CHEM 131 - ADVANCED GENERAL CHEMISTRY I Chirality and optical activity Question Bank Set 4

CHEM 131 - ADVANCED GENERAL

CHEMISTRY I - Chirality and optical

activity

Question Bank - Set 4

Liberty University

Question 1

Question

Explain the concept of chirality and how it relates to optical activity in molecules.

Solution

Step 1: Chirality Chirality is a property of a molecule that cannot be super-

imposed on its mirror image. In other words, a chiral molecule exists in two non-

superimposable mirror image forms known as enantiomers. These enantiomers

have the same physical and chemical properties except for their interaction with

plane-polarized light.

Step 2: Optical Activity When a chiral molecule is placed in a beam of

plane-polarized light, it will rotate the plane of polarization either to the left

(counter-clockwise - levorotatory) or to the right (clockwise - dextrorotatory).

The extent of this rotation is unique for each enantiomer and is measured using

a polarimeter. The specific rotation, denoted by [α], is a constant value for

each enantiomer at a given temperature, concentration, and wavelength of light.

The observed rotation depends on the concentration of the compound, the path

length of the sample tube, and the temperature.

Step 3: Relationship between Chirality and Optical Activity The

ability of a chiral molecule to rotate the plane of polarized light is due to the

asymmetric arrangement of atoms or groups around a chiral center within the

molecule. This asymmetry causes one enantiomer to rotate light in one direction

and the other enantiomer to rotate light in the opposite direction. Therefore,

the optical activity of a compound is directly related to its chirality.

Overall, chirality is a fundamental concept in organic chemistry that has

significant implications in determining the behavior and properties of molecules,

especially in the field of pharmaceuticals and biochemistry.

Question 2

Question

Determine whether the following molecules are chiral or achiral, and if chiral,

specify the number of chiral centers present:

a) 2-bromobutane

b) 2,3-dibromobutane

Solution

a) 2-bromobutane: Step 1: Determine the number of chiral centers in the

molecule. - A chiral center is a carbon atom bonded to four different groups. -

In 2-bromobutane, the carbon center bonded to the bromine atom is not chiral

because it is bonded to two hydrogen atoms (which are the same) and the two

carbon atoms. - Therefore, 2-bromobutane does not have any chiral centers.

Step 2: Determine if the molecule is chiral or achiral. - Since 2-bromobutane

does not have any chiral centers, it is an achiral molecule.

b) 2,3-dibromobutane: Step 1: Determine the number of chiral centers in

the molecule. - In 2,3-dibromobutane, both carbon atoms located in the middle

of the chain are chiral centers since each is bonded to four different groups (two

bromine atoms and two other carbon atoms). - Therefore, 2,3-dibromobutane

has 2 chiral centers.

Step 2: Determine if the molecule is chiral or achiral. - Since 2,3-dibromobutane

has chiral centers, it is a chiral molecule.

Question 3

Question

Explain the concept of chirality in chemistry and how it relates to optical ac-

tivity.

Solution

Step 1: Chirality in Chemistry A molecule is considered chiral if it cannot

be superimposed on its mirror image. In other words, a chiral molecule is

not identical to its mirror image. Chirality arises when a molecule contains

an asymmetric carbon atom, which is a carbon atom bonded to four different

groups or atoms.

Step 2: Optical Activity When a chiral molecule is placed in a plane-

polarized light, it can rotate the plane of polarization. This phenomenon is

known as optical activity. The two enantiomers of a chiral molecule (mirror

images) will rotate the plane of light in opposite directions - one clockwise (dex-

trorotatory or +) and one counterclockwise (levorotatory or -).

Step 3: Enantiopurity If a sample contains only one enantiomer (either all

+ or all -), it is said to be enantiopure. Enantiopure compounds are optically

pure and exhibit maximum optical activity.

Step 4: Racemic Mixtures A racemic mixture is one that contains equal

amounts of both enantiomers (+ and -). Racemic mixtures do not exhibit

optical activity since the rotations from the two enantiomers cancel each other

out.

Step 5: Importance of Chirality Chirality is important in various fields such

as pharmacology, as enantiomers of a drug can have different biological activ-

ities. This is why it is crucial to understand the concept of chirality and its

implications in chemistry and other scientific disciplines.

Question 4

Question

Explain the concept of chirality and optical activity in chemistry. Provide an

example of a chiral molecule and discuss how its optical activity arises.

Solution

Step 1: Chirality A molecule is chiral if it is not superimposable on its mirror

image. This means that the molecule and its mirror image are non-identical,

just like our hands are non-superimposable mirror images of each other.

Step 2: Optical Activity Chiral molecules can exhibit optical activity,

which is the ability to rotate the plane of polarized light. A solution of a chiral

compound will rotate the plane of polarized light either to the left (levorotary,

denoted by the prefix ”l-”) or to the right (dextrorotary, denoted by the prefix

”d-”).

Step 3: Example: Chiral Molecule - Lactic Acid One example of a

chiral molecule is lactic acid, which has the molecular formula C3H6O3. Lactic

acid exists in two enantiomeric forms: L(+) lactic acid and D(-) lactic acid.

These enantiomers are non-superimposable mirror images of each other.

Step 4: Optical Activity in Lactic Acid In the case of lactic acid, L(+)

lactic acid is levorotary and rotates plane-polarized light to the left, while D(-)

lactic acid is dextrorotary and rotates plane-polarized light to the right. This

optical activity arises from the asymmetric carbon atom present in lactic acid,

which results in two enantiomeric forms that interact with plane-polarized light

in different ways.

Question 5

Question

Explain why a molecule with a chiral carbon atom but lacking a stereocenter

cannot exhibit optical activity.

