CHEM 131 - ADVANCED GENERAL CHEMISTRY I Chirality and optical activity Question Bank Set 3

CHEM 131 - ADVANCED GENERAL

CHEMISTRY I - Chirality and optical

activity

Question Bank - Set 3

Liberty University

Question 1

Question

Explain the relationship between chirality and optical activity. How can one

determine whether a compound is optically active or not?

Solution

Step 1: Chirality refers to the property of a molecule that cannot be superim-

posed on its mirror image. A chiral molecule exists in two enantiomeric forms,

which are nonsuperimposable mirror images of each other.

Step 2: Optical activity is a physical property exhibited by chiral compounds

where they rotate the plane of polarized light. Enantiomers have equal but

opposite optical rotations, resulting in a net optical inactivity when present in

equal amounts.

Step 3: To determine whether a compound is optically active or not, one

can check if the molecule is chiral. This can be done by examining its symmetry

elements and determining if it possesses a chiral center (an atom with four

different groups attached).

Step 4: If a compound has a chiral center, it is considered chiral and can

exhibit optical activity. On the other hand, compounds without a chiral center,

such as meso compounds which have internal planes of symmetry, are optically

inactive despite being composed of chiral molecules.

Step 5: In summary, chirality and optical activity are closely related con-

cepts in organic chemistry. Chirality arises from lack of superimposability of

enantiomers, while optical activity results from the ability of chiral molecules to

rotate the plane of polarized light. Determining the presence of a chiral center

is crucial in identifying whether a compound is optically active or not.

Question 2

Question

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

activity. Provide an example of a chiral molecule and explain why it is optically

active.

Solution

Step 1: Chirality in Organic Molecules A molecule is chiral if it does not contain

a plane of symmetry, meaning that it cannot be superimposed on its mirror

image. Chirality arises from having four different substituents (atoms or groups)

attached to a central carbon atom, creating a non-superimposable mirror image

pair called enantiomers.

Step 2: Optical Activity Optical activity refers to the ability of a chiral

compound to rotate the plane of polarized light. Enantiomers rotate plane-

polarized light in equal but opposite directions, a property known as optical

activity. This rotation is due to the asymmetry in the spatial arrangement of

the atoms or groups in the molecule.

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

Lactic acid (2-hydroxypropanoic acid), which exists in two enantiomeric forms:

L-lactic acid and D-lactic acid. These two forms are mirror images of each other

and are optically active.

Step 4: Explanation of Optical Activity in Lactic Acid Lactic acid is chiral

because the carbon atom bonded to the hydroxyl group (OH), hydrogen (H),

carboxyl group (COOH), and a methyl group (CH3) is chiral. Due to the

asymmetry in its arrangement, L-lactic acid and D-lactic acid are optically

active enantiomers, rotating plane-polarized light in opposite directions.

By understanding chirality in organic molecules and its relationship to opti-

cal activity, we can appreciate the significance of enantiomers in various chem-

ical, biological, and pharmaceutical contexts.

Question 3

Question

Explain why a molecule with a chiral center can exhibit optical activity, while

a molecule without a chiral center cannot.

Solution



A molecule is chiral if it is not superimposable on its mirror image. This

arises when a molecule contains at least one chiral center (also known as

a stereocenter).



A chiral center is an atom, usually carbon, that is bonded to four different

groups. This arrangement leads to non-superimposable mirror images

(enantiomers).



Enantiomers have identical physical properties except for their interaction

with plane-polarized light, a phenomenon called optical activity.

Step 1: Optical Activity of Molecules with Chiral Centers



When a molecule with a chiral center is placed in a beam of plane-polarized

light, the two enantiomers (mirror images) will rotate the plane of polar-

ized light in opposite directions.



One enantiomer rotates the plane to the right, termed dextrorotary (+),

while the other enantiomer rotates the plane to the left, termed levorotary

(-).



The extent of rotation is quantified using specific rotation values.



Hence, molecules with chiral centers exhibit optical activity due to the

presence of enantiomers that rotate plane-polarized light.

Step 2: Lack of Optical Activity in Molecules without Chiral Cen-

ters



Molecules without chiral centers, such as meso compounds or achiral

molecules, do not exhibit optical activity.



This is because they are superimposable on their mirror images and do

not have enantiomers that can rotate plane-polarized light.



Therefore, even if the molecule may contain polar bonds or be asymmetric

in structure, it will not be optically active if it lacks chiral centers.

In summary, the presence of a chiral center in a molecule leads to non-

superimposable mirror images (enantiomers) that exhibit optical activity by

rotating plane-polarized light. Molecules without chiral centers do not show

optical activity as they are superimposable on their mirror images.

Question 4

Question

Explain why a molecule with a chiral center can exhibit optical activity while a

molecule with an internal plane of symmetry cannot.

Solution

Step 1: First, let’s define what chirality and optical activity are. - A molecule

is chiral if it cannot be superimposed on its mirror image. - Optical activity

refers to the ability of a chiral molecule to rotate the plane of polarized light.

Step 2: A molecule with a chiral center has four different groups attached

to the central carbon atom, leading to non-superimposable mirror image struc-

tures (enantiomers). A chiral molecule will interact with plane-polarized light

differently for each enantiomer. This leads to optical activity.

Step 3: On the other hand, a molecule with an internal plane of symmetry

can be separated into two symmetric halves that are mirror images of each other.

In this case, the molecule is achiral, as it is superimposable on its mirror image.

Step 4: Since an achiral molecule can be superimposed on its mirror image,

it will not exhibit optical activity. The plane-polarized light passing through an

achiral molecule will not be rotated.

Therefore, a molecule with a chiral center can exhibit optical activity due

to its non-superimposable mirror image structures, while a molecule with an

internal plane of symmetry cannot exhibit optical activity as it is achiral and

superimposable on its mirror image.

