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Organic Chemistry:
The study of carbon-containing
compounds and their properties.
Organic chemistry is the chemistry of carbon
containing compounds.
Homologous series:
 A homologous series is a group of organic
compounds having similar chemical
properties and a gradation of physical
properties..
 They have a common generic formula.
 Neighboring members differ by –CH2-
 eg alkanes, alkenes, alkynes
10.1.1 Describe the features of a homologous series
Are these molecules part of a
homologous group?
Are these molecules part of a
homologous group?
Yes - They have a common
generic formula.
- Neighboring members differ
by –CH2-
 Which statement about neighbouring
members of all homologous series is correct?
A. They have the same empirical formula.
B. They differ by a CH2 group.
C. They possess different functional groups.
D. They differ in their degree of unsaturation.
 Which statement about neighbouring
members of all homologous series is correct?
A. They have the same empirical formula.
B. They differ by a CH2 group.
C. They possess different functional groups.
D. They differ in their degree of unsaturation.
Hydrocarbons
. . . compounds composed of carbon and
hydrogen.
Saturated: carbon-carbon bonds are all
single bonds - alkanes [CnH2n+2]
H C
H
H
C
H
H
H
Ethane
The alkanes are a homologous series of saturated
hydrocarbons.
State the meaning of each of the following terms.
 (i) homologous series
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
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 (ii) hydrocarbon
.....................................................................................................................................
.....................................................................................................................................
 (iii) nomenclature
.....................................................................................................................................
.....................................................................................................................................
Physical properties of Alkanes:
10.1.2 Predict and explain the trends in boiling points of members of a
homologous series
Alkanes Boiling
Point (O
C)
State
Methane CH4 -164
Gases
at RTEthane C2H6 -89
Propane C3H8 -42
Butane C4H10 -0.5
Pentane C5H12 36
Liquids
at RTHexane C6H14 69
Heptane C7H16 98
Octane C8H18 125
Boiling point increases with increasing
carbon numbers (increasing chain
length) due to increasing temporary
dipoles causing stronger van der
Waals’ forces between the molecules
as their size increases
Increase is not linear due to the
increase in chain length being
proportionally greater in smaller
molecules
Other physical properties that show
predictable trends with increasing chain
length are density and viscosity
Physical properties of Alkanes:
1. Name each compound
2. Use the idea of Van der Waals forces
to predict the comparative boiling
points and melting points of these
compounds.
A
D
CB
Question
State and explain the trend in the boiling points of
the first five alkanes.
 boiling point increases as number of carbons
increases
increased surface area / greater Van der Waals’
forces / increased Mr
increased intermolecular forces
Examiners report
 generally done well, although boiling points were
sometimes explained in terms of an increase in the
number of bonds with no distinction between
intermolecular and intramolecular forces.
Which compound has the lowest boiling point?
A. CH3COOH
B. (CH3)4C
C. CH3CH2CH2CH2CH3
D. CH3CH2CH2CH2OH
Which compound has the lowest boiling point?
A. CH3COOH
B. (CH3)4C
C. CH3CH2CH2CH2CH3
D. CH3CH2CH2CH2OH
Empirical, molecular and structural
formula
Ethane
Empirical formula CH3
Molecular formula C2H6
Full structural formula
H H
H-C-C-H
H H
Condensed structural formula CH3CH3
10.1.3 Distinguish between empirical, molecular and structural formulas
Representing organic compounds.
 An empirical formula gives the lowest whole number ratio of atoms
present in a compound eg: CH3
 A molecular formula gives the actual number of atoms present in one
molecule of the substance eg: C2H6
 A structural formula shows how the atoms are arranged in the
molecule.
1. A full structural formula displays every atom and bond eg:
2. A condensed structural formula omits bonds and can show identical
groups bracketed together eg:
CH3CH3 OR CH3(CH2)2CH3
( abbreviations such as R and can be used)
3. A skeleton formula for more complex structures can also be used.
( Do NOT use these in your exam answers) eg:
Would this represent C2H6 OR C4H10????
