Stereochemistry and Molecular Models Organic Chemistry Lab Report

Translated from French to English – www.onlinedoctranslator.comCHM1321 Stereochemistry and Molecular Models
The following sections contain a great deal of background information. For more details, see
the course manual and notes.
Introduction
The examination and analysis of the three-dimensional structure of molecules are vital
aspects of modern organic chemistry. Since enzymes, proteins, and other molecular
machines in the human body are three-dimensional entities, the shape and stereochemical
configuration of molecules is extremely important in therapeutic chemistry and
biochemistry. All interactions between biological molecules or between biological molecules
and drugs are based on global geometry. Stereochemistry also plays a very important role in
chemical reactivity. Chemical reactions require the interaction of molecules and atoms in
such a way as to create a favorable orbital overlap. This is only possible if the orbitals have
the correct geometry.
It is very difficult to understand the three-dimensional structure of organic compounds
using two-dimensional drawings. The three-dimensional structure of molecules must be
taken into account to fully understand their structure and reactivity. Nowadays this can be
done using a computer, but the best way to work in three dimensions is with plastic
molecular models. They are strong, inexpensive and provide better visualization of
molecules.
In Scheme 1 the differences in reactivity which
OH
can be expected according to the different
base
O
stereochemical configurations are illustrated. The top
2
1 Br
reaction is a cyclization in which an epoxide is formed
OH
base
by the displacement of a halogen in an S processNOT2.
O
Br
Two-dimensional view, the reaction is simple. The
4
3
transformation of1in2is an example of a planar
OH
representation of a reaction. Stereochemistry
O
Br
complicates matters. Please note that the molecule
6
5
base
trans3is already oriented to produce the epoxy4,
Diagram 1:responsiveness and configuration
while the moleculecis5does not react can not form
the cycle. That
is explained by the fact that, in the structure5, the nucleophile and the leaving group cannot
adopt a conformationantiperiplanar due to their own configurations. In this reaction, an
antiperiplanar orientation is essential for the interaction of the unpaired electrons of the
nucleophile with the orbitals*of the leaving group to produce a new bond. This relationship
is hard to see on paper, but becomes evident using molecular models.
Darling Molecular Models
In this homework, you will use your molecular models to build a set of compounds from
a list provided by your teacher. Darling molecular models produce better results, but other
models may also be allowed. If you wish to use another set of molecular models, please
obtain permission from your professor.
ELEMENT
COLOR
Carbon
Black
Oxygen
Red
Nitrogen
Blue
Sulfur
YELLOW
Hydrogen
small marker
spherical white or
“empty” link
Table 1:Common colors
The models you will use are made up of plastic
parts of different colors. The table below presents the
color of the most common “organic” elements. The
other elements can be in different colors as required.
It is assumed that a hydrogen atom occupies each
“empty” bond of a carbon atom. In the case of
heteroatoms (O, N,etc.), it is usually convenient to use
a small spherical marker to indicate the presence of a
hydrogen atom. Open bonds on heteroatoms are
usually interpreted as lone electron pairs. When you
build your structures, you need to make sure that the
links fit together easily. Do not force the models and
do not try to make small cycles (3 or 4 atoms). Consult
the instruction booklet included with your set.
You will use structural models. In this type of molecular model, the “rods” represent
bonds between atoms and pairs of lone electrons. The atoms are assumed to be at the
meeting points of the rods (vertices) or at the ends of the “free rods”. Most of the pieces in
your sets are designed to form tetrahedral atoms consisting of a central atom surrounded
by 4 equally distributed bonds. It is important to always construct these tetrahedral atoms in
order to be able to visualize the molecular geometry g
my. THE
bonds converge at the vertices of a
(To)
Figure 1:Two pieces (a), one faith
two v-shaped pieces labeled sp3(Figure right
angle so that the V-shaped openings
are directed towards each other. Pinch the pieces together until you hear two clicks. The
double bonds (Figure 2) are constructed using the parts marked sp2, which look like
elongated oval rings. Triple bonds are represented by long arrow-shaped pieces.
(To)
(b)
(vs)
Figure 2:(a) two pieces marked sp2, and an elongated oval ring, form a double
bond, (b). (c) A triple bond, with linear geometry.
The easiest way to assemble molecules is usually to prepare the necessary number of
tetrahedral atoms and then join them. To make building molecules easier, try as much as
possible to make your molecules look like your drawings. Similarly, draw pictures that
resemble your molecules to make it easier to transfer the information to paper.
Structural representations
Figure 2 illustrates some molecular
projections. They are some of the many ways to
represent two-dimensional compound
structures. Each of these representations is
designed to provide structural information for
different purposes. The course manual provides
more details on these representations.
