1. Chapter 12 Unsaturated Hydrocarbons
12.1
Alkenes and Alkynes
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2. Unsaturated Hydrocarbons
Have fewer hydrogen atoms attached to the carbon chain than alkanes.
Are alkenes with double bonds.
Are alkynes with triple bonds.
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3. Bond Angles in Alkenes and Alkynes
According to VSEPR theory:
Three groups in a double bond are bonded at 120° angles.
Alkenes are flat, because the atoms in a double bond lie in the same plane. (trigonal planar)
The groups attached to a triple bond are at 180° angles. (linear)
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4. Naming Alkenes The names of alkenes Use the corresponding alkane name.
Change the ending to ­ene.
Alkene IUPAC Common
H2C=CH2 ethene ethylene
H2C=CH─CH3 propene propylene
cyclohexene

5. Ethene (ethylene) Ethene or ethylene Is an alkene C2H4.
Has two carbon atoms connected by a double bond.
Has two H atoms bonded to each C atom.
Is flat with all the C and H atoms in the same plane.
Is used to accelerate the ripening of fruits. Used to make polyethylene
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6. Naming Alkynes The names of alkynes Use the corresponding alkane name.
Change the ending to ­yne.
Alkyne IUPAC Common
HC≡CH ethyne acetylene
HC≡C─CH3 propyne

7. Guide to Naming Alkenes and Alkynes
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8. Naming Alkenes Write the IUPAC name for CH2=CH─CH2─CH3 and
STEP 1 Name the longest carbon chain
butene
STEP 2 Number the chain from the double bond
CH2=CH─CH2─CH3 1-butene
CH3─CH=CH─CH3 2-butene

9. Comparing Names of Alkanes, Alkenes, and Alkynes
TABLE 12.1
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10. Naming Alkenes with Substituents
CH3 │
Write the IUPAC name for CH3─CH─CH=CH─CH3
STEP 1 Name the longest carbon chain pentene
STEP 2 Number the chain from the double bond
CH3─CH─CH=CH─CH pentene
STEP 3 Give the location of each substituent
4-methyl- 2-pentene

11. Naming Alkynes with Substituents
CH3 │
Write the IUPAC name for HC≡C─CH─CH3
STEP 1 Name the longest carbon chain butyne
STEP 2 Number the chain from the double bond
1-butyne
STEP 3 Give the location of each substituent
3-methyl-1-butyne

12. Learning Check Write the IUPAC name for each of the following:
1. CH2=CH─CH2─CH3
2. CH3─CH=CH─CH3
CH3 |
3. CH3─CH=C─CH3
4. CH3─CC─CH3

13. Solution Write the IUPAC name for each of the following:
1. CH2=CH─CH2─CH butene
2. CH3─CH=CH─CH butene
CH3 |
3. CH3─CH=C─CH methyl-2-butene
4. CH3─CC─CH butyne

14. Learning Check Write the structural formula for each of the following:
A. 2-pentyne
B. 3-methyl-2-pentene

15. Solution Write the structural formula for each of the following:
A. CH3─CH2─C≡C─CH pentyne
CH3 
B. CH3─CH2─C=CH─CH3 3-methyl-2-pentene

16. Chapter 12 Unsaturated Hydrocarbons
12.2
Cis-Trans Isomers
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17. Cis and Trans Isomers In an alkene, cis and trans isomers are possible
because the double bond
Is rigid.
Cannot rotate.
Has groups attached to the carbons of the double bond that are fixed relative to each other.
CH3 CH CH3
CH = CH CH = CH
cis trans CH3

18. Cis-Trans Isomers Cis-trans isomers
Can be modeled by making a “double bond” with your fingers with both thumbs on the same side or opposite from each other.
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19. Cis-Trans Isomers Cis-trans isomers occur when
different groups are attached to
the double bond.
In a cis isomer, groups are attached on the same side of the double bond.
In the trans isomer, the groups are attached on opposite sides.
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20. Cis-Trans Isomerism
Cis-trans isomers do not occur if a carbon atom in the double bond is attached to identical groups.
Identical Identical
2-bromopropene ,1-dibromoethene
(not cis or trans) (not cis or trans) H Br H H

21. Naming Cis-Trans Isomers
The prefixes cis or trans are placed in front of the alkene name when there are cis-trans isomers.
cis trans
cis-1,2-dibromoethene trans-1,2-dibromoethene

22. Pheromones A pheromone
Is a chemical messenger emitted in tiny quantities.
Bombykol emitted by the silkworm moth to attract other moths has one cis and one trans double bond.
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23. Learning Check
Name each, using cis-trans prefixes when needed.