Summary of structural formula:
Which type of formula is used in each of the following:
CH2
C2H6
1. 2.
7.
4. 5. 6.
3.
Nomenclature:
Many organic compounds are known by a variety of names, and
many of these names give no indication of the nature of the
molecule.
e.g. for compounds from nature, the names may indicate the
source:
""-pinene" from pine trees or "citral" from citrus trees ...
These names are described as "Common" or "trivial" names (they
are not related to the chemical structure) – there are no rules
governing these names
A more systematic method was developed:
 Unique name for each organic compound
 Structure can be derived from name
 RULES are used to derive these names
10.1.10 Apply IUPAC rules for naming compounds containing up to 6 carbon
atoms with one of the following functional groups: alcohol, aldehyde,
ketone, carboxylic acid and halide
Rules for Naming Alkanes(CnH2n+2)
1. For alkanes beyond butane, add -ane to
the Greek root for the number of carbons.
C-C-C-C-C-C = hexane
2. Alkyl substituents (ie groups joined onto
another group): drop the -ane and add -yl.
-C2H5 is ethyl
Rules for Naming Alkanes (cont)
3. Positions of substituent groups are
specified by numbering the longest chain
sequentially( from the end which produces
the smallest numbers!!).
C
|
C-C-C-C-C-C
3-methylhexane(not 4-methylhexane)
4. Location and name are followed by root
alkane name. Substituents are given in
alphabetical order with the use of di-, tri-, etc.
Given that the suffix –ol indicates the presence of an –OH group, draw
the structural diagram for the compound named above.
 What is the IUPAC name for
CH3CH2CH(CH3)2?
 (Hint: draw the structure first!!)
A. 1,1-dimethylpropane
B. 2-methylbutane
C. isopentane
D. ethyldimethylmethane
 What is the IUPAC name for
CH3CH2CH(CH3)2?
 (Hint: draw the structure first!!)
A. 1,1-dimethylpropane
B. 2-methylbutane
C. isopentane
D. ethyldimethylmethane
 Which compound is a member of the same
homologous series as 1-chloropropane?
(remember – they must differ by –CH2- units)
A. 1-chloropropene
B. 1-chlorobutane
C. 1-bromopropane
D. 1,1-dichloropropane
 Which compound is a member of the same
homologous series as 1-chloropropane?
(remember – they must differ by –CH2- units)
A. 1-chloropropene
B. 1-chlorobutane
C. 1-bromopropane
D. 1,1-dichloropropane
 Which formula is a correct representation of
pentane?
A. CH3CH2CHCH2CH3
B. (CH3CH2)2CH3
C. CH3(CH2)3CH3
D. CH3(CH3)3CH3
 Which formula is a correct representation of
pentane?
A. CH3CH2CHCH2CH3
B. (CH3CH2)2CH3
C. CH3(CH2)3CH3
D. CH3(CH3)3CH3
Extra rules:
And finally:
What are isomers?:
10.1.4 Describe structural isomers as compounds with the same molecular
formula but different arrangements of atoms
10.1.5 Deduce structural formulas for the isomers of the non-cyclic alkanes up
to C6
10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up
to C6
10.1.7 Deduce structural formulas for the isomers of the straight chain
alkanes up to C6
10.1.8 Apply IUPAC rules for naming the isomers of the straight chain alkanes
up to C6
What are isomers?:
 Isomers are compounds with the same
molecular formula but different arrangements
of atoms.
 They will have the same molecular formula
but different structural formulae.
10.1.4 Describe structural isomers as compounds with the same molecular
formula but different arrangements of atoms
10.1.5 Deduce structural formulas for the isomers of the non-cyclic alkanes up
to C6
Suggest reasons for the differing boiling points?
Which formulas represent butane or its isomer?
I. CH3(CH2)2CH3
II. CH3CH(CH3)CH3
III (CH3)3CH
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
Which formulas represent butane or its isomer?
