CHEM 131 - ADVANCED GENERAL CHEMISTRY I Chirality and optical activity Question Bank Set 1

CHEM 131 - ADVANCED GENERAL

CHEMISTRY I - Chirality and optical

activity

Question Bank - Set 1

Liberty University

Question 1

Question

Determine the R/S configuration for each chiral carbon in the molecule below,

and then predict whether the molecule is optically active or not.

H−C(H)(CH3)−C(H3)(CH3)−C(H3)(CH3)−C(H3)(CH3)−C(H3)(CH3)−C(H3)(CH3)

Solution

Step 1: Assign priorities to the substituents attached to each chiral carbon:

H<CH3<C(H3)(CH3)<C(H3)(CH3)<C(H3)(CH3)

Step 2: Determine the configuration for each chiral carbon: - First chiral

carbon:

C(H)(CH3)−C(H3)(CH3)−C(H3)(CH3)

This is in a clockwise direction, so this chiral carbon is R.

- Second chiral carbon:

C(H3)(CH3)−C(H3)(CH3)−C(H3)(CH3)

This is in a counterclockwise direction, so this chiral carbon is S.

- Third chiral carbon:

C(H3)(CH3)−C(H3)(CH3)−C(H3)(CH3)

This is in a clockwise direction, so this chiral carbon is R.

- Fourth chiral carbon:

C(H3)(CH3)−C(H3)(CH3)−C(H3)(CH3)

This is in a counterclockwise direction, so this chiral carbon is S.

Step 3: The molecule has two chiral carbons with opposite configurations,

so the molecule is optically active.

Question 2

Question

Determine the configuration (R or S) and optical activity (dextrorotatory or

levorotatory) of the following compound:

chiral_compound.png

Solution

Step 1: Determine the priorities of the substituents attached to the chiral car-

bon.

Substituent Priority

Cl 1

CH2OH 2

OCH33

H 4

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

away from you, then trace from 1 to 2 to 3 in a counterclockwise direction.

The configuration is S(sinister).

Step 3: Determine the optical activity. In an S configuration, the compound

is levorotatory because the molecule rotates plane-polarized light anticlock-

wise. So the compound is S-levorotatory.

Question 3

Question

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

compounds.

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). An asymmetric carbon atom is an sp3 hybridized

carbon atom that has four different groups attached to it.

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

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

optical activity. Optical activity is related to the asymmetry of the chiral

molecule.

Step 3: Enantiomers Chiral molecules exist as a pair of enantiomers, which

are non-superimposable mirror images of each other. Enantiomers have identical

physical properties except for their interaction with other chiral molecules and

plane-polarized light.

Step 4: Specific Rotation The extent to which a chiral compound rotates

the plane of polarized light is given by its specific rotation ([α]). The specific

rotation is defined by the equation:

[α] = α·l

where αis the observed rotation in degrees, lis the path length in decimeters,

and cis the concentration in g/mL.

Step 5: Dextrorotatory and Levorotatory If an enantiomer rotates

plane-polarized light clockwise (to the right), it is called dextrorotatory (des-

ignated as +), while if it rotates the light counterclockwise (to the left), it is

called levorotatory (designated as -).

Step 6: Racemic Mixture A racemic mixture is a 50:50 mixture of enan-

tiomers in which the rotation of one enantiomer cancels out the rotation of the

other, resulting in no net optical activity. In a racemic mixture, the optical

rotations of the enantiomers are equal in magnitude but opposite in direction.

Question 4

Question

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

molecule with a meso compound containing two chiral centers is achiral. Provide

examples to illustrate your explanation.

Solution

Step 1: Chirality and Optical Activity - A molecule is chiral if it lacks an

internal plane of symmetry. A chiral molecule exists in two non-superimposable

mirror-image forms called enantiomers. - Optical activity refers to the ability of

a compound to rotate the plane of polarized light. Enantiomers exhibit optical

activity if they are not racemic (in equal amounts).

Step 2: Molecule with a Single Chiral Center is Optically Active

- A molecule with a single chiral center is optically active because it exists in

two enantiomeric forms that rotate plane-polarized light in opposite directions.

- For example, consider the molecule 2-chlorobutane. Its enantiomers are (R)-

2-chlorobutane and (S)-2-chlorobutane, which rotate polarized light in opposite

directions.

Step 3: Molecule with a Meso Compound containing Two Chiral

Centers is Achiral - A meso compound is a molecule with multiple chiral

centers but is achiral due to an internal plane of symmetry. - A molecule with a

meso compound containing two chiral centers is achiral because its internal plane

of symmetry causes the enantiomers to cancel out each other’s optical activity.

- For example, consider meso-tartaric acid. It contains two chiral centers, but

the internal plane of symmetry results in optical activity cancellation, making

the compound achiral.

Question 5

Question

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

their optical activity if applicable: 1. 2-methylbutan-2-ol 2. 3-methylpentan-3-

ol 3. 2,3-dimethylbutan-2-ol

Solution

1. 2-methylbutan-2-ol: Step 1: Identify the chiral center. In 2-methylbutan-

2-ol, the carbon atom bonded to four different groups (the hydroxyl group, a

hydrogen atom, a methyl group, and an ethyl group) is the chiral center.

Step 2: Determine chirality. Since 2-methylbutan-2-ol has a chiral center, it

is chiral.

2. 3-methylpentan-3-ol: Step 1: Identify the chiral center. In 3-methylpentan-

3-ol, the carbon atom bonded to four different groups (the hydroxyl group, a

hydrogen atom, a methyl group, and an ethyl group) is the chiral center.

Step 2: Determine chirality. Since 3-methylpentan-3-ol has a chiral center,

it is chiral.

3. 2,3-dimethylbutan-2-ol: Step 1: Identify the chiral center. In 2,3-

dimethylbutan-2-ol, there are no carbon atoms bonded to four different groups,

so there is no chiral center.

Step 2: Determine chirality. Since 2,3-dimethylbutan-2-ol does not have a

chiral center, it is achiral.

Question 6

Question

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

chiral molecule and discuss its optical activity.

Solution

Step 1: Chirality

Chirality is a geometric property of certain molecules where the molecule is not

superimposable on its mirror image. This means that a chiral molecule has a

non-superimposable mirror image, similar to how our left and right hands are

non-superimposable mirror images of each other.

