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Transcript of New Way Chemistry for Hong Kong A-Level Book 3A1 Isomerism 27.1Structural Isomerism...
![Page 1: New Way Chemistry for Hong Kong A-Level Book 3A1 Isomerism 27.1Structural Isomerism 27.2Stereoisomerism Chapter 27.](https://reader033.fdocuments.net/reader033/viewer/2022061501/56649f065503460f94c1b883/html5/thumbnails/1.jpg)
New Way Chemistry for Hong Kong A-Level Book 3A
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Isomerism
27.127.1 Structural IsomerismStructural Isomerism
27.227.2 StereoisomerismStereoisomerism
Chapter 27Chapter 27
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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.
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27.1 Structural Isomerism (SB p.61)
Chain isomers are isomers that have different carbon skeletons.e.g.
Structural Isomers with the Same Functional GroupStructural Isomers with the Same Functional Group
Chain Isomerism
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27.1 Structural Isomerism (SB p.62)
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.
Position Isomerism
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27.1 Structural Isomerism (SB p.62)
Metamers are those isomers with the functional group interrupting the carbon skeleton at different positions.e.g.
Metamerism
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27.1 Structural Isomerism (SB p.63)
Tautomers are those isomers with structures differing in arrangement of atoms. They are in dynamic equilibrium with each other.e.g.
Tautomerism
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27.1 Structural Isomerism (SB p.63)
Functional group isomers are isomers that have the same molecular formula but contain different functional groups.e.g.
Structural Isomers with Different Functional GroupsStructural Isomers with Different Functional Groups
Functional Group Isomerism
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27.1 Structural Isomerism (SB p.63)
More examples of functional group isomers:
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Example 27-1Example 27-1Draw the structural formulae for all the isomers with the molecular formula C3H6.O
Answer
27.1 Structural Isomerism (SB p.64)
Solution:
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Check Point 27-1 Check Point 27-1
State whether the compounds having the following molecular formulae exhibit structural isomerism.
(a) CH4
(b) C3H6
(c) C4H10
(d) C2H6OAnswer(a) No
(b) Yes
(c) Yes
(d) Yes
27.1 Structural Isomerism (SB p.64)
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27.2 Stereoisomerism (SB p.65)
Geometrical isomers are stereoisomers that have different arrangements of their atoms in space due to restricted rotation about a covalent bond.e.g.
Geometrical IsomerismGeometrical Isomerism
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27.2 Stereoisomerism (SB p.66)
Rotation of a carbon atom of a C = C bond through an angle of 90° causes the breaking of the bond.
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27.2 Stereoisomerism (SB p.66)
More examples of geometrical isomers:
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27.2 Stereoisomerism (SB p.66)
Geometrical isomers have different physical and chemical properties.
Physical properties
cis-But-2-ene trans-But-2-ene
Melting point (°C)
–139 –106
Boiling point (°C) 4 1trans-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
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27.2 Stereoisomerism (SB p.67)
cis-But-2-ene has higher boiling point ∵ net dipole moment
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
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27.2 Stereoisomerism (SB p.67)
Another example:
cis-butenedioic acid and trans-butenedioic acid
Physical properties
cis-Butenedioic acid
trans-Butenedioic acid
Melting point (°C) 130 286
Solubility in water (gram of solution per 100 g of water at 25 C)
78.8 0.7
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27.2 Stereoisomerism (SB p.68)
trans-butenedioic acid has higher melting point
form ∵ more extensive intermolecular hydrogen bonding
While the cis-isomer forms intramolecular hydrogen bonding less extensive intermolecular hydrogen bonding molecules are less strongly held lower melting point
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27.2 Stereoisomerism (SB p.68)
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.
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27.2 Stereoisomerism (SB p.68)
cis- and trans-butenedioic acids have different chemical properties
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Example 27-2Example 27-21,2-Dichloroethene has two geometrical isomers with different melting points and boiling points. Give explanations for these differences.