Question 5

Question

Determine the chirality of the following molecules and state whether they are

optically active:

1. 2-chloro-3-methylpentane

2. R-2,3-dibromobutane

3. (S)-2-aminopropanoic acid

Solution

1. For 2-chloro-3-methylpentane, we first need to identify the chiral centers in

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

In this case, the molecule has no chiral centers, so it is not chiral and not

optically active.

2. For R-2,3-dibromobutane, we need to determine the configuration at the

chiral centers. The R-configuration indicates that the highest priority group is

on the right side when the molecule is oriented with the lowest priority group

facing away from you. If the molecule has chiral centers and is optically active,

we must then determine if it is R or S.

3. For (S)-2-aminopropanoic acid, the (S)-configuration indicates that the

highest priority group is on the left side when the molecule is oriented with the

lowest priority group facing away from you. If the molecule has chiral centers

and is optically active, we must then determine if it is R or S.

Question 6

Question

Explain why a compound containing a chiral carbon atom may be optically

active, even if it is not a pure enantiomer.

Solution

To understand this concept, we need to consider the nature of chirality in

molecules and how it relates to optical activity.

Step 1: Understanding Chirality A molecule is chiral if it cannot be su-

perimposed on its mirror image. Chirality arises from the presence of an asym-

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

to four different groups, leading to non-superimposability of the molecule and

its mirror image.

Step 2: Enantiomers Enantiomers are a pair of molecules that are mir-

ror images of each other and non-superimposable. They have opposite optical

activity, meaning one is dextrorotatory (+) and the other is levorotatory (-).

Step 3: Racemic Mixtures When a compound contains a chiral carbon

atom but is not a pure enantiomer, it may exist as a racemic mixture. A racemic

mixture contains equal amounts of both enantiomers, resulting in cancelation

of their optical activities.

Step 4: Optically Active Mixtures However, if the compound is not a

pure racemic mixture (i.e., unequal amounts of enantiomers are present), the

optical activities of the enantiomers do not completely cancel out. As a result,

the mixture can still exhibit optical activity.

Step 5: Conclusion In summary, a compound containing a chiral carbon

atom may be optically active even if it is not a pure enantiomer due to the

presence of unequal amounts of enantiomers that lead to a net optical activity

in the mixture.

Question 7

Question

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

whether they are optically active or inactive:

I. (R)-2-bromobutane II. meso-2,3-dibromobutane

Solution

To determine chirality, we need to examine the presence of chiral centers and

symmetry. For optical activity, we need to determine if a molecule is chiral and

lacks a plane of symmetry. Let’s analyze each molecule separately:

I. (R)-2-bromobutane:



This molecule has a chiral center, as carbon is bonded to four different

groups.



Hence, it is chiral.



Since it is chiral and lacks a plane of symmetry, (R)-2-bromobutane is

optically active.

II. meso-2,3-dibromobutane:



In meso compounds, the molecule may have chiral centers, but overall it

is a achiral due to the presence of an internal plane of symmetry.



meso-2,3-dibromobutane has a plane of symmetry passing through the

central carbon, which divides the molecule into two equal halves.



Hence, meso-2,3-dibromobutane is achiral.

Therefore, (R)-2-bromobutane is chiral and optically active, while meso-2,3-

dibromobutane is achiral and optically inactive.

Question 8

Question

An unknown compound is found to be optically active. When a sample of this

compound is dissolved in a suitable solvent and passed through a polarimeter,

it is observed that the sample rotates the plane of polarized light by +35

. De-

termine whether the compound is chiral or achiral, and explain your reasoning.

Solution

Step 1: Recall that a compound is chiral if it lacks an internal plane of symmetry.

This means that the compound cannot be superimposed on its mirror image.

Step 2: Given that the compound is optically active and rotates the plane

of polarized light, we can conclude that it is chiral. This is because only chiral

compounds have the ability to rotate plane-polarized light.

Step 3: Since the compound is chiral, it must lack an internal plane of sym-

metry. This means that the compound’s mirror image cannot be superimposed

onto the original compound.

Step 4: Therefore, based on the optical activity and the definition of chirality,

we can determine that the unknown compound is chiral.

Question 9

Question

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

active.

chirality_molecule.png

Solution

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

non-superimposable mirror image. This molecule has a carbon atom (marked

with an asterisk) with four different substituents, making it a chiral center.

Step 2: Let’s draw the mirror image of the molecule:

chirality_mirror_image.png

Step 3: Comparing the original molecule to its mirror image, we see that

they are not superimposable. Therefore, the molecule is chiral.

Step 4: To determine if the molecule is optically active, we need to check if

it lacks an internal plane of symmetry. In this case, there is no internal plane

of symmetry, so the molecule is optically active.

Therefore, the molecule shown is chiral and optically active.

Question 10

Question

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

example of a chiral molecule and describe how its optical activity is determined.

Solution

Step 1: Chirality and Optical Activity A molecule is chiral if it cannot be

superimposed on its mirror image. Chirality is an important concept in chem-

istry because chiral molecules often exhibit optical activity. Optical activity

refers to the ability of a substance to rotate the plane of polarized light.

Step 2: Example of a Chiral Molecule One common example of a chiral

molecule is Limonene, a compound found in the peels of citrus fruits. Limonene

has a chiral center at the carbon atom highlighted below:

CH3−CH(CH3)−CH2−C(∗)(CH3)−CH2−CH = CH2

Step 3: Determining Optical Activity To determine the optical activity

of a chiral molecule like Limonene, we use a polarimeter. A polarimeter mea-

sures the rotation of plane-polarized light as it passes through a sample. When

polarized light passes through a solution of a chiral compound, the plane of

polarization rotates either clockwise (dextrorotatory, labeled as ”+”) or coun-

terclockwise (levorotatory, labeled as ”-”).

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

plane of polarized light is quantified by a property called specific rotation ([α]).

The specific rotation is unique for each enantiomer of a chiral molecule and is

dependent on factors like concentration, path length, and wavelength of light.