Step 2: Optical Activity

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

light. When plane-polarized light passes through a solution of a chiral molecule,

the plane of polarization will rotate. The direction and extent of rotation depend

on the specific chiral molecule and experimental conditions.

Step 3: Example of a Chiral Molecule - Lactic Acid

One example of a chiral molecule is lactic acid. Lactic acid exists in two

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

non-superimposable mirror images of each other, making lactic acid a chiral

molecule.

Step 4: Optical Activity of Lactic Acid

Lactic acid exhibits optical activity because of its chirality. When plane-polarized

light passes through a solution of lactic acid, the plane of polarization will rotate

in opposite directions for L-lactic acid and D-lactic acid. The specific angle of

rotation will depend on factors such as concentration, path length, and wave-

length of light.

In conclusion, chirality refers to the non-superimposable mirror image prop-

erty of certain molecules, while optical activity is the ability of chiral molecules

to rotate the plane of polarized light.

Question 7

Question

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

whether they are optically active or inactive:

I. 2 −chlorobutane II. 2 −bromobutane III. 2 −iodobutane

Solution

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

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

I. 2-chlorobutane: The carbon marked with * is the chiral center.

Carbon Atom Groups bonded to carbon atom

Chiral Center * C, H, Cl, C

Step 2: For a compound to be chiral, it must not have a plane of symmetry.

If a molecule is achiral, it will not rotate plane-polarized light and is optically

inactive. If a molecule is chiral and lacks a plane of symmetry, it is optically

active.

II. 2-bromobutane: The carbon marked with * is the chiral center.

Carbon Atom Groups bonded to carbon atom

Chiral Center * C, H, Br, C

Step 3:

III. 2-iodobutane: The carbon marked with * is the chiral center.

Carbon Atom Groups bonded to carbon atom

Chiral Center * C, H, I, C

Step 4:

I. 2-chlorobutane: Chiral and optically inactive.

II. 2-bromobutane: Chiral and optically active.

III. 2-iodobutane: Chiral and optically active.

Question 8

Question

Determine the relationship between the following pairs of compounds regarding

chirality and optical activity:

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

2. (R)-fluorochlorobromomethane and (S)-fluorochlorobromomethane

Solution

1. For the pair (R)-2-bromohexane and (S)-2-bromohexane, the compounds are

enantiomers of each other. This means that they are non-superimposable mirror

images of each other. Each compound has a chiral center due to the presence

of the carbon atom bonded to different groups. Since these two compounds

are enantiomers, they have opposite optical activities. One will rotate plane-

polarized light clockwise (dextrorotatory or +) and the other will rotate plane-

polarized light counterclockwise (levorotatory or -).

2. For the pair (R)-fluorochlorobromomethane and (S)-fluorochlorobromomethane,

the compounds are diastereomers of each other. This means that they are not

mirror images of each other and have different physical and chemical properties.

In this case, one compound has the (R) configuration at one chiral center and

the (S) configuration at the other chiral center, while the other compound has

the (S) configuration at the first chiral center and the (R) configuration at the

second chiral center. Since these compounds are diastereomers, they do not

have the same optical activity and their relationships with light rotation cannot

be predicted solely based on their configurations.

Question 9

Question

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

an example of a chiral compound and discuss why it exhibits optical activity.

Solution

1. Chirality: A molecule is considered chiral if it is not superimposable on its

mirror image. This means that chiral molecules exist in two non-superimposable

forms, known as enantiomers. Chirality arises when a molecule contains at least

one chiral center, where four different groups are bonded to a central carbon

atom.

2. Optical Activity: Optical activity refers to the ability of a substance

to rotate the plane of plane-polarized light. Chiral molecules exhibit optical

activity because they rotate the plane of polarized light in opposite directions.

One enantiomer will rotate the light to the right, which is labeled as (+) or dex-

trorotatory, while the other enantiomer will rotate the light to the left, labeled

as (-) or levorotatory.

3. Example of a Chiral Compound: One example of a chiral compound

is 2-chlorobutane. The carbon atom at the chiral center has four different sub-

stituents: a hydrogen atom, a methyl group, an ethyl group, and a chlorine

atom. These four groups result in two non-superimposable mirror image struc-

tures. These mirror images are enantiomers and one is levorotatory while the

other is dextrorotatory.

Therefore, 2-chlorobutane is a chiral compound that exhibits optical activity

because its enantiomers rotate the plane of polarized light in opposite directions

due to their non-superimposable nature.

Question 10

Question

Determine the chirality of the following molecule and predict whether it is op-

tically active or inactive:

chiral_molecule.png

Solution

Step 1: To determine the chirality of a molecule, we 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 given molecule, carbon atom C* is the only chiral center.

Step 2: Next, we need to determine the configuration at the chiral center.

Assign priorities to the four groups attached to the chiral center based on the

atomic number of the atoms directly bonded to it. The higher the atomic

number, the higher the priority.

Step 3: For carbon atom C*, the fluorine atom (F) has the highest atomic

number, followed by the chlorine atom (Cl), then the bromine atom (Br), and

finally the hydrogen atom (H).

Step 4: Orient the molecule so that the lowest priority group (H) points

away from you. Following the priority order, if you trace a path from the

highest priority group (F) to the second priority group (Cl) to the third priority

group (Br) and it appears clockwise, then the configuration is R (Latin: rectus,

meaning right-handed). If it appears counterclockwise, then the configuration

is S (Latin: sinister, meaning left-handed).

Step 5: In this case, after tracing the path from F to Cl to Br, the path

appears to be counterclockwise. Therefore, the configuration at carbon atom

C* is S.

Step 6: A chiral molecule is optically active if it lacks an internal plane of

symmetry. In this case, the given molecule is chiral because it has a chiral center

and lacks an internal plane of symmetry. Therefore, the molecule is optically

active.

Question 11

Question

For a compound to exhibit optical activity, it must be chiral. Explain why a

molecule with a chiral center is optically active.

Solution

Step 1: A chiral molecule is one that is not superimposable on its mirror image.

This means that a chiral molecule and its mirror image (enantiomer) are non-

superimposable.