I. CH3(CH2)2CH3
II. CH3CH(CH3)CH3
III (CH3)3CH
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
 They will have different IUPAC names.
Draw and name all of the isomers of C6H14
NB – remember all of the four bonds of a C atom are equivalent!
10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up
to C6
Cyclic Alkanes
Carbon atoms can form rings
containing only carbon-carbon single
bonds.
C3H6, C4H8, C6H12
Do cyclic alkanes have the same generic formula as straight
chain alkanes. Explain your answer.
Isomers of alkenes
 A different type of structural isomer is formed
when the double bond is formed in different
positions
 C4H8 has the following straight chain isomers
10.1.7 Deduce structural formulas for the isomers of the straight chain
alkenes up to C6
But-1-ene trans-but-2-ene cis-but-2-ene
10.1.8 Apply IUPAC rules for
naming the isomers
of the straight chain
alkenes up to C6
Note – molecules are
named using the smallest
numbered carbon that is
part of the double bond
Volatility and Solubility
 Structure of organic compounds can be
thought of in terms of:
1. a hydrocarbon skeleton
2. a functional group
 Both influence the physical properties of an
organic compound, such as volatility and solubility
10.1.13 Discuss the volatility and solubility in water of compounds containing
the functional groups: alcohol, aldehyde, ketone, carboxylic acid and
halide
Volatility
 Is a measure of how easily a substance
changes into the gaseous state
 The higher the volatility the lower the BP.
(remember that BP / volatility depends on
over coming the forces between the
molecules – topic 4)
 The stronger the intermolecular forces, the
higher the BP
Influence of the hydrocarbon skeleton
 The larger the molecule the stronger the van der
Waals’ (intermolecular) forces between the
molecules
 Increasing the number of carbons in the
hydrocarbon skeleton increases the BP /
decreases volatility
 Lower members of a homologous series are
usually gases or liquids at RT. Higher members
are mostly solids
Influence of the hydrocarbon skeleton
 Branching decreases the ability of the molecules
to align (line up next to each other) thus increases
the distance between the molecules and
decreases the intermolecular forces between
them
 Increased branching, increases volatility /
decreases BP
 Foe example
 Butane BP = - 0.5 o
C
 2-methylpropane BP = - 11.7o
C (branched isomer of
butane)
Influence of the functional group
 In addition to van der Waals’ forces,
functional groups that are polar contribute to
intermolecular forces and thus increase BP /
decrease volatility
 Polar groups develop dipole-dipole
interactions between molecules, which
increase BP
 Groups that form hydrogen bonds form even
stronger intermolecular forces between
molecules
Summary - volatility
 When comparing BP in different homologous
series, it’s important to compare molecules
with similar molecular masses. Which means
that the number of carbon is often different
For example
ethanol (C2H5OH), Mr = 46, BP = 78o
C
propane (C3H8), Mr = 44, BP = - 42o
C
Why does ethanol have a higher BP?
Summary - volatility
Most volatile least volatile
Alkane halogenoalkanes aldehyde ketone˃ ˃ ˃ ˃
alcohol carboxylic acid˃
Van der Waals’ dipole-dipole hydrogen bonding
Increasing strength of molecular attraction
Increasing boiling point
Solubility in water
 Largely determined by the extent to which the
solute molecules can interact to form
hydrogen bonds with water
Influence of the hydrocarbon skeleton
 Non-polar, therefore unable to form hydrogen
bonds and thus does not contribute to the
solubility of the molecule
 Therefore, higher members of a homologous
series are less soluble that lower members
Solubility in water
Influence of the functional group
Molecules with functional groups that can form
hydrogen bonds with water are soluble.
Alcohol, carboxylic acid and amines are
soluble in water.
Aldehydes, ketones, amides and esters are
less soluble
halogenoalkanes., alkanes and alkenes are
insoluble

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10.1 Introduction to Organic Chemistry

  • 1. Organic Chemistry: The study of carbon-containing compounds and their properties.
