Estereoquímica

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Estereoquímica	
  
	
  
Quiralidade	
  
•  No	
   inicio	
   do	
   século	
   XIX,	
   o	
  mineralogista	
   francês	
  
René	
  Hauy	
  observou	
  a	
  existência	
  de	
  dois	
  Apos	
  de	
  
cristais	
  de	
  quartzo.	
  
Cristais	
  de	
  quartzo	
  
ü  Enan0omorfos	
  (do	
  grego:	
  enán$os,	
  "opostos”,	
  +	
  morfo,	
  “forma”)	
  
(a)  Convolvulus	
  arvensis	
  
(b)	
  Lonicera	
  sempervirens	
  
(c)	
  Liguus	
  virgineus	
  
(d)	
  Bacillus	
  sub6lis.	
  
Quiralidade	
  
Quiralidade	
  
•  O	
  Termo	
  quiral	
  vem	
  da	
  palavra	
  grega	
  cheir,	
  
que	
  significa	
  “mão"	
  
Ø Todo	
  objeto	
  tem	
  uma	
  imagem	
  especular	
  
§  Objetos	
  quirais	
  têm	
  imagem	
  especular	
  não	
  sobreponível.	
  5.1 Chirality and Stereochemistry 187
Figure 5.2 Left and right hands are not
superposable.
Figure 5.1 The mirror image of a right
hand is a left hand.
Each of our hands is chiral. When you view your right hand in a mirror, the image that
you see in the mirror is a left hand (Fig. 5.1). However, as we see in Fig. 5.2, your left hand
and your right hand are not identical because they are not superposable. Your hands are
chiral. In fact, the word chiral comes from the Greek word cheir meaning hand. An object
such as a mug may or may not be chiral. If it has no markings on it, it is achiral. If the mug
has a logo or image on one side, it is chiral.
* For interesting reading, see Hegstrum, R. A.; Kondepudi, D. K. The Handedness of the Universe. Sci. Am.
1990, 262(1), 98–105, and Horgan, J. The Sinister Cosmos. Sci. Am. 1997, 276(5), 18–19.
Bindweed (top photo)
(Convolvulus sepium) winds in a
right-handed fashion, like the
right-handed helix of DNA.
5.1A The Biological Significance of Chirality
The human body is structurally chiral, with the heart lying to the left of center and the liver
to the right. Helical seashells are chiral and most are spiral, such as a right-handed screw.
Many plants show chirality in the way they wind around supporting structures. Honeysuckle
winds as a left-handed helix; bindweed winds in a right-handed way. DNA is a chiral
molecule. The double helical form of DNA turns in a right-handed way.
Chirality in molecules, however, involves more than the fact that some molecules adopt left-
or right-handed conformations. As we shall see in this chapter, it is the nature of groups bonded
at specific atoms that can bestow chirality on a molecule. Indeed, all but one of the 20 amino
acids that make up naturally occurring proteins are chiral, and all of these are classified as being
left-handed. The molecules of natural sugars are almost all classified as being right-handed. In
fact, most of the molecules of life are chiral, and most are found in only one mirror image form.*
This mug is chiral.
solom_c05_186-229hr.qxd 28-09-2009 14:28 Page 187
Quiralidade	
  
§  Objetos	
   aquirais	
   têm	
   imagem	
   especular	
  
sobreponível.	
  
Quiralidade	
  
Exercícios...	
  
Quiralidade	
  
Moléculas	
  também	
  
podem	
  ser	
  quirais!	
  
C	
  
Quiralidade	
  
Quiralidade	
  
A	
  descoberta	
  da	
  quiralidade	
  em	
  moléculas	
  
•  Em	
   1848,	
   Louis	
   Pasteur,	
   observou	
   que	
   dois	
  
Apos	
   de	
   cristais	
   do	
   tartarato	
   de	
   amônio	
   e	
  
sódio	
   eram	
   depositados	
   em	
   barris	
   de	
   vinho	
  
durante	
  a	
  fermentação.	
  
“A	
  oportunidade	
  favorece	
  a	
  mente	
  preparada"	
  
Importância	
  da	
  Quiralidade	
  nas	
  
Moléculas	
  
186
Stereochemistry
Chiral Molecules
5
We are all aware of the fact that certain everyday objects such as gloves and shoes possess the quality of “hand-
edness.” A right-handed glove only fits a right hand; a left-handed shoe only fits a left foot. Objects that can
exist in right-handed and left-handed forms are said to be chiral. In this chapter we shall find that molecules
can also be chiral and can exist in right- and left-handed forms. For example, one chiral form of the molecule
shown above is a painkiller (Darvon), and the other, a cough suppressant (Novrad)! It is easy to see why it is
important to understand chirality in molecules.
Me2N NMe2
Me
O
O
Me
O
O
DARVON
(painkiller)
NOVRAD
(anticough agent)
5.1 Chirality and Stereochemistry
Chirality is a phenomenon that pervades the universe. How can we know whether a par-
ticular object is chiral or achiral (not chiral)?
! We can tell if an object has chirality by examining the object and its mirror image.
Every object has a mirror image. Many objects are achiral. By this we mean that the
object and its mirror image are identical, that is, the object and its mirror image are super-
posable one on the other.* Superposable means that one can, in one’s mind’s eye, place
one object on the other so that all parts of each coincide. Simple geometrical objects such
as a sphere or a cube are achiral. So is an object like a water glass.
! A chiral object is one that cannot be superposed on its mirror image.
*To be superposable is different than to be superimposable. Any two objects can be superimposed simply by
putting one object on top of the other, whether or not the objects are the same. To superpose two objects (as
in the property of superposition) means, on the other hand, that all parts of each object must coincide.
The condition of superposability must be met for two things to be identical.
The glass and its mirror
image are superposable.
solom_c05_186-229hr.qxd 28-09-2009 14:28 Page 186
Importância	
  da	
  Quiralidade	
  nas	
  Moléculas	
  
CO2H
CH2CONH2
H
H2N
O
(S)-carvona
(odor de menta)
CO2H
H2NOCH2C
H
NH2
O
(odor de alcavaria)
(R)-carvona 
(R)-asparagina
(sabor amargo)
(S)-asparagina
(sabor doce)
Importância	
  da	
  Quiralidade	
  nas	
  Moléculas	
  
Isômeros	
  
•  Compostos que têm a mesma fórmula 
molecular, mas não têm estruturas idênticas, 
são chamados de isômeros. 
Enan0ômeros	
  mm	
  
aaaaaaaaaaaaaa	
  
Diastereoisômeros	
  	
  
	
  Isômeros	
  ConsAtucionais	
  
Isômeros	
  
Diastereoisômeros	
  
	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  Diastereoisômeros	
  
Isômeros	
  
•  Exercícios…	
  
1.  Desenhe	
  os	
  isômeros	
  cis	
  e	
  trans	
  para	
  as	
  
seguintes	
  substâncias:	
  
a.  1-­‐eAl-­‐3-­‐meAlciclobutano	
  
b.  2-­‐meAl-­‐hept-­‐3-­‐eno	
  
c.  1-­‐bromo-­‐4-­‐clorociclo-­‐hexano	
  
d.  1,3-­‐dibromociclobutano	
  
EnanAômeros	
  
•  São	
   estereoisômeros	
   cujas	
   moléculas	
   são	
  
imagens	
  especulares	
  não	
  superponíveis.	
  
186 C H A P T E R 5 Stereochemistry
A chiral molecule has a nonsuperimpos-
able mirror image.
An achiral molecule has a superimpos-
able mirror image.
Take a break and convince yourself that the two 2-bromobutane isomers are not iden-
tical, by building ball-and-stick models using four different-colored balls to represent the
four different groups bonded to the asymmetric carbon. Try to superimpose them.
Nonsuperimposable mirror-image molecules are called enantiomers (from the
Greek enantion, which means “opposite”). The two stereoisomers of 2-bromobu-
tane are enantiomers. A molecule that has a nonsuperimposable mirror image, like
an object that has a nonsuperimposable mirror image, is chiral. Each of the
enantiomers is chiral. A molecule that has a superimposable mirror image, like an
object that has a superimposable mirror image, is achiral. To see that the achiral
moleule is superimposable onits mirror image (i.e., they are identical molecules),
mentally rotate the achiral molecule clockwise. Notice that chirality is a property of
the entire molecule.
PROBLEM 6!
Which of the compounds in Problem 4 can exist as enantiomers?
5.5 Drawing Enantiomers
Chemists draw enantiomers using either perspective formulas or Fischer projections.
This book has been written in a way that allows you to use either perspective formulas or
Fischer projections. Most chemists use perspective formulas. If you choose to use perspec-
tive formulas, you can ignore all the Fischer projections in the book.
Perspective formulas show two of the bonds to the asymmetric carbon in the plane
of the paper, one bond as a solid wedge protruding out of the paper, and the fourth
bond as a hatched wedge extending behind the paper. You can draw the first enan-
tiomer by putting the four groups bonded to the asymmetric carbon in any order. Draw
the second enantiomer by drawing the mirror image of the first enantiomer.
an achiral
molecule
a chiral
molecule
superimposable
mirror image
nonsuperimposable
mirror image
enantiomers identical molecules
C H
Br
CH3
CH3CH2
CH
Br
H3C
CH2CH3
C
Br
H3C
H3C CH2CH3
C CH3
Br
CH3
CH3CH2
CH3CH2 CH3 CH3
CH2CH3
CH3CHCH2CH3
Br
Br
H
2-bromobutane
the two isomers of 2-bromobutane
enantiomers
mirror
Br
H
*
C C
Movie:
Nonsuperimposable mirror
image
BRUI05-182_237r4 20-03-2003 3:36 PM Page 186
Um centro assimétrico é a causa da 
quiralidade em uma molécula	
  
•  A causa usual da quiralidade em uma 
molécula é a presença de um centro 
assimétrico. 
•  Um centro assimétrico é um átomo que está 
ligado a quatro grupos diferentes. 
•  Esses	
   carbonos	
   são	
   chamados	
   de	
   centros	
   de	
  
quiralidade,	
  estereocentro,	
  centro	
  assimétrico	
  
e	
  centro	
  estereogênico.	
  
– Um	
   estereocentro	
   (centro	
   estereogênico)	
   é	
   um	
  
átomo	
   no	
   qual	
   o	
   intercâmbio	
   de	
   dois	
   grupos	
  
produz	
  um	
  estereoisômero.	
  
