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CHM!2201!
Organic!Chemistry!Lab!I!
Fall!2009!
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Department!of!Chemistry!
Villanova!University
Learning to Use the Molecular Model Kit and the
Basics of Isomerism
If you haven't already done so, buy a "Molecular Visions" molecular model kit
from the stockroom. It is the kit that comes in the green plastic box.
Here is what we'll cover in today's lab:
1) The Common 2nd Row Element Presentations (pp 10-11 of the model kit
manual)
With the help of your professor and TA's, together we are going to go
through some of the contents of your kit. We will follow what is on pages 10-11 of
the manual that came with your kit.
Note: try to learn which pieces are needed to make organic compounds;
also note how to put all the pieces back in the box as it is a tight fit and every
piece has a particular place!
2) Assembly and Disassembly of Atoms "With Bonds" (pp 15-17)
Carefully follow the descriptions on making an sp3 hybridized atom, and
making atoms joined by a double bond. It is important to learn how to put these
pieces together properly and securely, and learn how to take them apart
efficiently.
3) Constructing a variety of molecules
We'll begin by keeping things simple: putting together molecules
containing only carbon and hydrogen, frequently called hydrocarbons.
And we'll start off with saturated acyclic hydrocarbons. What does this mean?
3
Saturated means all the carbons are sp hybridized (or maybe a better way to think of it
is that the molecule is saturated with hydrogens, thus all carbon- carbon bonds present are
single bonds). Acyclic means "non-cyclic", meaning there are no rings of carbon atoms
present, which says that from any carbon atom in the molecule, you cannot traverse
along the carbon chain and get back to the same carbon you started from.
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CHM!2201!
Organic!Chemistry!Lab!I!
Fall!2009!
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Department!of!Chemistry!
Villanova!University
Exercise #1
Try putting together these molecules with your models, given the molecular
formula (name in parentheses):
CH4 (methane)
C2H6 (ethane)
C3H8 (propane)
Note how three-dimensional these systems are, and how many different shapes
(what organic chemists call conformations or conformers, the latter is short for
'conformational isomers') that ethane and propane can have. Various conformers of a
hydrocarbon are obtained by rotating about the C-C single bonds.
You will learn from lecture that there is usually considered to be "free rotation"
about C-C single bonds. Note that free rotation is in quotes. That is because there
are barriers to the rotation about C-C single bonds, but these barriers are usually very
small and the thermal energy present at ambient conditions (i.e. ~20 °C) is
sufficient that the barriers are easily traversed.
Question #1: What is the general formula for a saturated acyclic hydrocarbon?
Exercise #2
Now put together a molecule with molecular formula C4H10.
You probably put together this one, didn't you?
CH3CH2CH2CH3
The molecule above is known as n-butane (where the 'n' stands for normal,
meaning somehow it is 'normal' in organic chemistry when all the carbons in a
molecule are bonded together one after the other, in one continuous chain). Note that nbutane is also commonly written on paper as the following carbon skeletal diagram (also
called a bond-line formulas):
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CHM!2201!
Organic!Chemistry!Lab!I!
Fall!2009!
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Department!of!Chemistry!
Villanova!University
Note that in this notation a carbon atom is implied at the end of each line (unless some
other atom is explicitly written), and that none of the hydrogens are drawn. In this type of
drawing, the only atoms explicitly shown are those that are not carbon or a hydrogen
bonded to carbons (but hydrogens bonded to other atoms like N or O are shown).
Maybe some of you put together this one for C4H10:
This structure is also a C4H10, but it is not the same as n-butane. This C4H10 is called
isobutane (or 2-methylpropane using the IUPAC systematic nomenclature).
Since it has the same molecular formula as n-butane, but it is a different compound,
they are isomers. N-butane and isobutane are more accurately called
constitutional isomers because the atoms in each are bonded in a different order.
Exercise #3
As the molecular formula becomes larger, the number of isomers rises rapidly.
As an exercise, put together all the isomers of formula C5H12. You should come up with
three total.
Now that you know a bit about bond-line drawings and what they mean, put
together a model corresponding to the following bond-line formula:
This five-carbon molecule is cyclic and is known as cyclopentane.
Question #2: What is the molecular formula of cyclopentane? Does it fit the general
formula you derived for saturated acyclic hydrocarbons? If not, derive a general formula for
a saturated monocyclic (one ring) hydrocarbon.
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CHM!2201!
Organic!Chemistry!Lab!I!
Fall!2009!
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Department!of!Chemistry!
Villanova!University
Note: although it is easiest to write cyclopentane on paper as a simple pentagon,
this drawing doesn't represent well the three-dimensional shape of the molecule. Play
around with the cyclopentane molecule to see if you can guess what might be the
preferred shape (i.e. conformation).