Step 5: Conclusion In conclusion, chirality is a property of molecules that

gives rise to optical activity. Chiral molecules like Limonene can rotate the

plane of polarized light, and this rotation is quantified by the specific rotation

value.

Question 11

Question

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

an example of a chiral molecule and explain why it exhibits optical activity.

Solution

Step 1: Chirality and Optical Activity A molecule is considered chiral if it

cannot be superimposed on its mirror image. Chirality arises when a molecule

contains an asymmetric carbon atom - a carbon atom bonded to four different

groups or atoms. Chiral molecules exist in two enantiomeric forms (mirror

images), known as the R- and S-enantiomers.

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

ized light. Enantiomers exhibit optical activity because they interact differently

with plane-polarized light due to their unique asymmetric structures.

Step 2: Example of a Chiral Molecule - Lactic Acid Lactic acid, with

the chemical formula CH3CH(OH)COOH, is an example of a chiral molecule.

In lactic acid, the central carbon atom (the one bonded to the hydroxyl group

-OH) is chiral because it is bonded to four different groups: a hydrogen atom,

a hydroxyl group, a methyl group, and a carboxyl group.

Step 3: Explanation of Optical Activity in Lactic Acid The R- and S-

enantiomers of lactic acid are non-superimposable mirror images of each other.

When lactic acid is dissolved in a solvent and a beam of plane-polarized light is

passed through the solution, each enantiomer will rotate the plane of polariza-

tion in opposite directions. This phenomenon is known as optical activity.

In the case of lactic acid, the two enantiomers will rotate the plane of po-

larization in opposite directions due to their different spatial arrangements of

atoms around the chiral carbon. This property allows optical devices such as

polarimeters to differentiate between the two enantiomers based on their optical

activity.

Question 12

Question

Identify the following molecules as chiral or achiral, and determine whether each

is optically active or inactive:

a) 2-chlorobutane

b) 2-aminopropanoic acid

c) 2,3-dichlorobutane

Solution

a) 2-chlorobutane: - This molecule has a chiral center because the carbon atom

bonded to four different groups (C, H, Cl, CH3). - Therefore, 2-chlorobutane is

chiral. - Since it is chiral, it is expected to be optically active.

b) 2-aminopropanoic acid: - This molecule contains a chiral center due to

the presence of the nitrogen atom bonded to different groups (H, NH2, COOH,

CH3). - Consequently, 2-aminopropanoic acid is chiral. - As it is chiral, it is

optically active.

c) 2,3-dichlorobutane: - In this molecule, both carbon atoms bonded to

chlorine are equivalent as they have the same four substituents around them.

- Hence, 2,3-dichlorobutane is achiral. - Since it is achiral, it is optically

inactive.

Question 13

Question

A compound with molecular formula C9H10O exhibits optical activity and has

a chiral center. Upon reaction with PCC, it forms a single chiral product. Draw

the structures of the reactant and product.

Solution

Step 1: Determine the possible structures of the compound with the given

molecular formula C9H10O. The molecular formula C9H10O implies 9 carbons,

10 hydrogens, and 1 oxygen. One way to arrange these atoms is by considering

a cyclic structure due to the limited number of hydrogen atoms. Let’s start by

drawing a cyclic compound with a chiral center:

C = OC−C−C−C−C−C−C−C

The chiral center can be located at any of the carbon atoms in the ring.

Step 2: Determine the product that forms upon reaction with PCC. PCC

(pyridinium chlorochromate) is commonly used in organic chemistry to oxidize

primary alcohols to aldehydes. Since a single chiral product is formed, this

indicates that the product retains the original chiral center. The product may

be an aldehyde, such as benzaldehyde. Let’s draw the product structure based

on the information provided:

C = O C −H

The hydrogen is bound to the carbon next to the chiral center.

Therefore, the reactant is a cyclic compound with a chiral center and the

product is an aldehyde with a chiral center.

Question 14

Question

Determine the chirality (R/S) of the following compound and state whether it

is expected to be optically active:

CH3−C−CH3

Solution

To determine the chirality of the compound, we need to assign priorities to the

substituents attached to the chiral center based on their atomic numbers. The

substituents will be prioritized as follows: OH (highest priority, Priority 1),

CH3(second highest priority, Priority 2), H(lowest priority, Priority 3).

Step 1: Position the molecule so that the lowest priority group, H, is

pointing away from you.

Step 2: Trace a path from Priority 1 to Priority 2 to Priority 3. If the path is

clockwise, the configuration is R (Latin: rectus). If the path is counterclockwise,

the configuration is S (Latin: sinister).

Since the path goes clockwise from Priority 1 (OH ) to Priority 2 (CH3) to

Priority 3 (H), the chirality at the chiral center is R.

Step 3: To determine if the compound is optically active, we need to check

if it has a plane of symmetry. For a molecule to be optically inactive, it must

have a plane of symmetry.

In this case, the compound does not have a plane of symmetry. Since it is

chiral (R configuration), it is expected to be optically active.

Question 15

Question

Explain why 2,3-dibromobutane is chiral and determine its optical activity.

Solution

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

center. A stereocenter is a carbon atom with four different groups attached to

it.

Step 2: The molecular formula for 2,3-dibromobutane is C4H8Br2. The

carbon atom at position 2 in the molecule (between the two bromine atoms) is

a stereocenter because it has four different groups attached to it: H, Br, C2H5,

and Br.

Step 3: Due to the presence of a stereocenter, 2,3-dibromobutane is chiral.

Step 4: To determine the optical activity of a chiral molecule, we need to

consider the spatial arrangement of the groups attached to the stereocenter.

Step 5: In 2,3-dibromobutane, the spatial arrangement around the stere-

ocenter is such that it does not have a plane of symmetry. Therefore, it is

optically active.

Step 6: Since 2,3-dibromobutane is chiral and lacks a plane of symmetry, it

will exhibit optical activity.

Question 16

Question

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

optically active.

