is more
stable than the trans isomer
because its more stable
conformer has two groups equatorial.
The trans isomer is more
stable than the cis isomer
because its more stable
conformer has two groups equatorial.
4.47
Only the more stable conformer of each compound is drawn.
CH3
CH3
CH3
CH3
CH3
or
a.
CH3
CH3
CH3
or
b.
CH3
CH3
CH3
CH3
redraw to see
axial and equatorial
redraw to see
axial and equatorial
CH3
CH3
CH3
CH3
CH3
CH3
CH3
CH3
CH3
CH3
CH3
CH3
more stable
can all be equatorial
more stable
substituents on C1, C3, C5=
all equatorial
4.48
a.
OH
O
OH
HO
HO
HO
most stable
all groups are equatorial
4.49
HO
b.
ax
(CH3)2CH
CH2CH3
H
CH2CH3 eq
trans
(b) 1-ethyl-3-isopropylcyclohexane
ax
ax
CH2CH3 ax
O
HO
OH
HO
OH
I cunei indicano gruppi “sopra” alla pagina, e i trattini sono “sotto” dietro alla pagina.
I gruppi cis sono dalla stessa parte dell’anello, i gruppi trans da parti opposte.
ax
CH2CH3
H
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
trans-1-ethyl-2-methylcyclopentane
cis-1,2-dimethylcyclopropane
CH3
or
CH3
CH3
or
CH3
CH3CH2
cis = same side of the ring
both groups on wedges or
both on dashes
CH3
CH3CH2
CH3
trans = opposite sides of the ring
one group on a wedge,
one group on a dash
4.50
Per classificare un composto come isomero cis o trans, classificare ogni gruppo diverso
dall’idrogeno come su o giù. Gruppi dalla stessa parte = isomero cis, gruppi da parti opposte
= isomero trans.
down bond (up)
(equatorial) HH(up)
HO
a.
down bond
(up) (equatorial)
H
up bond
(axial)
H
Cl
H
HO
Cl
Cl
b.
OH
OH
H
H
up bond
H
(down) (equatorial)
down bond
(equatorial)
both groups down =
cis isomer
(up)
Br
H
c. H
(down)
H
Br
Br
down bond
H
(equatorial)
one group up, one down =
trans isomer
Cl
one group up, one down =
trans isomer
Br
4.51
ax
ax H
H
CH3
CH3
eq H
eq
eq
H eq
CH3
CH3
ax
ax
both groups equatorial
more stable
4.52
CH3
a.
trans: CH3
CH3
c.
CH3
groups on same side
cis isomer
CH3
cis:
H
CH3
b.
H
CH3
groups on opposite sides
trans isomer
CH3
CH3
H
H
two chair conformers for the cis isomer
Same stability since they are identical groups with
one equatorial, one axial.
H
CH3
H
H
CH3
H
CH3
both groups equatorial
more stable
two chair conformers for the trans isomer
d. The trans isomer is more stable because
it can have both methyl groups in the more roomy
equatorial position.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
4.53
a.
and
CH3
H
f.
same molecular formula C4H8
different connectivity
constitutional isomers
CH
CH3
CH3
CH3CH2
H
and
H
CH2CH3
CH3
different arrangement in three dimensions
stereoisomers
CH3
c.
CH3
H
and
H
CH3
CH3
CH3
1 down, 1 up =
trans
H
CH2CH3
d.
CH2CH3
redraw
CH3
CH2 CH3
CH
H
CH3
1 down, 1 up =
trans
CH
CH2CH3
CH2CH3
3-ethyl-2-methylpentane
3-ethyl-2-methylpentane
same molecular formula
same name
identical molecules
H
=
same arrangement in three dimensions
identical
CH3
CH3 CH2CH3
CH3 CH CH
H
H
CH3
and
b.
H
CH2CH3
up
CH3
eq
and CH3
H
CH3CH2
up
ax
ax
g.
eq
H
up
both up = cis
H
H
both up = cis
CH2CH3
and
same molecular formula C10H20
different connectivity
constitutional isomers
and
e.
same arrangement in three dimensions
identical
CH2CH3
CH2CH3
CH3
h.
and
3,4-dimethylhexane
2,4-dimethylhexane
same molecular formula C8H18
molecular formula: C6H10
molecular formula: C6H12
different IUPAC names
constitutional isomers
different molecular formulas
not isomers
4.54
up
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
constitutional isomer
One possibility:
stereoisomer
a.
trans
cis
H
H
b.
