Organic Chemistry I & II
Organic Chemistry I & II

Organic Chemistry I & II

Lead Author(s): Steven Forsey

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Organic Chemistry I & II is designed for instructors who want an active, dynamic, and understandable approach to support their own efforts in the classroom. This ever-evolving textbook includes auto-graded questions, videos and approachable language in order to make difficult concepts easier to understand and implement.

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Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

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Hardcover print text only

$301

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Always up-to-date content, constantly revised by community of professors

Content meets standard for Introduction to Organic Chemistry course, and is updated with the latest content

In-Book Interactivity

Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook

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Only available with supplementary resources at additional cost

Only available with supplementary resources at additional cost

Customizable

Ability to revise, adjust and adapt content to meet needs of course and instructor

All-in-one Platform

Access to additional questions, test banks, and slides available within one platform

Pricing

Average price of textbook across most common format

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

Up to 40-60% more affordable

Lifetime access on any device

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

$219

Hardcover print text only

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

$301

Hardcover print text only

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

$301

Hardcover print text only

Always up-to-date content, constantly revised by community of professors

Constantly revised and updated by a community of professors with the latest content

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

In-book Interactivity

Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

Customizable

Ability to revise, adjust and adapt content to meet needs of course and instructor

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

All-in-one Platform

Access to additional questions, test banks, and slides available within one platform

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

About this textbook

Lead Authors

Dr. Steven Forsey, Ph.D.University of Waterloo

Steven Forsey is currently a Professor at University of Waterloo, teaching a variety of organic chemistry courses to Chemistry, Science, Chemical Engineering, Nanotechnology and distance education students. He received his Ph.D. (2004) for Synthetic Organic Chemistry from University of Waterloo, Ontario. He is a recipient of the Excellence of Science Teaching Award and has acted as the Teaching Fellow for the Department of Chemistry since 2016.

Contributing Authors

Felix NgassaGrand Valley State University

Neil GargUCLA

Jennifer ChaytorSaginaw Valley State University

Greg DomskiAugustana College

Christian E. MaduCollin Community College

Christopher NicholsonUniversity of West Florida

Franklin OwEast Los Angeles College, UCLA

Robert S. PhillipsUniversity of Georgia

Grigoriy SeredaUniversity of South Dakota

Simon E. LopezUniversity of Florida

Brannon McCulloughNorthern Arizona University

Jason JonesKennesaw State University

José BoquinAugustana College

Stephanie BrouetSaginaw Valley State University

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Chapter 6: Stereochemistry

Understanding stereochemistry is key in the formation of pharmaceuticals. Many drugs with medical benefits have enantiomers that can cause adverse effects in patients. Chemical orientation is an important factor in how our bodies react with molecules. [1]​

Contents

Learning Objectives

  • Recognize the difference between conformers, structural isomers, and stereoisomers.
  • Recognize and name E and Z isomers of alkenes.
  • Recognize the stereochemical relationship between two or more molecules: enantiomers, diastereomers, structural, conformers or identical.
  • Distinguish between chiral and achiral molecules.
  • Recognize an achiral meso compound and determine its stereochemical relationship to other stereoisomers (diastereomers).
  • Identify asymmetric chiral centers and determine their R and S configurations.
  • Use wedged and dashed drawings and Fischer projections to determine the stereochemical configurations of compounds.
  • Use wedged and dashed drawings and Fischer projections to determine the stereochemical relationships between compounds.
  • Understand optical activity and the interaction of a chiral molecule with plane polarized light.
  • Understand optical activity, enantiomeric excess and the lack of optical activity of racemic mixtures.

6.1 Overview of Isomers

Figure 6.1: Overview of Isomers

Isomers

The word “isomer” comes from the Greek words isos and meros, meaning “made of the same parts”. Isomers are two or more different compounds that have the same molecular formula, but are not the same compound because they are not superposable in 3-D space.

There are two types of isomers, structural (or constitutional) isomers and stereoisomers.

6.1.1 Structural (or constitutional) isomers

Structural (or constitutional) isomers have the same molecular formula, but their atoms are bonded in a different order. These are different compounds and may have very different chemical and physical properties.

Figure 6.2​: Examples of sets of constitutional isomers and their boiling points.

Terminology Note: Constitutional isomerism is very different from conformational changes of a molecule. A single molecule will have different conformations due to rotation around a σ bond.

Figure 6.3: Constitutional isomers (top) versus changes in conformation (bottom).

6.1.2. Stereoisomers

Figure 6.4: Examples of stereoisomers, non-identical chiral molecules.

Stereoisomers have the same molecular formulas and connectivity, but differ in the arrangement of their atoms in space: for example, cis and trans isomers represent stereoisomers. The cis stereoisomer has groups on the same side of the molecule and the trans on the opposite side, as seen with cis-, and trans-1,2 dimethylcyclohexane; cis- and trans-but-2-ene; and  (S)-and (R)-butan-2-ol,  shown below (Figure 6.4). Both pairs of molecules have the same connectivity (bonding of atoms in the same order) but their atoms differ spatially. (S)-butan-2-ol and (R)-butan-2-ol are non-identical mirror images and are different molecules. This will be discussed in detail in later sections. Watch the video for visual clarification.

Before we can discuss different stereoisomers we must learn some nomenclature.

6.2 Alkene E and Z system of nomenclature

The double bond of an alkene prevents free rotation around the C=C bond. Consequently, some alkenes exhibit stereoisomerism. As stated above, alkenes show cis–trans stereoisomerism, but what do we call an alkene that has more than two different groups in the vinylic positions? For example, how would you name 1-bromo-2-chloro-2-fluoro-1-iodoethene?

Figure 6.5: Structure of 1-bromo-2-chloro-2-fluoro-1-iodoethene.

For alkenes that cannot be named using cis or trans nomenclature, the organic chemist follows the Cahn–Ingold–Prelog convention. This procedure was created by Robert Cahn, Christopher Ingold, and Vladimir Prelog, and is used to name stereoisomers and double bonds that are found in molecules. The following procedure outlines the set of rules for assigning an E or Z descriptor to double bonds.