Solution

Step 1: To understand this concept, we need to first clarify the definitions of a

chiral carbon atom and a stereocenter. - A chiral carbon atom is a carbon atom

bonded to four distinct groups. - A stereocenter is an atom, usually a carbon,

which is connected to four different substituents and is a center of chirality.

Step 2: When a molecule has a chiral carbon atom, it implies that it is a

molecule with a non-superimposable mirror image, making it optically active.

Step 3: However, if a molecule has a chiral carbon atom but lacks a stereo-

center, it means that the chiral carbon is not the only atom in the molecule with

four different substituents. This implies that the molecule lacks a true center of

chirality.

Step 4: Optical activity arises from the presence of a chiral center (stere-

ocenter) in a molecule because the molecule and its mirror image cannot be

superimposed.

Step 5: In the case where a molecule has a chiral carbon atom but lacks a

stereocenter, the molecule can be superimposed on its mirror image by another

arrangement within the molecule. This prevents the molecule from exhibiting

optical activity.

Therefore, a molecule with a chiral carbon atom but lacking a stereocenter

cannot exhibit optical activity.

Question 6

Question

Determine whether the following molecules are chiral or achiral, and specify if

they are optically active or inactive:

a) 2 −bromobutane b) 2 −chlorobutane

Solution

Step 1: To determine chirality, we need to examine if the molecule has a chiral

center. A chiral center is a carbon atom that is bonded to four different groups.

a) 2-bromobutane: The carbon attached to the bromine atom has two

hydrogen atoms, the bromine atom, and an ethyl group bonded to it. Since the

carbon atom has four different groups (H, Br, H, ethyl), it is a chiral center and

the molecule is chiral.

b) 2-chlorobutane: The carbon attached to the chlorine atom has two

hydrogen atoms, the chlorine atom, and a methyl group bonded to it. Since the

carbon atom does not have four different groups (H, Cl, H, methyl), it is not a

chiral center and the molecule is achiral.

Step 2: To determine optical activity, we need to check if the chiral molecule

has a non-superimposable mirror image (enantiomer).

a) 2-bromobutane: Since 2-bromobutane is chiral, it will have a non-

superimposable mirror image. Thus, 2-bromobutane is optically active.

b) 2-chlorobutane: Since 2-chlorobutane is achiral, it does not have a non-

superimposable mirror image. Hence, 2-chlorobutane is optically inactive.

Question 7

Question

Determine whether the following compounds are chiral or achiral, and if chiral,

state whether the compound is optically active:

a) 2,3-dibromobutane

b) 2,3-dibromopentane

c) 2-bromopentane

d) 2-chlorobutan-2-ol

Solution

Step 1: To determine chirality, we need to examine the presence of a chiral

center in each compound. A chiral carbon is a carbon atom bonded to four

different groups.

a) 2,3-dibromobutane: The carbon at position 2 is bonded to two bromine atoms

and two hydrogen atoms, making it a chiral center.

b) 2,3-dibromopentane: The carbon at position 2 is bonded to two bromine

atoms and two methyl groups, making it a chiral center.

c) 2-bromopentane: The carbon at position 2 is bonded to two hydrogen atoms,

a methyl group, and a bromine atom, making it a chiral center.

d) 2-chlorobutan-2-ol: The carbon at position 2 is bonded to a hydrogen atom, a

hydroxyl group, a chlorine atom, and a methyl group, making it a chiral center.

Step 2: Next, we determine if the chiral compounds are optically active. For

a compound to be optically active, it must be chiral and lack an internal plane

of symmetry.

a) 2,3-dibromobutane: This compound is chiral and optically active because

there is no internal plane of symmetry.

b) 2,3-dibromopentane: This compound is chiral and optically active because

there is no internal plane of symmetry.

c) 2-bromopentane: This compound is chiral and optically active because there

is no internal plane of symmetry.

d) 2-chlorobutan-2-ol: This compound is chiral and optically active because

there is no internal plane of symmetry.

Therefore, compounds a), b), c), and d) are all chiral, and they are optically

active.

Question 8

Question

Determine the optical activity of each of the following compounds:

1. 2,3-dibromobutane

2. 1,2-dichlorocyclopentane

3. 2-chloro-3-methylpentane

Solution

1. For 2,3-dibromobutane, we need to determine if the molecule is chiral or

achiral.



The molecule has a plane of symmetry, which means it is achiral.



Therefore, 2,3-dibromobutane is achiral and does not exhibit optical ac-

tivity.

2. For 1,2-dichlorocyclopentane, we need to determine if the molecule is

chiral or achiral.



The molecule does not have a plane of symmetry, so it is chiral.



Therefore, 1,2-dichlorocyclopentane is chiral and exhibits optical activity.

3. For 2-chloro-3-methylpentane, we need to determine if the molecule is

chiral or achiral.



The molecule does not have a plane of symmetry, so it is chiral.



Therefore, 2-chloro-3-methylpentane is chiral and exhibits optical activity.

Question 9

Question

Define chirality and explain how chirality is related to optical activity in molecules.

Solution

Step 1: Chirality: A molecule is chiral if it cannot be superimposed on its

mirror image. In other words, a molecule is chiral if it lacks an internal plane of

symmetry. Chirality arises when a molecule contains at least one stereocenter,

which is a carbon atom with four different substituents.

Step 2: Optical Activity: Chiral molecules often exhibit a property called

optical activity, where they can rotate the plane of polarized light. This rotation

occurs because the two enantiomers (mirror-image forms) of a chiral compound

interact differently with plane-polarized light due to their non-superimposable

nature.