  • 2. Organic chemistry is the chemistry of carbon containing compounds.
  • 3. Homologous series:  A homologous series is a group of organic compounds having similar chemical properties and a gradation of physical properties..  They have a common generic formula.  Neighboring members differ by –CH2-  eg alkanes, alkenes, alkynes 10.1.1 Describe the features of a homologous series
  • 4. Are these molecules part of a homologous group?
  • 5. Are these molecules part of a homologous group? Yes - They have a common generic formula. - Neighboring members differ by –CH2-
  • 6.  Which statement about neighbouring members of all homologous series is correct? A. They have the same empirical formula. B. They differ by a CH2 group. C. They possess different functional groups. D. They differ in their degree of unsaturation.
  • 7.  Which statement about neighbouring members of all homologous series is correct? A. They have the same empirical formula. B. They differ by a CH2 group. C. They possess different functional groups. D. They differ in their degree of unsaturation.
  • 8. Hydrocarbons . . . compounds composed of carbon and hydrogen. Saturated: carbon-carbon bonds are all single bonds - alkanes [CnH2n+2] H C H H C H H H Ethane
  • 9.
  • 10. The alkanes are a homologous series of saturated hydrocarbons. State the meaning of each of the following terms.  (i) homologous series ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... .....................................................................................................................................  (ii) hydrocarbon ..................................................................................................................................... .....................................................................................................................................  (iii) nomenclature ..................................................................................................................................... .....................................................................................................................................
  • 11. Physical properties of Alkanes: 10.1.2 Predict and explain the trends in boiling points of members of a homologous series Alkanes Boiling Point (O C) State Methane CH4 -164 Gases at RTEthane C2H6 -89 Propane C3H8 -42 Butane C4H10 -0.5 Pentane C5H12 36 Liquids at RTHexane C6H14 69 Heptane C7H16 98 Octane C8H18 125 Boiling point increases with increasing carbon numbers (increasing chain length) due to increasing temporary dipoles causing stronger van der Waals’ forces between the molecules as their size increases Increase is not linear due to the increase in chain length being proportionally greater in smaller molecules Other physical properties that show predictable trends with increasing chain length are density and viscosity
  • 12. Physical properties of Alkanes: 1. Name each compound 2. Use the idea of Van der Waals forces to predict the comparative boiling points and melting points of these compounds. A D CB
  • 13. Question State and explain the trend in the boiling points of the first five alkanes.
  • 14.  boiling point increases as number of carbons increases increased surface area / greater Van der Waals’ forces / increased Mr increased intermolecular forces Examiners report  generally done well, although boiling points were sometimes explained in terms of an increase in the number of bonds with no distinction between intermolecular and intramolecular forces.
  • 15. Which compound has the lowest boiling point? A. CH3COOH B. (CH3)4C C. CH3CH2CH2CH2CH3 D. CH3CH2CH2CH2OH
  • 16. Which compound has the lowest boiling point? A. CH3COOH B. (CH3)4C C. CH3CH2CH2CH2CH3 D. CH3CH2CH2CH2OH
  • 17. Empirical, molecular and structural formula Ethane Empirical formula CH3 Molecular formula C2H6 Full structural formula H H H-C-C-H H H Condensed structural formula CH3CH3 10.1.3 Distinguish between empirical, molecular and structural formulas
  • 18. Representing organic compounds.  An empirical formula gives the lowest whole number ratio of atoms present in a compound eg: CH3  A molecular formula gives the actual number of atoms present in one molecule of the substance eg: C2H6  A structural formula shows how the atoms are arranged in the molecule. 1. A full structural formula displays every atom and bond eg: 2. A condensed structural formula omits bonds and can show identical groups bracketed together eg: CH3CH3 OR CH3(CH2)2CH3 ( abbreviations such as R and can be used) 3. A skeleton formula for more complex structures can also be used. ( Do NOT use these in your exam answers) eg: Would this represent C2H6 OR C4H10????