Procedure: Using Cahn–Ingold–Prelog convention.

Step 1: Split the alkene in half.

Figure 6.6: Structure of 1-bromo-2-chloro-2-fluoro-1-iodoethene with red dashed line splitting the double bond into halves.

Step 2: Consider both sides of the double bond. The vinylic atom that has the higher atomic number receives a higher priority.

If the atoms are isotopes, then the higher mass number receives a higher priority. For example, tritium (3H or T) has a higher priority than deuterium (2H or D), followed by hydrogen (1H or H).

In this example, iodine has a higher atomic number than bromine, thus the iodine atom has the higher priority. On the other side of the double bond, chlorine has the higher atomic number and has a higher priority than fluorine.

Figure 6.7: Structure of 1-bromo-2-chloro-2-fluoro-1-iodoethene with red dashed line splitting the double bond into halves. Iodine and chlorine are highest priority on their respective carbons.

Step 3: If the higher priority atoms or groups are on the same side of the double bond, then the molecule is designated as Z (German: zusammen, or “together”). If the higher priority atoms or groups are on the opposite sides of the double bond, then the molecule is designated as E (German: entgegen, or “opposite”).

Thus, the above molecule is called (Z)-1-bromo-2-chloro-2-fluoro-1-iodoethene.

3D Molecule*: (Z)-1-bromo-2-chloro-2-fluoro-1-iodoethene



6.2.1 What happens if the atoms at the vinylic positions are the same?

If the atoms directly attached are the same, then explore outward until the first point of difference is observed. The atom with the higher atomic number receives a higher priority. Do not use the sums of the atomic numbers or the size of the groups. Look for the single atom of highest priority.

Example: Does 4-(tert-butyl)-3,5,5-trimethyl-hept-3-ene have an E or Z designation?

Figure 6.8: Structure of 4-(tert-butyl)-3,5,5-trimethyl-hept-3-ene.

Step 1: Split the alkene in half.

Step 2: Consider both sides of the double bond. The vinylic atom that has the higher atomic number receives a higher priority. In this case, both are carbons. Thus, you need work outward atom by atom until there is a point of difference.

Figure 6.9: Structure of 4-(tert-butyl)-3,5,5-trimethyl-hept-3-ene with red dashed line splitting the double bond into halves. Allylic carbons on left half are highlighted.

Look at the atoms directly bonded to the yellow circled carbons, and list them in decreasing atomic number. Carbon (a) has three hydrogens attached to it and carbon (b) has one carbon and two hydrogens. Since carbon has a higher atomic number than hydrogen, branch (b) has the higher priority.

Figure 6.10: Structure of 4-(tert-butyl)-3,5,5-trimethyl-hept-3-ene with red dashed line splitting the double bond into halves. Atomic priority list is given for highlighted atoms on the left.

Now do the same for the other side of the double bond.

The first set of atoms are carbons and have the same atomic number.

Figure 6.11: Structure of 4-(tert-butyl)-3,5,5-trimethyl-hept-3-ene with red dashed line splitting the double bond into halves. Allylic carbons on the right are now highlighted.

The next set of atoms are all carbons and have the same atomic number. Keep moving outward until you find a point of difference.

Figure 6.12: Structure of 4-(tert-butyl)-3,5,5-trimethyl-hept-3-ene with red dashed line splitting the double bond into halves. Atomic priority list is given for highlighted atoms on the right.

In the next set of atoms, the carbon in branch (c) has the highest atomic number and, thus, this branch has a higher priority.

Figure 6.13: Structure of 4-(tert-butyl)-3,5,5-trimethyl-hept-3-ene with red dashed line splitting the double bond into halves. Atomic priority list is given for next carbons along the chains on the right.

Step 3: Higher priority atoms or groups on the same side → Z. Higher priority atoms or groups on opposite sides → E.

Thus, the molecule is called (E)-4-(tert-butyl)-3,5,5-trimethyl-hept-3-ene.

6.2.2 Double bonds, triple bonds, and E, Z determination

If the atoms being compared are sp2 or sp, then they must be “artificially” transformed into sp3 atoms. Thus, they are treated as single bonds.

Figure 6.14: Transforming double and triple bonds into the equivalent number of single bonds for assigning priority.

Example: Does 3-(cyanomethyl)-2-(methylamino)-5-oxohex-2-enoic acid have an E or Z designation?

Figure 6.15​: Structure of 3-(cyanomethyl)-2-(methylamino)-5-oxohex-2-enoic acid.

Step 1: Split the alkene in half.

Figure 6.16​: Structure of 3-(cyanomethyl)-2-(methylamino)-5-oxohex-2-enoic acid with red dashed line splitting the double bond into halves.

Step 2: Work outward atom by atom until there is a point of difference. Click on the video to show the steps.

Figure 6.17: Assigning priority using the first point of difference for the structure of 3-(cyanomethyl)-2-(methylamino)-5-oxohex-2-enoic acid with red dashed line splitting the double bond into halves.


Higher priority atoms or groups on the same side → Z. Higher priority atoms or groups on opposite sides → E.

Thus, the molecule is called (E)-3-(cyanomethyl)-2-(methylamino)-5-oxohex-2-enoic acid.


Q6.1 - Level 2

In which alkene is (E)/(Z) isomerism possible?

question description
A

1)

B

2)

C

3)

D

4)


Q6.2 - Level 1

Identify each double bond, labeled 1 and 2, in the following structures as (E) or (Z).

question description
Premise
Response
1

1)

A

Z\textit{Z}

2

2)

B

E\textit{E}

C

E\textit{E}

D

Z\textit{Z}


Q6.3 - Level 2

Designate each double bond in the following structures, labeled 1 and 2, as (E) or (Z).

question description
Premise
Response
1

1)

A

Z\textit{Z}

2

2)

B

Z\textit{Z}

C

E\textit{E}

D

E\textit{E}


Q6.4 - Level 1

Designate each double bond in the following molecule, labeled 1 and 2, as (E) or (Z).

question description
Premise
Response
1

1)

A

Z\textit{Z}

2

2)

B

E\textit{E}


6.3 Achiral compounds and planes of symmetry

Another type of stereoisomerism is called mirror-image stereoisomerism. Mirror-image-related stereoisomers occur when a molecule and its mirror image are non-superposable.