Step 2: When a beam of plane-polarized light passes through a chiral molecule,

it interacts differently with the two enantiomers of the molecule due to their

non-superimposable nature.

Step 3: The interaction with the enantiomers causes a difference in the

rotation of the plane of polarized light. One enantiomer will rotate the plane of

polarized light clockwise (dextrorotary), while the other enantiomer will rotate

it counterclockwise (levorotary).

Step 4: This difference in the rotation of plane-polarized light by the two

enantiomers of a chiral molecule is what makes the molecule optically active.

Therefore, a molecule with a chiral center is optically active because its

two non-superimposable mirror images interact differently with plane-polarized

light, causing a rotation of the light in opposite directions.

Question 12

Question

A compound A with the molecular formula C5H11Cl shows optical activity.

When compound A is treated with a strong base, it forms two products: an

alkene B and an alkyl chloride C. Compound B does not show optical activity,

while compound C does. Write the structural formulas for compounds A, B,

and C. Explain the observed optical activities in each case.

Solution

Step 1: Determine the structures of compounds A, B, and C. Let’s start by

determining the possible structures of compound A with the molecular formula

C5H11Cl. Since it shows optical activity and contains a chiral center, let’s con-

sider a chiral carbon atom to form a unique enantiomer. The possible structure

for compound A is 2-chloropentane, which has a chiral center at carbon 2.

The structural formulas for compounds A, B, and C are as follows: Com-

pound A: 2-chloropentane (CH3CHClCH2CH2CH3) Compound B: 2-pentene

(CH3CH=CHCH2CH3) Compound C: 2-chloropentane (CH3CHClCH2CH2CH3)

Step 2: Explain the observed optical activities. Compound A (2-chloropentane)

shows optical activity because of its chiral center, which results in the presence

of enantiomers that rotate plane-polarized light in opposite directions.

Compound B (2-pentene) does not show optical activity because it does not

have a chiral center. The absence of a chiral center means that compound B is

achiral, and its mirror image is superimposable.

Compound C (2-chloropentane) shows optical activity because it has a chiral

center at carbon 2. The presence of a chiral center leads to the formation

of enantiomers, resulting in optical activity as the enantiomers rotate plane-

polarized light in opposite directions.

Question 13

Question

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

active.

chiral_compound.png

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 compound given, the carbon atom labeled with a star (*) is a

chiral center as it is bonded to four different groups: a hydrogen atom (H), a

bromine atom (Br), a chlorine atom (Cl), and a methyl group (CH3).

Step 2: Next, we need to check if the molecule is optically active. For a

molecule to be optically active, it must be chiral and lack a plane of symmetry.

A molecule has a plane of symmetry if it can be divided into two equal halves

such that one half is the mirror image of the other. In the given compound,

there is no plane of symmetry; therefore, it is optically active.

Thus, the compound shown is chiral and optically active.

Question 14

Question

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

ral, state whether they are optically active: 1. 2,3-dibromopentane 2. 2,3-

dibromobutane 3. 2,3-dichlorobutane

Solution

1. 2,3-dibromopentane:

Step 1: Determine if the compound has a chiral center. A chiral center is a

carbon atom bonded to four different groups.

In 2,3-dibromopentane, the carbon atom at the 3rd position is bonded to two

bromine atoms, a hydrogen atom, and an ethyl group. Therefore, it has a chiral

center.

Step 2: Check for a plane of symmetry in the molecule. If a compound has a

plane of symmetry, it is achiral.

2,3-dibromopentane does not have a plane of symmetry; therefore, it is chiral.

Step 3: Determine if the compound is optically active. To be optically active,

a chiral compound must lack internal symmetry and must not have a superim-

posable mirror image.

In the case of 2,3-dibromopentane, the mirror image of the molecule is not

superimposable, so it is optically active.

2. 2,3-dibromobutane:

Step 1: Check for a chiral center. In 2,3-dibromobutane, the carbon atom

at the 2nd position is bonded to two bromine atoms, a hydrogen atom, and a

methyl group. This carbon atom is a chiral center.

Step 2: Look for a plane of symmetry. There is no plane of symmetry in 2,3-

dibromobutane; hence, it is chiral.

Step 3: Determine if the compound is optically active. Since 2,3-dibromobutane

is chiral and lacks internal symmetry, it is optically active.

3. 2,3-dichlorobutane:

Step 1: Find any chiral centers in the compound. In 2,3-dichlorobutane, the

carbon atom at the 3rd position is bonded to two chlorine atoms, a hydrogen

atom, and a methyl group. This carbon atom is a chiral center.

Step 2: Check for a plane of symmetry. There is no plane of symmetry in

2,3-dichlorobutane, making it chiral.

Step 3: Determine if the compound is optically active. As 2,3-dichlorobutane

is chiral and lacks internal symmetry, it is optically active.

Question 15

Question

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

it is optically active.

Solution

Step 1: To determine the chirality of a compound, we examine its stereocenters.

A stereocenter is a carbon atom bonded to four different groups.

Step 2: The compound (2R,3S)-butan-2,3-diol has two stereocenters, located

at carbon atoms 2 and 3, as indicated by the (2R,3S) designation.

Step 3: At carbon atom 2, we have two different substituents - a hydrogen

atom and an -OH group. At carbon atom 3, we have two different substituents

- an -OH group and an -CH3 group.

Step 4: Thus, both carbon atoms 2 and 3 are stereocenters because they are

both bonded to four different groups, making the compound chiral.

Step 5: To determine if the chiral compound is optically active, we need to

check if it has a plane of symmetry. If there is a plane of symmetry present, the

compound is not optically active.

Step 6: The compound (2R,3S)-butan-2,3-diol lacks a plane of symmetry

because exchanging any two groups on one or both stereocenters will result in

a different arrangement. Therefore, the compound is optically active.

Therefore, the compound (2R,3S)-butan-2,3-diol is chiral and optically ac-

tive.

Question 16

Question

Draw the (R)-2-bromobutane molecule and determine if it is chiral. If it is

chiral, indicate whether it is optically active or inactive.

Solution

Step 1: Draw the (R)-2-bromobutane molecule.

CH3

C−H

C−Br

CH3

Step 2: Assign priorities to the groups attached to the chiral carbon (C-2).