  • 20. Which type of formula is used in each of the following: CH2 C2H6 1. 2. 7. 4. 5. 6. 3.
  • 21. Nomenclature: Many organic compounds are known by a variety of names, and many of these names give no indication of the nature of the molecule. e.g. for compounds from nature, the names may indicate the source: ""-pinene" from pine trees or "citral" from citrus trees ... These names are described as "Common" or "trivial" names (they are not related to the chemical structure) – there are no rules governing these names A more systematic method was developed:  Unique name for each organic compound  Structure can be derived from name  RULES are used to derive these names 10.1.10 Apply IUPAC rules for naming compounds containing up to 6 carbon atoms with one of the following functional groups: alcohol, aldehyde, ketone, carboxylic acid and halide
  • 22. Rules for Naming Alkanes(CnH2n+2) 1. For alkanes beyond butane, add -ane to the Greek root for the number of carbons. C-C-C-C-C-C = hexane 2. Alkyl substituents (ie groups joined onto another group): drop the -ane and add -yl. -C2H5 is ethyl
  • 23. Rules for Naming Alkanes (cont) 3. Positions of substituent groups are specified by numbering the longest chain sequentially( from the end which produces the smallest numbers!!). C | C-C-C-C-C-C 3-methylhexane(not 4-methylhexane) 4. Location and name are followed by root alkane name. Substituents are given in alphabetical order with the use of di-, tri-, etc.
  • 24. Given that the suffix –ol indicates the presence of an –OH group, draw the structural diagram for the compound named above.
  • 25.  What is the IUPAC name for CH3CH2CH(CH3)2?  (Hint: draw the structure first!!) A. 1,1-dimethylpropane B. 2-methylbutane C. isopentane D. ethyldimethylmethane
  • 26.  What is the IUPAC name for CH3CH2CH(CH3)2?  (Hint: draw the structure first!!) A. 1,1-dimethylpropane B. 2-methylbutane C. isopentane D. ethyldimethylmethane
  • 27.  Which compound is a member of the same homologous series as 1-chloropropane? (remember – they must differ by –CH2- units) A. 1-chloropropene B. 1-chlorobutane C. 1-bromopropane D. 1,1-dichloropropane
  • 28.  Which compound is a member of the same homologous series as 1-chloropropane? (remember – they must differ by –CH2- units) A. 1-chloropropene B. 1-chlorobutane C. 1-bromopropane D. 1,1-dichloropropane
  • 29.  Which formula is a correct representation of pentane? A. CH3CH2CHCH2CH3 B. (CH3CH2)2CH3 C. CH3(CH2)3CH3 D. CH3(CH3)3CH3
  • 30.  Which formula is a correct representation of pentane? A. CH3CH2CHCH2CH3 B. (CH3CH2)2CH3 C. CH3(CH2)3CH3 D. CH3(CH3)3CH3
  • 33.
  • 34.
  • 35. What are isomers?: 10.1.4 Describe structural isomers as compounds with the same molecular formula but different arrangements of atoms 10.1.5 Deduce structural formulas for the isomers of the non-cyclic alkanes up to C6 10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up to C6 10.1.7 Deduce structural formulas for the isomers of the straight chain alkanes up to C6 10.1.8 Apply IUPAC rules for naming the isomers of the straight chain alkanes up to C6
  • 36. What are isomers?:  Isomers are compounds with the same molecular formula but different arrangements of atoms.  They will have the same molecular formula but different structural formulae. 10.1.4 Describe structural isomers as compounds with the same molecular formula but different arrangements of atoms 10.1.5 Deduce structural formulas for the isomers of the non-cyclic alkanes up to C6
  • 37. Suggest reasons for the differing boiling points?