H3C

CH3−C(NH2)2

Solution

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

of symmetry. A molecule is chiral if it does not superimpose on its mirror image.

Step 2: The molecule shown has a central carbon atom bonded to four differ-

ent groups - methyl, ethylamine group, amino group, and hydrogen. Therefore,

it is chiral as it does not have a plane of symmetry.

Step 3: To check if the molecule is optically active, we need to see if it has a

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

Step 4: Since the central carbon atom in the molecule is bonded to four

different groups, it is a chiral center. Therefore, the molecule is optically active.

Step 5: The molecule is optically active because it is chiral and lacks a plane

of symmetry. It will rotate the plane of polarized light as it passes through it.

Question 17

Question

For a compound with the molecular formula C5H10O, four different stereoiso-

mers are possible. Which of these stereoisomers are chiral? For those that are

chiral, identify which ones are optically active.

Solution

Step 1: To determine the possible stereoisomers of a compound with formula

C5H10O, we need to consider the different ways the atoms can be arranged

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

groups. In C5H10O, there are 5 carbon atoms, and only those with all four

different groups will be chiral. Let’s first draw the structural formulas of the

four possible stereoisomers.

Step 2: Consider the isomers: 1. 3-pentanone (CH3-CH2-CO-CH3) 2. 2-

pentanone (CH3-CO-CH2-CH3) 3. 2-pentanol (CH3-CHOH-CH2-CH3) 4. 3-

pentanol (CH3-CH2-CH2-CHOH-CH3)

Step 3: Among the isomers considered, only 2-pentanol and 3-pentanol have

a chiral center, as they both have an asymmetric carbon atom (marked with an

asterisk). Let’s identify which of these chiral isomers are optically active.

Step 4: Optically active compounds must have chiral centers and not possess

a plane of symmetry. A compound is optically active if it rotates plane-polarized

light. To determine whether a compound is optically active, we need to assess

whether it has a plane of symmetry.

Step 5: In the case of 2-pentanol (CH3-CHOH-CH2-CH3), there is no plane

of symmetry present, so this compound is optically active. However, in the

case of 3-pentanol (CH3-CH2-CH2-CHOH-CH3), there is a plane of symmetry

through the central carbon atom, so this compound is not optically active.

Step 6: Therefore, the chiral isomer that is optically active among the pos-

sible stereoisomers of C5H10O is 2-pentanol.

Question 18

Question

Determine the relationship between the following pairs of compounds in terms

of chirality, and identify which pairs exhibit optical activity:

(i) (2S,3R)-2,3-dibromobutane and (2S,3S)-2,3-dibromobutane

(ii) (R)-3-bromohexane and (S)-3-bromohexane

(iii) (1S,2R)-1,2-dibromocyclobutane and (1R,2S)-1,2-dibromocyclobutane

Solution

Step 1: Chirality Relationship (i) For the pair (2S,3R)-2,3-dibromobutane and

(2S,3S)-2,3-dibromobutane: Both compounds have the same configuration at

the chiral carbon (2S), but different configurations at the second chiral car-

bon (3R versus 3S). Therefore, the relationship between these compounds is

diastereomers.

(ii) For the pair (R)-3-bromohexane and (S)-3-bromohexane: These com-

pounds have opposite configurations at the chiral carbon (R versus S). There-

fore, the relationship between these compounds is enantiomers.

(iii) For the pair (1S,2R)-1,2-dibromocyclobutane and (1R,2S)-1,2-dibromocyclobutane:

Both compounds have opposite configurations at both chiral carbons, making

them enantiomers.

Step 2: Optical Activity (i) A mixture of diastereomers does not exhibit

optical activity because they do not cancel each other out.

(ii) Enantiomers exhibit optical activity because they are non-superimposable

mirror images of each other.

(iii) Enantiomers exhibit optical activity for the same reason as in (ii).

Therefore, only pairs (ii) and (iii) exhibit optical activity.

Question 19

Question

Is the molecule shown below chiral? If so, identify all chiral centers and assign

the R/S configuration to each chiral center.

H−C−C−C−H

|||

HHH

Solution

Step 1: To determine if the molecule is chiral, we need to identify all chiral

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

There are three carbon atoms in the molecule shown, but only one of them is a

chiral center.

Step 2: Let’s assign priorities to the four groups attached to the chiral center.

The higher the atomic number of the atom directly bonded to the chiral center,

the higher the priority. In this case, the priorities are:



Priority 1: Carbon attached to two hydrogens and one carbon



Priority 2: Carbon attached to two hydrogens and one hydrogen



Priority 3: Hydrogen



Priority 4: Hydrogen

Step 3: Next, we need to position the molecule so that the lowest priority

group (H) is pointing away from us. Now, we look at the remaining three groups

and determine the order of priority (1-2-3) by rotating from 1 to 2 to 3. In this

case, the order is clockwise, therefore it is an R configuration.

Step 4: Therefore, the molecule is chiral and the chiral center has an R

configuration.

Question 20

Question

Determine whether each of the following compounds is chiral. If a compound is

chiral, indicate the number of chiral centers present.

1. 2-bromobutane

2. 2-chloro-3-methylpentane

3. 2,3-dibromobutane

Solution

1. For a compound to be chiral, it must not have a plane of symmetry. This

means that it cannot be superimposed on its mirror image. Therefore, a com-

pound with a chiral center is chiral. A chiral center is a carbon atom that is

bonded to four different groups. Let’s analyze each compound:

1. 2-bromobutane: The carbon bonded to the bromine atom (the second

carbon from the left) has three different substituents (methyl, ethyl, and

hydrogen). Therefore, 2-bromobutane is chiral with one chiral center.

2. 2-chloro-3-methylpentane: The carbon bonded to the chlorine atom

(the second carbon from the left) has three different substituents (methyl,

ethyl, and hydrogen). Therefore, 2-chloro-3-methylpentane is chiral with

one chiral center.