A shortcut—called a Fischer projection—for showing the three-dimensional
arrangement of groups bonded to an asymmetric carbon was devised in the late 1800s
by Emil Fischer. A Fischer projecton represents an asymmetric carbon as the point of
intersection of two perpendicular lines; horizontal lines represent the bonds that pro-
ject out of the plane of the paper toward the viewer, and vertical lines represent the
bonds that extend back from the plane of the paper away from the viewer. The carbon
chain always is drawn vertically with C-1 at the top of the chain.
To draw enantiomers using a Fischer projection, draw the first enantiomer by ar-
ranging the four atoms or groups bonded to the asymmetric carbon in any order. Draw
the second enantiomer by interchanging two of the atoms or groups. It does not matter
which two you interchange. (Make models to convince yourself that this is true.) It is
best to interchange the groups on the two horizontal bonds because the enantiomers
then look like mirror images on paper.
Note that interchanging two atoms or groups gives you the enantiomer—whether
you are drawing perspective formulas or Fischer projections. Interchanging two atoms
or groups a second time, brings you back to the original molecule.
A stereocenter (or stereogenic center) is an atom at which the interchange of two
groups produces a stereoisomer. Therefore, both asymmetric carbons—where the in-
terchange of two groups produces an enantiomer and the carbons where the inter-
change of two groups converts a cis isomer to a trans isomer (or a Z isomer to an E
isomer)—are stereocenters.
PROBLEM 7
Draw enantiomers for each of the following compounds using:
a. perspective formulas
b. Fischer projections
Br
1. CH3CHCH2OH
CH3
2.
CH3
OH
3.ClCH2CH2CHCH2CH3 CH3CHCHCH3
H
H
Br
CH3C H
CH3CH2
CC
HH
CH3CH2 CH3
Cl
Br
a stereocenter a stereocenter
a stereocenter
Br
CH2CH3
H
Fischer projections of the enantiomers 
of 2-bromobutane
asymmetric carbonCH3 CH3
CH2CH3
H Br
perspective formulas of the enantiomers of 2-bromobutane
CH
Br
CH3
CH2CH3
CH3C
Br
H
CH3CH2
Section 5.5 Drawing Enantiomers 187
The solid wedges represent bonds that
point out of the plane of the paper to-
ward the viewer.
The hatched wedges represent bonds
that point back from the plane of the
paper away from the viewer.
Make certain when you draw a perspec-
tive formula that the two bonds in the
plane of the paper are adjacent to one
another; neither the solid wedge nor
the hatched wedge should be drawn
between them.
In a Fischer projection horizontal lines
project out of the plane of the paper
toward the viewer and vertical lines ex-
tend back from the plane of the paper
away from the viewer.
Emil Fischer (1852–1919) was born
in a village near Cologne, Germany.
He became a chemist against the
wishes of his father, a successful mer-
chant, who wanted him to enter the
family business. He was a professor
of chemistry at the Universities of
Erlangen, Würzburg, and Berlin. In
1902 he received the Nobel Prize in
chemistry for his work on sugars.
During World War I, he organized
German chemical production. Two of
his three sons died in that war.
BRUI05-182_237r4 20-03-2003 3:36 PM Page 187
Centro	
  de	
  Quiralidade	
  
•  Exercício...	
  
Como desenhar enantiômeros	
  
•  Uma cunha sólida representa uma ligação que se estende para 
fora do plano do papel em direção ao observador. 
•  Uma cunha tracejada representa uma ligação que aponta para 
trás do plano do papel, para longe do observador. 
Fórmula	
  em	
  perspec0va	
  
A shortcut—called a Fischer projection—for showing the three-dimensional
arrangement of groups bonded to an asymmetric carbon was devised in the late 1800s
by Emil Fischer. A Fischer projecton represents an asymmetric carbon as the point of
intersection of two perpendicular lines; horizontal lines represent the bonds that pro-
ject out of the plane of the paper toward the viewer, and vertical lines represent the
bonds that extend back from the plane of the paper away from the viewer. The carbon
chain always is drawn vertically with C-1 at the top of the chain.
To draw enantiomers using a Fischer projection, draw the first enantiomer by ar-
ranging the four atoms or groups bonded to the asymmetric carbon in any order. Draw
the second enantiomer by interchanging two of the atoms or groups. It does not matter
which two you interchange. (Make models to convince yourself that this is true.) It is
best to interchange the groups on the two horizontal bonds because the enantiomers
then look like mirror images on paper.
Note that interchanging two atoms or groups gives you the enantiomer—whether
you are drawing perspective formulas or Fischer projections. Interchanging two atoms
or groups a second time, brings you back to the original molecule.
A stereocenter (or stereogenic center) is an atom at which the interchange of two
groups produces a stereoisomer. Therefore, both asymmetric carbons—where the in-
terchange of two groups produces an enantiomer and the carbons where the inter-
change of two groups converts a cis isomer to a trans isomer (or a Z isomer to an E
isomer)—are stereocenters.
PROBLEM 7
Draw enantiomers for each of the following compounds using:
a. perspective formulas
b. Fischer projections
Br
1. CH3CHCH2OH
CH3
2.
CH3
OH
3.ClCH2CH2CHCH2CH3 CH3CHCHCH3
H
H
Br
CH3C H
CH3CH2
CC
HH
CH3CH2 CH3
Cl
Br
a stereocenter a stereocenter
a stereocenter
Br
CH2CH3
H
Fischer projections of the enantiomers 
of 2-bromobutane
asymmetric carbonCH3 CH3
CH2CH3
H Br
perspective formulasof the enantiomers of 2-bromobutane
CH
Br
CH3
CH2CH3
CH3C
Br
H
CH3CH2
Section 5.5 Drawing Enantiomers 187
The solid wedges represent bonds that
point out of the plane of the paper to-
ward the viewer.
The hatched wedges represent bonds
that point back from the plane of the
paper away from the viewer.
Make certain when you draw a perspec-
tive formula that the two bonds in the
plane of the paper are adjacent to one
another; neither the solid wedge nor
the hatched wedge should be drawn
between them.
In a Fischer projection horizontal lines
project out of the plane of the paper
toward the viewer and vertical lines ex-
tend back from the plane of the paper
away from the viewer.
Emil Fischer (1852–1919) was born
in a village near Cologne, Germany.
He became a chemist against the
wishes of his father, a successful mer-
chant, who wanted him to enter the
family business. He was a professor
of chemistry at the Universities of
Erlangen, Würzburg, and Berlin. In
1902 he received the Nobel Prize in
chemistry for his work on sugars.
During World War I, he organized
German chemical production. Two of
his three sons died in that war.
BRUI05-182_237r4 20-03-2003 3:36 PM Page 187
•  Quando você desenhar uma fórmula em perspectiva, 
certifique‑se de que as duas ligações no plano do 
papel sejam adjacentes uma à outra; nem a cunha 
sólida nem a tracejada devem ser desenhadas entre 
elas. 
Como desenhar enantiômeros	
  
Fórmula	
  em	
  perspec0va	
  
Projeção	
  de	
  Fischer	
  	
  
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delete the image and then insert it again.
Como desenhar enantiômeros	
  
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•  Exercício...	
  
Nomeando	
  EnanAômeros	
  
•  O	
  sistema	
  de	
  nomenclatura	
  R,S	
  
–  R	
  do	
  laAm	
  rectus,	
  direita	
  e	
  S	
  do	
  laAm	
  sinister,	
  esquerda	
  
Nomeando	
  Enan0ômeros	
  
Exemplicando	
  
Nomeando	
  Enan0ômeros	
  
Etapas	
  para	
  a	
  determinação	
  da	
  configuração	
  R	
  ou	
  S:	
  
	
  	
  
1.  Numerar	
  os	
  átomos	
  (ou	
  grupos)	
  ligados	
  ao	
  carbono	
  
assimétrico	
  em	
  ordem	
  de	
  prioridade	
  	
  
•  maior	
  número	
  atômico,	
  maior	
  prioridade	
  (1);	
  
2.  Traçar	
  uma	
  seta	
  do	
  grupo	
  de	
  prioridade	
  1,	
  passando	
  
pelo	
  2	
  e,	
  então,	
  pelo	
  3.	
  	
  
	
  
Configuração S 
1
2 3
4
1
2
3
4
Configuração R
Sen0do	
  
horário	
  	
  R	
  
(do	
  laAm	
  
rectus,	
  direito)	
  
Sen0do	
  an0-­‐
horário	
  	
  S	
  (do	
  
laAm	
  sinister,	
  
esquerdo)	
  
3.  	
  Pode-­‐se	
  desenhar	
  a	
  seta	
  do	
  grupo	
  1	
  para	
  o	
  grupo	
  
2,	
  passando	
  pelo	
  grupo	
  4,	
  mas	
  nunca	
  passando	
  pelo	
  
grupo	
  3.	
  
	
  
190 C H A P T E R 5 Stereochemistry
4. In drawing the arrow from group 1 to group 2, you can draw past the group with the
lowest priority (4), but never draw past the group with the next lowest priority (3).
Now let’s see how to determine the configuration of a compound drawn as a Fischer
projection.
1. Rank the groups (or atoms) that are bonded to the asymmetric carbon in order of
priority.
2. Draw an arrow from the group (or atom) with the highest priority (1) to the group
(or atom) with the next highest priority (2). If the arrow points clockwise, the enan-
tiomer has the R configuration; if it points counterclockwise, the enantiomer has the S
configuration, provided that the group with the lowest priority (4) is on a vertical bond.
3. If the group (or atom) with the lowest priority is on a horizontal bond, the answer
you get from the direction of the arrow will be the opposite of the correct answer. For
example, if the arrow points clockwise, suggesting that the asymmetric carbon has the
R configuration, it actually has the S configuration; if the arrow points counterclock-
wise, suggesting that the asymmetric carbon has the S configuration, it actually has the
R configuration. In the following example, the group with the lowest priority is on a
horizontal bond, so clockwise signifies the S configuration, not the R configuration.
4. In drawing the arrow from group 1 to group 2, you can draw past the group (or
atom) with the lowest priority (4), but never draw past the group (or atom) with the
next lowest priority (3).
H CH3
COH
OH
O
3
1
(S)-lactic acid
CH3 H
COH
OH
O
3 44
1
2
(R)-lactic acid
2
H OH
CH3
CH2CH3
3 3
2
1
(S)-2-butanol
HO H
CH3
CH2CH3
2
44
1
(R)-2-butanol
CH3CH2 CH2CH2CH3
Cl
H
1
2
4
(R)-3-chlorohexane
CH3CH2CH2 CH2CH3
Cl
H
1
32
4
(S)-3-chlorohexane
3
(R)-1-bromo-3-pentanol
C
CH3CH2
OH
H
CH2CH2Br
1
3
2
4 C
CH3CH2
OH
H
CH2CH2Br
1
3
2
4
2 2
14
3
3
4
1
 CH3 and H
what is its configuration?
switch 
this molecule has the R configuration;
therefore, it had the S configuration 
before the groups were switched 
HO
CH2CH3
CH3
H
CC
HO
CH2CH3
H
CH3
Clockwise specifies R if the lowest prior-
ity substituent is on a vertical bond.
Clockwise specifies S if the lowest priori-
ty substituent is on a horizontal bond.
BRUI05-182_237r4 20-03-2003 3:36 PM Page 190
Nomeando	
  Enan0ômeros	
  