  • 38. Which formulas represent butane or its isomer? I. CH3(CH2)2CH3 II. CH3CH(CH3)CH3 III (CH3)3CH A. I and II only B. I and III only C. II and III only D. I, II and III
  • 39. Which formulas represent butane or its isomer? I. CH3(CH2)2CH3 II. CH3CH(CH3)CH3 III (CH3)3CH A. I and II only B. I and III only C. II and III only D. I, II and III
  • 40.  They will have different IUPAC names. Draw and name all of the isomers of C6H14 NB – remember all of the four bonds of a C atom are equivalent! 10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up to C6
  • 41.
  • 42. Cyclic Alkanes Carbon atoms can form rings containing only carbon-carbon single bonds. C3H6, C4H8, C6H12 Do cyclic alkanes have the same generic formula as straight chain alkanes. Explain your answer.
  • 43. Isomers of alkenes  A different type of structural isomer is formed when the double bond is formed in different positions  C4H8 has the following straight chain isomers 10.1.7 Deduce structural formulas for the isomers of the straight chain alkenes up to C6 But-1-ene trans-but-2-ene cis-but-2-ene
  • 44. 10.1.8 Apply IUPAC rules for naming the isomers of the straight chain alkenes up to C6 Note – molecules are named using the smallest numbered carbon that is part of the double bond
  • 45. Volatility and Solubility  Structure of organic compounds can be thought of in terms of: 1. a hydrocarbon skeleton 2. a functional group  Both influence the physical properties of an organic compound, such as volatility and solubility 10.1.13 Discuss the volatility and solubility in water of compounds containing the functional groups: alcohol, aldehyde, ketone, carboxylic acid and halide
  • 46. Volatility  Is a measure of how easily a substance changes into the gaseous state  The higher the volatility the lower the BP. (remember that BP / volatility depends on over coming the forces between the molecules – topic 4)  The stronger the intermolecular forces, the higher the BP
  • 47. Influence of the hydrocarbon skeleton  The larger the molecule the stronger the van der Waals’ (intermolecular) forces between the molecules  Increasing the number of carbons in the hydrocarbon skeleton increases the BP / decreases volatility  Lower members of a homologous series are usually gases or liquids at RT. Higher members are mostly solids
  • 48. Influence of the hydrocarbon skeleton  Branching decreases the ability of the molecules to align (line up next to each other) thus increases the distance between the molecules and decreases the intermolecular forces between them  Increased branching, increases volatility / decreases BP  Foe example  Butane BP = - 0.5 o C  2-methylpropane BP = - 11.7o C (branched isomer of butane)
  • 49. Influence of the functional group  In addition to van der Waals’ forces, functional groups that are polar contribute to intermolecular forces and thus increase BP / decrease volatility  Polar groups develop dipole-dipole interactions between molecules, which increase BP  Groups that form hydrogen bonds form even stronger intermolecular forces between molecules
  • 50. Summary - volatility  When comparing BP in different homologous series, it’s important to compare molecules with similar molecular masses. Which means that the number of carbon is often different For example ethanol (C2H5OH), Mr = 46, BP = 78o C propane (C3H8), Mr = 44, BP = - 42o C Why does ethanol have a higher BP?
  • 51. Summary - volatility Most volatile least volatile Alkane halogenoalkanes aldehyde ketone˃ ˃ ˃ ˃ alcohol carboxylic acid˃ Van der Waals’ dipole-dipole hydrogen bonding Increasing strength of molecular attraction Increasing boiling point
  • 52. Solubility in water  Largely determined by the extent to which the solute molecules can interact to form hydrogen bonds with water Influence of the hydrocarbon skeleton  Non-polar, therefore unable to form hydrogen bonds and thus does not contribute to the solubility of the molecule  Therefore, higher members of a homologous series are less soluble that lower members
  • 53. Solubility in water Influence of the functional group Molecules with functional groups that can form hydrogen bonds with water are soluble. Alcohol, carboxylic acid and amines are soluble in water. Aldehydes, ketones, amides and esters are less soluble halogenoalkanes., alkanes and alkenes are insoluble