Step 3: Relationship between Chirality and Optical Activity: The

relationship between chirality and optical activity can be understood through

the concept of enantiomeric pairs. Enantiomers are chiral molecules that are

non-superimposable mirror images of each other. When a sample contains equal

amounts of both enantiomers (racemic mixture), the optical activity is canceled

out, resulting in no net rotation of polarized light.

Step 4: Specific Rotation: The extent of optical rotation for a particular

enantiomer is quantified by a property known as specific rotation. The specific

rotation, denoted as [α], is a measure of how much a compound rotates plane-

polarized light and is dependent on factors such as concentration, path length,

and temperature.

Step 5: Application in Chemistry: Chirality and optical activity play

crucial roles in various fields of chemistry, especially in pharmaceuticals and

biochemistry. The biological activity of chiral drugs is often dependent on their

specific chirality, as different enantiomers can exhibit varying effects on the

human body.

In conclusion, chirality in molecules is intimately connected to optical ac-

tivity, where chiral compounds can interact differently with polarized light due

to their non-superimposable nature, leading to a phenomenon known as optical

rotation.

Question 10

Question

An organic compound has the following structure:

R−C(H)(H) −C(H3)(H) −C(H3)(H3)

Determine whether this compound is chiral or achiral. If it is chiral, identify

the chiral center(s). State whether the compound is optically active or inactive.

Solution

Step 1: Identify the Chiral Center(s)

A chiral center is a carbon atom bonded to four unique substituents. In

the given compound, the second carbon (C(H)(H)) is the only carbon that is

bonded to four unique substituents. Therefore, the compound is chiral and the

second carbon is the chiral center.

Step 2: Determine Optical Activity

To determine if the compound is optically active, we need to examine if it

has a plane of symmetry. A molecule is optically inactive if it has a plane of

symmetry.

In this case, if we draw an imaginary plane cutting through the molecule

at the chiral center (second carbon), the two halves of the molecule are not

mirror images of each other. Therefore, the compound does not have a plane of

symmetry and is optically active.

Thus, the compound is chiral with the second carbon as the chiral center,

and it is optically active.

Question 11

Question

Explain why a molecule with a stereocenter can be chiral, but a molecule with

an axis of symmetry cannot be chiral. Provide an example of each type of

molecule.

Solution

Step 1: A molecule is chiral if it does not have a plane of symmetry and is not

superimposable on its mirror image. A stereocenter is a carbon atom with four

different substituents, allowing for two non-superimposable mirror image forms

(enantiomers). This leads to chirality in a molecule.

Step 2: Let’s consider an example of a chiral molecule with a stereocenter:

2-chlorobutane. The carbon atom labeled with an asterisk (*) is a stereocenter

because it has four different substituents - a hydrogen atom, a methyl group, an

ethyl group, and a chlorine atom. The two possible forms of 2-chlorobutane are

non-superimposable mirror images of each other, making it a chiral molecule.

Step 3: On the other hand, a molecule with an axis of symmetry is not chiral

because it possesses a plane (or center) of symmetry. If a molecule has an axis

of symmetry, it can be rotated 180 degrees about that axis to superimpose with

its mirror image.

Step 4: An example of a molecule with an axis of symmetry that is not chiral

is trans-1,2-dichloroethylene. This molecule can be folded along its central axis

to superimpose with its mirror image, demonstrating lack of chirality.

In conclusion, a molecule with a stereocenter can be chiral due to the pres-

ence of non-superimposable mirror images (enantiomers), while a molecule with

an axis of symmetry cannot be chiral because it can be superimposed with its

mirror image.

Question 12

Question

Explain the concept of chirality and optical activity. Provide an example of a

molecule that exhibits chirality and explain how its chirality affects its optical

activity.

Solution

Step 1: Chirality

Chirality is a property of a molecule that cannot be superimposed on its mirror

image. In other words, a molecule is chiral if it lacks an internal plane of

symmetry. Chiral molecules are often referred to as ”handed” molecules because

they exist in two nonsuperimposable mirror image forms (enantiomers).

Step 2: Optical Activity

Optical activity is the ability of a chiral molecule to rotate the plane of polarized

light that passes through it. This phenomenon is observed when a beam of

polarized light passes through a solution of chiral molecules, causing the plane

of polarization to rotate either clockwise (dextrorotatory) or counterclockwise

(levorotatory).

Step 3: Example: Lactic Acid

One example of a chiral molecule is lactic acid (CH3CH(OH)COOH). Lactic

acid has two enantiomeric forms: L-lactic acid and D-lactic acid. These forms

are nonsuperimposable mirror images of each other and exhibit optical activity

due to their chirality.

Step 4: Effect on Optical Activity

L-lactic acid is levorotatory, meaning it rotates the plane of polarized light

counterclockwise. On the other hand, D-lactic acid is dextrorotatory, meaning

it rotates the plane of polarized light clockwise. This difference in optical activity

between enantiomers is a direct result of their chirality.

Question 13

Question

Explain why the compound shown below is chiral and determine if it is optically

active.

CH3−C H2−CH OH −COOH

Solution

Step 1: To determine if a compound is chiral, we must first identify if it has a

chiral center. A chiral center is a carbon atom bonded to four different groups.

In the compound given, the carbon atom bonded to the hydroxyl (OH) group

is a chiral center.

Step 2: Now, let’s determine if the compound is optically active. For a

compound to exhibit optical activity, it must be chiral and not have a plane of

symmetry.

Step 3: The compound CH3−C H2−CH OH −COOH is chiral because it

has a chiral center. To determine if it is optically active, we need to check if it

has a plane of symmetry.

Step 4: If we try to draw a plane of symmetry through the chiral carbon atom

in the compound, we will find that it is not possible. Therefore, the compound

does not have a plane of symmetry.