Answer
27.2 Stereoisomerism (SB p.69)
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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
27.2 Stereoisomerism (SB p.69)
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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.
27.2 Stereoisomerism (SB p.70)
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Check Point 27-2 Check Point 27-2
State whether the following compounds exhibit geometrical isomerism.
(a) 1,2-Dibromoethene
(b) 1,1-Dibromopropene
(c) Ethene-1,2-diolAnswer
(a) Yes
(b) No
(c) Yes
27.2 Stereoisomerism (SB p.70)
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27.2 Stereoisomerism (SB p.70)
• 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.
EnantiomerismEnantiomerism
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27.2 Stereoisomerism (SB p.71)
Not superimposable
Mirror
Mirror image of a left hand is a right hand
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27.2 Stereoisomerism (SB p.72)
sp3-hybridized carbon atom with two or more identical groups attached is achiral and contains a plane of symmetry
Trichloromethane Propan-2-ol
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27.2 Stereoisomerism (SB p.72)
sp3-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
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Example 27-3Example 27-3Draw 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 3-dimensional structure for both enantiomers showing their relationship.
(a) 2,3,3-Trichloropentane
(b) 3-Methylpentane
(c) 2-Phenylbutane
(d) trans-But-2-ene
Answer
27.2 Stereoisomerism (SB p.73)
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Solution:
(a) 2,3,3-Trichloropentane is a chiral compound. The chiral carbon atom is shown by an asterisk.
Their enantiomers are:
27.2 Stereoisomerism (SB p.73)
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Solution:
(b) There is no chiral carbon atom in the 3-methylpentane molecule. The compound has no enantiomers.
Note that carbon-3 is not a chiral centre because it has two identical groups (–CH2CH3) bonded to it.
27.2 Stereoisomerism (SB p.74)
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Solution:
(c) 2-Phenylbutane is a chiral compound. The chiral carbon atom is shown by an asterisk.
Their enantiomers are:
27.2 Stereoisomerism (SB p.74)
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Solution:
(d) There is no chiral carbon atom in the trans-but-2-ene molecule. The compound has no enantiomers.
27.2 Stereoisomerism (SB p.74)
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27.2 Stereoisomerism (SB p.74)
Enantiomers have identical physical properties
Properties of Enantiomers: Optical Activity
Physical properties (+) Butan-2-ol (–) Butan-2-ol
Boiling point (°C) 99.5 99.5
Density (g cm–3 at 20°C)
0.808 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.
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27.2 Stereoisomerism (SB p.74)
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.
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Example 27-4Example 27-43-Methylpent-1-ene is found to be an optically active compound. However, when the compound reacts with excess hydrogen, the product of the reaction is found to be optically inactive. Explain why.
Answer
27.2 Stereoisomerism (SB p.75)
Solution:
There is a chiral carbon atom which is denoted by an asterisk (*) in 3-methylpent-1-ene. Thus, the compound is optically active.
When it reacts with excess hydrogen, the product of the reaction is 3-methylpentane. The original chiral carbon atom becomes an achiral one as it links up with two identical groups of (–CH2CH3). As the compound is symmetrical, it is optically ina
ctive.
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Check Point 27-3 Check Point 27-3
Draw the structural formula for the following compound and decide whether it will show optical activity.
(a) 1-Chloro-2-methylbutane
Answer
27.2 Stereoisomerism (SB p.76)
(a)
The carbon atom marked with an asterisk (*) is a chiral cen
tre as it is attached to four different groups.
Therefore, the compound is optically active.
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Check Point 27-3 Check Point 27-3
Draw the structural formula for the following compound and decide whether it will show optical activity.
(b) 2-Chloro-2-methylbutaneAnswer
27.2 Stereoisomerism (SB p.76)
(b)
This compound is optically inactive be
cause there is no chiral centre.
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Check Point 27-3 Check Point 27-3
Draw the structural formula for the following compound and decide whether it will show optical activity.
(c) 1-Chloro-3-methylbutane
Answer
27.2 Stereoisomerism (SB p.76)
(c)
This compound is optically inactive
because there is no chiral centre.