24. Solution trans-2-butene 1,1-dichloropropene cis-1,2-dibromoethene
Identical atoms on one C; no cis or trans

25. Chapter 12 Unsaturated Hydrocarbons
12.3
Addition Reactions
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26. Addition Reactions In addition reactions,
Reactants add to the carbon atoms in double or triple bonds.
A double or triple bond is easily broken, which makes them very reactive.
TABLE 12.2
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27. Hydrogenation In hydrogenation,
Hydrogen atoms add to the carbon atoms of a double bond or triple bond.
A catalyst such as Pt, Pd or Ni is used to speed up the reaction.

28. Hydrogenation of Oils Adding H2 to double bonds in vegetable
oils produces
Compounds with higher melting points.
Solids at room temperature such as margarine,
soft margarine,
and shortening.
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29. Learning Check
Write an equation for the hydrogenation of 1-butene using a platinum catalyst.

30. Solution Write an equation for the hydrogenation of 1-butene
using a platinum catalyst. Pt
CH2=CH─CH2─CH3 + H2 CH3─CH2─CH2─CH3

31. Trans Fats In vegetable oils, the unsaturated fats usually contain
cis double bonds.
During hydrogenation, some cis double bonds are converted to trans double bonds (more stable) causing a change in the fatty acid structure
If a label states “partially” or “fully hydrogenated”, the fats contain trans fatty acids.

32. Learning Check Write the product of each the following reactions: Pt
CH3─CH=CH─CH3 + H2
+ H2

33. Solution Pt
CH3─CH=CH─CH3 + H CH3─CH2─CH2─CH3
+ H Pt

34. Halogenation
In halogenation, halogen atoms add to the carbon atoms of a double bond or triple bond.

35. Learning Check Write the product of the following addition reactions:
1. CH3─CH=CH─CH3 + Cl2 2.

36. Solution Write the product of the following addition reactions: Pt
1. CH3─CH=CH─CH3 + Cl2
Cl Cl
l l
CH3─CH─CH─CH3 2.

37. Testing for Unsaturation
Br2
When bromine (Br2) is added to an alkane, the red color of bromine persists.
When bromine (Br2) is added to an alkene or alkyne, the red color of bromine disappears immediately.
Br2
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38. Hydrohalogenation
In hydrohalogenation, the atoms of a hydrogen halide add to the carbon atoms of a double bond or triple bond.

39. Markovnikov’s Rule
When an unsymmetrical alkene undergoes hydrohalogenation, the H in HX adds to the carbon in the double bond that has the greater number of H atoms .

40. Mechanism of Hydrohalogenation
The the electron pair in the π bond (double bond) act a nucleophile attacking the proton (electrophile) to form a cation.
The halide ion acts as a nucleophile attacking the cation (electrophile) to form the product.

41. Markovnikov’s Rule Revisited
Like radicals, organic cations are generally hard to form
Tertiary are easier to form than secondary which are easier to form than primary
Cation forms on most substituted carbon and that’s where the halogen or water (next reaction) goes

42. Hydration In the addition reaction called hydration
An acid H+ catalyst is required.
Water (HOH) adds to a double bond.
An H atom bonds to one C in the double bond.
An OH bonds to the other C.
H OH
H │ │
CH3─CH=CH─CH3 + H─OH CH3─CH─CH─CH3

43. Hydration When hydration occurs with a double bond that has an
unequal number of H atoms,
The H atom bonds to the C in the double bond with the most H.
The OH bonds to the C in the double bond with the fewest H atoms.
OH H
H │ │
CH3─CH=CH2 + H─OH CH3─CH─CH2
(1H) (2H)

44. Mechanism of Hydrohalogenation
The the electron pair in the π bond (double bond) act a nucleophile attacking the proton (electrophile) to form a cation.
The water acts as a nucleophile attacking the cation (electrophile) to form the product and eject a proton.