Step 5: Since the compound is chiral and lacks a plane of symmetry, it

is optically active. When a beam of plane-polarized light passes through an

optically active compound, the plane of polarization will rotate in one direction.

Question 14

Question

Determine whether the following molecules are chiral or achiral, and if chiral,

state whether they are optically active:

a) CHClBrF b) CHFClBr

Solution

Step 1: To determine chirality, we need to identify if the molecule has a chiral

center, which is a carbon atom bonded to four different groups.

a) For molecule CHClBrF, the central carbon is bonded to H, Cl, Br, and

F. Since all four groups bonded to the central carbon are different, CHClBrF is

chiral.

Step 2: To determine optical activity, we need to check if the molecule lacks

a plane of symmetry. A molecule is optically active if it is chiral and lacks a

plane of symmetry.

In the case of CHClBrF, it is chiral and does not have a plane of symmetry,

so it is optically active.

b) For molecule CHFClBr, the central carbon is bonded to H, F, Cl, and

Br. Since all four groups bonded to the central carbon are different, CHFClBr

is chiral.

Step 3: To determine optical activity, we need to check if the molecule lacks

a plane of symmetry.

In the case of CHFClBr, it is chiral but does have a plane of symmetry

(if you rotate the molecule 180 degrees around the central C-F bond), so it is

achiral and not optically active.

Question 15

Question

Explain why a molecule with a chiral center is optically active, while a molecule

with a plane of symmetry is not optically active.

Solution

1. A molecule is considered chiral if it cannot be superimposed on its mirror

image. This typically arises when a molecule has a chiral center, which is

a carbon atom bonded to four different groups.

2. When light interacts with a chiral molecule, the two enantiomers (mirror

image forms) will rotate the plane of polarized light in opposite directions.

This phenomenon is known as optical activity.

3. The optical activity of a chiral molecule is due to its lack of internal

symmetry, leading to different interactions with right-handed and left-

handed circularly polarized light.

4. Conversely, a molecule with a plane of symmetry can be superimposed

on its mirror image. This results in no difference in the interactions with

right-handed versus left-handed circularly polarized light.

5. Since a molecule with a plane of symmetry does not have the property of

enantiomerism and does not produce differing interactions with polarized

light, it is optically inactive.

Question 16

Question

Draw the enantiomers of the compound shown below and determine which enan-

tiomer is optically active. Explain your reasoning.

CH3−CH(−CH3)−CH2−OH

Solution

Step 1: Draw the enantiomers of the compound.

CH3−C(−CH3)H −CH2−OH

Step 2: Determine the chirality of the compound. The central carbon atom

is chiral because it has four different groups attached to it (CH3, -CH3, H, OH).

This chiral center gives rise to enantiomers.

Step 3: Determine which enantiomer is optically active. Since the molecule

is chiral and its mirror image is non-superimposable, the molecule is optically

active. One enantiomer will rotate plane-polarized light in a clockwise direction

(dextrorotatory), while the other enantiomer will rotate plane-polarized light in

a counterclockwise direction (levorotatory).

Question 17

Question

Explain the concept of chirality and how it relates to optical activity in organic

molecules. Provide an example of a chiral molecule and identify its enantiomers.

Solution

Step 1: Chirality in Organic Molecules Chirality refers to the property of

a molecule that cannot be superimposed on its mirror image. This property

arises when a molecule has a stereogenic center, also known as a chiral center,

which is a carbon atom bonded to four different groups.

Step 2: Optical Activity Chiral molecules are optically active, meaning

they rotate the plane of polarized light. This rotation can be either clockwise

(dextrorotatory, labeled as +) or counterclockwise (levorotatory, labeled as -).

The amount of rotation is quantified using a specific unit called specific rotation.

Step 3: Example of a Chiral Molecule One common example of a chiral

molecule is 2-chlorobutane, which has a chiral center at the second carbon atom.

Step 4: Enantiomers of 2-Chlorobutane The enantiomers of 2-chlorobutane

are non-superimposable mirror images of each other. By labeling the groups

attached to the chiral carbon, we can identify the two enantiomers: - (R)-2-

chlorobutane: In this enantiomer, the priority groups (from highest to lowest)

on the chiral carbon are arranged in a clockwise manner. - (S)-2-chlorobutane:

In this enantiomer, the priority groups on the chiral carbon are arranged in a

counterclockwise manner. These enantiomers have the same physical properties

(melting point, boiling point, etc.), but they interact differently with plane-

polarized light.

Question 18

Question

How does chirality relate to optical activity? Provide an example of a chiral

molecule and explain how its chirality affects its optical activity.

Solution

Chirality refers to the property of asymmetry in a molecule, where the molecule

is not superimposable on its mirror image. Chiral molecules exist as a pair of

enantiomers, which are non-superimposable mirror images of each other. One

of the key consequences of chirality is optical activity.

Step 1: Optical activity is the ability of a chiral molecule to rotate the plane

of polarized light. Each enantiomer of a chiral molecule will rotate polarized

light in equal but opposite directions.

Step 2: An example of a chiral molecule is Limonene, which is found in

citrus fruit peels. Limonene has a chiral center at the carbon atom bound to

the methyl and the ethyl groups.

Step 3: Due to its chirality, limonene exists as two enantiomers: (+)-

limonene and (-)-limonene. These enantiomers will rotate polarized light in

opposite directions.

Step 4: The specific rotation of a compound quantifies its ability to rotate

the plane of polarized light. The specific rotation of (+)-limonene is about +125

degrees, while that of (-)-limonene is about -125 degrees.

Step 5: In a mixture of equal amounts of (+)-limonene and (-)-limonene, the

optical rotations will cancel each other out, resulting in no net optical rotation.