Chiral: A chiral molecule is a molecule that is not superimposable on its mirror image and is a different molecule.

Achiral: A molecule that can be superposed onto its mirror image and is the same molecule. 

First, let us look at molecules that are achiral (where the mirror images are the same molecule), and then discuss the concept of mirror-image stereoisomers. Consider bromochloromethane. Is its mirror image the same molecule or different?

Figure 6.18: Bond-line and ball and stick representations of the stereoisomers of bromochloromethane.

3D Molecule*: bromochloromethane


To determine if the mirror images are the same, in your mind try to transpose one drawing onto the other without breaking bonds. You may move the drawing in your mind any way you wish. Click on the video below to see an example. In the above molecule, the mirror image is the same compound, thus, the molecule is achiral. Notice also that a plane of symmetry, or mirror plane, is located in the molecule. This leads to another way of determining if a molecule is achiral, i.e., has an internal plane of symmetry.

Figure 6.19: Fischer projection and bond-line representations of the stereoisomers of bromochloromethane with plane of symmetry shown.


A plane of symmetry is defined as an imaginary plane that bisects a molecule in such a way that the two halves of the molecule are mirror images of each other. All molecules with a plane of symmetry are achiral. The plane of symmetry can cut through atoms or bonds. The following are all achiral molecules. The video below does not feature audio. 


Figure 6.20: Several examples of molecules with planes of symmetry shown.


Q6.5 - Level 1

Are the following molecules chiral or achiral? Match the molecules with the terms chiral and achiral.

question description
Premise
Response
1

1)

A

Chiral

2

2)

B

Chiral

3

3)

C

Achiral

D

Achiral

E

Chiral

F

Achiral


6.4 Chiral molecules and enantiomers

A molecule that cannot be superposed on its mirror image is said to be chiral. Bromochlorofluoromethane is chiral because the mirror image cannot be superposed. Thus, these isomers are different molecules.

Enantiomers: Two stereoisomers that are non-superposable mirror images.

Figure 6.21: Bond-line and ball and stick models of the stereoisomers of bromochlorofluoromethane.


Notice that the carbons in both structures are connected to four different atoms. This carbon atom is called an asymmetric atom or chirality center. Often, asymmetric atoms are marked with an asterisk. Molecules having only one chirality center are always chiral, and no plane of symmetry exists in the molecule.

Enantiomers are different molecules. They have the same physical properties, such as boiling and melting points, however, they may act totally different biologically. 

6.5 Nomenclature: R and S system using the Cahn–Ingold–Prelog convention

Example: Does butan-2-ol have an R or S configuration?

Figure 6.22: Structure of butan-2-ol.

Terminology Note: A change in configuration always means breaking a bond or a change in the orientation of the atoms in space. A change in configuration is different from a change in conformation.

Figure 6.23a. Change in conformation. The molecule on the left is the same molecule as the molecule on the right but with different conformations.


Figure 6.23b. ​A change in configuration. These are different molecules

Procedure:

Step 1: Identify the asymmetric atom or, in this case, the asymmetric carbon atom. An asymmetric carbon will have four different groups attached to it.

Figure 6.24: Assigning the chirality center in butan-2-ol.

Step 2: Prioritize the groups attached to the chirality center (this is similar to the E, Z nomenclature of alkenes).

Priority is placed on the atomic number of the atom that is directly attached to the asymmetric carbon atom. The group with the highest atomic number is given the highest priority and so on.

If the atoms are isotopes, then the higher mass number receives a higher priority; for example, tritium (3H or T) has a higher priority than deuterium (2H or D), followed by hydrogen (1Hor H).

The asymmetric carbon atom is bonded to an oxygen, two carbons, and a hydrogen atom. Since oxygen has the highest atomic number, it has the highest priority. Label this as priority 1. The hydrogen attached to the asymmetric carbon atom will have the lowest priority. Label this as priority 4. The two carbons attached to the chirality center have the same atomic number; therefore, go to the next set of atoms and make a list of atoms to which they are bonded. From this list, we see, at the first point of difference, that the bottom carbon is bonded to a carbon that has a higher atomic number than the hydrogen in the top branch. This branch would have the second highest priority and is labeled priority 2, leaving the top branch as priority 3.

When a priority cannot be assigned on the basis of the atomic number of the atoms that are directly attached to the chirality center, then the next set of atoms in the unassigned groups are examined. Assign priority at the first point of difference.

Figure 6.25: Assigning priority about chirality centre in butan-2-ol.

Step 3: Now that you have prioritized the atoms bonded to the chirality center, rotate the molecule so that the lowest priority group is at the back. Watch video 3.4 below to help you visualize what is going on.

Step 4: Trace a path from the highest to the lowest priority.

Figure 6.26: Assigning R configuration about chirality center in butan-2-ol.


The given structure has an R configuration and is called (R)-butan-2-ol.

Example: What is the configuration of the mirror image of (R)-butan-2-ol?

Step 1: Identify the asymmetric atom or, in this case, the asymmetric carbon atom . An asymmetric carbon will have four different groups attached to it.

6.27_v2.png
Figure 6.27: Assigning priority about chirality centre in mirror image of (R)-butan-2-ol.

Step 2: Prioritize the groups attached to the chirality center from highest to lowest atomic number. When a priority cannot be assigned on the basis of the atomic number of the atoms that are directly attached to the chirality center, then the next set of atoms in the unassigned groups is examined. Assign priority at the first point of difference.

Step 3: Rotate the molecule so that the lowest priority group is at the back. Watch the video to help you visualize what is going on.

Step 4: Trace a path from the highest to lowest priority. If the path is clockwise, then the enantiomer is designated as R. If the path is counterclockwise, then the enantiomer is designated as S.

Figure 6.28​: Assigning S configuration about chirality center in butan-2-ol.


The mirror image of (R)-butan-2-ol has an S configuration and is called (S)-butan-2-ol.

