Lab Exercise: Visualizing and drawing Organic Molecules in 3-Dimensions
The simplest organic molecule is methane, and in a previous lab, we considered its
structure and the fact that the four hydrogens point towards the corners of a tetrahedron. Thus,
methane is not planar, and in order to represent it in 3-dimensions, we draw two of the hydrogens
with normal lines, one with a bold line, and one with a dashed line.
Exercise 1: Make a model of methane and orient it as shown in Figure 1, then draw it to the
right of the model in such a way as to represent the orientation of the molecule using bold and
dashed lines as needed, and with all the hydrogens labeled.
Figure 1 Draw the molecule above with the hydrogens labeled (Ha, Hb, etc.)
Exercise 2: Now, rotate the model of methane by 180º along the axis shown in Figure 2 and
draw it to the right of the model in such a way as to represent the orientation of the molecule
using bold and dashed lines as needed, and with all the hydrogens labeled.
Figure 2 Draw the molecule above with the hydrogens labeled (Ha, Hb, etc.)
Ethane and Eclipsed Versus Staggered Conformations
Ethane is the simplest organic molecule which can potentially exist in more than one
conformation. The C-C bond is, of course, a σ-bond, and as such, it can undergo rotation. The
lowest energy conformation of ethane is the staggered conformation as shown in the model
below. Note that Ha, the two carbons, and Hf are in the same plane and that they have a dihedral
angle of 180º.
Exercise 3: Make a model of ethane in the conformation shown in Figure 3, then draw it to the
right of the model in such a way as to represent the orientation of the molecule using bold and
dashed lines as needed, and with all the hydrogens labeled.
Figure 3 Draw the molecule above with the hydrogens labeled (Ha, Hb, etc.)
If we rotate about the C-C bond by 60º, we obtained the eclipsed conformation of ethane.
As we saw in the first molecular models exercise, eclipsed ethane is about 3 kcal/mol higher in
energy than staggered ethane. The model below was created by rotating the methyl group on the
right clockwise by 60º. Note that Ha, the two carbons, and He are in the same plane and that they
have a dihedral angle of 0º. This is not a stable conformation, it is an energy maximum, and the
molecule does not exist in this conformer for any length of time, it is simply accessed as the
molecule rotates from one staggered orientation to another. This is true of almost all eclipsed
conformations, they are energy maxima and are only accessed fleetingly.
Exercise 4: Rotate about the C-C bond of your model of ethane to place it in the conformation
shown in Figure 4 by holding the methyl group on the left steady and rotating the methyl group
on the right. Draw the resulting conformer being sure to label the hydrogens as indicated in the
Figure 4 Draw the molecule above with the hydrogens labeled (Ha, Hb, etc.)
Butane and Anti vs. Gauche Conformers
As we saw in the Conformational Analysis of Alkanes, there are three staggered
conformations for butane, one anti and two gauche conformers. The two gauche conformers for
butane are gauche-(+) and gauche-(-). These are named after the right-hand rule: in the (+)
compound, when sighting from the front carbon to the rear carbon, one travels in a clockwise
direction from CH3(1) to CH3(2). In the (-) compound, when sighting from the front carbon to
the rear carbon one travels in counterclockwise direction from CH3(1) to CH3(2). These are
shown in Figure 5. These are chiral conformers, the (+) conformer is the mirror image of the (-)
conformer, and these two conformers are not superimposable. However, the two conformers
undergo rapid rotation about the central C-C bond, and as such, the chiral conformers cannot be
isolated at room temperature. Thus, butane is NOT a chiral molecule, even though it has chiral
Exercise 5 (provide answers above each model shown in Figure 13 as appropriate)
a) Build a model of anti-butane, and draw a Newman projection of your model with the carbon
labeled "front carbon" in front. Label the substituents (note that CH3(1) is in front, and CH3(2) is
in back) and orient your drawing such that Hd points towards the bottom of the page.
b) Convert your model of anti-butane to the gauche-(+)-butane conformation (Figure 13b) by
rotating the substituents on the front carbon by 120º. Draw a Newman projection of your model
with the carbon labeled "front carbon" in front. Label the substituents (note that CH3(1) is in
front, and CH3(2) is in back) and orient your drawing such that Hd points towards the bottom of
c) Convert your model of gauche-(+)-butane to the gauche-(-)-butane conformation (Figure 13c)
by rotating the substituents on the front carbon by 120º. Draw a Newman projection of your
model with the carbon labeled "front carbon" in front. Label the substituents (note that CH3(1) is
in front, and CH3(2) is in back) and orient your drawing such that Hd points towards the bottom
of the page.
Stereochemical designation (R,S and E,Z)
We saw earlier that molecules can exist with different stereoisomeric structures, and in
order to be able to refer to molecules unambiguously, we need a system of stereochemical
designation. Some time ago, organic chemists decided they would use the Cahn-Ingold-Prelog
(CIP) system to do this. In this system, groups are assigned priorities based on atomic number as
described in your text book. The relationship of the groups to each other can then be examined
and the stereocenter can be unambiguously assigned a descriptor. For tetrahedral stereocenters
(i.e., "chiral centers") the descriptor is either R or S, while for planar stereocenters (i.e., alkenes
and related planar functional groups), it is E or Z.