Exercise #4
Now put together the one carbon larger saturated cyclic hydrocarbon, known as
cyclohexane:
Notice how difficult it is to get this structure in the shape of a regular hexagon with all
carbons in the same plane. Here we again demonstrate the three- dimensional
nature of organic molecules and why they can be difficult to represent on a twodimensional piece of paper! After playing around with cyclohexane you may have
discovered that the preferred conformation for cyclohexane is commonly called the
"chair" form, as it somewhat resembles the shape of a reclining chair. An attempt to draw
this on paper using a bond-line formula looks like the following:
Note that the twelve hydrogens in the planar conformation of cyclohexane are equivalent,
but in the preferred chair conformation they are not, with there being two sets of 6
equivalent hydrogens. These two types of hydrogens are called axial
equatorial (He):
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(Ha) and
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CHM!2201!
Organic!Chemistry!Lab!I!
Fall!2009!
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Department!of!Chemistry!
Villanova!University
Cyclohexane can also exist in the boat conformation which is less stable than the chair
conformation. Convert your cyclohexane chair conformer model to the boat conformer by
flipping up one end of the chair as illustrated below.
The decrease in stability of the boat conformer is due to eclipsing of some of the bonds,
giving it torsional strain. Further destabilization of the boat conformer is caused by the
close proximity of the flagpole hydrogens (the hydrogens in the stern and bow of the boat)
which imparts steric strain.
Question #31. Re-examine the cyclohexane model that you constructed to answer the
following questions:
a) Is the cyclohexane ring rigid or flexible, that is, can the ring of atoms move and take
various arrangements in space, or is the ring of atoms locked into only one
configuration?
b) Of the many configurations, which appears best for the ring: a planar or a puckered
arrangement?
c) Arrange the ring atoms into the chair and then the boat conformations and compare
to a picture of a lounge chair and a boat illustrated below. Do the terms fit the
pictures?
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CHM!2201!
Organic!Chemistry!Lab!I!
Fall!2009!
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Department!of!Chemistry!
Villanova!University
Question #41. With the model in the chair conformation, rest it on your bench top.
a) How many hydrogens are in contact with the bench top?
b) How many hydrogens point in a direction 180° opposite to these?
c) Take your pencil and place it into the center of the ring perpendicular to the bench
top. Now rotate the ring around the pencil; this is called an axis of rotation. How
many hydrogens are on bonds parallel to this axis? These hydrogens are called the
axial hydrogens and the bonds are called axial bonds.
d) If you look at the perimeter of the cyclohexane system, the remaining hydrogens lie
roughly in a ring perpendicular to the axis through the center of the molecule. How
many hydrogens are on bonds lying in this ring? These hydrogens are called
equatorial hydrogens and the bonds are called equatorial bonds.
e) Replace one of the axial hydrogens with a colored atom. Do a “ring flip” by
moving the carbon with the colored atom up and moving the carbon furthest away
from it down. In what position is the colored component after the ring flip? (1) Do
another ring flip. In what position is the colored component now? (2) Observe all
axial positions and follow them through a ring flip.
up
down
A "Ring Flip"
f) Using the following drawing, replace both positions labeled A and A1 by colored
component atoms. Are they cis or trans? (1) Do a ring flip. Are the two colored
components cis or trans? (2) Does the geometry change for the two components as
the ring undergoes a ring flip? (3) Repeat the exercise replacing atoms in positions
labeled A and B1 and answer the same three questions for this model.
g) Replace one of the colored components with a methyl (CH3) group. Manipulate the
model so that the methyl group is in an axial position; examine the model. Do a
ring flip placing the –CH3 group in an equatorial position; examine the model.
Which of the chair conformations, -CH3 axial or -CH3 equatorial, is more crowded?
(1) What would account for one of the conformations being more crowded that the
other? (2) Which would be of higher energy, and thus, less stable? (3) In your
notebook, draw the two conformations and connect with equilibrium arrows.
Given your answers to 4g1, 4g2, and 4g3, towards which conformation will the
equilibrium lie (indicate by drawing one arrow bigger and thicker than the other)?
h) A substituent group in the equatorial position of a chair conformation is more
stable that the same substituent group in the axial position. Do you agree or
disagree? Explain your answer.
i) Remembering that “trans” refers to substituents on the opposite side of a ring and
“cis” refers to substituents of the same side of the ring, construct the chair
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CHM!2201!
Organic!Chemistry!Lab!I!
Fall!2009!
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Department!of!Chemistry!