CH3(4)

C−H(3)

C−H(2)

C−Br(1)

CH3

Step 3: Orient the molecule so that the lowest priority group (in this case,

C-Br) is pointing away from you.

CH3

C−H

C−Br

C−H

CH3

Step 4: Determine the configuration of the molecule by looking at the direc-

tion of priority groups from 1 to 3 (clockwise or counterclockwise). Since the

direction is clockwise (R), the molecule is (R)-2-bromobutane, and it is chiral.

Step 5: Optical activity in a molecule arises when it is chiral. Since (R)-2-

bromobutane is chiral, it is optically active.

Therefore, (R)-2-bromobutane is a chiral molecule that is optically active.

Question 17

Question

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

a molecule that exhibits optical activity and determine whether it is optically

active or not.

Solution

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

imposed on its mirror image. Chiral molecules are non-superimposable mirror

images of each other, much like our left and right hands. A molecule that

possesses a chiral center is called a chiral molecule.

Step 2: Optical Activity Optical activity is the ability of a substance

to rotate the plane of polarized light. Chiral molecules show optical activity

because they interact differently with left- and right-handed circularly polarized

light.

Step 3: Example and Determination Let’s consider the molecule 2,3-

dibromobutane (C4H8Br2) as an example. It has a chiral center (the carbon

atom bonded to the two bromine atoms) and therefore can exhibit optical ac-

tivity.

To determine if the molecule is optically active, we need to check if it is in

a chiral configuration. If the molecule has an internal plane of symmetry, it is

not chiral and therefore not optically active.

In the case of 2,3-dibromobutane, if we rotate the molecule so that the

bromine atoms are on the same side, we can see that it has an internal plane of

symmetry. Therefore, 2,3-dibromobutane is not optically active.

Question 18

Question

A compound is found to be optically inactive, even though it contains a stereo-

center. Explain why this is possible and provide an example.

Solution

Step 1: A compound with a stereocenter can be optically inactive if it exists as

a racemic mixture, where the two enantiomers cancel out each other’s optical

activity. This occurs when there is an equal amount of both enantiomers present.

Step 2: Let’s consider the example of 2,3-dichlorobutane. This compound

has a stereocenter at the second carbon atom, which gives rise to two enan-

tiomers. However, if an equal amount of each enantiomer is present, the com-

pound will be optically inactive.

Step 3: The chemical structure of 2,3-dichlorobutane is:

H−C−C−CH3

| |

Cl Cl

Step 4: In this case, the two enantiomers of 2,3-dichlorobutane will have

equal and opposite rotations of plane-polarized light, resulting in no net optical

activity.

Thus, a compound with a stereocenter can be optically inactive when it

exists as a racemic mixture containing equal amounts of both enantiomers.

Question 19

Question

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

molecule that exhibits chirality and discuss how its chiral nature influences its

optical activity.

Solution

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

perimposed on its mirror image. A chiral molecule has a non-superimposable

mirror image, known as its enantiomer. Chirality arises when a molecule has

an asymmetric center, such as a carbon atom bonded to four different groups.

Step 2: Optical Activity Optical activity is the ability of a chiral molecule

to rotate the plane of polarized light. Enantiomers of a chiral molecule will

rotate the plane of polarized light in equal but opposite directions, one clockwise

(dextrorotatory) and the other counterclockwise (levorotatory). The magnitude

of this rotation is quantified by the specific rotation, [α], which is dependent on

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

nature of the substance.

Step 3: Example: Lactic Acid Lactic acid (2-hydroxypropionic acid) is an

example of a molecule that exhibits chirality. Lactic acid has a chiral center at

the carbon atom bonded to the hydroxyl group, methyl group, and carboxylic

acid group. It exists as two enantiomers: L-lactic acid and D-lactic acid.

Step 4: Influence on Optical Activity Lactic acid’s enantiomers, L-lactic

acid, and D-lactic acid, exhibit optical activity. L-lactic acid is levorotatory,

meaning it rotates plane-polarized light counterclockwise, while D-lactic acid

is dextrorotatory, rotating light clockwise. The specific rotation of each enan-

tiomer differs due to the arrangement of atoms around the chiral center.

Question 20

Question

Consider a molecule with the following structure:

CH3−CH2−CH(Cl)−CH3

Determine whether the molecule is chiral or achiral, and if chiral, indicate

whether it is optically active or inactive.

Solution

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

chiral center (an atom bonded to four different groups). In this case, the carbon

atom bonded to the chlorine atom serves as a chiral center.

Step 2: Now, let’s examine the molecule’s mirror image. To do this, we

create a mirror image of the molecule and attempt to overlay it with the original

molecule.

Step 3: In this case, when we attempt to overlay the mirror image onto the

original molecule, we find that the chlorine atom can never overlay the carbon

atom, indicating that the molecule is chiral.

Step 4: Next, we determine if the molecule is optically active. For that,

we check if the molecule’s mirror image is non-superimposable on the original

molecule.

Step 5: Since we found that the mirror image cannot be superimposed on

the original molecule, the molecule is optically active.

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

Question 21

Question

Draw the 3D structure of a chiral molecule, label each chiral center, and deter-

mine if the molecule is optically active.

Solution

To determine the optical activity of a molecule, we must first identify if the

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

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

If a molecule has at least one chiral center, it is chiral.

Step 1: Draw the 3D structure of the chiral molecule.

H3C−[: 30](−[: −30]CH3)−[: 150](−[: 90]H)−[: −150](−[: 90]H)(−[: −210]OH)

Step 2: Label each chiral center.

In the given molecule, the carbon atom bonded to OH, CH3, H, and another

H is a chiral center.

Step 3: Determine if the molecule is optically active.

Since the molecule has one chiral center, it is chiral and thus optically active.

Question 22

Question

Explain why the compound 2-chlorobutane is chiral, while the compound 2,3-

dichlorobutane is not chiral. Additionally, determine which compound, if any,

exhibits optical activity.

Solution

Step 1: To determine if a compound is chiral, we need to examine its molecular

structure and look for a chiral center. A chiral center is a carbon atom attached

to four different groups. In 2-chlorobutane, the carbon atom bonded to the

chlorine atom is a chiral center as it is attached to four different groups: a

hydrogen atom, a methyl group, an ethyl group, and a chlorine atom. Therefore,

2-chlorobutane is a chiral compound.