(R)-butan-2-ol and (S)-butan-2-ol are enantiomers. They are non-superposable mirror images and are different molecules. Click on the video for a comparison of the enantiomers

Figure 6.29​: Enantiomers (R)-butan-2-ol and (S)-butan-2-ol.


Example: Does the 2-amino-3-hydroxypropanoic acid have an R or S configuration?

Figure 6.30: Structure of 2-amino-3-hydroxypropanoic acid.

Step 1: Identify the asymmetric carbon atom . An asymmetric carbon will have four different groups attached to it.

Step 2: Prioritize the groups attached to the asymmetric carbon atom from highest to lowest atomic number. When a priority cannot be assigned on the basis of the atomic number of the atoms that are directly attached to the chirality center, then the next set of atoms in the unassigned groups are examined. Assign priority at the first point of difference.

​As seen in Figure 6.31, the first set of atoms attached to the asymmetric carbon are two carbons, a nitrogen and a hydrogen. Nitrogen has the highest priority because it has the highest atomic number and hydrogen would have the lowest priority. Since there are two carbons attached to the chirality center we need to look at the next set of atoms. Double bonds are treated as single bonds; thus, the carbonyl carbon is attached to three oxygens and the bottom carbon is attached to one oxygen and two hydrogens. Listing the atoms in decreasing atomic number shows that oxygen has a higher atomic number than hydrogen, this gives the carboxylic acid side chain has the second highest priority.

Figure 6.31: Prioritizing groups of 2-amino-3-hydroxypropanoic acid around the chiral center.

Step 3: Rotate the molecule so that the lowest priority group is in the back. Watch the video to help you visualize what is going on. Since the lowest priority is already in the back, we do not need to rotate or change the drawing (Figure 6.32). 

Figure 6.32: Assigning R configuration to 2-amino-3-hydroxypropanoic acid.

Step 4: Trace a path from the highest to the lowest priority. Notice in Figure 6.32 the lowest priority, hydrogen, looks as if it comes before the third priority. This is because of the way it is drawn. If you rotate the figure slightly the hydrogen will go between the first and third priority groups. Tracing a path from highest to lowest priority follows a clockwise path, thus 2-amino-3-hydrosypropanoic acid has an R configuration.

Watch the video for a stepwise explanation.

Many students confuse R and S determination with the direction the drawing is rotated to place the lowest priority atom or group at the back. How you rotate the drawing does not matter. The R and S determination is made after you have moved the lowest priority to the back, as you'll see by working through the example below. 

Example: Does the following molecule have an R or S configuration?

Figure 6.33: Structure of 2-chlorobutane.

Step 1: Identify the asymmetric atom.

Figure 6.34: Rotating 2-chlorobutane to facilitate assigning R or S.

Step 2: Prioritize the groups attached to the chirality center.

Step 3: Rotate the molecule so that the lowest priority group is at the back. This is not the step that determines R or S. The easiest way to move the lowest priority to the back is to rotate the molecule around a sigma bond that is in the plane of the page. In Figure 6.34 we see that ethyl chain stayed in the plane of the page while the other three groups are rotated.

Step 4: Trace a path from the highest to the lowest priority. Tracing a path from highest to lowest priority follows a clockwise path, thus 2-chlorobutane has an R configuration.

To stress this point, 12 different drawings of (S)-butan-2-ol are presented below. No matter which way your draw (S)-butan-2-ol, it will always be (S)-butan-2-ol. For each drawing, rotate the lowest priority hydrogen to the back and you will see that they all have an S configuration.

Figure 6.35​: Twelve different ways to represent (S)-butan-2-ol.


Q6.6 - Level 2

Are the following pair of molecules enantiomers or the same molecule drawn differently?

question description
A

Enantiomers

B

Same molecule, drawn differently

Watch the following video to see the explanation to Q6.6.


Q6.7 - Level 2

Are the following pair of molecules enantiomers or the same molecule drawn differently?

question description
A

Enantiomers

B

Same molecule, drawn differently



Stereochemistry

Although enantiomers have the same physical properties (e.g.. melting point, boiling point, density, and solubility), they will usually interact differently with other chiral molecules. A common example of this involves the taste and smell receptors in the tongue and nose, respectively.

Two famous examples involve the compounds limonene and carvone. The (+) enantiomer of limonene is recognized as an orange smell, while the (–) enantiomer is recognized as a lemon smell. Also, while (–)-carvone has a spearmint smell, (+)-carvone has a dill or caraway smell.


Keeping it Real Q6.1

Assign the absolute configuration in (–)-limonene, the enantiomer that smells like lemons.

A

R

B

S


One of the most famous examples of the importance of stereochemistry and medicines is the story of thalidomide. In 1957, racemic thalidomide was marketed as a drug for the treatment of morning sickness and was viewed as a ‘wonder drug’ for pregnant women. However, reports soon surfaced of children being born with abnormalities. The drug was quickly recalled, but over 10,000 babies were born with birth defects, including mental illnesses and badly deformed or completely missing limbs. The drug is thought to have claimed the lives of over 2,000 children.

What went wrong? Although the (R)-enantiomer of thalidomide was useful for the treatment of morning sickness, it turns out that the (S)-enantiomer is a teratogen and caused the birth defects.

You might wonder if one can use the (R)-enantiomer of thalidomide as a drug. Unfortunately, this does not work either because thalidomide readily racemizes under physiological conditions to produce an equal mixture of R and S enantiomers. Even recently, scientists have been trying to get around this stereochemistry problem.

6.6 Shortcut for determining R or S if lowest priority atom or group is in the front

If you understand this rule, then it will make life a lot easier when determining R and configurations from a drawing.

An asymmetric carbon atom can only be R or S. If the lowest priority is at the back, as it should be, then all you need to do is look at the molecule and prioritize the groups or atoms to determine if the numbering is clockwise or counterclockwise.

If the lowest priority is in the front, prioritize the groups, determine if it is clockwise or counterclockwise; whatever you determine, it to be must be the opposite since you are looking at the molecule from the wrong side. Click on the video for a visual explanation.