Villanova!University
conformation of trans-1,2-dimethylcyclohexane. By means of ring flips examine
the model with the two –CH3 groups axial and then equatorial. Which is the more
stable conformation? Explain your answer.
j) Construct cis-1,2-dimethylcyclohexane. Do ring flips and examine the two chair
conformations. Which is the more stable conformation? (1) Explain your answer.
Which of the two isomers, trans-1,2-dimethylcyclohexane or cis-1,2dimethylcyclohexane, is the more stable isomer? (2) Explain your answer.
Exercise #5
Let's now put together the simplest example of a hydrocarbon containing one
carbon-carbon double bond, ethene, or more commonly called ethylene (C2H4, or
CH2CH2). Hydrocarbons with C=C double bonds are usually called alkenes (sometimes
you'll hear them called olefins as well). They are considered to be unsaturated
(unsaturated in the sense that they don't have the maximum number of hydrogens like an
alkane has). Of note is how the ethylene molecule is planar, and that is due to the ! bond
2
2
formed from the sp hybridized carbons. Normally all six atoms associated with the sp
carbon atoms of a C=C double bond are in the same plane.
Now put together as many structures as you can that correspond to an acyclic
alkene with four carbon atoms and one double bond.
You should have put together the following four isomers drawn as bond-line structures:
The last two isomers above, the 2-butenes, deserve special mention. The trans
designation refers to the relationship of the two methyl groups, which are on opposite sides of
the double bond. The cis isomer has the methyl groups on the same side of the double
bond.
If you have a suspicion that these two isomers of 2-butene are not
constitutional isomers, you are correct! They fall into a different class of isomers known as
stereoisomers, which are isomers that have the same constitution but differ in the
arrangement of their atoms in space. More on these two later.
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CHM!2201!
Organic!Chemistry!Lab!I!
Fall!2009!
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Department!of!Chemistry!
Villanova!University
Question #5: What is the general formula for acyclic alkenes with one C=C
double bond?
Exercise #6
Note that C=C double bonds can also be incorporated within a ring. As an example
put together cyclohexene, which has bond-line structure given below:
The cyclohexene above is a stable and isolable compound (translation: it can be made
and stored in a bottle!). The name more accurately would be ciscyclohexene, analogous to cis-2-butene from the earlier exercise.
Now put together a model of trans-cyclohexene. It is actually sort of hard to
manipulate the pieces of the model kit to make this isomer. Perhaps not surprisingly,
trans-cyclohexene is not a stable and isolable compound.
trans-cyclohexene
Question #6: Why do you think this trans isomer of cyclohexene is not a known and
isolable compound? Why is the cis isomer more stable?
Question #7: Put together trans-cycloheptene (a 7-membered ring) and transcyclooctene (an 8-membered ring) and compare them to their cis isomer counterparts.
What conclusion(s) might you draw about the effect of increasing the ring size has on the
likeliness of isolating one of these trans-cycloalkenes?
trans-cycloheptene
trans-cyclooctene
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CHM!2201!
Organic!Chemistry!Lab!I!
Fall!2009!
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Department!of!Chemistry!
Villanova!University
Exercise #7
To demonstrate another topic to be eventually covered in lecture, put together
the molecule bromochlorofluoromethane (CHFClBr, simply a system with one carbon
with the four substituents Br, CI, F, and H). Use the colored spheres in your model kit to
represent the substituents (hydrogen = white, bromine = blue, fluorine = red, chlorine =
green). Compare the one you made with those of your neighbors.
So, did all of you put together the exact same molecule? Make sure to take a
very close look! The answer almost certainly is that all of you did not put together the exact
same molecule. Two different structures are possible with this formula. To obtain the second
one from the one you put together, simply switch the positions of any two of the
substituents.
These two isomers of bromochlorofluoromethane, which differ from each other only in the
way their atoms are arranged in space, also belong to that special class of isomers called
stereoisomers. Specifically, these two structures are enantiomers, which are defined as
non-identical mirror images (some textbooks say enantiomers are non-superimposable
mirror images, which is pretty much the same definition). Enantiomers are analogous to a
pair of hands.
Nonsuperimposable mirror images
Right hand Left hand
Question #81: (a) Does the enantiomer you just constructed have a plane of
symmetry? A plane of symmetry can be described as a cutting plane – a plane that when
passed through the model or object divides it into two equivalent halves; the elements on
one side of the plane are the exact reflection of the elements on the other side.
Molecules without a plane of symmetry are chiral. In the model you constructed, the
tetrahedral carbon is the chiral center; the molecule is also chiral. A simple test for a
chiral center in a molecule is to look for a carbon center with four different atoms or
groups attached to it; this molecule will have no plane of symmetry. (b) Copy the three
structures below into your notebook and label the chiral center in each structure with an
asterisk (*).