3. 2,3-dibromobutane: The second and third carbons from the left are

both bonded to two bromine atoms and two hydrogen atoms each, so they

do not have a chiral center. Therefore, 2,3-dibromobutane is achiral.

Question 21

Question

Explain why the following compound is chiral:

2−bromobutane

Solution

To determine if a compound is chiral, we need to examine its molecular structure

and look for the presence of an asymmetric carbon atom (also known as a chiral

center or stereocenter). An asymmetric carbon atom is a carbon atom that is

bonded to four different groups.

Step 1: Draw the structural formula of 2-bromobutane:

CH3CHBrCH2CH3

Step 2: Identify the carbon atom bonded to four different groups, which is

the second carbon atom in the chain (the carbon atom bonded to the bromine

atom).

Step 3: Determine if the four groups bonded to the asymmetric carbon

are different. In 2-bromobutane, the four groups attached to the asymmetric

carbon atom are: a hydrogen atom, a methyl group, an ethyl group, and a

bromine atom. Since each group is different, this carbon atom is a chiral center.

Therefore, 2-bromobutane is a chiral compound due to the presence of an

asymmetric carbon atom with four different groups attached to it.

Question 22

Question

Determine the absolute configuration (R or S) at the chiral center of the com-

pound shown below. Indicate whether the compound is optically active or in-

active.

chiral_compound.png

Solution

Step 1: Identify the priority of the substituents attached to the chiral center

based on atomic number (higher atomic number = higher priority).

Cl >C>H>C

Step 2: Orient the molecule so that the lowest priority group (H in this case)

is pointing away from you.

chiral_compound_oriented.png

Step 3: Trace a path from the highest priority group to the second highest

priority group going in a clockwise direction. If the path goes in a clockwise

direction, the configuration is R (Latin: rectus, right-handed); if it goes in a

counterclockwise direction, the configuration is S (Latin: sinister, left-handed).

chiral_compound_labeled.png

The path from Cl to C to the right side C goes in a counterclockwise direc-

tion, so the configuration at the chiral center is S.

Step 4: Determine if the compound is optically active or inactive. If a

compound has a chiral center (i.e., it is not a meso compound), then it is

optically active and can rotate plane-polarized light.

Therefore, the compound shown is S-configured and optically active.

Question 23

Question

Explain why 2,3-dibromobutane is chiral while 2,3-dichlorobutane is not chiral.

Solution

To determine chirality, we need to examine whether a molecule has a non-

superimposable mirror image. A chiral molecule is optically active, meaning

that it can rotate the plane of polarized light.

Step 1: Determine if 2,3-dibromobutane is chiral

2,3-dibromobutane has the chemical formula CH3CHBrCHBrCH3. To de-

termine chirality, we look at the carbon atom bonded to the two bromine atoms

(the chiral center):

CH3−C−H

Br −C−Br

The carbon atom bonded to the two bromine atoms has four different groups

attached to it (methyl group, hydrogen atom, and two bromine atoms). There-

fore, 2,3-dibromobutane is chiral.

Step 2: Determine if 2,3-dichlorobutane is chiral

2,3-dichlorobutane has the chemical formula CH3CHClCHClCH3. Look at

the carbon atom bonded to the two chlorine atoms:

CH3−C−H

Cl −C−Cl

The carbon atom bonded to the two chlorine atoms has two identical groups

attached to it (two chlorine atoms), making it achiral and not optically active.

Therefore, 2,3-dibromobutane is chiral while 2,3-dichlorobutane is not chiral.

Question 24

Question

Explain why a molecule with a chiral center is optically active. Provide an

example of a molecule with a chiral center, and discuss whether it is optically

active or not.

Solution

Step 1: Chirality and Optical Activity A molecule is chiral if it is not

superimposable on its mirror image. Chirality arises from a molecule having a

chiral center (also known as a stereocenter), which is an atom that is attached to

four different groups. Chirality is important in the context of optical activity,

which refers to the ability of certain chiral molecules to rotate the plane of

polarized light.

Step 2: Optical Activity of Chiral Molecules Chiral molecules are op-

tically active because they interact with plane-polarized light in a unique way

due to their asymmetry. When plane-polarized light passes through a solu-

tion of chiral molecules, the light is rotated either clockwise (dextrorotatory) or

counterclockwise (levorotatory) by an angle specific to that molecule.

Step 3: Example of a Molecule with a Chiral Center Consider the

molecule 2-chlorobutane (CH3CHClCH2CH3). The carbon atom bonded to the

chlorine atom is a chiral center since it is connected to four different groups: a

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

Step 4: Optical Activity Discussion Due to the presence of the chiral

center (the carbon atom bonded to the chlorine atom), 2-chlorobutane is a

chiral molecule. As a result, 2-chlorobutane is optically active because it has a

chiral center that can interact with plane-polarized light by rotating its plane.

In conclusion, molecules with a chiral center are optically active because

of their asymmetry, allowing them to interact with plane-polarized light in a

unique way.

Question 25

Question

Explain the concept of chirality and optical activity in organic chemistry. Pro-

vide an example of a chiral molecule and explain how its enantiomers exhibit

optical activity.

Solution

Step 1: Chirality in Organic Chemistry Chirality is a property of a molecule

that results from its inability to be superimposed on its mirror image. This

means that a chiral molecule and its mirror image are non-identical, similar to

how our left and right hands are not superimposable. Chirality is an important

concept in organic chemistry as it plays a crucial role in various chemical and

biological processes.

Step 2: Optical Activity Optical activity refers to the ability of a chiral

molecule to rotate the plane of polarized light. When a beam of plane-polarized

light passes through a solution of chiral molecules, the plane of polarization

rotates either to the left (levorotatory, denoted as ”-”) or to the right (dex-

trorotatory, denoted as ”+”). The extent of rotation is determined by the

concentration of the chiral molecules, the path length of the light through the

solution, and a constant characteristic of the specific compound called specific

rotation.