Figure 6.36. Assigning R or S for bromochlorofluorine. ​



Q6.8 - Level 2

Identify the chirality center as R, S, or the molecule is not chiral.

question description
A

RR

B

SS

C

molecule not chiral


Q6.9 - Level 2

Identify the chirality center as R, S, or the molecule is not chiral.

question description
A

RR

B

SS

C

molecule not chiral


Q6.10 - Level 2

Which of these groups has the highest priority in R/S configuration determination?

question description
A

1)

B

2)

C

3)

D

4)

E

5)


Q6.11 - Level 2

Identify the labeled chirality centers as R or S in the molecule shown below.

question description
A

2S\textit{S}, 6S\textit{S}

B

2S\textit{S}, 6R\textit{R}

C

2R\textit{R}, 6S\textit{S}

D

2R\textit{R}, 6R\textit{R}


Q6.12 - Level 2

Identify the labeled chirality centers as R or S in the molecule shown below.

question description
A

1S\textit{S}, 2S\textit{S}

B

1S\textit{S}, 2R\textit{R}

C

1R\textit{R}, 2S\textit{S}

D

1R\textit{R}, 2R\textit{R}


6.7 Optical activity

When plane-polarized light passes through a solution containing a chiral molecule, the plane-polarized light will be deflected to the left or right. If the plane-polarized light is rotated clockwise, then the rotation is (+) and the molecule is dextrorotatory. If the plane-polarized light is deflected counterclockwise, then the rotation is (and the molecule is levorotatory.

The measured or the observed optical rotation “α” of the sample depends upon the structure of the chiral molecule, concentration of the chiral molecule, length of the sample cell, wavelength of the light, the solvent used, and temperature. As the value is solvent-dependent, the specific rotation [α] of a sample is defined for each chiral molecule as follows.

Figure 6.37: Equation for specific rotation.

Specific rotation is a physical constant for a substance as are melting and boiling point, density, etc. There is no straightforward correlation between the R,S designation of an enantiomer and the direction (+) or (−) in which it rotates plane-polarized light. For example, you do not know if (R)-2-bromobutane will bend the plane-polarized light to the right (+) or the left (−) just by looking at its structure. However, once you determine experimentally the optical activity of (R)-2-bromobutane, you know the S configuration will have an opposite value of equal magnitude.

Figure 6.38: Enantiomers of 2-bromobutane having equal but opposite specific rotations.

6.8 Racemic mixture

Racemic Mixture: A solution containing equal concentrations of dextro- and levorotatory enantiomers will be optically inactive. This is called a racemic mixture, or racemate. A solution that is optically inactive does not rotate plane-polarized light. A racemic mixture can be designated by using the symbol (±)-butan-2-ol, which indicates you have an equal molar mixture of the enantiomers (S)-(+)-butan-2-ol and (R)-(−)-butan-2-ol.


Q6.13 - Level 2

A racemic mixture will rotate plane-polarized light?

A

Zero degrees

B

45 degrees

C

90 degrees

D

180 degrees


Q6.14 - Level 2

Which of the following compounds would be optically active?

question description
A

1)

B

2)

C

3)


Q6.15 - Level 2

Which of the following is true of any (S) – enantiomer?

A

It rotates plane-polarized light to the right

B

It rotates plane-polarized light to the left

C

It is the mirror image would be the (R\textit{R}) enantiomer

D

It has the highest priority group on the right


Q6.16 - Level 2

Which of the following is an alcohol that can exhibit optical activity?

question description
A

1)

B

2)

C

3)

D

4)


Q6.17 - Level 2

If a solution has an optical activity of + 25 degrees, which of the following statements can be said with certainty?

A

The compound in the solution has an (R\textit{R}) configuration

B

The compound in the solution has an (S\textit{S}) configuration

C

The compound in the solution has at least one stereogenic unit

D

The solution contains only one optically active compound


6.9 Enantiomeric excess and racemic mixtures

Not all solutions have only one enantiomer, and not all reactions produce only one enantiomer. If a solution contains only one enantiomer, then it is said to be optically, or enantiomerically, pure.

What happens to the optical activity if a solution contains both enantiomers? Since the optical activity of a pair of enantiomers is equal in magnitude but opposite in rotation to plane-polarized light, the optical activity of each dextrorotatory (+) molecule will cancel the optical activity of each levorotatory (−) molecule.

Enantiomeric Excess

A solution containing unequal amounts of each enantiomer will have an excess of one, or enantiomeric excess, which can be calculated in two ways.

Figure 6.39: Equation for determining % enantiomeric excess.

Let’s go through a thought experiment to understand what information these two equations provide. Assume we have a pure solution of (S)-butan-2-ol with an optical activity of +13.52. Using the two formulas, calculate the enantiomeric excess (%ee) and the observed specific rotation as the amount of (R)-butan-2-ol is increased. To make numbers easy let’s keep the total number of moles constant at 20 moles. This is an easy number to visualize and graphically display.

Start off with the enantiomeric pure solution of (S)-(+)-butan-2-ol, which has a specific rotation of +13.52. Now add three moles of (R)-(−)-butan-2-ol to the solution. Every mole of (R)-(−)-butan-2-ol will cancel the optical rotation of plane-polarized light as an equal number of moles of (S)-(+)-butan-2-ol, as shown in the diagram. The observed optical activity decreases to 9.46. As we add more (R)-(−)-butan-2-ol, the observed specific rotation will continue to decrease. Notice that a 50% enantiomeric excess does not mean 50% of the molecules are (+) and 50% are (−). It means that 50% of the molecules are (S)-(+)-butan-2-ol and 50% of the molecules are an equal mixture of (S)-(+)-butan-2-ol and (R)-(−)-butan-2-ol. The enantiomeric excess is 50%. 

When an equal number of moles of (S)-(+)-butan-2-ol and (R)-(−)-butan-2-ol are in the solution, the enantiomer excess is zero and the solution will have an optical activity of zero. This is called a racemic mixture, where neither enantiomer is in excess. As we continue to add (R)-(−)-butan-2-ol there will be an excess of the (−) enantiomer with a negative observed specific rotation that will become more negative until we have a pure solution of (R)-(−)-butan-2-ol, with a specific rotation of −13.25. (In reality, you can never obtain a 100% pure solution of R since there will always be S present in the solution; however, for this thought exercise, let us assume there is no S present).

