27.2

Chapter 27

Isomerism

27.1

Structural Isomerism

27.2

Stereoisomerism

1 New Way Chemistry for Hong Kong A-Level Book 3A

27.1 Structural Isomerism (SB p.60)

Isomers are different compounds that have the same molecular formula . They have different linkages or spatial arrangements of atoms .



Structural Isomers with the Same Functional Group

Chain Isomerism

Chain isomers are isomers that have different carbon skeletons .

e.g.



Position Isomerism

Position isomers are isomers that have the same carbon skeleton and functional group . They differ only in the position of the functional group .

e.g.



Metamerism

Metamers are those isomers with the functional group interrupting the carbon skeleton at different positions .

e.g.



Tautomerism

Tautomers are those isomers with structures differing in arrangement of atoms . They are in dynamic equilibrium with each other.

e.g.



Structural Isomers with Different Functional Groups

Functional Group Isomerism

Functional group isomers are isomers that have the same molecular formula but contain different functional groups .

e.g.



More examples of functional group isomers:



Example 27-1

Solution:

Draw the structural formulae for all the isomers with the molecular formula C

3

H

6.

O

Answer


10


Check Point 27-1

State whether the compounds having the following molecular formulae exhibit structural isomerism.

(a) CH

4

(b) C

3

H

6

(c) C

4

H

10

(d) C

2

H

6

O

(a) No

Answer

(b) Yes

(c) Yes

(d) Yes

New Way Chemistry for Hong Kong A-Level Book 3A

27.2 Stereoisomerism (SB p.65)

Geometrical Isomerism

Geometrical isomers are stereoisomers that have different arrangements of their atoms in space due to restricted rotation about a covalent bond .

e.g.



Rotation of a carbon atom of a C = C bond through an angle of 90 ° causes the breaking of the

 bond .



More examples of geometrical isomers:



Geometrical isomers have different physical and chemical properties .

Physical properties

Melting point ( ° C)

Boiling point ( ° C)

cis-But-2-ene

–139

4

trans-But-2-ene

–106

1

trans-But-2-ene has higher melting point

 more regular and symmetrical structure

 molecules pack more compactly in crystal lattice

 difficult to break the lattice

 higher melting point



cis-But-2-ene has higher boiling point

∵ net dipole moment

15

 molecules are held together by dipole-dipole interactions



trans-isomer has no net dipole moment , their molecules are held by instantaneous dipole-induced dipole interactions

 dipole-dipole interactions are stronger

 cis -but-2-ene has higher boiling point



Another example:

cis-butenedioic acid and trans-butenedioic acid

16

Physical properties

Melting point ( ° C)

Solubility in water

(gram of solution per 100 g of water at 25



C)

cis-Butenedioic acid

130

78.8


trans-Butenedioic acid

286

0.7


trans-butenedioic acid has higher melting point

∵ form more extensive intermolecular hydrogen bonding

17

While the cis -isomer forms intramolecular hydrogen bonding

 less extensive intermolecular hydrogen bonding

 molecules are less strongly held

 lower melting point



Although trans -butenedioic acid can form more extensive hydrogen bonds with water molecules, cis-butenedioic acid is more soluble in water than the trans -isomer because of the greater dipole moment .


27.2 Stereoisomerism (SB p.68) cis - and trans -butenedioic acids have different chemical properties



Example 27-2

1,2-Dichloroethene has two geometrical isomers with different melting points and boiling points. Give explanations for these differences.


Answer


Solution:

• trans -1,2-dichloroethene has a more regular and symmetrical structure than the cis-isomer

• Molecules of trans -1,2-dichloroethene can pack more compactly in the crystal lattice

• The crystal lattice of the trans -isomer is more difficult to break compared with that of the cis-isomer

• trans -1,2-dichloroethene has a higher melting point than cis -1,2-dichloroethene



Solution:

• cis -1,2-dichloroethene has a net dipole moment

• Molecules are held together by dipole-dipole interactions

• Molecules of trans -1,2-dichloroethene are held together by instantaneous dipole-induced dipole interactions

• Instantaneous dipole-induced dipole interactions are weaker than dipole-dipole interactions

• Less energy is required to separate the trans -1,2-dichloroethene molecules in the process of boiling.

• trans -1,2-dichloroethene has a lower boiling point than cis -1,2-dichloroethene .


23



State whether the following compounds exhibit geometrical isomerism.

(a) 1,2-Dibromoethene

(b) 1,1-Dibromopropene

(c) Ethene-1,2-diol

Answer

(a) Yes

(b) No

(c) Yes



Enantiomerism

• Enantiomerism occurs in those compounds whose molecules are chiral .

• A chiral molecule is one that is not superimposable with its mirror image .

• The chiral molecule and its mirror image are enantiomers .



Mirror

25

Mirror image of a left hand is a right hand

Not superimposable


27.2 Stereoisomerism (SB p.72) sp 3 -hybridized carbon atom with two or more identical groups attached is achiral and contains a plane of symmetry

26

Trichloromethane Propan-2-ol


27.2 Stereoisomerism (SB p.72) sp 3 -hybridized carbon atom with four different groups attached is chiral and do not contain a plane of symmetry e.g. butan-2-ol

Chiral centres


28


Example 27-3

Draw the structural formulae for the following compounds.

Determine whether the compounds are chiral or not. If they are, denote the chiral carbon atom by an asterisk (*). Draw the 3dimensional structure for both enantiomers showing their relationship.