H
OH
H
OH
cis
HO
OH
H
OH
c.
OH
H
trans
Cl
cis
Cl
Cl
Cl
Cl
Cl
trans
4.55
Three constitutional isomers of C7H14:
1,1-dimethylcyclopentane
1,2-dimethylcyclopentane
1,3-dimethylcyclopentane
or
or
trans
cis
trans
cis
4.56 L’ ossidazione provoca un aumento del numero di legami C−Z, o in una diminuzione del numero
di legami C−H.
La riduzione provoca una diminuzione del numero di legami C−Z, o un aumento del numero di
legami C−H.
a.
CH3
O
O
C
C
H
CH3
O
OH
c.
Decrease in the number of C–H bonds.
Increase in the number of C–O bonds.
Oxidation
CH3
C
CH3
HO OH
C
CH3
CH3
No change in the number of C–O
or C-H bonds. Neither
O
b.
CH3
C
CH3
CH3CH2CH3
Decrease in the number of C–O bonds.
Increase in the number of C–H bonds.
Reduction
d.
O
OH
Decrease in the number of C–O bonds.
Increase in the number of C–H bonds.
Reduction
4.57 I prodotti della combustione di un idrocarburo sono sempre gli stessi: CO2 e H2O.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
4.58
a.
CH3CH2CH3
b.
+
+
flame
5 O2
9 O2
flame
3 CO2 +
6 CO2 +
4 H2O + heat
6 H2O + heat
Usare le definizioni della risposta 4.56 per classificare le reazioni.
O
a.
=
CH3CHO
CH3
C
CH3CH2OH
H
d.
CH2
CH2
Decrease in the number of C–O
bonds. Reduction
H C C H
Decrease in the number of C–H
bonds. Oxidation
CH3
b.
Increase in the number of C–Z
bonds. Oxidation
Increase in the number of C–O
bonds. Oxidation
c.
CH2
CH2
f.
HOCH2CH2OH
CH3CH2OH
CH2
CH2
Loss of one C–O
bond and one C–H
bond. Neither
Two new C–O
bonds. Oxidation
4.59
CH2Br
e.
O
Usare la regola della risposta 4.57.
flame
a. CH3CH2CH2CH2CH(CH3)2
7 CO2 + 8 H2O + heat
11 O2
flame
b.
4 CO2 + 5 H2O + heat
(13/2) O2
4.60
1 C–O bond
2 C–O bonds
H
O
a.
OH
2 C–H bonds
H
benzene
an arene oxide
increase in C–O bonds
oxidation reaction
H
1 C–H bond
phenol
loss of 1 C–O bond,
loss of 1 C–H bond
neither
b. Phenol is more water soluble than benzene because it is polar (contains an O–H group)
and can hydrogen bond with water, whereas benzene is nonpolar and cannot hydrogen bond.
4.61
L’ammide del ciclo a quattro termini ha angoli di legame a 90° che originano tensione angolare,
e quindi è più reattiva.
Fondamenti di chimica organica
Janice Gorzynski Smith
Copyright © 2009 – The McGraw-Hill Companies srl
amide
H
N
penicillin G
S
N
O
O
CH3
CH3
COOH
strained amide
more reactive
4.62
CH3
CH3
CH3
CH3
cis-1,4-dimethylcyclohexane
trans-1,4-dimethylcyclohexane
more symmetrical
better packing
higher melting point
4.63
Cl H
Example:
HH
IH
HH
H
Cl
C
H
H
HH
H
HH
I
H
C
C
H
Although I is a much bigger atom than Cl, the C–I bond
is also much longer than the C–Cl bond. As a result the
eclipsing interaction of the H and I atoms is not very much
different from the H,Cl eclipsing interaction in magnitude.
H
C
H
H
H
longer bond
4.64
H
H
H
decalin
H
trans-decalin
cis-decalin
H
H
H
H
H
trans
The trans isomer is more stable since
the carbon groups at the ring junction are
both in the favorable equatorial position.
H
1,3-diaxial interaction
cis
This bond is axial, creating unfavorable
1,3-diaxial interactions.

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