Step 2: On the other hand, in 2,3-dichlorobutane, there is no chiral center

present. Both carbon atoms bonded to chlorine atoms have two hydrogen atoms

attached to them, making them achiral. Therefore, 2,3-dichlorobutane is not

chiral.

Step 3: For a compound to exhibit optical activity, it must be chiral. Since

2-chlorobutane is chiral, it has the potential to exhibit optical activity. However,

optical activity also depends on the specific spatial arrangement of the molecule,

which can only be determined experimentally. In contrast, 2,3-dichlorobutane

is not chiral and therefore cannot exhibit optical activity.

Question 23

Question

Explain why some molecules are chiral while others are not. Provide an example

of a chiral molecule and explain how its chirality leads to optical activity.

Solution

Step 1: Chirality in Molecules

Chirality in molecules arises from the presence of an asymmetric carbon

atom, also known as a chiral center. A carbon atom is considered asymmetric

when its four substituents are different from one another. This lack of symmetry

results in a non-superimposable mirror image, making the molecule chiral.

Step 2: Chiral vs. Achiral Molecules

Molecules that contain at least one chiral center are considered chiral, while

molecules without any chiral centers are achiral. Achiral molecules can have

symmetry elements that allow their mirror images to be superimposed.

Step 3: Example of a Chiral Molecule: Limonene

One example of a chiral molecule is limonene, a compound responsible for

the citrusy aroma in fruits like lemons and oranges. Limonene contains a chiral

center at its carbon 1 position.

Step 4: Optical Activity

Due to their chiral nature, chiral molecules like limonene exhibit optical

activity. This means they rotate plane-polarized light passing through them.

The direction and magnitude of rotation depend on the molecule’s chirality and

concentration.

Step 5: Conclusion

In conclusion, chirality in molecules is determined by the presence of an

asymmetric carbon atom, leading to non-superimposable mirror images. Chi-

ral molecules display optical activity, demonstrating their unique property in

interacting with plane-polarized light.

Question 24

Question

Determine whether the following compound is chiral or achiral, and if chiral,

indicate whether it is optically active:

CHBrClF

Solution

Step 1: Determine if the molecule is chiral.

A molecule is chiral if it lacks an internal plane of symmetry. Looking at the

structure of CHBrClF, we see that there is no internal plane of symmetry. Each

atom bonded to the central carbon is different, making the molecule chiral.

Step 2: Determine if the molecule is optically active.

In order for a chiral molecule to be optically active, it must be able to rotate

plane-polarized light. To determine the optical activity of CHBrClF, we need

to consider whether the molecule is overall chiral due to the presence of a chiral

center. Since CHBrClF contains a single chiral center (the carbon atom), it will

be optically active.

Therefore, CHBrClF is a chiral molecule and is optically active.

Question 25

Question

An unknown compound is found to be optically active. It contains a chiral

center and has the molecular formula C7H14O. Upon analysis, it is determined

that the compound has only two possible stereoisomers. Draw the structural

formulas for both stereoisomers and determine which one is the levorotatory

form.

Solution

Step 1: Determine the possible structural formulas for the two stereoisomers.

Since the compound contains a chiral center, there are two possible stereoiso-

mers: the enantiomer pairs.

Step 2: Draw the structural formulas for the two stereoisomers. Let’s denote

the chiral center as an asterisk (*) to represent the carbon atom with four

different substituents.

Structural formulas for the stereoisomers:

Structure 1 - Dextrorotatory(R−enantiomer) : CH3−CH2−CH(∗)(CH3)−CH2−CH2−OH

Structure 2 - Levorotatory(S−enantiomer) : CH3−CH2−CH(∗)(CH3)−CH2−CH2−OH

Step 3: Determine the levorotatory form. To determine which stereoisomer is

levorotatory, we need to analyze them based on the Cahn-Ingold-Prelog priority

rules. In this case, the most important thing to consider is the priority of the

substituents attached to the chiral center.

In the given structures, the substituents are the same (two methyl groups

and one hydrogen atom), so we need to compare the fourth substituents. The

R-enantiomer has the higher priority substituent on the fourth position, making

it the levorotatory form.

Therefore, the levorotatory form is the Structure 1 - Dextrorotatory (R-

enantiomer).

Question 26

Question

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

molecules.

Solution

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

image. In other words, a molecule is chiral if it has a non-superimposable mirror

image. Chiral molecules have the property of optical activity, which means they

rotate the plane of polarized light. This phenomenon occurs because chiral

molecules exist in two forms - enantiomers - that are mirror images of each

other and have different interaction with plane-polarized light.

Step 1: Definition of Chirality A molecule is chiral if it lacks an internal

plane of symmetry, resulting in a non-superimposable mirror image. Chirality

can be identified by the presence of a chiral center, which is an atom bonded to

four unique substituents.

Step 2: Enantiomers Enantiomers are pairs of molecules that are non-

superimposable mirror images of each other. They have identical physical and

chemical properties except for their interaction with plane-polarized light.

Step 3: Optical Activity Optical activity refers to the rotation of plane-

polarized light by a chiral molecule. One enantiomer will rotate light clockwise

(dextrorotatory, labeled as ”+”) while the other will rotate it counterclockwise

(levorotatory, labeled as ”-”). The magnitude of rotation can be measured using

a polarimeter.

Step 4: Specific Rotation The specific rotation ([α]) of a compound is a

measure of its ability to rotate plane-polarized light at a specific concentration,

path length, and wavelength. It is calculated using the formula:

[α] = α/(c·l)

where: - [α] is the specific rotation in degrees per decimeter per gram (deg

dm−1g−1),−αis the observed rotation in degrees, - cis the concentration of

the compound in grams per milliliter, - lis the path length in decimeters.

Step 5: Racemic Mixtures A racemic mixture is a 50:50 mixture of two

enantiomers, which cancels out their optical activities. As a result, racemic

mixtures do not rotate plane-polarized light.

Understanding the concept of chirality is essential in organic chemistry as it

plays a significant role in the reactivity and properties of molecules.