4.  O	
  grupo	
  de	
  menor	
  prioridade	
  fica	
  para	
  trás.	
  
Nomeando	
  a	
  par0r	
  da	
  Projeção	
  de	
  Fischer	
  
Nomeando	
  Enan0ômeros	
  
190 C H A P T E R 5 Stereochemistry
4. In drawing the arrow from group 1 to group 2, you can draw past the group with the
lowest priority (4), but never draw past the group with the next lowest priority (3).
Now let’s see how to determine the configuration of a compound drawn as a Fischer
projection.
1. Rank the groups (or atoms) that are bonded to the asymmetric carbon in order of
priority.
2. Draw an arrow from the group (or atom) with the highest priority (1) to the group
(or atom) with the next highest priority (2). If the arrow points clockwise, the enan-
tiomer has the R configuration; if it points counterclockwise, the enantiomer has the S
configuration, provided that the group with the lowest priority (4) is on a vertical bond.
3. If the group (or atom) with the lowest priority is on a horizontal bond, the answer
you get from the direction of the arrow will be the opposite of the correct answer. For
example, if the arrow points clockwise, suggesting that the asymmetric carbon has the
R configuration, it actually has the S configuration; if the arrow points counterclock-
wise, suggesting that the asymmetric carbon has the S configuration, it actually has the
R configuration. In the following example, the group with the lowest priorityis on a
horizontal bond, so clockwise signifies the S configuration, not the R configuration.
4. In drawing the arrow from group 1 to group 2, you can draw past the group (or
atom) with the lowest priority (4), but never draw past the group (or atom) with the
next lowest priority (3).
H CH3
COH
OH
O
3
1
(S)-lactic acid
CH3 H
COH
OH
O
3 44
1
2
(R)-lactic acid
2
H OH
CH3
CH2CH3
3 3
2
1
(S)-2-butanol
HO H
CH3
CH2CH3
2
44
1
(R)-2-butanol
CH3CH2 CH2CH2CH3
Cl
H
1
2
4
(R)-3-chlorohexane
CH3CH2CH2 CH2CH3
Cl
H
1
32
4
(S)-3-chlorohexane
3
(R)-1-bromo-3-pentanol
C
CH3CH2
OH
H
CH2CH2Br
1
3
2
4 C
CH3CH2
OH
H
CH2CH2Br
1
3
2
4
2 2
14
3
3
4
1
 CH3 and H
what is its configuration?
switch 
this molecule has the R configuration;
therefore, it had the S configuration 
before the groups were switched 
HO
CH2CH3
CH3
H
CC
HO
CH2CH3
H
CH3
Clockwise specifies R if the lowest prior-
ity substituent is on a vertical bond.
Clockwise specifies S if the lowest priori-
ty substituent is on a horizontal bond.
BRUI05-182_237r4 20-03-2003 3:36 PM Page 190
1.  O	
  grupo	
  de	
  menor	
  prioridade	
  fica	
  para	
  trás.	
  
2.  Se	
  o	
  grupo	
  com	
  menor	
  prioridade	
  esAver	
  na	
  
ligação	
  horizontal,	
  a	
  nomeação	
  será	
  oposta	
  à	
  
direção	
  da	
  seta.	
  
190 C H A P T E R 5 Stereochemistry
4. In drawing the arrow from group 1 to group 2, you can draw past the group with the
lowest priority (4), but never draw past the group with the next lowest priority (3).
Now let’s see how to determine the configuration of a compound drawn as a Fischer
projection.
1. Rank the groups (or atoms) that are bonded to the asymmetric carbon in order of
priority.
2. Draw an arrow from the group (or atom) with the highest priority (1) to the group
(or atom) with the next highest priority (2). If the arrow points clockwise, the enan-
tiomer has the R configuration; if it points counterclockwise, the enantiomer has the S
configuration, provided that the group with the lowest priority (4) is on a vertical bond.
3. If the group (or atom) with the lowest priority is on a horizontal bond, the answer
you get from the direction of the arrow will be the opposite of the correct answer. For
example, if the arrow points clockwise, suggesting that the asymmetric carbon has the
R configuration, it actually has the S configuration; if the arrow points counterclock-
wise, suggesting that the asymmetric carbon has the S configuration, it actually has the
R configuration. In the following example, the group with the lowest priority is on a
horizontal bond, so clockwise signifies the S configuration, not the R configuration.
4. In drawing the arrow from group 1 to group 2, you can draw past the group (or
atom) with the lowest priority (4), but never draw past the group (or atom) with the
next lowest priority (3).
H CH3
COH
OH
O
3
1
(S)-lactic acid
CH3 H
COH
OH
O
3 44
1
2
(R)-lactic acid
2
H OH
CH3
CH2CH3
3 3
2
1
(S)-2-butanol
HO H
CH3
CH2CH3
2
44
1
(R)-2-butanol
CH3CH2 CH2CH2CH3
Cl
H
1
2
4
(R)-3-chlorohexane
CH3CH2CH2 CH2CH3
Cl
H
1
32
4
(S)-3-chlorohexane
3
(R)-1-bromo-3-pentanol
C
CH3CH2
OH
H
CH2CH2Br
1
3
2
4 C
CH3CH2
OH
H
CH2CH2Br
1
3
2
4
2 2
14
3
3
4
1
 CH3 and H
what is its configuration?
switch 
this molecule has the R configuration;
therefore, it had the S configuration 
before the groups were switched 
HO
CH2CH3
CH3
H
CC
HO
CH2CH3
H
CH3
Clockwise specifies R if the lowest prior-
ity substituent is on a vertical bond.
Clockwise specifies S if the lowest priori-
ty substituent is on a horizontal bond.
BRUI05-182_237r4 20-03-2003 3:36 PM Page 190
Nomeando	
  Enan0ômeros	
  
3.  A	
  seta	
  pode	
  ir	
  do	
  grupo	
  1	
  ao	
  2,	
  passando	
  pelo	
  
grupo	
  4,	
  mas	
  nunca	
  pelo	
  grupo	
  3.	
  
CO2H
OH
CH3H (S)-­‐ácido	
  láAco	
  
Nomeando	
  Enan0ômeros	
  
ü  Uma	
  projeção	
  de	
  Fischer	
  pode	
  somente	
  ser	
  rodada	
  180°	
  	
  
no	
  plano	
  do	
  papel,	
  obtendo-­‐se	
  a	
  mesma	
  molécula.	
  
	
  
Atribuição	
  de	
  prioridades	
  
§  	
   Em	
   caso	
   de	
   empate	
   nos	
   átomos	
   diretamente	
   ligados	
   ao	
  
carbono	
   assimétrico,	
   compara-­‐se	
   os	
   átomos	
   ligados	
   a	
   eles,	
  
uAlizando	
  os	
  mesmos	
  critérios.	
  
Nomeando	
  Enan0ômeros	
  
(R)-1-bromopropan-2-ol 
H
CH2BrH3C
OH
4
2
1
3
C
H
H
H
C
H
H
Br (Br,H,H)
(H,H,H)
§ Grupos	
  com	
  ligações	
  duplas	
  e	
  triplas	
  são	
  “desdobrados”	
  em	
  
duas	
  ou	
  três	
  ligações	
  simples.	
  
The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and 
then open the file again. If the red x still appears, you may have to delete the image and then insert it again.
Atribuição	
  de	
  prioridades	
  
Nomeando	
  Enan0ômeros	
  
Exercício...	
  
1.  Liste	
  os	
  subsAtuintes	
  em	
  ordem	
  decrescente	
  
de	
  prioridade	
  para	
  cada	
  um	
  dos	
  seguintes	
  
conjuntos:	
  
The image part with relationship ID rId2 was not found in the file.
•  Exercício...	
  