R, S Designations
Tetrahedral stereocenters (i.e., "chiral centers") are designated as R or S. To determine if
a tetrahedral stereocenter is R or S, we first assign a CIP priority to each group bound to the
stereocenter. The highest priority group is 1, the lowest is 4. The groups are then oriented such
that group 4 points away from the observer. Groups 1, 2, and 3 will then be oriented towards the
observer. We then trace an arc from group 1 to group 2, then to group 3. This arc will be either
clockwise, or counter clockwise. If it's clockwise, the stereocenter is designated as R, if it's
counterclockwise, it is S.
E, Z Designations
Planar stereocenters (i.e., alkenes and related planar functional groups), are designated as
E or Z. To determine if an alkene is E or Z, we first assign a CIP priority to the two groups
bound to each carbon. So, each carbon will have a group labeled as 1 (higher priority) and 2
(lower priority). If the two groups labeled as 1 are on the same side of the molecule, the
molecule is designated az Z (also called cis) If they are on the opposite sides, it is designated as
E (also called trans).
2-Chlorobutane is one of the simplest chiral molecules, yet it can be depicted in different
ways. This exercise is intended to stress this point and illustrate that simple reorientation of a
molecule and bond rotation provide structures which can look different on paper. It also
illustrates the point the interchanging the two groups on a stereocenter provides the opposite
configuration of the stereocenter.
Exercise 6: Shown in Figure 6 are the two enantiomers of 2-Chlorobutane drawn in the zigzag
conformation with the chlorine on the 2nd carbon from the left. First, convince yourself that
enantiomers are not identical. Make models of the two molecules shown below and then align
any two of the substituents of the carbon bearing the chlorine atom and then attempt to align the
other two. You won't be able to, so stop trying (but not before you've tried all possible pairings
of two substituents).
Figure 6 2-chlorobutane
Now, assign a CIP priority to the four different groups (Cl=1, Et=2, Me=3, H=4) bound
to the stereocenter and orient your models such that the lowest priority group (H) is pointing
away from you. Examine your models and indicate if each molecule is R or S. HINT: Note that
the hydrogen and chlorine have been exchanged and that the R / S configuration is different in
the two structures.
R or S R or S
Several other conformations and orientations of these molecules are depicted below.
Make a model of each by rotating about the σ-bond of either the R- or S- model and holding it in
the same orientation as in the drawing. Then:
i) Use your model to help you write a line drawing to the right of each conformation. Note that
in some cases, you will have to rotate about a C-C bond, while in others, you can simply reorient
the molecule in space.
ii) Indicate if each molecule is the R- or S- enantiomer (the answer to the first one is provided as
Visualizing σ-Bond rotations - Two Types of Carbon
Students often find visualization of rotations about C-C bonds difficult without the aid of
models, but there is an informal yet systematic way of doing this that can help. Consider
compound A shown in Figure 7 which is drawn in a zig-zag conformation wherein the line of
carbons is parallel to the bottom of the paper. Notice that there are two orientations of the
carbons, the substituents on some point towards the top of the page, whereas in others, they point
towards the bottom. For example, in C-2 (see numbering above / below each carbon), the
chlorine points towards the top of the page (as does the H, even though it is not shown), while in
C-3, the CH3 points towards the bottom (again, as does the H, even though it is not shown).
Both substituents are bold indicating that they are coming out of the plane of the paper. To the
right of A, compound B is drawn. In this compound, the chlorine points towards the bottom of
the page while the CH3 group points towards the top, and both substituents are dashed, indicating
that they are going back behind the plane of the paper. These are, in fact, the same molecule;
they are just drawn differently. They appear to be stereochemically different, but this is because
the molecule has been essentially flipped like a pancake, and as such, all the substituents that
pointed out of the plane of the page now point behind the plane of the page. If this visualization
is difficult for you, prepare a model of compound A and flip it such that it adopts the orientation
of compound B and examine the position of the substituents.