Question 27

Question

Determine whether the following molecules are chiral, achiral, or meso com-

pounds:

a) 2,3-dichloro-2-butene b) 2,3-dibromobutane c) tartaric acid d) meso-tartaric acid

Solution

a) 2,3-dichloro-2-butene

A molecule is chiral if it does not have an internal plane of symmetry. Let’s

examine the structure of 2,3-dichloro-2-butene:

- C ≡C H Cl Cl

In this molecule, the two chlorine atoms are different, and each group bonded

to the double-bonded carbons is different. Therefore, 2,3-dichloro-2-butene is

chiral.

b) 2,3-dibromobutane

Now let’s consider 2,3-dibromobutane:

HCHBrCBrH

In this molecule, there is a plane of symmetry that divides the molecule into

two identical halves. Therefore, 2,3-dibromobutane is achiral.

c) Tartaric acid

Tartaric acid is a molecule with two chiral centers. The structure can be

represented as:

HOOC C OH CHOH CHOH C

1 2 3 4 5

↑ ↑ ↑ ↑ ↑

OH H OH H COOH

Since tartaric acid has two chiral centers and is not superimposable on its

mirror image, it is chiral.

d) Meso-tartaric acid

Meso compounds are molecules that contain chiral centers but are achiral

overall due to the presence of an internal plane of symmetry. The structure of

meso-tartaric acid is:

HOOC C OH CHOH CHOH COOH

1 2 3 4

↑ ↑ ↑ ↑

OH H OH COOH

In this molecule, there is an internal plane of symmetry that divides the

compound into two equal halves. Therefore, meso-tartaric acid is achiral.

Question 28

Question

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

chiral_molecule.png

Solution

Step 1: Assign priorities to the four substituents attached to the chiral center

based on the atomic number of the atoms directly bonded to the chiral center.

The higher the atomic number, the higher the priority. In this molecule, the

priorities are assigned as follows:

Substituent Priority

Cl (chlorine) 1

O (oxygen) 2

CH3 (methyl group) 3

H (hydrogen) 4

Step 2: Orient the molecule so that the lowest priority substituent (H) is

pointing away from you. Then, trace a path from priority 1 to 2 to 3. If the

path is clockwise, the configuration is R. If the path is counterclockwise, the

configuration is S.

In the given molecule, the path from 1 to 2 to 3 is counterclockwise, so the

configuration of the chiral center is S.

Question 29

Question

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

certain chiral compounds exhibit optical activity while others do not.

Solution

Step 1: Chirality in molecules refers to the property of having a non-superimposable

mirror image. This is often caused by having a carbon atom bonded to four

different groups, known as a chiral center.

Step 2: Optical activity is the ability of a compound to rotate the plane of

polarized light. Chiral molecules exhibit optical activity because their mirror

images are nonsuperimposable, leading to different interactions with polarized

light.

Step 3: The presence of a chiral center in a molecule allows for different

spatial arrangements of the molecule which interact differently with polarized

light, resulting in optical activity.

Step 4: However, not all chiral compounds exhibit optical activity. One

reason for this is when the chiral molecule is in a symmetric environment that

cancels out the overall optical activity.

Step 5: For example, if a chiral molecule is in a perfect symmetric envi-

ronment, the rotations caused by the two enantiomers (mirror image molecules)

may cancel each other out, resulting in no overall optical activity.

Step 6: Therefore, the relationship between chirality and optical activity

lies in the asymmetric nature of chiral molecules which allows for different in-

teractions with polarized light. While most chiral compounds exhibit optical

activity, the presence of symmetry in the molecule can lead to a lack of optical

activity.

Question 30

Question

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

active.

chiral_compound.png

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 bonded to four different groups.

In the given compound, the carbon atom circled below is a chiral center as it is

bonded to four different groups:

chiral_center.png

Step 2: A chiral molecule is one that is not superimposable on its mirror

image. In this case, the presence of the chiral center makes the molecule chiral.

Since the compound is chiral, we need to determine if it is optically active.

Step 3: To determine if a chiral molecule is optically active, we need to check

if it has a plane of symmetry. If a molecule has a plane of symmetry, it will not

exhibit optical activity. In the given compound, there is no plane of symmetry

that can divide it into two equal halves.

Step 4: Therefore, the compound shown is chiral and optically active. It will

rotate the plane of polarized light and exhibit optical activity.

Question 31

Question

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

tivity. Provide an example of a chiral molecule and describe how its optical

activity is determined.

Solution

Step 1: Chirality in Chemistry

Chirality in chemistry refers to the property of a molecule that cannot be

superimposed on its mirror image. This means that chiral molecules exist in two

non-superimposable forms called enantiomers. Enantiomers have the same phys-

ical and chemical properties except for their interaction with plane-polarized

light, known as optical activity.

Step 2: Optical Activity

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

ized light. This phenomenon is measured using a polarimeter, which measures

the angle of rotation caused by passing polarized light through the sample. The

direction and extent of rotation are specific to each enantiomer.

Step 3: Example of a Chiral Molecule: Lactic Acid

One example of a chiral molecule is lactic acid, which has two enantiomeric

forms: L-lactic acid and D-lactic acid. These enantiomers are non-superimposable

mirror images of each other.

Step 4: Determining Optical Activity of Lactic Acid

To determine the optical activity of lactic acid, we would pass polarized

light through a sample of each enantiomer using a polarimeter. The degree and

direction of rotation of the plane of polarized light would differ for L-lactic acid

and D-lactic acid, allowing us to distinguish between the two enantiomers based

on their optical activity.

Therefore, chirality in chemistry is closely related to optical activity, as chiral

molecules exhibit optical activity due to their asymmetric carbon centers, which

results in their ability to rotate plane-polarized light.