CH3CH2CH2CH2CH3
Condensed formula
Zigzag representation
H
H
VS2H5
H
H
Newman’s projection
CH3
Diagram 2:Structural representations
THEcondensed formulasdo not provide
no information on the three-dimensional structure. They are used when the composition of
a molecule is important or to represent a structure using word processing software. These
structures provide information about the bonds between the atoms of a compound, based
on assumptions about the number of atoms that can be joined to a given atom.
In order to understand these formulas, you need to know how many bonds of each type
an atom can form. Carbons generally form four bonds, nitrogen three, oxygen two, and
halogens one. All hydrogen atoms bonded to an atom are shown immediately following that
atom. An atom following the hydrogens is assumed to be bonded to the preceding nonhydrogen atom, up to the number
maximum number of bonds that this atom can form. Groups bonded to a given atom are often
indicated in parentheses. Here are some examples :
a) CH3CH2CH2CH3
b) CH3CHClCH2CH3
c) CH3CH2OH
d) CH3OCH3
e) HOCH(CH3)2
In the first example (a), the four carbon atoms form a linear chain. The first and fourth
carbon are surrounded by three hydrogen atoms, while the second and third are
surrounded by two hydrogen atoms. Please note that each carbon is bonded to four other
atoms. In the second example (b), the first carbon is bonded to three hydrogen atoms as
well as the second carbon. The second carbon is bonded to a hydrogen, to a chlorine as well
as to the third carbon. The third carbon is bonded to two other hydrogens and the last
carbon, itself bonded to three hydrogen atoms.
Compounds (c) and (d) illustrate the bonding of heteroatoms. In the third molecule, the
first carbon is bonded to three hydrogen atoms and the second carbon, itself bonded to two
hydrogens and one oxygen. The last hydrogen is bonded to oxygen. In the fourth structure,
the first carbon is bonded to three hydrogens and oxygen. Oxygen is bonded to the second
carbon, itself bonded to three hydrogens.
The fifth compound is an example of groups. In this case, the oxygen is bonded to the
hydrogen and the first carbon. This carbon is bonded to a hydrogen and two CH groups3.
The bonds to these groups are all carbon-carbon bonds.
CH3CHCHCH3
H3VS
H
VS
VS
H
condensed
CH3
structure
Diagram 3:an alkene
The structure in Scheme 3 is an example of a condensed
formula that includes a double bond. The first carbon is
bonded to three hydrogens and the second carbon. The
second carbon is bonded to a hydrogen and the third
carbon. The third carbon is bonded to a hydrogen and the
last group CH3. Drawing this structure with the bonds
clearly indicates that the second and third carbons now
have three bonds each. Since a carbon normally forms four
bonds, we deduce that these two carbon atoms are
necessarily linked by a double bond.
The performances inzigzagprovide stereochemical
information in compact form. In these structures, the carbon
and hydrogen atoms are not indicated explicitly. Only bonds
that connect carbon atoms are drawn. The carbons are
assumed to be at each vertex and end of the zigzag. The atomic
symbols of the heteroatoms are indicated. Hydrogen atoms are
only explicitly shown if they are bonded to a drawn atom. In
these structures, it is important to remember that a carbon
always forms a maximum of four bonds. If a vertex or end has
less than four bonds, the missing bonds (up to
CH3CH2CH3
H2
VS
H2VS CH2
H2VS CH2
OH
CH3CH2COOH
O
Diagram 4:Representations
zigzag
four) are formed with hydrogen atoms. You will find in Diagram 4 some condensed formulas
and the corresponding zigzag representations.
In the first example, three carbon atoms form a linear chain. This chain is drawn in the
form of a zigzag. The ends and vertices of this chain represent carbon atoms. The first and
third carbons are assumed to be bonded with three hydrogens each, while the central
carbon is bonded with two hydrogens (total of four bonds).
In the second example, five carbon atoms form a ring. Each vertex represents a carbon
atom and each line represents a bond. The number of hydrogen atoms bonded to each
carbon can be deduced keeping in mind that each carbon atom can be bonded to a
maximum of four other atoms.
In the last example, three carbon atoms form a chain. The third carbon is bonded to two
oxygen atoms. Please note that the oxygen symbol is explicitly drawn since oxygens are
heteroatoms. The last hydrogen atom is listed after an oxygen; it is therefore bound to
oxygen. Since the first oxygen is not followed by a hydrogen, it forms a double bond. This
double bond can only bind the oxygen to the carbon that precedes it.
Your molecular models naturally form zigzags due to the tetrahedral geometry of atoms.
This can be useful for drawing and building structures since your structures and drawings
look alike. In the sections below, you will be asked to construct molecules from condensed
formulas and then draw the zigzag representations. Build each molecule and place it on the
desk in a zigzag pattern. All you have to do is draw what you see. This method avoids errors
when going from a molecular model to a drawing or vice versa.