1.  Se	
  cerAfique	
  se	
  cada	
  uma	
  das	
  seguintes	
  
estruturas	
  tem	
  configuração	
  R	
  ou	
  S	
  
Section 5.6 Naming Enantiomers: The R,S System of Nomenclature 191
It is easy to tell whether two molecules are enantiomers (nonsuperimposable) or
identical molecules (superimposable) if you have molecular models of the molecules—
just see whether the models superimpose. If, however, you are working with structures
on a two-dimensional piece of paper, the easiest way to determine whether two mole-
cules are enantiomers or identical molecules is by determining their configurations. If
one has the R configuration and the other has the S configuration, they are enan-
tiomers. If they both have the R configuration or both have the S configuration, they
are identical molecules.
When comparing two Fischer projections to see if they are the same or different,
never rotate one 90° or turn one over, because this is a quick way to get a wrong an-
swer. A Fischer projection can be rotated 180° in the plane of the paper, but this is the
only way to move it without risking an incorrect answer.
PROBLEM 8!
Indicate whether each of the following structures has the R or the S configuration:
a. c.
b. d.
PROBLEM 9! SOLVED
Do the following structures represent identical molecules or a pair of enantiomers?
a.
b.
c. d.
SOLUTION TO 9a The first structure shown in part (a) has the S configuration, and the
second structure has the R configuration. Because they have opposite configurations, the
structures represent a pair of enantiomers.
PROBLEM 10!
Assign relative priorities to the following groups:
a.
b.
c.
d. CH CH2CH3CH2 CH3
¬CH2CH2CH2Br¬Cl¬CH2CH2Br¬CH(CH3)2
¬CH2OH¬CH3¬OH¬CH“O
¬H¬CH2CH2OH¬CH3¬CH2OH
CH3 CH2CH3 and
Cl
H
H Cl
CH3
CH2CH3
C
H
CH2Br
OH and
CH3
C
HO
H
CH3
CH2Br
C
CH3
CH2Br
Cl and
CH2CH3
C
CH3CH2
Cl
CH3
CH2Br
C
HO
CH3
H and
CH2CH2CH3
C
CH3CH2CH2
OH
CH3
H
Cl
C
CH3CH2
CH2Br
CH2CH2Cl
OH
HO
C
CH3
CH(CH3)2
CH2CH3
CH2Br
BRUI05-182_237r4 20-03-2003 3:36 PM Page191
5.8 Properties of Enantiomers: Optical Activity 201
Tell whether the two structures in each pair represent enantiomers or two molecules of the
same compound in different orientations.
Review Problem 5.13
Tartaric acid is found naturally in
grapes and many other plants.
Crystals of tartaric acid can be
sometimes be found with wine.
Mixtures of the enantiomers of a compound have different properties than pure samples
of each, however. The data in Table 5.1 illustrate this for tartaric acid. The natural isomer,
(!)-tartaric acid, has a melting point of 168–170°C, as does its unnatural enantiomer,
(")-tartaric acid. An equal mixture tartaric acid enantiomers, (!/")-tartaric acid, has a
melting point of 210–212°C, however.
Enantiomers show different behavior only when they interact with other chiral sub-
stances, including their own enantiomer, as shown by the melting point data above.
Enantiomers show different rates of reaction toward other chiral molecules, that is, toward
reagents that consist of a single enantiomer or an excess of a single enantiomer. Enantiomers
also show different solubilities in solvents that consist of a single enantiomer or an excess
of a single enantiomer.
One easily observable way in which enantiomers differ is in their behavior toward plane-
polarized light. When a beam of plane-polarized light passes through an enantiomer, the
plane of polarization rotates. Moreover, separate enantiomers rotate the plane of plane-
polarized light equal amounts but in opposite directions. Because of their effect on plane-
polarized light, separate enantiomers are said to be optically active compounds.
In order to understand this behavior of enantiomers, we need to understand the nature
of plane-polarized light. We also need to understand how an instrument called a polarime-
ter operates.
TABLE 5.1 Physical Properties of 2-Butanol and Tartaric Acid Enantiomers
Compound Boiling Point (bp) or Melting Point (mp)
(R)-2-Butanol 99.5°C (bp)
(S)-2-Butanol 99.5°C (bp)
(!)-(R,R)-Tartaric acid 168–170°C (mp)
(")-(S,S)-Tartaric acid 168–170°C (mp)
(!/")-Tartaric acid 210–212°C (mp)
5.8 Properties of Enantiomers: Optical Activity
The molecules of enantiomers are not superposable and, on this basis alone, we have con-
cluded that enantiomers are different compounds. How are they different? Do enantiomers
resemble constitutional isomers and diastereomers in having different melting and boiling
points? The answer is no. Pure enantiomers have identical melting and boiling points. Do
pure enantiomers have different indexes of refraction, different solubilities in common sol-
vents, different infrared spectra, and different rates of reaction with achiral reagents? The
answer to each of these questions is also no.
Many of these properties (e.g., boiling points, melting points, and solubilities) are depen-
dent on the magnitude of the intermolecular forces operating between the molecules
(Section 2.13), and for molecules that are mirror images of each other these forces will be
identical. We can see an example of this if we examine Table 5.1, where boiling points of
the 2-butanol enantiomers are listed.
(b) and H Cl
CH3
F
CH3F
Cl
H
(c) and
H OH
OHH
(a) andCl F
H
Br
H
Br
F
Cl
solom_c05_186-229hr.qxd 28-09-2009 14:28 Page 2012.  Determine	
  se	
  as	
  duas	
  estruturas,	
  em	
  cada	
  
par	
  visto	
  a	
  seguir,	
  representam	
  enanAômeros	
  
ou	
  moléculas	
  domesmo	
  composto.	
  
Section 5.6 Naming Enantiomers: The R,S System of Nomenclature 191
It is easy to tell whether two molecules are enantiomers (nonsuperimposable) or
identical molecules (superimposable) if you have molecular models of the molecules—
just see whether the models superimpose. If, however, you are working with structures
on a two-dimensional piece of paper, the easiest way to determine whether two mole-
cules are enantiomers or identical molecules is by determining their configurations. If
one has the R configuration and the other has the S configuration, they are enan-
tiomers. If they both have the R configuration or both have the S configuration, they
are identical molecules.
When comparing two Fischer projections to see if they are the same or different,
never rotate one 90° or turn one over, because this is a quick way to get a wrong an-
swer. A Fischer projection can be rotated 180° in the plane of the paper, but this is the
only way to move it without risking an incorrect answer.
PROBLEM 8!
Indicate whether each of the following structures has the R or the S configuration:
a. c.
b. d.
PROBLEM 9! SOLVED
Do the following structures represent identical molecules or a pair of enantiomers?
a.
b.
c. d.
SOLUTION TO 9a The first structure shown in part (a) has the S configuration, and the
second structure has the R configuration. Because they have opposite configurations, the
structures represent a pair of enantiomers.
PROBLEM 10!
Assign relative priorities to the following groups:
a.
b.
c.
d. CH CH2CH3CH2 CH3
¬CH2CH2CH2Br¬Cl¬CH2CH2Br¬CH(CH3)2
¬CH2OH¬CH3¬OH¬CH“O
¬H¬CH2CH2OH¬CH3¬CH2OH
CH3 CH2CH3 and
Cl
H
H Cl
CH3
CH2CH3
C
H
CH2Br
OH and
CH3
C
HO
H
CH3
CH2Br
C
CH3
CH2Br
Cl and
CH2CH3
C
CH3CH2
Cl
CH3
CH2Br
C
HO
CH3
H and
CH2CH2CH3
C
CH3CH2CH2
OH
CH3
H
Cl
C
CH3CH2
CH2Br
CH2CH2Cl
OH
HO
C
CH3
CH(CH3)2
CH2CH3
CH2Br
BRUI05-182_237r4 20-03-2003 3:36 PM Page 191
3.  Considere	
  as	
  duas	
  estruturas	
  a	
  seguir	
  e	
  estabeleça	
  se	
  elas	
  
representam	
  enanAômeros	
  ou	
  duas	
  moléculas	
  do	
  mesmo	
  
composto.	
  	
  
! When light is filtered through
two polarized lenses at a 90° angle
to one other, no light is transmitted
through them.
192 C H A P T E R 5 Stereochemistry
PROBLEM 11"
Indicate whether each of the following structures has the R or the S configuration:
a. c.
b. d.
5.7 Optical Activity
Enantiomers share many of the same properties—they have the same boiling points,
the same melting points, and the same solubilities. In fact, all the physical properties
of enantiomers are the same except those that stem from how groups bonded to the
asymmetric carbon are arranged in space. One of the properties that enantiomers do
not share is the way they interact with polarized light.
What is polarized light? Normal light consists of electromagnetic waves that oscil-
late in all directions. Plane-polarized light (or simply polarized light), in contrast, os-
cillates only in a single plane passing through the path of propagation. Polarized light
is produced by passing normal light through a polarizer such as a polarized lens or a
Nicol prism.
You experience the effect of a polarized lens with polarized sunglasses. Polarized
sunglasses allow only light oscillating in a single plane to pass through them, so they
block reflections (glare) more effectively than nonpolarized sunglasses.
In 1815, the physicist Jean-Baptiste Biot discovered that certain naturally occurring
organic substances such as camphor and oil of turpentine are able to rotate the plane of
polarization. He noted that some compounds rotated the plane of polarization clock-
wise and others counterclockwise, while some did not rotate the plane of polarization
at all. He predicted that the ability to rotate the plane of polarization was attributable to
some asymmetry in the molecules. Van’t Hoff and Le Bel later determined that the
molecular asymmetry was associated with compounds having one or more asymmet-
ric carbons.
When polarized light passes through a solution of achiral molecules, the light
emerges from the solution with its plane of polarization unchanged. An achiral com-
pound does not rotate the plane of polarization.It is optically inactive.
direction of light propagation
polarizernormal
light
light
source
plane-polarized
light
light waves oscillate
in a single plane
light waves oscillate
in all directions
CH3 CH2CH2CH3
CH2CH2CH2CH3
CH2CH3
HO H
CH2CH2CH3
CH2OH
CH3 H
Br
CH2CH3
CH3CH2 CH2Br
CH(CH3)2
CH3
Born in Scotland, William Nicol
(1768–1851) was a professor at the
University of Edinburgh. He devel-
oped the first prism that produced
plane-polarized light. He also devel-
oped methods to produce thin slices
of materials for use in microscopic
studies.
BRUI05-182_237r4 20-03-2003 3:36 PM Page 192
S,	
  S	
  R	
  ,S	
  
Enan0ômeros	
  
§ 	
  Mesmas	
  propriedades	
  \sicas:	
  
	
  	
  -­‐	
  temperatura	
  de	
  fusão	
  
	
  	
  -­‐	
  temperatura	
  de	
  ebulição	
  
	
  	
  -­‐	
  densidade,	
  	
  
	
  	
  -­‐	
  índice	
  de	
  refração,	
  etc.	
  
§ 	
  Diferentes	
  propriedades	
  biológicas.	
  
§ 	
  Diferentes	
  propriedades	
  óp0cas	
  	
  
• 	
  diferente	
  interação	
  com	
  a	
  luz	
  polarizada.	
  
Compostos	
  quirais	
  são	
  
op6camente	
  a6vos	
  
	
  
	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  A0vidade	
  Ó0ca	
  
O	
  Polarímetro	
  e	
  a	
  luz	
  polarizada	
  
Luz	
  normal:	
  composta	
  por	
  ondas	
  
eletromagnéAcas	
  que	
  oscilam	
  em	
  
todas	
  as	
  direções.	
  	
  
Luz	
  plano-­‐polarizada:	
  	
  oscila	
  em	
  
apenas	
  uma	
  direção.	
  É	
  obAda	
  a	
  
parAr	
  da	
  luz	
  comum	
  -­‐	
  	
  prisma	
  de	
  
Nicol.	
  
O	
  Polarímetro	
  e	
  a	
  luz	
  polarizada	
  
Polarímetro:	
  aparelho	
  para	
  medida	
  do	
  desvio	
  da	
  
luz	
  polarizada.	
  	
  
Moléculas quirais: capacidade de desviar o plano da luz polarizada. 
 desvio para direita  dextrógira  (+) 
desvio para esquerda  levógira  (-) 
•  Se	
  uma	
  substância	
  gira	
  o	
  plano	
  de	
  polarização	
  
n o	
   s e n 0 d o	
   h o r á r i o , 	
   é 	
   c h a m a d a	
  
dextrorrotatória	
  (+).	
  