Figure 6.40​

Example: Ephedrine is a bronchodilator and is used as a decongestant. The isomer that is used in many cold remedies is (–)-(1R,2S)-ephedrine. You are trying to purify a 50:50 mixture of ephedrine and its enantiomer. A solution of enantiomerically pure (–)-(1R,2S)-ephedrine in a solution of ethanol has a specific rotation of −6.3. After the purification, you measure the observed specific rotation to be −6.1. What is the %ee of ephedrine and what percentage of the enantiomer of ephedrine is still in the sample?

Figure 6.41: Determination of % enantiomeric excess ephedrine.

An enantiomeric excess of 97% indicates that 97% of the solution is (–)-ephedrine and 3% is a mixture of (–)-ephedrine and its enantiomer, (+)-ephedrine. This means that 98.5% of your sample is (–)-ephedrine and 1.5% is the enantiomer, (+)-ephedrine.


Q6.18 - Level 2

The specific rotation of L-dopa in water is -39.5. The specific rotation of a solution of L-dopa and its enantiomer was determined to be -19.75. Calculate the % ee of this mixture.

A

15%

B

25%

C

50%

D

75%

E

80%


Q6.19 - Level 3

The specific rotation of L-dopa in water is -39.5. The specific rotation of a solution of L-dopa and its enantiomer was determined to be -19.75. Calculate the % of L-dopa in the solution.

A

15%

B

25%

C

50%

D

75%

E

80%


6.10 Fischer projections

Fischer projections are generally used to help determine the stereochemistry of straight-chained carbohydrates, such as sugars, which have multiple chirality centers. They are also very helpful when comparing stereoisomers.

Fischer projection:

Each intersection of the horizontal and vertical lines represents a carbon. Each horizontal line represents a bond coming toward the viewer (wedged line). Each vertical line represents bonds going back away from the viewer (dashed line). The video linked further down on this page will provide a visual guide.

Figure 6.42: Converting from bond-line to Fischer projection for (R)-glyceraldehyde.​


Fischer projections of multiple carbons have all the bonds eclipsed. This allows you to find planes of symmetry quickly to determine if a molecule is chiral or achiral. Note: this is not a stable conformation of the molecule. However, this does not matter because a Fischer projection helps you look at the stereochemistry of a molecule or the stereochemical relationships between molecules.

Example: Is (2R,3R)-butane-2,3-diol a chiral molecule?

Figure 6.44: Converting from bond-line to Fischer projection for (2R,3R)-butane-2,3-diol.​


6.11 Shortcut for determining R and S configurations of Asymmetric Carbon atoms in a Fischer projection

As mentioned above, an asymmetric carbon atom can only be R or S. If the lowest priority is at the back, then you can determine R and S configuration quickly by prioritizing the groups or atoms to determine if the numbering is clockwise or counterclockwise.

With Fischer projections, the horizontal lines are sticking out of the page, which means that if this line is bonded to a hydrogen (the lowest priority), then you will need to rotate the hydrogen to the back. This process can become confusing. A shortcut exists with the Fischer projections. If the lowest priority is sticking out of the page on a horizontal line, then prioritize the groups, determine the direction of rotation (clockwise or counterclockwise) and whatever you determine it to be, it must be the opposite since you are looking at the molecule from the wrong side.

Example: Determine the stereochemistry of the chirality centers of hexane-2,3,4-triol.

Figure 6.45: Fischer projection of hexane-2,3,4-triol.

Answer: (2R,3S,4R)-hexane-2,3,4-triol

Click on the video for further explanation.


Figure 6.46a. Determination of R or S for carbon 2 in Fischer projection of hexane-2,3,4-triol.​


Figure 6.46b. Determination of R or S for carbon 3 in Fischer projection of hexane-2,3,4-triol.​


Figure 6.46c. Determination of R or S for carbon 4 in Fischer projection of hexane-2,3,4-triol.​


Figure 6.46d. Determination of R or S for all chirality centers in Fischer projection of hexane-2,3,4-triol, using the shortcut method.​


Q6.20 - Level 2

Which of the following compounds represents (S)-2-bromobutane?

question description
A

1)

B

2)

C

3)

3D Molecule*: (S)-2-bromobutane



Q6.21 - Level 3

Identify the chirality centers as (R) or (S) in the Fischer projection of 2,3-dichlorobutanal given below.

question description
A

1S\textit{S}, 2S\textit{S}

B

1S\textit{S}, 2R\textit{R}

C

1R\textit{R}, 2S\textit{S}

D

1R\textit{R}, 2R\textit{R}


Q6.22 - Level 3

Identify the chirality centers as R or S in the Fischer projection of 4-chlorobutane-1,2,3-triol given below.

question description
A

(2S,3S)(2S,3S)

B

(2S,3R)(2S,3R)

C

(2R,3S)(2R,3S)

D

(2R,3R)(2R,3R)


6.12 Enantiomers and diastereomers

For each asymmetric carbon atom there are two configurations, R or S; thus, the maximum number of stereoisomers that a molecule can have is 2n, where n = # of chirality centers (carbons with four different groups attached to it).

2,3-Dibromopentane has two stereogenic units or chirality centers and has 22 = 4 possible stereoisomers. The combination of stereoisomers would be (2R,3R), (2S,3S), (2R,3S), and (2S,3R).

Drawing them as Fischer projections, we see that (2R,3R) and (2S,3S) are enantiomers and (2R,3S) and (2S,3R) are enantiomers:

Figure 6.47: Structure of 2,3-dibromopentane.​


Figure 6.48: Fischer projections of stereoisomers of 2,3-dibromopentane.​


What stereochemical relationship do (2R,3R)- and (2S,3R)-dibromopentane have with each other? They are stereoisomers because they have the same connectivity. However, they are not mirror images of each other. These types of stereoisomers are called diastereomers.