Step 6: Therefore, the chirality of a molecule, such as in the case of

limonene, directly affects its optical activity by determining the direction and

magnitude of rotation of polarized light.

Question 19

Question

A compound has the molecular formula C5H11Cl and is optically active. It

reacts with sodium metal to form a colorless gas and a compound with the

molecular formula C5H10. Propose a structure for the compound and explain

why it is optically active.

Solution

Step 1: First, let’s determine the number of degrees of unsaturation in the

compound C5H10. The formula for degrees of unsaturation is given by

DOU = 2C+ 2 −H+N−X

where C is the number of carbon atoms, H is the number of hydrogen atoms,

N is the number of nitrogen atoms, and X is the number of halogen atoms. In

this case, we have:

DOU = 2(5) + 2 −10

2= 1

This means that the compound C5H10 has one degree of unsaturation, which

suggests the presence of a double bond or a ring in the structure.

Step 2: Now, let’s propose a structure for the compound C5H10. One pos-

sible structure is pent-1-ene, which has a double bond and is chiral.

Step 3: The compound C5H11Cl could be 2-chloropentane. When 2-chloropentane

reacts with sodium metal, it undergoes dehalogenation to form pent-1-ene and

a colorless gas (sodium chloride).

Step 4: Pent-1-ene is optically active because it has a chiral center at the

carbon atom bonded to two different groups (the methyl group and the hy-

drogen atom). This chiral carbon gives rise to two enantiomers that are non-

superimposable mirror images of each other.

Therefore, the compound C5H11Cl is likely 2-chloropentane, which upon

reaction with sodium metal forms the optically active compound pent-1-ene.

Question 20

Question

Explain why the compound shown below is chiral and determine whether it is

optically active.

chiral_compound.png

Solution

Step 1: To determine if a compound is chiral, we need to check if it has a

non-superimposable mirror image (enantiomer).

Step 2: In the compound given, the carbon atom labeled with * has four

different groups attached to it, making it a chiral center.

Step 3: The mirror image of this compound will not be superimposable, so

the compound is chiral.

Step 4: For a chiral compound to be optically active, it must be able to

rotate plane-polarized light. This depends on the presence of a chiral center

and no internal plane of symmetry.

Step 5: In this case, the compound is chiral and does not have an internal

plane of symmetry, so it is optically active.

Step 6: Therefore, the given compound is chiral and optically active.

Question 21

Question

Explain the concept of chirality and optical activity. Provide examples to illus-

trate your explanation.

Solution

Chirality refers to the property of asymmetry in a molecule, where the molecule

and its mirror image cannot be superimposed on each other. A chiral molecule

exists in two forms: enantiomers. Enantiomers are non-superimposable mirror

images of each other. The presence of a chiral center or asymmetric carbon

atom is necessary for chirality to occur. Examples of chiral molecules include

amino acids, sugars, and some drugs.

Optical activity is the property of chiral molecules to rotate the plane of

polarized light. Enantiomers have the ability to rotate polarized light in opposite

directions. One enantiomer rotates light to the right (dextrorotatory or +) and

the other to the left (levorotatory or -). The degree of rotation is determined

by the concentration of the sample, path length of the polarized light, and the

specific rotation constant for the compound. This property is important in the

field of pharmaceuticals and food industries to distinguish between enantiomers

with different biological activities.

Example: Consider the molecule 2-chlorobutane. This molecule contains

a chiral center at the second carbon atom. The two possible enantiomers of

2-chlorobutane are:

- (R)-2-chlorobutane (dextrorotatory) - (S)-2-chlorobutane (levorotatory)

These two enantiomers are non-superimposable mirror images of each other,

making 2-chlorobutane a chiral molecule.

Question 22

Question

A compound with the molecular formula C6H12O2exhibits optical activity.

It is found that it has 4 chiral carbons and a plane of symmetry. Draw the

possible structures for this compound and determine the total number of possible

stereoisomers.

Solution

Step 1: Determine the maximum number of stereoisomers based on the number

of chiral carbons. The formula 2n, where nis the number of chiral carbons, gives

the maximum number of stereoisomers. In this case, n= 4, so the maximum

number of stereoisomers is 24= 16.

Step 2: However, we are told that the compound has a plane of symmetry.

This implies that half of the stereoisomers will be a mirror image of the other

half. Therefore, the total number of unique stereoisomers will be 16

2= 8.

Step 3: Let’s draw the possible structures for this compound. Since the

compound has 4 chiral carbons, we can have 2 stereoisomers for each chiral

center.

The possible structures are: 1. All chiral centers in R configuration 2. One

chiral center in S configuration, the rest in R configuration 3. Two chiral centers

in S configuration, the rest in R configuration 4. Three chiral centers in S

configuration, the rest in R configuration 5. All chiral centers in S configuration

6. One chiral center in R configuration, the rest in S configuration 7. Two chiral

centers in R configuration, the rest in S configuration 8. Three chiral centers in

R configuration, the rest in S configuration

So, there are 8 possible stereoisomers for a compound with the given condi-

tions.

Question 23

Question

Determine the configurations (R or S) of the following chiral molecules and

identify whether they are optically active or not:

1-bromo-1-chloro-1-fluoro-ethane

Solution

Step 1: Identify the priority of substituents attached to the chiral carbon based

on the atomic number of the atoms directly bonded to it.

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Bromine (Br) has the highest atomic number, so it gets the highest prior-

ity.



Chlorine (Cl) comes next in priority.



Fluorine (F) has the lowest atomic number, giving it the lowest priority.

Step 2: Orient the molecule so that the lowest priority group (F) is pointing

away from you.