Question 32

Question

Determine the relationship between the following pairs of compounds with re-

gard to chirality and optical activity:

i) (R)-2-bromobutane and (S)-2-bromobutane

ii) (R)-2-chlorobutane and (S)-2-chlorobutane

iii) (R)-2-iodobutane and (S)-2-iodobutane

Solution

i) For a compound to exhibit optical activity, it must be chiral, meaning it

lacks a plane of symmetry. A compound with a chiral center, such as a carbon

atom bonded to four different groups, is optically active. (R)-2-bromobutane

and (S)-2-bromobutane are enantiomers, meaning they are non-superimposable

mirror images of each other, making them chiral and optically active. Therefore,

(R)-2-bromobutane is the enantiomer of (S)-2-bromobutane.

ii) Similarly, (R)-2-chlorobutane and (S)-2-chlorobutane are also enantiomers,

making them chiral and optically active. Therefore, (R)-2-chlorobutane is the

enantiomer of (S)-2-chlorobutane.

iii) On the other hand, (R)-2-iodobutane and (S)-2-iodobutane are mirror

images of each other but not enantiomers. They are diastereomers because they

differ at one, but not all, chiral centers. Diastereomers are not superimposable,

non-mirror images of each other, meaning they are also chiral. Therefore, (R)-

2-iodobutane is the diastereomer of (S)-2-iodobutane.

Question 33

Question

Explain why the compound (2S)-2-bromobutane is chiral while the compound

(RS)-2-bromobutane is not chiral. Additionally, determine whether each com-

pound is optically active or inactive.

Solution

Step 1: Chirality of (2S)-2-bromobutane

A molecule is chiral if it lacks an internal plane of symmetry. In the case of

(2S)-2-bromobutane, it has a chiral center at the second carbon atom, as the

carbon is bonded to four different groups: a bromine atom, a hydrogen atom,

a methyl group, and an ethyl group. The mirror image of the molecule cannot

be superimposed on the original molecule, making it chiral.

Step 2: Optical activity of (2S)-2-bromobutane

Since (2S)-2-bromobutane is chiral, it is optically active. It rotates the plane of

polarized light and exhibits optical activity.

Step 3: Chirality of (RS)-2-bromobutane

On the other hand, (RS)-2-bromobutane is not chiral because if we interchange

the positions of the groups attached to the chiral carbon, we get the same

molecule. This means it has an internal plane of symmetry and is achiral.

Step 4: Optical activity of (RS)-2-bromobutane

Since (RS)-2-bromobutane is not chiral, it is optically inactive. It does not

rotate the plane of polarized light and exhibits no optical activity.

Question 34

Question

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

chiral molecule and discuss how it exhibits optical activity.

Solution

Step 1: Chirality A molecule is considered chiral if it lacks an internal plane

of symmetry. This means that the molecule and its mirror image are not super-

imposable. Chirality arises when a molecule contains at least one chiral center,

which is a carbon atom bonded to four different groups.

Step 2: Optical Activity Optical activity is the ability of a chiral molecule

to rotate the plane of polarized light. Chiral molecules that rotate the plane

of polarized light to the right (clockwise) are referred to as dextrorotatory (d-)

while those that rotate to the left (counterclockwise) are levorotatory (l-).

Step 3: Example of Chiral Molecule - Lactic Acid Lactic acid, with

the chemical formula C3H6O3, is a common example of a chiral molecule. It has

two enantiomers: L-lactic acid (levorotatory) and D-lactic acid (dextrorotatory).

These enantiomers are nonsuperimposable mirror images of each other due to

the chiral carbon atom in the molecule.

Step 4: Optical Activity of Lactic Acid When a beam of polarized light

passes through a solution of L-lactic acid, it will rotate the plane of polarization

to the left (levorotatory). On the other hand, a solution of D-lactic acid will

rotate the plane to the right (dextrorotatory). This optical activity is a result of

the asymmetric arrangement of atoms around the chiral carbon atom in lactic

acid.

Question 35

Question

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

molecule that exhibits chirality and describe how its optical activity is deter-

mined.

Solution

Step 1: Chirality Chirality is a property of a molecule that results from its

three-dimensional structure. A molecule is chiral if it cannot be superimposed

on its mirror image. In other words, a chiral molecule is non-superimposable on

its mirror image, much like a right hand and a left hand are non-superimposable.

Chiral molecules have a unique spatial arrangement of atoms that gives them

distinct physical and chemical properties.

Step 2: Optical Activity Optical activity is a phenomenon displayed by

chiral molecules when they interact with plane-polarized light. Chiral molecules

can rotate the plane of polarized light either to the left (levorotatory) or to the

right (dextrorotatory) as it passes through a sample of the compound. This

rotation of the plane of polarized light is measured using a polarimeter, and the

extent of rotation is quantified as specific rotation.

Step 3: Example: Lactic Acid An example of a chiral molecule is lactic

acid, which exists in two enantiomeric forms: L-lactic acid and D-lactic acid.

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

optical activity of lactic acid is determined by the spatial arrangement of its

atoms, specifically the arrangement around the chiral carbon atom.

Step 4: Determination of Optical Activity To determine the optical

activity of lactic acid, a sample of the compound is placed in a polarimeter.

If the plane-polarized light rotates to the right, the sample is dextrorotatory

and labeled as (+). If the light rotates to the left, the sample is levorotatory

and labeled as (-). The specific rotation value obtained from the polarimeter

provides quantitative information about the degree of optical activity of the

compound.

In conclusion, chirality and optical activity are important concepts in organic

chemistry that describe the unique spatial arrangement and interaction of chiral

molecules with plane-polarized light, respectively.

- Third chiral carbon:

C(H3)(CH3)−C(H3)(CH3)−C(H3)(CH3)

This is in a clockwise direction, so this chiral carbon is R.

- Fourth chiral carbon:

C(H3)(CH3)−C(H3)(CH3)−C(H3)(CH3)

This is in a counterclockwise direction, so this chiral carbon is S.

Step 3: The molecule has two chiral carbons with opposite configurations,

so the molecule is optically active.

Question 2

Question

Determine the configuration (R or S) and optical activity (dextrorotatory or

levorotatory) of the following compound:

chiral_compound.png

Solution

Step 1: Determine the priorities of the substituents attached to the chiral car-

bon.

Substituent Priority

Cl 1

CH2OH 2

OCH33

H 4

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

away from you, then trace from 1 to 2 to 3 in a counterclockwise direction.

The configuration is S(sinister).

Step 3: Determine the optical activity. In an S configuration, the compound

is levorotatory because the molecule rotates plane-polarized light anticlock-

wise. So the compound is S-levorotatory.

Question 3

Question

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

compounds.

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). An asymmetric carbon atom is an sp3 hybridized

carbon atom that has four different groups attached to it.