•  Se	
  uma	
  substância	
  gira	
  o	
  plano	
  de	
  polarização	
  
no	
   senAdo	
   an0-­‐horário ,	
   é	
   chamada	
  
levorrotatória	
  (-­‐).	
  
H
C2H5
CH3
OH
(R)-2-Butanol
rotation 
A0vidade	
  Ó0ca	
  
•  Não	
  há	
  correlação	
  entre	
  as	
  configurações	
  dos	
  
enanAômeros	
   	
   e	
   o	
   senAdo	
   (+)	
   ou	
   	
   (-­‐)	
   da	
  
rotação	
  da	
  luz	
  polarizada.	
  
A0vidade	
  Ó0ca	
  
A	
  rotação	
  depende:	
  
	
  
-­‐	
  da	
  natureza	
  da	
  amostra;	
  	
  
-­‐	
  do	
  comprimento	
  do	
  tubo;	
  	
  
-­‐	
  da	
  concentração	
  da	
  amostra;	
  
-­‐	
  da	
  temperatura;	
  	
  
-­‐	
  do	
  solvente;	
  
-­‐	
  da	
  fonte	
  de	
  luz	
  uAlizada.	
  	
  
A0vidade	
  Ó0ca	
  
	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  Polarímetro	
  
•  Rotação	
  específica	
  
–  [α]	
  é	
  a	
  rotação	
  específica	
  
– α	
  é	
  a	
  rotação	
  observada	
  
– c	
  concentração	
  em	
  g	
  cm-­‐3	
  
–  l	
  é	
  o	
  comprimento	
  da	
  célula	
  em	
  decímetros	
  (dm).	
  
A0vidade	
  Ó0ca	
  
	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  Polarímetro	
  
	
  
–  Isso	
   significa	
   que	
   um	
   desvio	
   no	
   senAdo	
   horário	
   de	
  
3,12°	
  foi	
  observado	
  com	
  uma	
  amostra	
  contendo	
  1,00	
  
g/mL	
  da	
  substância	
  opAcamente	
  aAva,	
  em	
  uma	
  célula	
  
de	
  1,00	
  dm,	
  uAlizando-­‐se	
   a	
   linha	
  D	
  de	
  uma	
   lâmpada	
  
de	
  sódio	
  (λ=	
  599,6	
  nm)	
  como	
  fonte	
  de	
  luz	
  e	
  mantendo	
  
uma	
  temperatura	
  de	
  25	
  °C	
  
A0vidade	
  Ó0ca	
  
•  Cada	
  substância	
  opAcamente	
  aAva	
  tem	
  uma	
  
rotação	
  específica.	
  
	
  
	
  
	
  
•  O	
  senAdo	
  da	
  rotação	
  da	
  luz	
  plano	
  polarizada	
  é	
  
incorporado	
  ao	
  nome	
  do	
  composto	
  
opAcamente	
  aAvo.	
  
A0vidade	
  Ó0ca	
  
•  Exercício…	
  
Para	
  uma	
  solução	
  de	
  uma	
  substância	
  orgânica	
  de	
  concentração	
  0,5	
  g	
  cm-­‐3	
  de	
  
um	
  composto	
  em	
  um	
  polarímetro	
  de	
  20	
  cm	
  de	
  comprimento	
  a	
  rotação	
  
observada	
  é	
  de	
  1,5	
  °.	
  Qual	
  a	
  rotação	
  observada?	
  
A0vidade	
  Ó0ca	
  
Estereoisômeros	
  com	
  mais	
  de	
  um	
  
centro	
  assimétrico	
  
•  r e g r a	
   g e r a l : 	
   n úme r o	
   m á x im o	
   d e	
  
estereoisômeros	
   é	
   2n,	
   onde	
   n	
   =	
   número	
   de	
  
carbonos	
  assimétricos.	
  
•  Por	
  exemplo:	
  2-­‐bromo-­‐3-­‐clorobutano	
  
H
C CH3C
H
H3C
Br Cl
1 2 3 4
(IV)(III)(II)(I)
H
CH3
Cl
H Br
CH3
Cl
CH3
H
Br H
CH3
Cl
CH3
H
H Br
CH3
H
CH3
Cl
Br H
CH3
H
C CH3C
H
H3C
Br Cl
1 2 3 4
§ 	
  I	
  e	
  II,	
  III	
  e	
  IV:	
  EnanAômeros.	
  
§  I	
  e	
  IV,	
  I	
  e	
  III,	
  II	
  e	
  IV,	
  II	
  e	
  III:	
  
Diastereoisômeros	
  
não	
  são	
  imagens	
  
especulares	
  
ü  Diastereoisômeros	
  propriedades	
  vsicas	
  diferentes	
  
H
CO2H
OH
H OH
CO2H
V
Iguais
H
CO2H
OH
H OH
CO2H
HO
CO2H
H
HO H
CO2H
VI
180º
H
CO2H
OH
HO H
CO2H
VII
HO
CO2H
H
H OH
CO2H
Diferentes
HO
CO2H
H
H OH
CO2H
VIII
180º
VII	
  e	
  VIII:	
  Enan0ômeros	
  
V	
  e	
  VI:	
  Compostos	
  meso	
  
V	
  e	
  VII,	
  V	
  e	
  VIII:	
  Diastereoisômeros	
  
Compostos	
  meso	
  
•  contêm	
  carbonos	
  assimétricos,	
  mas	
  imagens	
  
especulares	
  sobreponíveis	
  :	
  
–  possui	
  dois	
  ou	
  mais	
  carbonos	
  assimétricos;	
  
–  possui	
  plano	
  de	
  simetria;	
  
–  são	
  opAcamente	
  inaAvos.	
  
–  Elas	
  são	
  moléculas	
  aquirais.	
  
	
  
Compostos	
  meso	
  
Estereoisomerismo	
  de	
  Compostos	
  
Cíclicos	
  
•  Composto	
  meso	
  
Me
H Me
H Me
H Me
H
mirror
enantiomers
H
MeMe
H
Plane of 
symmetry 
a meso compound 
achiral 
Estereoisomerismo	
  em	
  compostos	
  
cíclicos	
  
Me
Me
Me
Me
Plane of 
symmetry 
v  1,4-Dimethylcyclohexane 
ü  As substâncias 1,4-dissubstituídas cis ou trans do 
ciclohexano são estereoisômeros mas não são 
quirais. 
v  1,3-Dimethylcyclohexane 
Me Me
**
cis-1,3-dimethyl
cyclohexane
Plane of 
symmetry 
•  Exercício…	
  
•  Quais	
  das	
  seguintes	
  substâncias	
  têm	
  um	
  estereoisômero	
  que	
  é	
  uma	
  
substância	
  meso?	
  
