Organic Chemistry I & II
Organic Chemistry I & II

Organic Chemistry I & II

Lead Author(s): Steven Forsey

Student Price: Contact us to learn more

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.

This content has been used by 15,690 students

What is a Top Hat Textbook?

Top Hat has reimagined the textbook – one that is designed to improve student readership through interactivity, is updated by a community of collaborating professors with the newest information, and accessed online from anywhere, at anytime.


  • Top Hat Textbooks are built full of embedded videos, interactive timelines, charts, graphs, and video lessons from the authors themselves
  • High-quality and affordable, at a significant fraction in cost vs traditional publisher textbooks
 

Key features in this textbook

Engage students with all different learning styles, offering plentiful visuals, instructional videos, interactive demonstrations, real world examples and auto-graded drawing questions.
Built-in assessment questions embedded throughout chapters so students can read a little, do a little, and test themselves to see what they know! “Keeping it Real”Sections connect theory to real-world examples and demonstrate the real-life applications of Organic Chemistry.
Questions are embedded right in the textbook and can be auto-graded, including chemical drawings.

Comparison of Organic Chemistry Textbooks

Consider adding Top Hat’s Organic Chemistry textbook to your upcoming course. We’ve put together a textbook comparison to make it easy for you in your upcoming evaluation.

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

Pricing

Average price of textbook across most common format

Up to 40-60% more affordable

Lifetime access on any device

$219

Hardcover print text only

$301

Hardcover print text only

$301

Hardcover print text only

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

Content meets standard for Introduction to Anatomy & Physiology 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

Only available with supplementary resources at additional cost

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

Explore this textbook

Read the fully unlocked textbook below, and if you’re interested in learning more, get in touch to see how you can use this textbook in your course today.

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.

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

E\textit{E}

2

2)

B

Z\textit{Z}

C

Z\textit{Z}

D

E\textit{E}


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.

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

Achiral

2

2)

B

Achiral

3

3)

C

Achiral

D

Chiral

E

Chiral

F

Chiral


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.

F6.27.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