Diastereomers: These stereoisomers are not mirror images of each other. They have different physical properties and, unlike enantiomers, can be more readily separated or isolated from each other. Take, for example, cis and trans 1,4-dibromocyclohexane.

Figure 6.49: Diastereomers cis-1,4-dibromocyclohexane and trans-1,4-dibromocyclohexane.

Make sure you understand the difference between diastereomers and structural or constitutional isomers. This difference can be demonstrated with alkenes that have the chemical formula C2FClBrI. (Z)-1-Bromo-1-chloro-2-fluoro-2-iodoethene and (E)-1-bromo-1-chloro-2-fluoro-2-iodoethene are diastereomers, whereas (Z)-1-bromo-1-chloro-2-fluoro-2-iodoethene and (E)-1-bromo-2-chloro-2-fluoro-1-iodoethene are constitutional isomers.

Figure 6.50: Isomeric relationships between alkenes of formula C2FClBrI.

In visualizing the stereoisomers of 2-bromo-3-chlorobutane, notice how the Fischer projections allow for a quick visual omparison between them.

Figure 6.51: Visualizing the stereoisomers of 2-bromo-3-chlorobutane with Fischer projections and dashed and wedge structures.”


6.13 How to recognize chirality centers

As the number of chirality centers increases so does the number of possible stereoisomers. This is very important in the synthesis of molecules. For every stereogenic center you create, you may produce two products: a pair of stereoisomers.

How many chirality centers and stereoisomers does the following molecule have? Click on the video below to see if you are correct. 

Figure 6.52​: Example structure for determining chirality centres and stereoisomers.​


How many chirality centers and stereoisomers does the following molecule have? Click on the video below to see if you are correct. 

Figure 6.53: Example structure for determining chirality centres and stereoisomers.​​



Q6.23 - Level 2

There are 8 chirality centers in the following molecule. Identify each asymmetric carbon atom by clicking on the circled carbons that are chiral.

Note: sp2 carbon atoms of a double bond cannot be chiral centers since they are planar. However, they can be a stereogenic unit as in cis-trans isomerism (see Figure 6.50).

Q6.24 - Level 2

There are 5 chirality centers in the following molecule. Identify each asymmetric carbon atom by clicking on the circled carbons that are chiral. Make sure you click on the center of the circle.


6.14 Meso Compounds

A meso compound is a molecule that has two or more stereogenic units; however, due to symmetry in the molecule, it is achiral. 

For example (1R,2S)-1,2-dibromocyclohexane has two chirality centers or asymmetric carbon atoms, but the whole molecule is achiral. View the video for a visual presentation.

Figure 6.54: Meso compound (1R,2S)-1,2-dibromocyclohexane with plane of symmetry shown, viewed as Haworth projection (left) and as bond-line diagrams (right).​

There are different types of symmetry, such as inversion (reflection about a point), which renders a molecule achiral. This chapter will only talk about planes of symmetry. This chapter will also not talk about conformational stereoisomers, such as cis-1,2-dimethylcyclohexane, which has two chair conformations that are enantiomeric. This textbook will not consider conformational stereoisomers. For more information on these topics, consult other references.

With this new knowledge of meso compounds, calculate how many stereoisomers tartaric acid has. There are two chirality centers, thus tartaric acid has potentially 4 stereoisomers: (R,R), (S,S), (R,S), and (S,R). First, draw them as Fischer projections for easy visualization.

Figure 6.55​​: Tartaric acid with chirality centers indicated with an asterisk.​

As you can see from the Fischer projections, (R,R)- and (S,S)-tartaric acid are enantiomers. They have the same melting point with opposite optical activity. However, (R,S)-tartaric acid is a meso compound since its mirror image is the same molecule. The optical activity of meso-(R,S)-tartaric acid is zero, as expected with an achiral molecule. Thus, there are 3 stereoisomers of tartaric acid, not 4, because one isomer is a meso compound. 

Figure 6.56: Stereoisomers of tartaric.​

In visualizing the stereoisomers of 2,3-dibromobutane, notice how the Fischer projections allow for a quick visual comparison between them.

Figure 6.57: Visualizing the stereoisomers of 2,3-dibromobutane with Fischer projections and dashed and wedge structures.


Q6.25 - Level 1

Is 1,1,4-trimethylcyclohexane a meso compound?

A

Yes

B

No

Watch the video below for a walkthrough of this problem:

Question: Determine whether each molecule below is chiral or achiral. Which molecules are meso? Click on the video below to see if you are correct.

Figure 6.58​​: Structures A - F


Question: Which pairs of the following compounds are enantiomers and diastereomers? Which compound is meso? Click on the video below to see if you are correct.

Figure 6.59​: Structures 1-3



Q6.26 - Level 2

Are the following molecules chiral or achiral? Match the molecules with the terms chiral and achiral.

question description
Premise
Response
1

1)

A

Chiral

2

2)

B

Chiral

3

3)

C

Achiral

D

Chiral

E

Achiral

F

Achiral


Q6.27 - Level 1

Are the following molecules chiral or achiral? Match the molecules with the terms chiral and achiral.

question description
Premise
Response
1

1)

A

Achiral

2

2)

B

Chiral

3

3)

C

Chiral

D

Achiral

E

Chiral

F

Achiral


6.15 Chiral molecules without asymmetric atoms

If a molecule has one chirality center, then the molecule will be chiral since it is asymmetric. If a molecule has two or more chirality centers, then the molecule will be chiral unless there is a plane of symmetry; i.e., a meso compound. The next question is; does a chiral molecule have to have a stereogenic atom. The answer is no. Allenes are a great example of this.

Allenes have the sequence of an sp2 carbon then an sp carbon followed by an sp2 carbon. This arrangement of the π bonds causes the groups at the end of the allene to be perpendicular. If an allene has different groups at each end, then a pair of enantiomers can form. They will be optically active and bend plane-polarized light in equal magnitude but in opposite directions. There is no cis–trans isomerism because the groups at either end are not in the same plane. Watch the video for a visual explanation.