a.  2,3-­‐dimeAlbutano	
  
b.  3,4-­‐dimeAl-­‐hexano	
  
c.  2-­‐bromo-­‐3-­‐meAlpentano	
  
d.  1,3-­‐dimeAlciclo-­‐hexano	
  
e.  1,4-­‐dimeAlciclo-­‐hexano	
  
f.  1,2-­‐dimeAlciclo-­‐hexano	
  
g.  1,2-­‐dimeAlciclo-­‐hexano	
  
h.  3,4-­‐dieAl-­‐hexano	
  
i.  1-­‐bromo-­‐2-­‐meAlciclo-­‐hexano	
  
Check each compound to see if it has the necessary requirements to have a stereoisomer
that is a meso compound. That is, does it have two asymmetric carbons with the same four
substituents attached to each of the asymmetric carbons?
Compounds A, E, and G do not have a stereoisomer that is a meso compound because
they don’t have any asymmetric carbons.
Compounds C and H each have two asymmetric carbons. They do not have a stereoisomer
that is a meso compound because each of the asymmetric carbons is not bonded to the
same four substituents.
Compounds B, D, and F have a stereoisomer that is a meso compound—they have two asym-
metric carbons and each asymmetric carbon is bonded to the same four atoms or groups.
The isomer that is the meso compound is the one with a plane of symmetry when an
acyclic compound is drawn in its eclipsed conformation (B), or when a cyclic compound is
drawn with a planar ring (D and F).
Now continue on to Problem 26.
PROBLEM 26!
Which of the following compounds has a stereoisomer that is a meso compound?
a. 2,4-dibromohexane d. 1,3-dichlorocyclohexane
b. 2,4-dibromopentane e. 1,4-dichlorocyclohexane
c. 2,4-dimethylpentane f. 1,2-dichlorocyclobutane
PROBLEM 27 SOLVED
Which of the following are chiral?
H3C CH3
H3C
CH3 H3C
CH3 Cl
H3C
CH3
H3C
CH3
H3C Cl Cl
Cl CH3
Cl
Cl CH3
H3C
CH3
CH3
H H
H3C
FD
H
CH3
H
H3C
B
or
CH2CH3
CH2CH3
CH3
CH3H
H
CH3CH2
H H
H3C
CH2CH3
CH3
CC
F
CH3
CH3
D
CH3
CH3
B
CH3CH2CHCHCH2CH3
CH3
CH3
H
CH3
Br
C
CH3CHCHCH2CH3
CH3
Br
G
CH3CH2CHCHCH2CH3
CH2CH3
CH2CH3
E
CH3
H3C
A
CH3CHCHCH3
CH3
CH3
Section 5.10 Meso Compounds 203
BRUI05-182_237r4 20-03-2003 3:36 PM Page 203
Check each compound to see if it has the necessary requirements to have a stereoisomer
that is a meso compound. That is, does it have two asymmetric carbons with the same four
substituents attached to each of the asymmetric carbons?
Compounds A, E, and G do not have a stereoisomer that is a meso compound because
they don’t have any asymmetric carbons.
Compounds C and H each have two asymmetric carbons. They do not have a stereoisomer
that is a meso compound because each of the asymmetric carbons is not bonded to the
same four substituents.
Compounds B, D, and F have a stereoisomer that is a meso compound—they have two asym-
metric carbons and each asymmetric carbon is bonded to the same four atoms or groups.
The isomer that is the meso compound is the one with a plane of symmetry when an
acyclic compound is drawn in its eclipsed conformation (B), or when a cyclic compound is
drawn with a planar ring (D and F).
Now continue on to Problem 26.
PROBLEM 26!
Which of the following compounds has a stereoisomer that is a meso compound?
a. 2,4-dibromohexane d. 1,3-dichlorocyclohexane
b. 2,4-dibromopentane e. 1,4-dichlorocyclohexane
c. 2,4-dimethylpentane f. 1,2-dichlorocyclobutane
PROBLEM 27 SOLVED
Which of the following are chiral?
H3C CH3
H3C
CH3 H3C
CH3 Cl
H3C
CH3
H3C
CH3
H3C Cl Cl
Cl CH3
Cl
Cl CH3
H3C
CH3
CH3
H H
H3C
FD
H
CH3
H
H3C
B
or
CH2CH3
CH2CH3
CH3
CH3H
H
CH3CH2
H H
H3C
CH2CH3
CH3
CC
F
CH3
CH3
D
CH3
CH3
B
CH3CH2CHCHCH2CH3
CH3
CH3
H
CH3
Br
C
CH3CHCHCH2CH3
CH3
Br
G
CH3CH2CHCHCH2CH3
CH2CH3
CH2CH3
E
CH3
H3C
A
CH3CHCHCH3
CH3
CH3
Section 5.10 Meso Compounds 203
BRUI05-182_237r4 20-03-2003 3:36 PM Page 203
Check each compound to see if it has the necessary requirements to have a stereoisomer
that is a meso compound. That is, does it have two asymmetric carbons with the same four
substituents attached to each of the asymmetric carbons?
Compounds A, E, and G do not have a stereoisomer that is a meso compound because
they don’t have any asymmetric carbons.
Compounds C and H each have two asymmetric carbons. They do not have a stereoisomer
that is a meso compound because each of the asymmetric carbons is not bonded to the
same four substituents.
Compounds B, D, and F have a stereoisomer that is a meso compound—they have two asym-
metric carbons and each asymmetric carbon is bonded to the same four atoms or groups.
The isomer that is the meso compound is the one with a plane of symmetry when an
acyclic compound is drawn in its eclipsed conformation (B), or when a cyclic compound is
drawn with a planar ring (D and F).
Now continue on to Problem 26.
PROBLEM 26!
Which of the following compounds has a stereoisomer that is a meso compound?
a. 2,4-dibromohexane d. 1,3-dichlorocyclohexane
b. 2,4-dibromopentane e. 1,4-dichlorocyclohexane
c. 2,4-dimethylpentane f. 1,2-dichlorocyclobutane
PROBLEM 27 SOLVED
Which of the following are chiral?
H3C CH3
H3C
CH3 H3C
CH3 Cl
H3C
CH3
H3C
CH3
H3C Cl Cl
Cl CH3
Cl
Cl CH3
H3C
CH3
CH3
H H
H3C
FD
H
CH3
H
H3C
B
or
CH2CH3
CH2CH3
CH3
CH3H
H
CH3CH2
H H
H3C
CH2CH3
CH3
CC
F
CH3
CH3
D
CH3
CH3
B
CH3CH2CHCHCH2CH3
CH3
CH3
H
CH3
Br
C
CH3CHCHCH2CH3
CH3
Br
G
CH3CH2CHCHCH2CH3
CH2CH3
CH2CH3
E
CH3
H3C
A
CH3CHCHCH3
CH3
CH3
Section 5.10 Meso Compounds 203
BRUI05-182_237r4 20-03-2003 3:36 PM Page 203
Check each compound to see if it has the necessary requirements to have a stereoisomer
that is a meso compound. That is, does it have two asymmetric carbons with the same four
substituents attached to each of the asymmetric carbons?
Compounds A, E, and G do not have a stereoisomer that is a meso compound because
they don’t have any asymmetric carbons.
Compounds C and H each have two asymmetric carbons. They do not have a stereoisomer
that is a meso compound because each of the asymmetric carbons is not bonded to the
same four substituents.
Compounds B, D, and F have a stereoisomer that is a meso compound—they have two asym-
metric carbons and each asymmetric carbon is bonded to the same four atoms or groups.
The isomer that is the meso compound is the one with a plane of symmetry when an
acyclic compound is drawn in its eclipsed conformation (B), or when a cyclic compound is
drawn with a planar ring (D and F).
Now continue on to Problem 26.
PROBLEM 26!
Which of the following compounds has a stereoisomer that is a meso compound?
a. 2,4-dibromohexane d. 1,3-dichlorocyclohexane
b. 2,4-dibromopentane e. 1,4-dichlorocyclohexane
c. 2,4-dimethylpentane f. 1,2-dichlorocyclobutane
PROBLEM 27 SOLVED
Which of the following are chiral?
H3C CH3
H3C
CH3 H3C
CH3 Cl
H3C
CH3
H3C
CH3
H3C Cl Cl
Cl CH3
Cl
Cl CH3
H3C
CH3
CH3
H H
H3C
FD
H
CH3
H
H3C
B
or
CH2CH3
CH2CH3
CH3
CH3H
H
CH3CH2
H H
H3C
CH2CH3
CH3
CC
F
CH3
CH3
D
CH3
CH3
B
CH3CH2CHCHCH2CH3CH3
CH3
H
CH3
Br
C
CH3CHCHCH2CH3
CH3
Br
G
CH3CH2CHCHCH2CH3
CH2CH3
CH2CH3
E
CH3
H3C
A
CH3CHCHCH3
CH3
CH3
Section 5.10 Meso Compounds 203
BRUI05-182_237r4 20-03-2003 3:36 PM Page 203
•  Exercício…	
  
A	
  substância	
  a	
  seguir	
  tem	
  tem	
  somente	
  um	
  carbono	
  assimétrico.Por	
  que	
  tem	
  
quatro	
  estereoisômeros?	
  
198 C H A P T E R 5 Stereochemistry
groups on the same side of the carbon chain are called the erythro enantiomers
(Section 22.3). Those with similar groups on opposite sides are called the threo enan-
tiomers. Therefore, 1 and 2 are the erythro enantiomers of 3-chloro-2-butanol (the hy-
drogens are on the same side), whereas 3 and 4 are the threo enantiomers. In each of
the Fischer projections shown here, the horizontal bonds project out of the paper to-
ward the viewer and the vertical bonds extend behind the paper away from the viewer.
Groups can rotate freely about the carbon–carbon single bonds, but Fischer projec-
tions show the stereoisomers in their eclipsed conformations.
A Fischer projection does not show the three-dimensional structure of the molecule,
and it represents the molecule in a relatively unstable eclipsed conformation. Most
chemists, therefore, prefer to use perspective formulas because they show the mole-
cule’s three-dimensional structure in a stable, staggered conformation, so they provide
a more accurate representation of structure. When perspective formulas are drawn to
show the stereoisomers in their less stable eclipsed conformations, it can easily be
seen—as the eclipsed Fischer projections show—that the erythro isomers have similar
groups on the same side. We will use both prespective formulas and Fischer projections
to depict the arrangement of groups bonded to an asymmetric carbon.
PROBLEM 18
The following compound has only one asymmetric carbon. Why then does it have four
stereoisomers?
PROBLEM 19!
a. Stereoisomers with two asymmetric carbons are called _____ if the configuration of
both asymmetric carbons in one isomer is the opposite of the configuration of the asym-
metric carbons in the other isomer.
b. Stereoisomers with two asymmetric carbons are called _____ if the configuration of
both asymmetric carbons in one isomer is the same as the configuration of the asym-
metric carbons in the other isomer.
c. Stereoisomers with two asymmetric carbons are called _____ if one of the asymmetric
carbons has the same configuration in both isomers and the other asymmetric carbon
has the opposite configuration in the two isomers.
PROBLEM 20!
a. How many asymmetric carbons does cholesterol have?
b. What is the maximum number of stereoisomers that cholesterol can have?
c. How many of these stereoisomers are found in nature?
cholesterol
CH3
CH3
H H
HH3C
H3C
H3C
HO
CH3CH2CHCH2CH CHCH3
Br
*
erythro enantiomers
perspective formulas of the stereoisomers of 3-chloro-2-butanol (eclipsed)
1 2
Cl
H
Cl
H 3 4
H3C
H
Cl H
OH
CH3
C
H
OH
CH3
C C
H3C
C
CH3
H
H
HO
Cl
C
CH3
C C
H3C
H
HO C
H3C
threo enantiomers
Tutorial:
Identification of 
asymmetric carbons
BRUI05-182_237r4 20-03-2003 3:36 PM Page 198
•  Exercício…	
  
Quntos	
  carbonos	
  assiméricos	
  tem	
  o	
  colesterol?	
  
Qual	
  é	
  o	
  número	
  máximo	
  de	
  estereoisômeros	
  que	
  o	
  colesterol	
  pode	
  ter?	
  
198 C H A P T E R 5 Stereochemistry
groups on the same side of the carbon chain are called the erythro enantiomers
(Section 22.3). Those with similar groups on opposite sides are called the threo enan-
tiomers. Therefore, 1 and 2 are the erythro enantiomers of 3-chloro-2-butanol (the hy-
drogens are on the same side), whereas 3 and 4 are the threo enantiomers. In each of
the Fischer projections shown here, the horizontal bonds project out of the paper to-
ward the viewer and the vertical bonds extend behind the paper away from the viewer.
Groups can rotate freely about the carbon–carbon single bonds, but Fischer projec-
tions show the stereoisomers in their eclipsed conformations.
A Fischer projection does not show the three-dimensional structure of the molecule,
and it represents the molecule in a relatively unstable eclipsed conformation. Most
chemists, therefore, prefer to use perspective formulas because they show the mole-
cule’s three-dimensional structure in a stable, staggered conformation, so they provide
a more accurate representation of structure. When perspective formulas are drawn to
show the stereoisomers in their less stable eclipsed conformations, it can easily be
seen—as the eclipsed Fischer projections show—that the erythro isomers have similar
groups on the same side. We will use both prespective formulas and Fischer projections
to depict the arrangement of groups bonded to an asymmetric carbon.
PROBLEM 18
The following compound has only one asymmetric carbon. Why then does it have four
stereoisomers?
PROBLEM 19!
a. Stereoisomers with two asymmetric carbons are called _____ if the configuration of
both asymmetric carbons in one isomer is the opposite of the configuration of the asym-
metric carbons in the other isomer.
b. Stereoisomers with two asymmetric carbons are called _____ if the configuration of
both asymmetric carbons in one isomer is the same as the configuration of the asym-
metric carbons in the other isomer.
c. Stereoisomers with two asymmetric carbons are called _____ if one of the asymmetric
carbons has the same configuration in both isomers and the other asymmetric carbon
has the opposite configuration in the two isomers.
PROBLEM 20!
a. How many asymmetric carbons does cholesterol have?
b. What is the maximum number of stereoisomers that cholesterol can have?
c. How many of these stereoisomers are found in nature?
cholesterol
CH3
CH3
H H
HH3C
H3C
H3C
HO
CH3CH2CHCH2CH CHCH3
Br
*
erythro enantiomers
perspective formulas of the stereoisomers of 3-chloro-2-butanol (eclipsed)
1 2
Cl
H
Cl
H 3 4
H3C
H
Cl H
OH
CH3
C
H
OH
CH3
C C
H3C
C
CH3
H
H
HO
Cl
C
CH3
C C
H3C
H
HO C
H3C
threo enantiomers
Tutorial:
Identification of 
asymmetric carbons
BRUI05-182_237r4 20-03-2003 3:36 PM Page 198
•  Nomenclatura	
  de	
  Compostos	
  com	
  mais	
  de	
  Um	
  