The molecule can be oriented as follows:

C(-[2]H)(-[4])(-[6])(-[0]F)

Step 3: Trace a circular path from the highest priority group (Br) to the

second-highest (Cl) to the third-highest (H) atom.



Going from Br to Cl to H in a circular path, the direction is clockwise.

Step 4: Determine the configuration of the chiral center based on the direc-

tion of the circular path traced in Step 3.



Since the path is clockwise, the configuration is R.

Step 5: Determine if the molecule is optically active.



A molecule is optically active if it lacks a plane of symmetry. In this case,

the molecule is optically active since it has different groups attached to

the chiral center (Br, Cl, H, and F) and lacks a plane of symmetry.

Therefore, the configuration of 1-bromo-1-chloro-1-fluoro-ethane is Rand it

is optically active.

Question 24

Question

Explain why the compound (2R,3S)-2,3-dibromobutane is chiral and determine

whether it is optically active.

Solution

Step 1: To determine if a compound is chiral, we must first identify whether

it has a chiral center. A chiral center is a carbon atom that is bonded to four

different substituents.

Step 2: In the compound (2R,3S)-2,3-dibromobutane, the second carbon

atom (labelled as 2) has a bromine atom, a methyl group, an ethyl group, and

a hydrogen atom bonded to it. Therefore, this carbon atom meets the criteria

of having four different substituents and is a chiral center.

Step 3: Next, we need to determine the chirality of the compound based on

the configuration of the chiral centers. The (2R,3S) designation indicates the

configuration around each chiral center.

Step 4: The designation (2R) means that the priority groups (based on

atomic number) follow a clockwise direction, while the designation (3S) means

that the priority groups follow a counterclockwise direction.

Step 5: Since the compound has two chiral centers and the designations are

different for each center, the compound is indeed chiral.

Step 6: Now, to determine if the chiral compound is optically active, we

need to check if it has a plane of symmetry. A molecule is optically active if it

lacks a plane of symmetry.

Step 7: In (2R,3S)-2,3-dibromobutane, there is no plane of symmetry that

can divide the molecule into two mirror-image halves. Therefore, the compound

is optically active.

Therefore, the compound (2R,3S)-2,3-dibromobutane is chiral and optically

active.

Question 25

Question

Explain the concept of chirality and optical activity. Give an example of a chiral

molecule and discuss its optical activity.

Solution

Step 1: Chirality A molecule is chiral if it cannot be superimposed on its

mirror image. Chirality arises in molecules that have an asymmetric carbon

atom, also known as a chiral center. This asymmetric carbon atom is bonded

to four different groups, leading to non-superimposable mirror images called

enantiomers.

Step 2: Optical Activity When a chiral molecule is placed in a plane-

polarized light, it rotates the plane of polarization. This phenomenon is known

as optical activity. Enantiomers will rotate the plane of polarized light in equal

but opposite directions. One enantiomer will rotate the light clockwise (dextro-

rotatory) while the other will rotate it counterclockwise (levorotatory).

Step 3: Example: Chiral Molecule - Lactic Acid Lactic acid is an

example of a chiral molecule. It has a chiral center at the carbon atom bonded to

the carboxyl group (COOH) and the hydroxyl group (OH). The two enantiomers

of lactic acid are L-lactic acid (levorotatory) and D-lactic acid (dextrorotatory).

Step 4: Optical Activity of Lactic Acid When a plane-polarized light is

passed through a solution of L-lactic acid, it will rotate the light counterclock-

wise. Conversely, a solution of D-lactic acid will rotate the light clockwise. This

optical activity is due to the presence of the chiral carbon atom in lactic acid

molecules.

Question 26

Question

Explain why the compound (2S,3S)-butanediol is chiral and show how to deter-

mine its optical activity.

Solution

Step 1: To determine if a molecule is chiral, we first need to identify if it has a

chiral center. A chiral center is a carbon atom that is bonded to four different

groups. In the case of (2S,3S)-butanediol, the two carbon atoms bonded to

alcohol groups are chiral centers, as each carbon atom is bonded to -OH, -H,

-CH3, and another carbon atom.

Step 2: Next, we determine the configuration of the chiral centers. In the

(2S,3S)-butanediol molecule, the configuration of the chiral centers is specified

as (S) for each. This means that the groups are arranged so that the lowest

priority group (H) is directed away from the viewer.

Step 3: Optical activity arises from the presence of chiral centers in a

molecule. Since (2S,3S)-butanediol has two chiral centers with configurations

(S) for each, the molecule is chiral and optically active.

Step 4: To determine the direction of optical activity, we can assign priorities

to the groups bonded to the chiral centers. For (2S,3S)-butanediol, the -OH

group has the highest priority, followed by -CH3, then -H, and finally the other

carbon atom.

Step 5: By using the Cahn-Ingold-Prelog priority rules, we can determine

that the (2S,3S)-butanediol molecule will rotate plane-polarized light in a spe-

cific direction. The exact direction (clockwise or counterclockwise) will depend

on the specific arrangement of the molecule in space.

Therefore, the compound (2S,3S)-butanediol is chiral and exhibits optical

activity due to its two chiral centers with (S) configurations.

Question 27

Question

An organic compound has the following structure:

CH3−CH(OH) −CH3

Is the compound chiral? If so, determine if it is dextrorotatory or levorota-

tory.

Solution

Step 1: To determine if the compound is chiral, we need to check if it has a chiral

center. A chiral center is a carbon atom bonded to four different groups. In the

given compound, the central carbon atom is bonded to two methyl groups, one

hydrogen atom, and one hydroxyl group. Since the hydroxyl group is different

from the other three groups, the compound has a chiral center and is chiral.