Zigzag representations often include slurs that look like dashes.bevelOrdotted. These
bonds are used to show the three-dimensional tetrahedral shape of an atom. It is assumed
that the bonds drawn as lines are in the plane of the paper, that the bonds in the form of
dashes inbevelproject themselvesabove of the plane of the paper and that the links in the
form of linesdottedproject themselvesunder the plan paper.
OH
Trait
dotted
OH
OH
Trait
Tetrahedron
bevel
Wrong
(see the model)
Diagram 5:How to properly draw 3D structures with zigzag representations.
When you draw beveled or dotted lines, you must respect the shape of the tetrahedron. A
tetrahedron is made up of two Vs, joined at their apex and placed at an angle of 90° to each
other. When drawing a stereochemical configuration (beveled or dashed lines), you should
always draw the atoms to show the two Vs joined at their tops (see above). Two of the bonds
should be in the plane of the paper (single lines), one should point up and the other down.
ANewman projectionis a drawing that illustrates the dihedral angle of a bond. The
dihedral angle is defined by 4 atoms, and the bond in question is assumed to be between
the second and third of these four atoms. The Newman projection is very useful in
determining the reactivity and overall shape of a molecule.
H
H
Carbon
back
Carbon
H
H
Before
H
H
90° rotation of the molecule
H
H
H
H
Carbon
back
H
Carbon
Before
H
Overlay
H
both
H
H
H
H
H
H
H
Projection
by Newman
H
H
H
H
Conformation
offbeat
f
H
H
HH
H
H
Conformation
eclipsed
(To)
(b)
(vs)
Figure 3:(a) Construction of a Newman projection of ethane. Newman’s
projections of the staggered (b) and eclipsed (c) conformation of ethane.
In Figure 3, ethane is shown with all hydrogen atoms in their stereochemical
configuration. If you look at the molecule in the direction of the carbon-carbon bond, you
see the projection of this bond and the spatial distribution of the hydrogen atoms. A
Newman projection is used to illustrate this spatial distribution of atoms.
In a Newman projection, the carbon atom at the back is represented by a circle. All the
hydrogen atoms bonded to this carbon are represented by lines at an angle of 120°. Notice
that these lines stop at the circle that represents the carbon atom behind.
The three carbon-hydrogen bonds bonded to the front carbon are shown as 120° angled lines.
It is assumed that the carbon is located at the vertex formed by these lines. The Newman
projection is formed by superimposing the representations of the forward carbon and the
backward carbon. It shows the conformation of the molecule. The angle between a hydrogen
and the front carbon as well as between a hydrogen and the back carbon is calledthe dihedral
angle (F). It is defined by the four atoms that form it (HCCH).
Your set of molecular models contains small plastic spheres designed to fit together on
the tetrahedral atoms. The sphere can be used to help visualize Newman projections. Place a
sphere on the atomback of the bond you observe. If you look in the angle of the bond, you
will be able to observe the Newman projection. The sphere represents the circle shown in
the Newman projection. The use of a sphere whose color contrasts with that of the front
atom makes it easier to visualize the Newman projection.
The dihedral angle defined by the atoms creates the particular conformations of a
molecule. A conformation is simply a shape that a compound can assume by means of
rotation of bonds. The example above shows two types of conformations that are of
particular importance. In the first example, the dihedral angle shown is 60°. This
conformation, in which the hydrogen atoms are separated as much as possible, is called
offbeat(Figure 3b). The other structure illustrates a conformation whose dihedral angle is 0°.
This conformation, in which the carbon-hydrogen bonds of the front and back carbons are
superimposed, is referred to aseclipsed(Figure 3c). In addition to these conformations, there
are two types of eclipsed conformations:antiAndLEFT. Descriptions of these conformations
can be found in your course manual.
Absolute configuration
The form of a chiral compound is indicated by a system of nomenclature which provides the
absolute configuration of each stereogenic center of the compound. This system, analogous to
the distinction between the left hand and the right hand, is called the Cahn-Ingold-Prelog system
or, more simply,RS system. The course manual gives more details on this. Here is the procedure
for giving the nomenclature of a stereogenic center using this system:
1. Give an order of priority (1, 2, 3, 4) to each of the four groups linked to a stereogenic
atom, according to the atomic number of the atoms directly linked to the stereogenic
atom. Priority noh1 is assigned to the atom with the largest atomic number.
2. In the event of a tie (same atomic number), the next directly linked atom of each of the
groups in question is examined. This process is continued, examining each of the
atoms at the branch point, until the tie is broken. Atoms are examined according to
atomic number, from largest to smallest.
3. In the case of a double or triple bond, we draw the group again by repeating the atoms at
each end of the bondp.It should be noted that only the atoms directly linked to the
stereogenic atom are repeated, and not the entire groups.