Figure 6.60: Stereoisomeric relationship of Allenes with molecular formula C3H2Br2.​


6.16 Review

Figure 6.61
Figure 6.62

6.17 Chapter Summary

Isomers are two or more different compounds that have the same molecular formula, a summary of isomers is shown below.

Figure 6.63

Structural or Constitutional Isomers

These isomers have the same molecular formula, but their atoms are bonded in a different order. These different isomers have very different physical properties. Some examples are shown below.

Figure 6.64

A constitutional isomer is not the same thing as a conformational change. A conformational change is the same molecule that has undergone a rotation about a sigma bond.

Stereoisomers

These isomers have the same molecular formula and the same connectivity, but their atoms are arranged differently in space. Example of this are cis and trans isomers, shown below.

Figure 6.65

Alkene E and Z Nomenclature

The double bond of alkene prevents free rotation of the C=C bond. The “cis-trans” terminology is reserved for alkenes that have one substituent on each alkene carbon atom. When an alkene has three or four substituents then it is named differently using the Cahn-Ingold-Prelog convention.

First, split the molecule in half and consider both sides of the double bond. The vinylic atom that has a higher atomic number receives a higher priority. If the atoms are isotopes the higher mass receives the higher priority. If the atoms at the vinylic positions are the same then explore outwardly until the first point of difference is observed. The atom with the higher atomic number receives the higher priority. Do not use sums of the atomic numbers or the size of the groups. 

Figure 6.66

When the substituents contain double or triple bonds they must be artificially transformed into sp3 atoms.

Figure 6.67

Chiral Compounds

Another type of stereoisomerism is called mirror-image stereoisomerism. Mirror-image stereoisomerism occurs when a molecule and its mirror image are non-superposable. A molecule whose mirror image is non-superposable is said to be chiral and is a different molecule. A molecule that can be superposed onto its mirror image is said to be achiral and is the same molecule. An easy way to determine if a molecule is chiral or achiral is to look for a plane of symmetry.

Figure 6.68

Other achiral molecules are shown below.

Figure 6.69

Enantiomers

Enantiomers are two stereoisomers that are non-superposable mirror images of each other. These molecules do not have a plane of symmetry. 

Figure 6.70

A carbon atom that is connected to four different substituents is called an asymmetric atom or a chirality center and is often marked with an asterisk. Molecules that have only one chirality center are always chiral and have no plane of symmetry. Enantiomers are different molecules that have the same physical properties but are distinguished in chiral environments, such as biological systems.

R-S Nomenclature

To identify a chirality center using nomenclature, the R-S system using the Cahn-Ingold-Prelog convention is used. Prioritize the substituents attached to the stereogenic unit as done with E, Z nomenclature in alkenes. Rotate the substituents around the chirality center such that the lowest priority is in the back. Considering the first, second and third priority groups trace a path from the highest to the lowest priority. If the path is clockwise the stereocenter is designated “R”; if it is counterclockwise the chirality center is designated “S”. See the example below.

Figure 6.71


Figure 6.72

A pair of enantiomers always has the opposite R-S designations, an example of butan-2-ol is shown below.

Figure 6.73

Optical Activity

Although enantiomers are different molecules they have most of the same physical properties (i.e. MP, BP, etc), except the direction in which they rotate plane-polarized light (optical activity). A pair of enantiomers will each rotate plane polarized light in the opposite direction but with equal magnitude. When light is rotated to the right it is designated (+) and to the left it is designated (-). There is no correlation between the direction the light is rotated and the “R-S” designation.

Figure 6.74

A solution that contains an equal mixture of enantiomers will be optically inactive and have an optical rotation of zero. This type of solution is referred to as a racemic mixture and can be written as “(+)”. A solution that contains an achiral molecule will also be optically inactive. 

Enantiomeric Excess

A solution containing only one enantiomer is referred to as “optically/enantiomerically pure”. When a solution contains an equal mixture of enantiomers it is racemic and not optically active. When a solution contains an unequal mixture of enantiomers then the optical activity of one enantiomer will cancel out the optical activity of the other enantiomer, however, since the mixture is unequal, there will be an excess of one enantiomer. This is referred to as the enantiomeric excess and is calculated as shown below.

Figure 6.75

Fisher Projections

Fisher projections are typically used to help determine the stereochemistry of straight chained carbohydrates such as sugars that have multiple chirality centers. They are also very helpful when comparing stereoisomers. When drawing a Fisher projection remember the following: each intersection of a horizontal and vertical line represents a carbon atom; horizontal lines represent a bond coming toward the viewer (wedge); vertical lines represent a bond coming away from the viewer (dash).

Figure 6.76

Diastereomers

For each chirality center in a molecule there are two possible configurations: R and S, thus the maximum number of stereoisomers that a molecule can have is 2n, where n = # of chirality centers. 2,3-Dibromopentane has two chirality centers and thus has 4 possible stereoisomers.

Figure 6.77

The pairs of molecules that are not enantiomers are called diastereomers, which is defined as stereoisomers that are not mirror images of each other (i.e. (2R, 3R) and (2S, 3R) are diastereomers). Unlike enantiomers, diastereomers have different physical properties such as MP, BP, etc. Examples of cis and trans diastereomers are shown below.

Figure 6.78

Meso Compounds

A molecule that has two or more chirality centers, that also has a plane of symmetry, is an achiral molecule and is referred to as a meso compound. Note: meso compounds have asymmetric atoms (chirality centers) but the plane of symmetry makes the molecule itself achiral.

Figure 6.79

Chiral Molecules That Do Not Have Chirality Centers

Most molecules that do not contain a chirality center are achiral, however an exception to this are allene molecules. Allenes have the sequence of an sp2 carbon then an sp carbon followed by an sp2 carbon. This arrangement of pi bonds causes the groups at the end of the allene to be perpendicular. If there are different groups at each end of an allene then a pair of enantiomers can form. They will be optically active. 

Figure 6.80


End of Chapter 6

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Image Credits

[1] Image courtesy of Rhoda Baer in the Public Domain.
[*] 3D Molecule: courtesy of QR Chem. QR Chem is a resource created by students and Professor Neil Garg at UCLA.