Carbono	
  Assimétrico	
  
Fischer 
Projection 
Nomenclatura	
  de	
  Compostos	
  com	
  mais	
  de	
  Um	
  Carbono	
  Assimétrico	
  
Nomenclatura	
  de	
  Compostos	
  com	
  mais	
  de	
  Um	
  Carbono	
  Assimétrico	
  
H3C
Ph
COOH
HO
Et Br
Et OH
Br Ph
CH3
COOH
Et OH
Br Ph
CH3
COOH
Fischer 
Projection 
Nomenclatura	
  de	
  Compostos	
  com	
  mais	
  de	
  Um	
  Carbono	
  Assimétrico	
  
H3C
Cl
CH3
H
Cl H
(I)
(2S, 3S)-Dichlorobutane
Cl H
H Cl
CH3
CH3
CH3
Cl
H3C
H
ClH
(II)
(2R, 3R)-Dichlorobutane
H Cl
Cl H
CH3
CH3
mirror
enantiomers 
Nomenclatura	
  de	
  Compostos	
  com	
  mais	
  de	
  Um	
  Carbono	
  Assimétrico	
  
H3C
Cl
CH3
Cl
H H
(III)
(2S, 3R)-Dichlorobutane
H Cl
H Cl
CH3
CH3
v  (III) is achiral (a meso compound) 
Plane of 
symmetry 
•  Exercício:	
  
1.  Escreva	
  a	
  fórmula	
  de	
  projeção	
  de	
  Fischer	
  ou	
  
a	
   fórmula	
   	
   tridimensional	
   do	
   composto	
  
abaixo.	
   Em	
   seguida	
   dê	
   os	
   nomes	
   dos	
  	
  
compostos,	
  Incluindo	
  as	
  designações	
  (R)	
  e	
  (S)	
  
•  Exercício:	
  
2.  Escreva	
   a	
   fórmula	
   de	
   projeção	
   de	
   Fischer	
   ou	
   a	
   fórmula	
  	
  
tridimensional	
  do2,3-­‐dibrobutano.	
  Em	
  seguida	
  dê	
  os	
  nomes	
  dos	
  	
  
compostos,	
  Incluindo	
  as	
  designações	
  (R)	
  e	
  (S).	
  
2S,3R	
   2S,3S	
   2R,3R	
  
Mistura	
  Racêmica	
  
•  Uma	
  mistura	
  equimolar	
  de	
  dois	
  enanAômeros	
  
é	
  chamada	
  de	
  mistura	
  racêmica	
  ou	
  racemato.	
  
•  Uma	
  mistura	
  racêmica	
  não	
  causa	
  rotação	
  
líquida	
  no	
  plano	
  da	
  luz	
  polarizada.	
  
H
C2H5
CH3
OH
(R)-2-Butanol
H
C2H5
H3C
HO
(S)-2-Butanol
(if present)
rotation 
equal & opposite 
rotation by the 
enantiomer 
Mistura	
  Racêmica	
  
	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  Pureza	
  óp0ca	
  e	
  excesso	
  enan0omérico	
  
•  Uma	
  amostra	
  de	
  uma	
  substância	
  opAcamente	
  
aAva	
  que	
  consiste	
  em	
  um	
  único	
  enanAômero	
  
é	
   denominada	
   enanAomericamente	
   pura	
   ou	
  
que	
  tem	
  um	
  excesso	
  enanAomérico	
  de	
  100%.	
  
•  O	
   excesso	
   enanAomérico	
   (ee)	
   pode	
   ser	
  
calculado	
  a	
  parAr	
  da	
  rotação	
  ópAca:	
  
	
  
	
  
%	
  ee	
  =	
  	
  
rotação	
  específica	
  observada	
  
rotação	
  específica	
  do	
  enanAômero	
  puro	
  
X	
  100	
  
•  1.	
  Exercício:	
  Uma	
  mistura	
  dos	
  enanAômeros	
  do	
  2-­‐
butanol	
  apresenta	
  rotação	
  específica	
  de	
  +6,76	
  (use	
  
os	
  dados	
  da	
  figura	
  abaixo).	
  O	
  ee	
  é:	
  
	
  
– Qual	
  a	
  composição	
  estereoisomérica	
  real	
  da	
  mistura?	
  
%	
  ee	
  =	
  	
  
rotação	
  específica	
  observada	
  
rotação	
  específica	
  do	
  enanAômero	
  puro	
  
X	
  100	
  
2.  Uma	
  amostra	
  de	
  2-­‐meAl-­‐1-­‐butanol	
  tem	
  rotação	
  
específica	
   de	
   +1,151.	
   Qual	
   a	
   percentagem	
   de	
  
excesso	
   enanAomérico	
   da	
   amostra?	
   Que	
  
enanAômero	
  está	
  em	
  excesso,	
  (R)	
  ou	
  (S)	
  ?	
  
–  use	
  os	
  dados	
  da	
  figura	
  abaixo	
  
Estereoisômeros	
  Conformacionais	
  
•  	
  	
  Estão	
  relacionados	
  um	
  ao	
  outro	
  por	
  rotações	
  de	
  torno	
  de	
  ligação	
  
•  Os	
   diferentes	
   arranjos	
   espaciais	
   de	
   átomos	
  
resultantes	
  da	
   rotação	
  em	
  torno	
  de	
  uma	
   ligação	
  
simples	
  são	
  chamados	
  conformações.	
  
•  Cada	
  estrutura	
  possível	
  é	
  chamada	
  confôrmero.	
  
•  A	
  invesAgação	
  de	
  várias	
  conformações	
  de	
  uma	
  
substância	
   e	
   as	
   respecAvas	
   estabilidades	
   é	
  
chamada	
  Análise	
  conformacional.	
  
	
  	
  	
  	
  	
  	
  	
  	
  	
  Fórmula	
  de	
  projeção	
  de	
  Newman	
  
4.8 Sigma Bonds and Bond Rotation 157
trol, a much more environmentally sensitive method than the
use of insecticides.
Research suggests there are roles for pheromones in the
lives of humans as well. For example, studies have shown
that the phenomenon of menstrual synchronization among
women who live or work with each other is likely caused by
pheromones. Olfactory sensitivity to musk, which includes
steroids such as androsterone, large cyclic ketones, and lac-
tones (cyclic esters), also varies cyclically in women, differs
between the sexes, and may influence our behavior. Some
of these compounds are used in perfumes, including cive-
tone, a natural product isolated from glands of the civet cat,
and pentalide, a synthetic musk.
Civetone Pentalide
O
O
O
Androsterone
H
HO
O
H3C
H3C
H
H
H
4.8 Sigma Bonds and Bond Rotation
Two groups bonded by only a single bond can undergo rotation about that bond with respect
to each other.
! The temporary molecular shapes that result from such a rotation are called con-
formations of the molecule.
! Each possible structure is called a conformer.
! An analysis of the energy changes that occur as a molecule undergoes rotations
about single bonds is called a conformational analysis.
4.8A Newman Projections and How to Draw Them
When we do conformational analysis, we will find that certain types of structural formu-
las are especially convenient to use. One of these types is called a Newman projection
formula and another type is a sawhorse formula. Sawhorse formulas are much like
dash–wedge three-dimensional formulas we have used so far. In conformational analyses,
we will make substantial use of Newman projections.
To write a Newman projection formula:
! We imagine ourselves taking a view from one atom (usually a carbon) directly
along a selected bond axis to the next atom (also usually a carbon atom).
! The front carbon and its other bonds are represented as .
! The back carbon and its bonds are represented as .
Newman projection
formula
Sawhorse formula
Learn to draw Newman projections
and sawhorse formulas. Build
handheld molecular models and
compare them with your drawings.
Helpful Hint
solom_c04_137-185hr.qxd 24-09-2009 11:22 Page 157
	
  	
  	
  	
  	
  	
  	
  	
  	
  Representação	
  em	
  cavalete	
  
•  Par	
   a	
   escrever	
   uma	
   formula	
   de	
   projeção	
   de	
  
Newman:	
  
– Nos	
   imaginamos	
   tendo	
   uma	
   visão	
   de	
   um	
   átomo	
  
(normalmente	
  de	
  carbono)	
  diretamente	
  ao	
   longo	
  
de	
   um	
   eixo	
   de	
   ligação	
   selecionado	
   a	
   parAr	
   do	
  
átomo	
   seguinte	
   (também	
   geralmente	
   um	
   átomo	
  
de	
  carbono);	
  
– O	
  carbono	
  da	
  frente	
  são	
  representados	
  por	
  
– O	
   carbono	
   de	
   trás	
   e	
   as	
   suas	
   ligações	
   são	
  
representadas	
  por	
  
Análise	
  conformacional	
  
•  Quando	
   a	
   rotação	
   ocorre	
   em	
   torno	
   de	
   uma	
  
ligação	
   carbono-­‐carbono	
   do	
   etano,	
   pode	
  
resultar	
  em	
  duas	
  conformações.	
  
– Uma	
  conformação	
  em	
  oposição	
  (ou	
  alternada)	
  
– Uma	
  conformação	
  eclipsada.	
  
Análise	
  conformacional	
  
Look from this
direction
Hc
H Hb
Ha
HH
staggered 
confirmation
of ethane
f1 = 60o 
f2 = 180o 
•  Por	
  exemplo	
  o	
  etano:	
  conformação	
  em	
  oposição	
  
Look from this
direction
eclipsed 
confirmation
of ethaneH H
H H
HH
f = 0o 
•  Por	
  exemplo	
  o	
  etano:	
  conformação	
  eclipsada	
  
	
  	
  	
  	
  Análise	
  conformacional	
  
ü  Barreira	
  torsional	
  da	
  ligação	
  simples	
  
	
  Análise	
  conformacional	
  
•  Um	
  confôrmero	
  em	
  oposição	
  é	
  mais	
  estável	
  que	
  
um	
  confôrmero	
  eclipsado.	
  