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Check Point 27-3 Check Point 27-3
Draw the structural formula for the following compound and decide whether it will show optical activity.
(d) 2-Chloro-3-methylbutaneAnswer
27.2 Stereoisomerism (SB p.76)
(d)
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.
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27.2 Stereoisomerism (SB p.76)
Light is an electromagnetic radiation.
Plane-polarized Light
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27.2 Stereoisomerism (SB p.76)
Some possible planes of electrical oscillations in a beam of ordinary light.
Polarizer
Plane-polarized light
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27.2 Stereoisomerism (SB p.77)
Polarimeter is a device used for measuring the effect of optically active compounds on plane-polarized light.
Polarimeter
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27.2 Stereoisomerism (SB p.78)
If the tube of polarimeter is empty, or if an optically inactive s
ubstance is present, the axes of the plane-polarized light and t
he analyzer will be exactly parallel.
If the tube contains an optically active substance, the plane-
polarized 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 (+).
If the rotation is anticlockwise, the rotation is said to be
negative (–).
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27.2 Stereoisomerism (SB p.78)
The substance that rotates the plane-polarized light in cl
ockwise direction is said to be dextrorotatory.
cm) 10 dm (1 dmin tube theoflength
cm gin solution theofion concentrat c
rotation observed
rotation specific ][ where c
][
3-
The substance that rotates the plane-polarized light in anticloc
kwise direction is said to be levorotatory.
Specific rotation is used to measure the rotations on a standar
d basis.
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27.2 Stereoisomerism (SB p.78)
The specific rotation depends on the temperature and the
wavelength of light that is employed.
A specific rotation might be given as follows:12.3][ 25
D
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
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New Way Chemistry for Hong Kong A-Level Book 3A
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Check Point 27-4 Check Point 27-4
(a) Bromination of propane (C3H8) yields two monosubstituted products with the same molecular formula, C3H7Br.
(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.
Answer
27.2 Stereoisomerism (SB p.78)
(a) (i)
(ii) They are structural isomers.
(iii) No, because they do not have chiral
centres.
![Page 47: New Way Chemistry for Hong Kong A-Level Book 3A1 Isomerism 27.1Structural Isomerism 27.2Stereoisomerism Chapter 27.](https://reader033.fdocuments.net/reader033/viewer/2022061501/56649f065503460f94c1b883/html5/thumbnails/47.jpg)
New Way Chemistry for Hong Kong A-Level Book 3A
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Check Point 27-4 Check Point 27-4
(b) Dehydration of butan-2-ol gives but-2-ene as the major 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? Explain your answer.
Answer
27.2 Stereoisomerism (SB p.78)
(b) (i)
(ii) They are geometrical isomers.
(iii) No, because they do not have chiral
centres.
![Page 48: New Way Chemistry for Hong Kong A-Level Book 3A1 Isomerism 27.1Structural Isomerism 27.2Stereoisomerism Chapter 27.](https://reader033.fdocuments.net/reader033/viewer/2022061501/56649f065503460f94c1b883/html5/thumbnails/48.jpg)
New Way Chemistry for Hong Kong A-Level Book 3A
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Check Point 27-4 Check Point 27-4
(c) Draw the structural formulae for all possible structural isomers of C4H8O2 containing the –COO group.
Answer
27.2 Stereoisomerism (SB p.78)
(c)
![Page 49: New Way Chemistry for Hong Kong A-Level Book 3A1 Isomerism 27.1Structural Isomerism 27.2Stereoisomerism Chapter 27.](https://reader033.fdocuments.net/reader033/viewer/2022061501/56649f065503460f94c1b883/html5/thumbnails/49.jpg)
New Way Chemistry for Hong Kong A-Level Book 3A
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Check Point 27-4 Check Point 27-4
(d) Draw the structural formulae for all possible geometrical isomers of hexa-2,4-diene.
Answer
27.2 Stereoisomerism (SB p.78)
(d)
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The END