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

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

optical activity. Optical activity is related to the asymmetry of the chiral

molecule.

Step 3: Enantiomers Chiral molecules exist as a pair of enantiomers, which

are non-superimposable mirror images of each other. Enantiomers have identical

physical properties except for their interaction with other chiral molecules and

plane-polarized light.

Step 4: Specific Rotation The extent to which a chiral compound rotates

the plane of polarized light is given by its specific rotation ([α]). The specific

rotation is defined by the equation:

[α] = α·l

where αis the observed rotation in degrees, lis the path length in decimeters,

and cis the concentration in g/mL.

Step 5: Dextrorotatory and Levorotatory If an enantiomer rotates

plane-polarized light clockwise (to the right), it is called dextrorotatory (des-

ignated as +), while if it rotates the light counterclockwise (to the left), it is

called levorotatory (designated as -).

Step 6: Racemic Mixture A racemic mixture is a 50:50 mixture of enan-

tiomers in which the rotation of one enantiomer cancels out the rotation of the

other, resulting in no net optical activity. In a racemic mixture, the optical

rotations of the enantiomers are equal in magnitude but opposite in direction.

Question 4

Question

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

molecule with a meso compound containing two chiral centers is achiral. Provide

examples to illustrate your explanation.

Solution

Step 1: Chirality and Optical Activity - A molecule is chiral if it lacks an

internal plane of symmetry. A chiral molecule exists in two non-superimposable

mirror-image forms called enantiomers. - Optical activity refers to the ability of

a compound to rotate the plane of polarized light. Enantiomers exhibit optical

activity if they are not racemic (in equal amounts).

Step 2: Molecule with a Single Chiral Center is Optically Active

- A molecule with a single chiral center is optically active because it exists in

two enantiomeric forms that rotate plane-polarized light in opposite directions.

- For example, consider the molecule 2-chlorobutane. Its enantiomers are (R)-

2-chlorobutane and (S)-2-chlorobutane, which rotate polarized light in opposite

directions.

Step 3: Molecule with a Meso Compound containing Two Chiral

Centers is Achiral - A meso compound is a molecule with multiple chiral

centers but is achiral due to an internal plane of symmetry. - A molecule with a

meso compound containing two chiral centers is achiral because its internal plane

of symmetry causes the enantiomers to cancel out each other’s optical activity.

- For example, consider meso-tartaric acid. It contains two chiral centers, but

the internal plane of symmetry results in optical activity cancellation, making

the compound achiral.

Question 5

Question

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

their optical activity if applicable: 1. 2-methylbutan-2-ol 2. 3-methylpentan-3-

ol 3. 2,3-dimethylbutan-2-ol

Solution

1. 2-methylbutan-2-ol: Step 1: Identify the chiral center. In 2-methylbutan-

2-ol, the carbon atom bonded to four different groups (the hydroxyl group, a

hydrogen atom, a methyl group, and an ethyl group) is the chiral center.

Step 2: Determine chirality. Since 2-methylbutan-2-ol has a chiral center, it

is chiral.

2. 3-methylpentan-3-ol: Step 1: Identify the chiral center. In 3-methylpentan-

3-ol, the carbon atom bonded to four different groups (the hydroxyl group, a

hydrogen atom, a methyl group, and an ethyl group) is the chiral center.

Step 2: Determine chirality. Since 3-methylpentan-3-ol has a chiral center,

it is chiral.

3. 2,3-dimethylbutan-2-ol: Step 1: Identify the chiral center. In 2,3-

dimethylbutan-2-ol, there are no carbon atoms bonded to four different groups,

so there is no chiral center.

Step 2: Determine chirality. Since 2,3-dimethylbutan-2-ol does not have a

chiral center, it is achiral.

Question 6

Question

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

chiral molecule and discuss its optical activity.

Solution

Step 1: Chirality

Chirality is a geometric property of certain molecules where the molecule is not

superimposable on its mirror image. This means that a chiral molecule has a

non-superimposable mirror image, similar to how our left and right hands are

non-superimposable mirror images of each other.

Step 2: Optical Activity

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

light. When plane-polarized light passes through a solution of a chiral molecule,

the plane of polarization will rotate. The direction and extent of rotation depend

on the specific chiral molecule and experimental conditions.

Step 3: Example of a Chiral Molecule - Lactic Acid

One example of a chiral molecule is lactic acid. Lactic acid exists in two

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

non-superimposable mirror images of each other, making lactic acid a chiral

molecule.

Step 4: Optical Activity of Lactic Acid

Lactic acid exhibits optical activity because of its chirality. When plane-polarized

light passes through a solution of lactic acid, the plane of polarization will rotate

in opposite directions for L-lactic acid and D-lactic acid. The specific angle of

rotation will depend on factors such as concentration, path length, and wave-

length of light.

In conclusion, chirality refers to the non-superimposable mirror image prop-

erty of certain molecules, while optical activity is the ability of chiral molecules

to rotate the plane of polarized light.

Question 7

Question

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

whether they are optically active or inactive:

I. 2 −chlorobutane II. 2 −bromobutane III. 2 −iodobutane

Solution

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

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

I. 2-chlorobutane: The carbon marked with * is the chiral center.

Carbon Atom Groups bonded to carbon atom

Chiral Center * C, H, Cl, C

Step 2: For a compound to be chiral, it must not have a plane of symmetry.

If a molecule is achiral, it will not rotate plane-polarized light and is optically

inactive. If a molecule is chiral and lacks a plane of symmetry, it is optically

active.

II. 2-bromobutane: The carbon marked with * is the chiral center.

Carbon Atom Groups bonded to carbon atom

Chiral Center * C, H, Br, C

Step 3:

III. 2-iodobutane: The carbon marked with * is the chiral center.

Carbon Atom Groups bonded to carbon atom

Chiral Center * C, H, I, C

Step 4:

I. 2-chlorobutane: Chiral and optically inactive.

II. 2-bromobutane: Chiral and optically active.

III. 2-iodobutane: Chiral and optically active.

Question 8

Question

Determine the relationship between the following pairs of compounds regarding

chirality and optical activity:

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

2. (R)-fluorochlorobromomethane and (S)-fluorochlorobromomethane