The difference between these two renditions of the same molecule is fundamentally
related to the "zig-zag" orientation of the carbons in each drawing; in A, carbon 2 "zigs" while in
B carbon 2 "zags". In other words, in compound A, the arc formed from following C-1 to C-2 to
C-3 runs clockwise while the same arc in compound B runs counterclockwise (see the arrows in
the drawings in Figure 15). Similarly, the arc traced from C-2 to C-3 to C-4 in compound A is
counterclockwise while in B it is clockwise. In general, if we have a substituent on a molecule
which is bold when tracing an arc clockwise, it will be dashed if it is rendered in such a way as
to trace a counterclockwise arc. Similarly, if we have a substituent on a molecule which is
dashed when tracing an arc clockwise, it will be bold if it is rendered in such a way as to trace a
Rotating about C-C bonds can have the effect of switching the arc from clockwise to
counterclockwise. This is illustrated in Figure 8. Compound C is drawn in the zig-zag
conformation, whereas in compound D the bond between C2 and C3 has been rotated by 180 º,
while in compound E the bond between C5 and C6 has been rotated by 180º. All three are the
same isomer, it's just that they have been drawn differently with C-C bonds rotated in different
orientations. In compound C, the arcs traced upon going from C1-C2-C3 and C5-C6-C7 are
both clockwise. In compound D the arc traced upon going from C1-C2-C3 is now counter
clockwise, and as a consequence, the OH at C2 is behind the plane of the paper and is drawn
dashed. Similarly, in compound E the arc traced upon going from C5-C6-C7 is now counter
clockwise, and as a consequence, the chlorine at C6 comes out of the plane of the paper and is
drawn bold. Prepare models of these compounds and rotate the bonds to get the molecules into
the conformations depicted, and convince yourself that these are all the same compound.
Exercise 7: Shown in Figure 9 are various isomers of 2-chloro-3-methylpentan-1-ol drawn in
different conformations and orientations. Using the template provided on top of each drawing,
add the chloro- and methyl-substituents with the correct stereochemistry. Try to do this by
seeing if the arc traced by each stereogenic carbon is the same as it is in the compounds shown
below the templates. If this is difficult, make models of the starting conformation and
orientation, and rotate the model to the same conformation and orientation as the template.
Exercise 8: Are the pairs of compounds shown Figure 10 isomers or are they the same
compound drawn differently?
isomers or same cmpd isomers or same cmpd isomers or same cmpd
Chirality, Enantiomers, and Diastereomers
We previously discussed the definition of chirality and enantiomers, and in this exercise,
we will practice determining if a molecule is chiral, and learn how to draw the enantiomer of a
chiral molecule. Remember that in order for a molecule to be chiral, it must have a non-
superimposable mirror image that can, in principle, be isolated at room temperature. We saw
earlier that even though butane has chiral conformations, it is not a chiral molecule because the
conformations are not isolable at room temperature.
Exercise 9: A number of substituted cyclohexanes are shown in Figure 11 drawn in the chair
conformation. Make a model of each compound and answer the following questions:
a) Do the molecules contain a mirror plane in their chair conformation (yes or no), and if so,
indicate where the mirror plane is?
b) Make a model of the mirror image of the chair conformation of each molecule. Are the two
models superimposable on each other (yes or no)?
c) Draw the flat representation below each of the molecules.
d) Do the molecules contain a mirror plane in their flat conformations (yes or no), and if so,
indicate where the mirror plane (or mirror planes, if more than one exist) should be?
e) You have bottles each labeled with one of these compounds on it, are the molecules inside the
bottle chiral or achiral?
a) yes or no a) yes or no a) yes or no a) yes or no
b) yes or no b) yes or no b) yes or no b) yes or no
c) draw flat above c) draw flat above c) draw flat above c) draw flat above
d) yes or no d) yes or no d) yes or no d) yes or no
e) chiral or achiral e) chiral or achiral e) chiral or achiral e) chiral or achiral
The simplest way to draw the enantiomer of a chiral molecule is to invert the
stereochemistry at each tetrahedral stereocenter as shown in the example in Figure 12. But keep
in mind that only chiral molecules have enantiomers. Recall that diastereomers are
stereoisomers which are not enantiomers. Several molecules which have diastereomeric
relationships to each other are shown in Figure 13. Note that they fit all the criteria of being
stereoisomers (identical formula and connectivity), but they are not enantiomers.
Exercise 10: Several compounds are shown in Figure 14.
A) For each, make two models, one model of the compounds as shown, the other where each
tetrahedral stereocenter is inverted. Compare the two models side-by-side to see if the model
with the inverted stereocenter(s) is the enantiomer (ent), or if it is identical (ident) to the original
(circle one). Remember, you can rotate around σ-bonds to place the molecule in the
conformation with the highest symmetry.
B) Note that some of the compounds shown above contain tetrahedral stereocenters, but are
achiral. Such compounds are called meso.1 Circle yes or no below each compound.
ent or ident ent or ident ent or ident ent or ident ent or ident
meso yes or no meso yes or no meso yes or no meso yes or no meso yes or no
Hopefully, this exercise has taught you how to recognize chiral molecules, isomers, and
how to draw molecules wherein σ-bonds have been rotated, or which are oriented differently in
space. Models are a helpful tool to visualize these concepts; however, for simple organic
structures, one should be able to solve these sorts of problems without the aid of models.
1 A more rigorous definition of meso is "an achiral molecule which has a diastereomeric relationship to a chiral
molecule." In other words, if an achiral molecule has a stereoisomer which is chiral, then the achiral molecule
is also meso. Most achiral molecules do not have chiral diastereomers. The definition provided in the text is
less rigorous, yet easier to apply, and it works for the vast majority of molecules one encounters, so it will
suffice for our purposes.