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CHM!2201!
Organic!Chemistry!Lab!I!
Fall!2009!
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Department!of!Chemistry!
Villanova!University
Question #91: Take the bromochlorofluoromethane model you constructed above
and pretend that you are observing it in a mirror. Construct the model of the image that
would be projected in the mirror. You now have two models. (a) If one is the object,
what could you call the other (with regard to the mirror)? (b) Do either have a plane of
symmetry? (c) Are both chiral? Now try to superimpose one model onto the other, that
is, to place one model on top of the other in such a way that all five elements (i.e. the
colored atoms) fall exactly one-on-top-of-the-other. (d) Can you superimpose one model
onto the other? Enantiomers are two molecules that are related to each other such that
they are nonsuperimposable mirror images of each other. (e) Are the two models you
have a pair of enantiomers?
Question #101: Molecules with a plane of symmetry are achiral. Replace the blue
substituent of one of your molecules with a second green one. The model should now
have three different substituents attached to the carbon. (a) Does the model now have a
plane of symmetry? (b) Passing the cutting plane through the model, what colored
elements does it cut in half? (c) What is on the left and right half of the cutting plane?
Set this model in front of your imaginary mirror and construct another model showing the
projected image. You now have two models – an object and its mirror image. (d) Are
these two models superimposable on each other? (e) Are the two models representative
of different molecules or identical molecules?
Each stereoisomer in a pair of enantiomers has the property of being able to rotate
monochromatic plane-polarized light. The instrument chemists use to demonstrate this
property is called a polarimeter (see Carey, pages 282 – 284 for a more detailed
description of the instrument). A pure solution of a single one of the enantiomers
(referred to as an optical isomer) can rotate the polarized light in either a clockwise
(dextrorotatory, + ) or a counterclockwise (levorotatory, - ) direction. Thus, those
molecules that are optically active possess a “handedness” or chirality. Achiral molecules
are optically inactive and do not rotate polarized light.
Exercise #81
R/S Convention for Chiral Centers
A system of assigning configuration to a chiral carbon was devised by the chemists R.S.
Cahn, C.K. Ingold and V. Prelog. The system uses the letters R and S to designate the
configuration at the chiral carbon: R, from the Latin rectus, or right; S, from the Latin
sinister, or left. The designation arises from the priority order assigned to the four groups
attached to the chiral carbon. Priority number is based on the atomic number of the atoms
directly attached to the chiral carbon. The higher the atomic number, the higher the
priority. In the case where two atoms have the same atomic number, you must move
along the bonds to the next atoms out from the chiral center until an atom of different
atomic number is reached. In order to apply the Cahn-Ingold-Prelog system, hold a model
of the molecule so that the atom or group with the lowest priority is pointed directly away
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CHM!2201!
Organic!Chemistry!Lab!I!
Fall!2009!
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Department!of!Chemistry!
Villanova!University
from you (see illustration below). Then examine the remaining three attachments in terms
of the order of their priorities:
If movement of your eye is from the highest to the lowest priority in a clockwise
direction, the configuration of the chiral carbon is R.
If movement of your eye is from the highest to the lowest priority in a counterclockwise direction, the configuration of the chiral carbon is S.
An Example of assignment using R/S priority rules: priority order 4 > 3 > 2 > 1
Question #111: Construct a model of 2-bromobutane, CH3CHBrCH2CH3. Answer
the following questions:
(a) What are the four groups attached to the chiral carbon? Assign priorities to these
groups: 4 = highest; 1 = lowest.
(b) View the model as shown in the above figure and draw the arrangement you see. Is it
R or S?
(c) Exchange the H and Br. Again view the model and draw the arrangement. Is it R or
S?
(d) Are the configurations opposite one another? Are they enantiomers?
We've covered a handful of different types of isomers today. A simplified "isomer family tree" to help
in determining a specific isomeric relationship is given on the next page.
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Isomers (dif f erent compounds that
have the same molecular f ormula)
Are the atoms bonded in the same
order f or the isomers being compared?
No
Yes
Constitutional
Isomers
Stereoisomers
No
Are the structures non-identical
mirror images?
Diastereoisomers
Yes
Enantiomers
Note that the cis and trans-2-butenes from earlier, which we stated are
stereoisomers, are classified further as diastereomers using the chart above.
You'll learn much more about all these topics in lecture soon, but now you hopefully
have an appreciation of the diverse structure of even simple organic molecules,
isomerism, and additionally know how to efficiently use your plastic model kit.
1. © 1997 Saunders College Publishing
8/31/2009
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