4. In the case of isotopes, the tie must be broken using the atomic mass (the isotope with
the highest atomic mass has the highest priority).
5. Rotate the structure so that the group with the lowest priority is farthest from the
observer. The nomenclature of a molecule is given according to the meaning
rotation of the three groups with the highest priority, in the order 1-2-3. The stereogenic
center is namedRin the case of a rotation in the directionhourly, AndSin the case of a rotation
in the directioncounterclockwise.
Chair conformation
Cyclohexane is a special structure in organic chemistry because it is the only cyclic
hydrocarbon with a tension-free conformation. This conformation is called the chair
conformation because its linear representation resembles a recliner. When looking at a
cyclohexane from above, the molecules have a hexagonal shape. However, a slight rotation
to look at it from the side reveals a three-dimensional shape called a chair. The course
manual provides more details about the chair conformation.
HOME DUTY – DEADLINE: February 17, 2023 TO BE
SUBMITTED TO THE DEDICATED FILE ON BRIGHTSPACE
-15 points if photos are not included; -15 points if the student card/identification is not
visible in the photos
Issues? Read the information at the end of this document carefully.
Part A: Enantiomers and diastereoisomers.
1. Build a model of the moleculeH2NCH(CH3)CH2CH3(compound i), and also that of the
mirror image of the compound. Take a photo of both models, with your student ID
visible. The photo must clearly illustrate that the two structures are mirror images.
2. Carefully draw the skeletal formulas (zigzag representation) of all possible stereoisomers
of the molecule using the appropriate stereochemical notation.
If possible, draw the mirror images side by side indicating the mirror plane. You can draw
your structures by hand or with ChemDraw(see information below).
3. Determine if the mirror images can be superimposed and if they are therefore
enantiomers and/or diastereoisomers (or if the molecule is achiral).
4. Identify the stereogenic center(s) of the molecule (if possible).
5. Determine the configuration (ROrS) of each stereogenic center (if possible).
6. If there are more than 2 isomers, do a full stereochemical analysis of your structures,
indicating which pairs are enantiomers and which are diastereoisomers.
7. Repeat the steps#2 to #6 for each of the following compounds:
compound ii
CH3CClBrCH(OH)CH3
compound iii
CH3CH2CH(CH2OH)CH2OH
compound iv
Cl
Cl
O
Part B: Newman Projections.
1. Build a model of the following molecule (make sure it lines up with the molecule perfectly)
and inspect it along the bond indicated by the eye:
H
H
HO
CH3
H
HO
2. Take a photo from this perspective, where your student card is also visible in the photo.
3. Using your structure, make an energy diagram to show the change in free energy when
the FORWARD atom is flipped CLOCKWISE, from 0º to 360º, in 60º increments. In your
diagram, you should clearly show therelative energies of each conformation.
4. Draw the Newman projections (by hand or with ChemDraw) of each conformation at the
bottom of your diagram.
Part C: Chair conformation.
1. Build a model of this molecule in a chair conformation:
Cl
Br
2. Take a photo of the model, with your student ID visible.
3. Draw (by hand or with ChemDraw) your structure carefully using the appropriate
convention for the chair conformation.
4. List all axial and equatorial substituents (including hydrogen atoms).
5. Without breaking any bonds, transform your structure into the other chair shape. Be careful
not to rotate the whole molecule.
6. Draw your structure appropriately indicating all axial and equatorial substituents
(including hydrogen atoms).
7. Indicate which of the two chair conformations would be more energetically stable.
Information :
Your responses must be uploaded as a digital document to Brightspace (we will only accept
pdf formats) by the deadline. It is your responsibility to ensure that the file is readable and
submitted correctly: late submissions or corrupt files will not be scored. Your document
must include both photos (of parts A and B) as well as your structures.
If you haven’t received your molecular model kit, you can use alternatives. Examples:
drops of gum and toothpicks
Plasticine and Q-tips
If you don’t have a student ID, you can use another photo ID in your images (name and
photo must be visible, but you can hide other personal information).
If you draw your structures and Newman projections by hand, they should be drawn in pen,
clear and legible – if we cannot understand what you have drawn, it will not be graded. To
submit your assignment, you can scan or take a photo of your written responses and attach
your photos to form a single document, using the software of your choice.
It is strongly recommended to use ChemDraw to draw your structures. On the lab’s
Brightspace there is a video that explains how you can download a free copy of this
software. If you are using ChemDraw for your structures, follow these steps:
•
•
open a new document
go inQueueToApply Document Settings from…ToACS Document 1996
This will apply the necessary presets so that your structures can be easily and perfectly
copied and pasted into most applications (eg Microsoft Word), along with your answers and
photos.
A warning:using ChemDraw generally provides much neater structures; however, drawing
molecules using this program takes much longer than drawing them by hand.