Step 2: Next, to determine if the chiral compound is dextrorotatory or

levorotatory, we need to look at the orientation of the groups attached to the

chiral center. Given that the compound is a chiral molecule, we can assign

priorities to the groups attached based on atomic number (the higher the atomic

number, the higher the priority).

Step 3: Assigning priorities to the groups attached to the chiral center: -

The hydroxyl group (OH) has the highest priority due to the oxygen atom. -

The two methyl groups (CH3) have the same priority, which is lower than the

hydroxyl group. - The hydrogen atom (H) has the lowest priority.

Step 4: Once we have assigned priorities to the groups, we need to orient

the molecule in 3D space so that the lowest priority group (hydrogen atom) is

pointing away from us. Now, we can visualize the rotation of the other groups

in a clockwise or counterclockwise manner.

Step 5: Since the two methyl groups are pointing away from us in the Fischer

projection, we can see that the rotation goes from the highest priority group

(OH) to the second-highest priority group (CH3) in a clockwise direction. This

indicates that the compound is dextrorotatory.

Therefore, the compound is chiral and dextrorotatory.

Question 28

Question

Determine the relationship between the following pairs of compounds when it

comes to chirality and optical activity:

1. (R)-2-bromobutane and (S)-2-bromobutane

2. (R)-2-chlorobutane and (R)-2-chloro-2-methylbutane

Solution

1. For the first pair of compounds:



(R)-2-bromobutane and (S)-2-bromobutane are enantiomers of each other.



Both compounds are chiral since they have a chiral center (carbon with

four different substituents).



Since they are enantiomers, they will exhibit optical activity and rotate

plane-polarized light in equal but opposite directions.

2. For the second pair of compounds:



(R)-2-chlorobutane and (R)-2-chloro-2-methylbutane are different com-

pounds.



Both compounds are chiral since they have a chiral center.



However, they are not enantiomers because they have different structures.



Each compound will exhibit optical activity, but the extent and direction

of rotation will depend on the specific compound.

Question 29

Question

What is the relationship between chirality and optical activity? Explain how

the chirality of a molecule can affect its ability to rotate plane-polarized light.

Solution

Step 1: Chirality and Optical Activity Chirality is a property of a molecule

that cannot be superimposed onto its mirror image. A molecule that is chiral

exists in two non-superimposable mirror image forms called enantiomers. Op-

tical activity is the ability of a chiral molecule to rotate the plane of polarized

light.

Step 2: Chirality of Molecules Chirality arises in molecules that have

an asymmetric carbon atom – a carbon atom that is bonded to four different

groups. These different groups create a spatial arrangement that cannot be

superimposed onto its mirror image.

Step 3: Formation of Enantiomers When a chiral molecule forms enan-

tiomers, they will have the same physical and chemical properties except for

their interaction with polarized light. One enantiomer will rotate the plane

of polarized light in one direction (clockwise or ”dextrorotatory”), while the

other enantiomer will rotate it in the opposite direction (counterclockwise or

”levorotatory”).

Step 4: Optical Activity The ability of a chiral molecule to rotate plane-

polarized light arises due to the interaction between the light’s electric field

vector and the asymmetric arrangement of the molecule’s atoms. This interac-

tion causes a phase shift in the electric field vector, resulting in the rotation of

the plane of polarized light.

Step 5: Effect of Chirality on Optical Activity The extent of rotation

of plane-polarized light by a chiral molecule depends on factors such as the

number of chiral centers, the nature of substituents attached to the chiral atom,

and the concentration of the enantiomeric mixture. Molecules with multiple

chiral centers or bulky substituents tend to exhibit greater optical activity.

Therefore, the chirality of a molecule directly influences its ability to ro-

tate plane-polarized light, with the enantiomers of a chiral molecule exhibiting

opposite directions of rotation.

Question 30

Question

Determine whether the following compounds are chiral or achiral, and specify

their optical activity if chiral:

1. 2-bromobutane

2. 2-chloropentane

Solution

1. For a molecule to be chiral, it must not have a plane of symmetry. Let’s

examine the structures:

1. 2-bromobutane: The structure of 2-bromobutane is CH3CHBrCH2CH3.

There is no plane of symmetry that can divide the molecule into two

identical halves. Therefore, 2-bromobutane is chiral. Since it is chiral, it

will exhibit optical activity.

2. 2-chloropentane: The structure of 2-chloropentane is CH3CHClCH2CH2CH3.

There is a plane of symmetry that can divide the molecule into two iden-

tical halves. Therefore, 2-chloropentane is achiral.

2. To determine the optical activity of a chiral compound, we need to ex-

amine its stereocenters.

1. 2-bromobutane: The carbon atom attached to the bromine (C-2) is a

stereocenter. Since it has 4 different groups attached to it (H, CH3, Br,

CH2CH3), it is chiral. The compound 2-bromobutane is therefore opti-

cally active.

2. 2-chloropentane: Since it is achiral, 2-chloropentane is optically inac-

tive.

Question 31

Question

Explain why a molecule with a chiral center is optically active, while a molecule

without a chiral center is not optically active.

Solution

To understand why a molecule with a chiral center is optically active, while a

molecule without a chiral center is not optically active, it is important to first

define chirality and optical activity.

Chirality: A molecule is chiral if it does not possess an internal plane of

symmetry. This means that the molecule and its mirror image (enantiomer) are

non-superimposable.

Optical activity: Optical activity refers to the ability of a chiral molecule

to rotate plane-polarized light. Enantiomers (mirror images) of a chiral molecule

will rotate plane-polarized light in equal but opposite directions.