Step 3: Example of a Chiral Molecule and Optical Activity One ex-

ample of a chiral molecule is lactic acid. Lactic acid exists as two enantiomers:

L-lactic acid and D-lactic acid. These enantiomers are mirror images of each

other but cannot be superimposed. When a polarized light is passed through

a solution of L-lactic acid, it rotates the plane of polarization to the left (lev-

orotatory), while D-lactic acid rotates the plane of polarization to the right

(dextrorotatory).

In summary, chirality in organic chemistry refers to the property of non-

superimposability of a molecule and its mirror image, leading to optical activity

where chiral molecules rotate the plane of polarized light. The enantiomers

of chiral molecules exhibit optical activity in opposite directions due to their

mirror-image relationship.

Question 26

Question

A compound has the molecular formula C8H8O2and exhibits optical activity.

Upon careful analysis, it is determined that the compound contains a single

chiral center. Its specific rotation in a 1:1 mixture of ethanol and water at

20◦C is +48.7◦. What is the configuration (R or S) of the chiral center in this

compound?

Solution

Step 1: Calculate the observed specific rotation using the formula:

Observed specific rotation = α

l×100

c

Where: α= observed rotation in degrees l= length of the tube in decimeters

c= concentration of the compound in g/mL

Given that the observed specific rotation is +48.7◦, we have:

48.7 = +48.7

1×100

c

Step 2: Calculate the concentration of the compound.

c= 2.05 g/mL

Step 3: Calculate the molar mass of the compound.

Molar mass = 8(12.01) + 8(1.008) + 2(16.00) = 152.16 g/mol

Step 4: Calculate the specific rotation using the formula:

Specific rotation = [α]obs ×l

c×Molar mass

Substituting the values:

Specific rotation = 48.7×1

2.05 ×152.16= +0.155 deg mL g−1dm3cm−1

Step 5: Determine the observed specific rotation of the enantiomer with the

opposite configuration using the formula:

Observed specific rotation of the enantiomer = −(Specific rotation)

Therefore, the observed specific rotation of the enantiomer with the opposite

configuration is −0.155◦.

Step 6: Determine the configuration of the chiral center from the observed

and enantiomeric specific rotations. Since the observed rotation is greater than

the enantiomeric rotation, the configuration of the chiral center is S.

Question 27

Question

Determine the R/S configurations for the following chiral molecules. Which of

the molecules are optically active?

CH3−CH(OH) −CH2−CH(NH2)−COOH

Solution

To determine the R/S configurations for the chiral molecules, we need to assign

priorities to the substituents attached to the chiral center based on the atomic

number of the atoms directly bonded to the chiral center.

1. Assign priorities to the substituents based on the atomic number of the

atoms directly bonded to the chiral center. The highest priority is assigned

to the group with the highest atomic number.

2. If two substituents have the same atom directly bonded to the chiral cen-

ter, look at the atoms bonded to those atoms. Continue this process until

there is a point of difference.

3. Rotate the molecule so that the lowest priority group is facing away from

you.

4. Trace a path from the group with the lowest priority (4) to the group with

the highest priority (1) through groups 2 and 3. If the path is clockwise,

the configuration is R (rectus); if it is counterclockwise, the configuration

is S (sinister).

Step 1: Assign priorities to the substituents based on the atomic number

of the atoms directly bonded to the chiral center. The atomic numbers of the

atoms directly bonded to the chiral center are: O (1 for COOH), N (7 for NH2),

C (6 for OH), and C (6 for CH3). The priorities are:

COOH (O) >NH2(N) >OH (C) >CH3(C)

Step 2: Rotate the molecule so that the lowest priority group, CH3, is facing

away from you.

Step 3: Trace a path from CH3(4) to COOH (1) through OH (2) and NH2

(3). The path is counterclockwise.

Step 4: Since the path is counterclockwise, the configuration is S.

The molecule described is optically active because it has one chiral center

and is a single enantiomer.

Question 28

Question

For each of the following molecules, determine if the molecule is chiral or achiral.

If the molecule is chiral, designate the stereocenter(s) or chiral center(s).

1. 3-chloro-2-butanol

2. 2,3-dibromopentane

3. 1,2-dichlorocyclohexane

Solution

1. 3-chloro-2-butanol:

Step 1: Identify the chiral center(s) in the molecule.

3-chloro-2-butanol has one chiral center at the carbon bonded to the hydroxyl

group. This carbon is attached to four different groups (H, OH, Cl, CH3).

Therefore, 3-chloro-2-butanol is chiral.

2. 2,3-dibromopentane:

Step 1: Identify the chiral center(s) in the molecule.

2,3-dibromopentane has two chiral centers, at the second and third carbons

from the left end. Each of these carbons is bonded to four different groups (H,

Br, Br, CH3).

Therefore, 2,3-dibromopentane is chiral.

3. 1,2-dichlorocyclohexane:

Step 1: Identify the chiral center(s) in the molecule.

1,2-dichlorocyclohexane does not have any chiral centers. The molecule pos-

sesses a plane of symmetry passing through the carbon atoms (1 and 2) bearing

the chlorine atoms. This makes the molecule achiral.

Therefore, 1,2-dichlorocyclohexane is achiral.

Question 29

Question

An unknown compound X has the molecular formula C5H10 O and exhibits op-

tical activity. When compound X is reacted with hot acidic potassium dichro-

mate, it is oxidized to a carboxylic acid, which is optically inactive. Draw two

different structures of compound X that satisfy these conditions and indicate

which one is chiral.

Solution

Step 1: Let’s first list the possible functional groups present based on the molecu-

lar formula and the given information: The molecular formula C5H10O suggests

that compound X may contain a ketone, aldehyde, ester, or alcohol functional

group. Since the compound is optically active and is oxidized to an optically

inactive carboxylic acid, it must contain a chiral center. We need to find two dif-

ferent structures for compound X that are chiral and can be oxidized to optically

inactive carboxylic acids.