(a) 2,3,3-Trichloropentane

(b) 3-Methylpentane

(c) 2-Phenylbutane

(d) trans -But-2-ene

Answer



Solution:

(a) 2,3,3-Trichloropentane is a chiral compound. The chiral carbon atom is shown by an asterisk.

Their enantiomers are:



Solution:

(b) There is no chiral carbon atom in the 3-methylpentane molecule. The compound has no enantiomers.

30

Note that carbon-3 is not a chiral centre because it has two identical groups ( –CH

2

CH

3

) bonded to it .



Solution:

(c) 2-Phenylbutane is a chiral compound. The chiral carbon atom is shown by an asterisk.

Their enantiomers are:



Solution:

(d) There is no chiral carbon atom in the trans -but-2-ene molecule. The compound has no enantiomers.



Properties of Enantiomers: Optical Activity

Enantiomers have identical physical properties

Physical properties (+) Butan-2-ol (–) Butan-2-ol

Boiling point ( ° C)

Density (g cm

–3 at 20 ° C)

99.5

0.808

99.5

0.808

Refractive index (at

20 ° C)

1.397

1.397

One easily observable difference of a pair of enantiomers is their properties towards plane-polarized light .



When a beam of plane-polarized light passes through a solution of an enantionmer , the plane of polarization rotates .

Two enantiomers that are mirror images of one another cause exactly the same extent of rotation , but in opposite directions , clockwise in one and anticlockwise in the other.

Enantiomers are also called optical isomers and said to be optically active .



Solution:

Example 27-4

There is a chiral carbon atom which is denoted by an asterisk

3-Methylpent-1-ene is found to be an optically active compound.

(*) in 3-methylpent-1-ene. Thus, the compound is optically

However, when the compound reacts with excess hydrogen, the product of the reaction is found to be optically inactive. Explain why.

is 3-methylpentane. The original chiral carbon atom becomes

Answer an achiral one as it links up with two identical groups of

( –CH

2

CH

3

). As the compound is symmetrical , it is optically inactive .


36



Draw the structural formula for the following compound and decide whether it will show optical activity.

(a) 1-Chloro-2-methylbutane

Answer

(a)

The carbon atom marked with an asterisk (*) is a chiral centre as it is attached to four different groups.

Therefore, the compound is optically active.


37




(b) 2-Chloro-2-methylbutane

Answer

(b)

This compound is optically inactive because there is no chiral centre.


38




(c) 1-Chloro-3-methylbutane

Answer

(c)

This compound is optically inactive because there is no chiral centre.


39




(d) 2-Chloro-3-methylbutane

(d)

Answer

The carbon atom marked with an asterisk (*) is a chiral centre as it is attached to four different groups. Therefore, the compound is optically active.



Plane-polarized Light

Light is an electromagnetic radiation .



Polarizer

41

Some possible planes of electrical oscillations in a beam of ordinary light.

Plane-polarized light



Polarimeter

Polarimeter is a device used for measuring the effect of optically active compounds on planepolarized light .



If the tube of polarimeter is empty, or if an optically inactive substance is present, the axes of the plane-polarized light and the analyzer will be exactly parallel .

If the tube contains an optically active substance, the planepolarized light will rotate as it passes through the tube.

If the analyzer is rotated in a clockwise direction, the rotation is said to be positive (+).

43

If the rotation is anticlockwise , the rotation is said to be negative (–).



The substance that rotates the plane-polarized light in clockwise direction is said to be dextrorotatory .

The substance that rotates the plane-polarized light in anticlockwise direction is said to be levorotatory .

44

Specific rotation is used to measure the rotations on a standard basis.

[



]

 where

 c

 

[



]

 specific rotation

  observed rotation c

 concentrat ion of the solution in g cm

3

  length of the tube in dm (1 dm



10 cm)



The specific rotation depends on the temperature and the wavelength of light that is employed.

A specific rotation might be given as follows:

[



]

25

D

 

3 .

12



D = D line of a sodium lamp that was used as light source

25 = temperature of 25 ° C was maintained

[



] = 1 g cm

–3 of the optically active substance was contained in a 1 dm tube

+3.12

° = rotation of 3.12

° in a clockwise direction


46



(a) (i)

(a) Bromination of propane (C

3

H

8

) yields two monosubstituted products with the same molecular formula, C

3

H

7

Br.

(i) Name the products and draw their structural formulae.

(ii) State the relationship between the two products.

(iii) Do you think the products are optically active?

Explain your answer.

(ii) They are structural isomers.

Answer

(iii) No, because they do not have chiral centres.


47


Check Point 27-4 product. A careful study shows that there are two forms of but-2-ene formed in the reaction.

(i) Name the products and draw their structural formulae.

(ii) State the relationship between the two products.

(iii) Do you think the products are optically active?

(iii) No, because they do not have chiral centres.

Answer




(c)

(c) Draw the structural formulae for all possible structural isomers of C

4

H

8

O

2 containing the –COO group.

Answer




(d) Draw the structural formulae for all possible geometrical isomers of hexa-2,4-diene.

(d)

Answer


The END


27.2

Chapter 27

Isomerism

Structural Isomerism

Stereoisomerism

Example 27-1

Example 27-2

Example 27-3

Example 27-4

The END

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27.2

Chapter 27

Isomerism

Structural Isomerism

Stereoisomerism

Example 27-1

Example 27-2

Example 27-3

Example 27-4

The END

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