Step 1: Molecule with a chiral center is optically active - A chiral center is a

carbon atom bonded to four different groups. - The arrangement of these groups

creates a non-superimposable mirror image. - As a result, the two enantiomers

of a molecule with a chiral center will rotate plane-polarized light in opposite

directions, making the molecule optically active.

Step 2: Molecule without a chiral center is not optically active - If a molecule

lacks a chiral center, it means the molecule possesses an internal plane of sym-

metry. - The presence of an internal plane of symmetry allows the molecule and

its mirror image to be superimposable. - Since the mirror image can be superim-

posed onto the original molecule, there will be no net rotation of plane-polarized

light. Therefore, the molecule is not optically active.

In conclusion, a molecule with a chiral center is optically active because its

enantiomers are non-superimposable, while a molecule without a chiral center

is not optically active due to the presence of an internal plane of symmetry.

Question 32

Question

Explain the concept of chirality in organic chemistry and how it relates to op-

tical activity. Provide an example of a chiral molecule and discuss how its

enantiomers interact with plane-polarized light.

Solution

Chirality in organic chemistry refers to molecules that are non-superimposable

mirror images of each other. These molecules are known as enantiomers. Chi-

rality arises when a molecule has an asymmetric carbon atom, also known as a

chiral center.

Step 1: Chirality and Enantiomers A chiral molecule is one that does

not possess an internal plane of symmetry. Enantiomers are a pair of chiral

molecules that are mirror images of each other but cannot be superimposed.

They have the same physical and chemical properties except for their interaction

with other chiral molecules and plane-polarized light.

Step 2: Optical Activity When a chiral molecule is placed in a beam

of plane-polarized light, it can rotate the plane of polarization either clockwise

or counterclockwise. This effect is known as optical activity. Enantiomers ex-

hibit optical activity in equal but opposite magnitudes, a property known as

enantiomeric excess.

Step 3: Example: Chiral Molecule One example of a chiral molecule

is 2 −chlorobutane,C H3CH (Cl)CH2CH3, which has a chiral center at the

carbon atom bonded to the chlorine atom.

Step 4: Interaction with Plane-Polarized Light The two enantiomers

of 2−chlorobutane will interact differently with plane-polarized light. One enan-

tiomer will rotate the plane of polarization clockwise, while the other will rotate

it counterclockwise, to the same extent, due to their mirror image relationship.

Understanding chirality and optical activity is crucial in the field of organic

chemistry, especially in areas such as drug development and asymmetric syn-

thesis.

Question 33

Question

Explain why the compound (1R,2S)-2-bromocyclohexan-1-ol is chiral and de-

termine its optical activity.

Solution

Step 1: The compound (1R,2S)-2-bromocyclohexan-1-ol is chiral because it does

not have a plane of symmetry and contains four different substituents attached

to the asymmetric carbon atom.

Step 2: To determine the optical activity of a chiral compound, we need to

examine the configuration of the substituents around the chiral center. In this

compound, the chiral center is the carbon atom labeled as 2.

Step 3: The configuration (1R,2S) indicates that the groups attached to

the chiral center are arranged in a counterclockwise manner, with the highest

priority group (bromine) on the dashed bond pointing away from the viewer.

Step 4: Since the compound is (1R,2S), it has an S configuration at the chiral

center, which means it rotates plane-polarized light counterclockwise. Therefore,

(1R,2S)-2-bromocyclohexan-1-ol is optically active.

Question 34

Question

Explain the concept of chirality in organic chemistry and how it relates to optical

activity.

Solution

Chirality in organic chemistry refers to the property of molecules that are

non-superimposable mirror images of each other. Such molecules are called

chiral molecules and exist in two forms: enantiomers. Enantiomers have the

same physical and chemical properties except for their interaction with plane-

polarized light, a property known as optical activity.

Step 1: Structure of Chiral Molecules Chiral molecules have an asym-

metric carbon atom, also known as a chiral center, which is bonded to four

different groups. This tetrahedral arrangement prevents the molecule from be-

ing superimposed on its mirror image.

Step 2: Enantiomers When two chiral molecules are non-superimposable

mirror images of each other, they are referred to as enantiomers. Enantiomers

have opposite configurations at each chiral center.

Step 3: Optical Activity Due to their different 3D structures, enantiomers

interact differently with plane-polarized light. One enantiomer will rotate the

plane of polarized light clockwise (+), while the other will rotate it counter-

clockwise (-).

Step 4: Specific Rotation The magnitude of the optical activity is quan-

tified using a specific rotation value, denoted as [α]. The formula to calculate

specific rotation is:

[α] = α

where: - [α] = specific rotation - α= observed rotation in degrees - l = path

length in decimeters - c = concentration in g/mL

Step 5: Racemic Mixture A racemic mixture is a 1:1 mixture of enan-

tiomers. Due to their opposite optical activities, the rotations of the enantiomers

cancel each other out, resulting in no net rotation of plane-polarized light.

Understanding chirality and its relationship to optical activity is crucial in

the study of organic chemistry, especially in the fields of drug development and

biologically active compounds.

Question 35

Question

Determine the absolute configuration (R or S) of the following chiral compound:

chiral_compound.png

Solution

1. Identify the priority of the four substituents attached to the chiral center

based on atomic number. The higher the atomic number, the higher the

priority.

2. Draw the molecule in a way that the lowest priority group (H) is away

from you and the remaining three groups are in the foreground.

3. Orient the molecule so that the fourth priority group (H) is pointing away

from you, and the remaining three groups are in a triangle pointing to-

wards you.

4. Identify the direction of the priority groups: 1) Br - clockwise (highest pri-

ority), 2) F - counterclockwise (second priority), 3) CH3 - counterclockwise

(third priority).

5. Since the direction is clockwise, the absolute configuration of the chiral

center is R.