Step 2: One possible structure for compound X is 2-pentanol (CH3CH2CH(OH)CH2CH3).

2-Pentanol contains a chiral center, the carbon atom bonded to the hydroxyl

group. This structure can be oxidized to pentanoic acid, which is optically

inactive.

Step 3: Another possible structure for compound X is 3-methyl-2-butanol

(CH3CH(OH)C(CH3)2CH3). 3-Methyl-2-butanol also contains a chiral center,

the carbon atom bonded to the hydroxyl group. This structure can be oxidized

to 3-methylbutanoic acid, which is optically inactive.

Hence, both 2-pentanol and 3-methyl-2-butanol are chiral structures of com-

pound X that can be oxidized to optically inactive carboxylic acids.

Question 30

Question

Explain why the molecule shown below is chiral and determine its optical ac-

tivity.

H−C(H)(H) −C(H)(H)(F) −C(H)(H)(Cl) −C(H)(H)(Br) −C(H)(H)(I) −H

Solution

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

has a chiral center, which is a carbon atom with four different groups attached

to it.

H−C(H)(H) −C(H)(H)(F) −C(H)(H)(Cl) −C(H)(H)(Br) −C(H)(H)(I) −H

In this molecule, the carbon atom with the H, F, Cl, Br, and I attached to

it is a chiral center because it has four different groups attached.

Step 2: A molecule with a chiral center is chiral and can exist in two enan-

tiomeric forms that are non-superimposable mirror images of each other.

Step 3: To determine the optical activity, we need to look at the configuration

of the chiral center. In this case, the configuration is R (clockwise order of

priority: I >Br >Cl >F) given by the Cahn-Ingold-Prelog rules.

Step 4: The molecule is optically active because it is chiral. Its optical

activity is determined by the configuration of the chiral center. Since the con-

figuration is R, the molecule is dextrorotatory (d-plus).

Therefore, the molecule is chiral and dextrorotatory.

Question 31

Question

Explain why the compound 2-chlorobutane is chiral and determine whether it

exhibits optical activity.

Solution

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

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

Step 2: The compound 2-chlorobutane has a chiral center because the carbon

atom bonded to the chlorine atom is also bonded to two different groups (ethyl

and methyl groups) and a hydrogen atom. These four groups are all different.

Step 3: Since 2-chlorobutane has a chiral center, it is chiral.

Step 4: To determine if a chiral compound exhibits optical activity, we need

to check if it is optically active. A compound is optically active if it rotates

plane-polarized light.

Step 5: The ability for a chiral compound to rotate plane-polarized light

depends on the spatial arrangement of the groups around the chiral center. In

2-chlorobutane, the ethyl and methyl groups are not significantly different in

size or shape, leading to cancellation of the optical activity.

Step 6: Therefore, even though 2-chlorobutane is chiral, it does not exhibit

optical activity due to the symmetrical arrangement of the ethyl and methyl

groups.

Question 32

Question

Determine whether the following compounds are chiral or achiral:

(R)−2−bromobutane and (S)−2−bromobutane

Explain your reasoning.

Solution

Step 1: To determine whether a compound is chiral or achiral, we need to

examine its symmetry. A molecule is chiral if it does not have an internal plane

of symmetry. If a molecule has an internal plane of symmetry, it is achiral.

Step 2: Let’s consider (R)-2-bromobutane first. This compound has a chiral

center due to the presence of the bromine atom bonded to the second carbon.

To determine the configuration at this chiral center, assign priorities to the four

substituents based on atomic number: higher atomic number = higher priority.

The lowest priority group (hydrogen in this case) is pointing away from the

viewer. The remaining three groups (bromine, carbon, carbon) are arranged in

a clockwise direction, which gives the configuration as R.

Step 3: Now, let’s consider (S)-2-bromobutane. By following the same pro-

cess as in Step 2, we find that the configuration at the chiral center in this

compound is S.

Step 4: Since (R)-2-bromobutane and (S)-2-bromobutane have different con-

figurations at the chiral center, they are enantiomers of each other. Enantiomers

are non-superimposable mirror images.

Step 5: Therefore, both (R)-2-bromobutane and (S)-2-bromobutane are chi-

ral compounds.

Step 6: In conclusion, the compounds (R)-2-bromobutane and (S)-2-bromobutane

are chiral due to the presence of a chiral center, and they are enantiomers of

each other.

Question 33

Question

Consider a molecule with the following chirality centers: A, B, and C. Draw the

structure of the molecule and determine whether it is chiral or achiral. If chiral,

classify the molecule as either dextrorotatory or levorotatory.

Solution

Step 1: Draw the structure of the molecule with the chirality centers A, B, and

Step 2: Determine whether the molecule is chiral or achiral. To determine

if the molecule is chiral, we need to check if it has a non-superimposable mirror

image. If the molecule has a plane of symmetry or a center of symmetry, it is

achiral.

Step 3: Determine the chirality of the molecule. If the molecule is chiral,

the next step is to determine if it is dextrorotatory or levorotatory. This can be

done by analyzing the arrangement of substituents around the chirality centers

A, B, and C.

Step 4: Analyze the substituents around chirality center A. Check the pri-

ority of the substituents around chirality center A based on the Cahn-Ingold-

Prelog rules. If the substituents are arranged in a clockwise direction, the chi-

rality center is R (rectus). If they are arranged in a counterclockwise direction,

the chirality center is S (sinister).

Step 5: Repeat the same process for chirality centers B and C. Check the

priority of the substituents around chirality centers B and C using the Cahn-

Ingold-Prelog rules. Determine if each center is R or S based on the arrangement

of substituents.

Step 6: Determine the overall chirality of the molecule. Once the chirality of

each center is determined, consider the overall arrangement of R and S centers

to classify the molecule as either dextrorotatory or levorotatory.