•  Tensão	
  torsional:	
  é	
  o	
  nome	
  dado	
  à	
  repulsão	
  senAda	
  
pelos	
  elétrons	
  ligantes	
  de	
  um	
  subsAtuinte	
  quando	
  
passam	
  perto	
  dos	
  elétrons	
  de	
  outro	
  subsAtuinte.	
  
Análise	
  conformacional	
  do	
  butano	
  
•  Se	
  considerarmos	
  a	
  rotação	
  sobre	
  a	
  ligação	
  
C2-­‐C3	
  do	
  butano,	
  descobriremos	
  que	
  existem	
  
seis	
  confôrmeros	
  importantes:	
  
Análise	
  conformacional	
  do	
  butano	
  
C2-­‐C3	
  
CH3
H
CH3
H
CH3
HH
H
CH3
H
H
HCH3
H H
CH3 H
H
anti conformer
(I)
(lowest energy)
eclipsed conformer
(II)
gauche conformer
(III)
CH3
H H
H H
H3C
eclipsed conformer
(IV)
(highest energy)
CH3
H H
H CH3
H
eclipsedconformer
(VI)
H
CH3
H
H
CH3H
gauche conformer
(V)
CH3 on front carbon
rotates 60o clockwise
=
P
P
P
P
P
P
Análise	
  conformacional	
  do	
  butano	
  
Análise	
  conformacional	
  do	
  butano	
  
Análise	
  conformacional	
  do	
  butano	
  
Análise	
  conformacional	
  
•  O	
  confôrmero	
  mais	
  estável	
  é	
  o	
  an6.	
  
•  Os	
   confôrmeros	
   an0	
   e	
   gauche	
   não	
   tem	
   a	
  
mesma	
  energia	
  devido	
  à	
  tensão	
  estérica.	
  
•  Tensão	
   estérica:	
   repulsão	
   entre	
   nuvens	
  
eletrônicas	
   de	
   átomos	
   ou	
   grupos.	
   Esse	
   0po	
   de	
  
tensão	
  estérica	
  é	
  chamada	
  interação	
  gauche.	
  
•  Todos	
   os	
   confôrmeros	
   eclipsados	
   tem	
   tanto	
  
tensão	
  torsional	
  quanto	
  tensão	
  estérica.	
  
Exercícios	
  
1.  Desenhe	
  todos	
  os	
  confôrmeros	
  em	
  oposição	
  e	
  
eclipsados	
  que	
  resultam	
  da	
  rotação	
  em	
  torno	
  da	
  
ligação	
  C-­‐2-­‐-­‐-­‐C-­‐3	
  do	
  pentano.	
  
2.  Desenhe	
  um	
  diagrama	
  de	
  energia	
  potencial	
  para	
  a	
  
rotação	
  da	
  ligação	
  C-­‐2-­‐-­‐-­‐C-­‐3	
  do	
  pentano	
  ao	
  longo	
  de	
  
360°,	
  iniciando	
  do	
  confôrmero	
  menos	
  estável.	
  
3.  Usando	
  projeções	
  de	
  Newman,	
  desenhe	
  os	
  
confôrmeros	
  mais	
  estáveis	
  para	
  os	
  seguintes	
  itens:	
  
a.  3-­‐meAlpentano,	
  considerando	
  a	
  rotação	
  em	
  torno	
  da	
  
ligação	
  C2-­‐C3	
  
b.  3-­‐meAl-­‐hexano	
  (C3-­‐C-­‐4)	
  
c.  3,3-­‐dimeAl-­‐hexano	
  (C3-­‐C4)	
  
Análise	
  conformacional	
  
•  Quanto	
   mais	
   estável	
   for	
   a	
   conformação,	
   maior	
   é	
   a	
  
fração	
  da	
  molécula	
  que	
  estará	
  naquela	
  conformação.	
  
Cicloalcanos:	
  tensão	
  no	
  anel	
  
•  Nem	
   todos	
   os	
   cicloalcanos	
   tem	
   a	
   mesma	
  
estabilidade	
  relaAva.	
  
•  O	
   ciclo-­‐hexano	
   é	
   o	
   cicloalcano	
   mais	
   estável,	
   o	
  
ciclopropano	
   e	
   o	
   ciclobutano	
   são	
   muito	
   menos	
  
estáveis.	
  Essa	
  diferença	
  de	
  estabilidade	
  é	
  devido	
  
à	
  tensão	
  de	
  anel,	
  que	
  pode	
  ser:	
  
–  Tensão	
  angular:	
   	
  é	
  o	
   resultado	
  do	
  desvio	
  de	
  ângulos	
  
de	
   ligação	
   ideais	
   provocado	
   por	
   limitações	
  
estrututurais	
   inerentes	
   (tais	
   como	
   o	
   tamanho	
   do	
  
anel).	
  
Tensão	
  de	
  torsão:	
  é	
  causada	
  pela	
  repulsão	
  entre	
  os	
  elétrons	
  ligantes	
  de	
  um	
  subsAtuinte	
  e	
  os	
  
elétrons	
  ligantes	
  de	
  um	
  subsAtuinte	
  próximo.	
  
Tensão	
  estérica:	
   é	
   causada	
  por	
  átomos	
  ou	
  grupos	
  de	
  átomos	
  muito	
  aproximados	
  uns	
  dos	
  
outros.	
  
	
  
	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  Ciclopropano	
  
•  Tem	
  tensão	
  angular	
  tensão	
  de	
  torsão	
  
	
  
Cicloalcanos:	
  tensão	
  no	
  anel	
  
H H
H H
H H
sp3 hybridized carbon 
(normal tetrahedral 
bond angle is 109.5o) 
v  Internal bond angle (q) ~60o (~49.5o 
deviated from the ideal tetrahedral angle) 
	
  	
  Ciclopropano	
  
Cicloalcanos:	
  tensão	
  no	
  anel	
  
Ciclobutano	
  
•  Tem	
  considerável	
  tensão	
  angular.	
  
H H
HH
H
H
H
H
v  Internal bond angle (q) ~88o (~19,5o deviated 
from the normal 109.5o tetrahedral angle) 
v  Se fosse planar a tensão angular seria menor 
(90°) 
Cicloalcanos:	
  tensão	
  no	
  anel	
  
•  Substâncias	
   cíclicas	
   se	
   torcem	
  e	
   se	
   curvam	
  a	
  fim	
  de	
  obter	
  
uma	
   estrutura	
   que	
   minimize	
   os	
   três	
   Apos	
   diferentes	
   de	
  
tensão	
  que	
  pode	
  desestabilizar	
  uma	
  substância	
  cíclica.	
  	
  
Cicloalcanos:	
  tensão	
  no	
  anel	
  
Ciclopentano	
  
•  Possui	
  tensão	
  de	
  torsão	
  e	
  tensão	
  angular	
  
pequenas,	
  ele	
  é	
  quase	
  tão	
  estável	
  quanto	
  o	
  ciclo-­‐
hexano.	
  
H
H
H
H
H HH
H H
H
v  If cyclopentane were planar, q ~108o, very close 
to the normal tetrahedral angle of 109.5o 
v  However, planarity would introduce considerable 
torsional strain (i.e. 10 C–H bonds eclipsed) 
v  Therefore cyclopentane has a slightly bent 
conformation 
Conformações	
  do	
  ciclo-­‐hexano	
  
•  O	
  ciclo-­‐hexano	
  é	
  mais	
  estável	
  do	
  os	
  outros	
  
cicloalcanos	
  e	
  ele	
  tem	
  várias	
  conformações	
  que	
  
são	
  importantes:	
  
–  	
  conformação	
  mais	
  estável	
  do	
  ciclo-­‐hexano	
  é	
  a	
  
conformação	
  em	
  cadeira.	
  
	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  Conformações	
  do	
  ciclo-­‐hexano	
  
•  Através	
   de	
   rotações	
   parciais	
   em	
   torno	
   das	
  
ligações	
   simples	
   carbono-­‐carbono	
   do	
   anel	
   a	
  
conformação	
   em	
   cadeira	
   pode	
   assumir	
   outra	
  
forma	
  chamada	
  de	
  conformação	
  em	
  barco.	
  164 Chapter 4 Nomenclature and Conformations of Alkanes and Cycloalkanes
(a) (b)
CH2
H
H H
H
H
H
H
H
H
CH2
H
H
H
H
4
1
2
365
Figure 4.12 (a) A Newman projection of the chair conformation of cyclohexane. (Comparisons
with an actual molecular model will make this formulation clearer and will show that similar
staggered arrangements are seen when other carbon–carbon bonds are chosen for sighting.) 
(b) Illustration of large separation between hydrogen atoms at opposite corners of the ring
(designated C1 and C4) when the ring is in the chair conformation.
as well. When viewed along any carbon–carbon bond (viewing the structure from an end,
Fig. 4.12), the bonds are seen to be perfectly staggered. Moreover, the hydrogen atoms at
opposite corners of the cyclohexane ring are maximally separated.
! By partial rotations about the carbon–carbon single bonds of the ring, the chair
conformation can assume another shape called the boat conformation (Fig. 4.13).
! The boat conformation has no angle strain, but it does have torsional strain.
(b)
(c)(a)
H
H H
H
H
H
H
HH
H
H
H
H
H
HH
H
H
H
H
H HH
H
Figure 4.13 (a) The boat conformation of cyclohexane is
formed by “flipping” one end of the chair form up (or down).
This flip requires only rotations about carbon–carbon single
bonds. (b) Ball-and-stick model of the boat conformation. 
(c) A space-filling model of the boat conformation.
(a) (b)
H
HH
HH
H H
H
CH2
CH2
H H
H H
41
Figure 4.14 (a) Illustration of the eclipsed conformation of the boat conformation of cyclohexane.
(b) Flagpole interaction of the C1 and C4 hydrogen atoms of the boat conformation. The C1–C4
flagpole interaction is also readily apparent in Fig. 4.13c.
When a model of the boat conformation is viewed down carbon–carbon bond axes along
either side (Fig. 4.14a), the C H bonds at those carbon atoms are found to be eclipsed,
causing torsional strain. Additionally, two of the hydrogen atoms on C1 and C4 are close
enough to each other to cause van der Waals repulsion (Fig. 4.14b). This latter effect has
9
You will best appreciate the
differences between the chair and
boat forms of cyclohexane by
building and manipulating
molecular models of each.
Helpful Hint
solom_c04_137-185hr.qxd