Step 7: Conclusion Based on the analysis of the chirality centers A, B,

and C, determine whether the molecule is chiral or achiral, and classify it as

dextrorotatory or levorotatory.

Question 34

Question

A compound with the molecular formula C6H14 exhibits optical activity. When

a sample of this compound is dissolved in an organic solvent, the solution rotates

plane-polarized light in a clockwise direction. Deduce the possible structures of

the compound and explain why it exhibits optical activity.

Solution

Step 1: Calculate the degree of unsaturation to determine the possible structures

of the compound. The degree of unsaturation (DU) can be calculated using the

formula:

DU = 2n+ 2 −X−H

where nis the number of carbon atoms in the molecular formula, Xis the

number of halogen atoms, and His the number of hydrogen atoms.

For C6H14 :

DU = 2×6+2−6−14

2= 0

Step 2: The absence of double bonds or rings suggests that the compound

is an alkane. Since the compound exhibits optical activity, it must be chiral. In

an alkane, chirality arises from the presence of asymmetrical carbon atoms. The

possible structural formula for the compound is therefore a chain of 6 carbon

atoms with one asymmetric (chiral) carbon atom.

Step 3: Define chirality in chemistry as the property of a molecule that is

not superimposable on its mirror image. In organic chemistry, chirality is often

exhibited by molecules with an asymmetric (chiral) carbon atom, which is a

carbon atom bonded to four different groups.

Step 4: Due to the presence of an asymmetric carbon atom, the compound

exists as a pair of enantiomers, which are non-superimposable mirror images of

each other. These enantiomers rotate plane-polarized light in equal but opposite

directions, causing the solution to exhibit optical activity.

Therefore, the compound with the molecular formula C6H14 that exhibits

optical activity is likely a chiral hexane compound with one asymmetric carbon

atom.

Question 35

Question

Draw the structure of a chiral molecule and explain why it is chiral. Additionally,

determine whether the molecule is optically active or inactive and justify your

answer.

Solution

Step 1: A chiral molecule is a molecule that cannot be superimposed on its

mirror image. Let’s consider the molecule (2-bromobutane) shown below:

H−C−C

| |

Structure: H H3C CH2Br

The second carbon atom in the molecule is attached to four different groups:

H, CH3, CH2Br. This asymmetric center makes the molecule chiral.

Step 2: To determine whether the molecule is optically active, we need to

check if the molecule has a plane of symmetry. If a molecule is achiral (has a

plane of symmetry), it will not exhibit optical activity.

In the case of (2-bromobutane), there is no plane of symmetry that can

divide the molecule into two equal halves that are mirror images of each other.

Therefore, the molecule is optically active.

Thus, the molecule (2-bromobutane) is chiral and optically active.

Question 2

Question

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

activity. Provide an example of a chiral molecule and explain why it is optically

active.

Solution

Step 1: Chirality in Organic Molecules A molecule is chiral if it does not contain

a plane of symmetry, meaning that it cannot be superimposed on its mirror

image. Chirality arises from having four different substituents (atoms or groups)

attached to a central carbon atom, creating a non-superimposable mirror image

pair called enantiomers.

Step 2: Optical Activity Optical activity refers to the ability of a chiral

compound to rotate the plane of polarized light. Enantiomers rotate plane-

polarized light in equal but opposite directions, a property known as optical

activity. This rotation is due to the asymmetry in the spatial arrangement of

the atoms or groups in the molecule.

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

Lactic acid (2-hydroxypropanoic acid), which exists in two enantiomeric forms:

L-lactic acid and D-lactic acid. These two forms are mirror images of each other

and are optically active.

Step 4: Explanation of Optical Activity in Lactic Acid Lactic acid is chiral

because the carbon atom bonded to the hydroxyl group (OH), hydrogen (H),

carboxyl group (COOH), and a methyl group (CH3) is chiral. Due to the

asymmetry in its arrangement, L-lactic acid and D-lactic acid are optically

active enantiomers, rotating plane-polarized light in opposite directions.

By understanding chirality in organic molecules and its relationship to opti-

cal activity, we can appreciate the significance of enantiomers in various chem-

ical, biological, and pharmaceutical contexts.

Question 3

Question

Explain why a molecule with a chiral center can exhibit optical activity, while

a molecule without a chiral center cannot.

Solution



A molecule is chiral if it is not superimposable on its mirror image. This

arises when a molecule contains at least one chiral center (also known as

a stereocenter).



A chiral center is an atom, usually carbon, that is bonded to four different

groups. This arrangement leads to non-superimposable mirror images

(enantiomers).



Enantiomers have identical physical properties except for their interaction

with plane-polarized light, a phenomenon called optical activity.

Step 1: Optical Activity of Molecules with Chiral Centers



When a molecule with a chiral center is placed in a beam of plane-polarized

light, the two enantiomers (mirror images) will rotate the plane of polar-

ized light in opposite directions.



One enantiomer rotates the plane to the right, termed dextrorotary (+),

while the other enantiomer rotates the plane to the left, termed levorotary

(-).



The extent of rotation is quantified using specific rotation values.



Hence, molecules with chiral centers exhibit optical activity due to the

presence of enantiomers that rotate plane-polarized light.

Step 2: Lack of Optical Activity in Molecules without Chiral Cen-

ters



Molecules without chiral centers, such as meso compounds or achiral

molecules, do not exhibit optical activity.



This is because they are superimposable on their mirror images and do

not have enantiomers that can rotate plane-polarized light.



Therefore, even if the molecule may contain polar bonds or be asymmetric

in structure, it will not be optically active if it lacks chiral centers.

In summary, the presence of a chiral center in a molecule leads to non-

superimposable mirror images (enantiomers) that exhibit optical activity by

rotating plane-polarized light. Molecules without chiral centers do not show

optical activity as they are superimposable on their mirror images.

Question 4

Question

Explain why a molecule with a chiral center can exhibit optical activity while a

molecule with an internal plane of symmetry cannot.