45. Learning Check Write the product for the hydration of each of the
following: H+
1. CH3─CH2─CH=CH─CH2─CH HOH
CH3
│ H+
2. CH3─C=CH─CH2─CH HOH
HOH

46. Solution H OH │ │ 1. CH3─CH2─CH─CH─CH2─CH3 CH3 │ 2. CH3─C─CH─CH2─CH3
│ │
1. CH3─CH2─CH─CH─CH2─CH3
CH3 │
2. CH3─C─CH─CH2─CH3
│ │
OH H OH 3. H

47. Learning Check
Write the products of each reaction.

48. Solution

49. Chapter 12 Unsaturated Hydrocarbons
12.4
Polymers of Alkenes
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50. Polymers Polymers are Large, long-chain molecules.
Found in nature, including cellulose in plants, starches in food, proteins and DNA in the body.
Also synthetic such as polyethylene and polystyrene, Teflon, and nylon.
Made up of small repeating units called monomers.
Made by reaction of small alkenes.

51. Polymerization
In polymerization, small repeating units called monomers join to form a long chain polymer.
monomer unit repeats n

52. Common Synthetic Polymers
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53. Polymers from Addition Reactions
TABLE 12.3
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54. More Monomers and Polymers
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55. Recycling Plastics Recycling is simplified by using codes found
on plastic items.
1 PETE Polyethyleneterephtalate
2 HDPE High-density polyethylene
3 PV Polyvinyl chloride
4 LDPE Low-density polyethylene
5 PP Polypropylene
6 PS Polystyrene
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56. Chapter 12 Unsaturated Hydrocarbons
12.5
Aromatic Compounds
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57. Benzene Structure Benzene
Has 6 electrons shared equally among the 6 C atoms.
Is also represented as a hexagon with a circle drawn inside.
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58. Aromatic Compounds in Nature and Health
Vanillin Aspirin
Ibuprofen Acetaminophen

59. Naming Aromatic Compounds
Aromatic compounds are named
With benzene as the parent chain.
With one side group named in front of benzene.
methylbenzene chlorobenzene

60. Some Common Names Some substituted benzene rings
Have common names used for many years.
With a single substituent use a common name or are named as a benzene derivative.
toluene aniline phenol
(methylbenzene) (aminobenzene) (hydroxybenzene)

61. Aromatic Compounds with Two Groups
Two naming systems are used when two groups are
attached to a benzene ring.
Number the ring to give the lowest numbers to the side groups.
Use prefixes to show the arrangement:
ortho(o-) for 1,2-
meta(m-) for 1,3-
para(p-) for 1,4-

62. Aromatic Compounds with Two Groups
3-chlorotoluene ,4-dichlorobenzene chlorophenol
m-chlorotoluene p-dichlorobenzene o-chlorophenol

63. Learning Check Select the correct name for each compound:
1) chlorocyclohexane
2) chlorobenzene
3) 1-chlorobenzene
1) 1,2-dimethylbenzene
2) m-xylene
3) 1,3-dimethylbenzene

64. Solution
2) chlorobenzene
2) m-xylene
3) 1,3-dimethylbenzene

65. Learning Check
Write the structural formulas for each of the following:
A. 1,3-dichlorobenzene
B. o-chlorotoluene

66. Solution
A. 1,3-dichlorobenzene
B. o-chlorotoluene

67. Learning Check Identify the organic family for each: A. CH3─CH2─CH=CH2B.
C. CH3─C≡CH D.

68. Solution Identify the organic family for each:
A. CH3─CH2─CH=CH alkene
B cycloalkane (alkane)
C. CH3─C≡CH alkyne
D aromatic

69. Chapter 12 Unsaturated Hydrocarbons
12.6
Properties of Aromatic
Compounds
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70. Properties of Aromatic Compounds
Have a stable aromatic bonding system.
Are resistant to many reactions.
Undergo substitution reactions, which retain the stability of the aromatic bonding system. Addition or elimination reactions would change the number of double bonds.

71. Substitution Reactions
In a substitution reaction, a hydrogen atom on a benzene ring is replaced by an atom or group of atoms.
Type of substitution H on benzene replaced by
Halogenation chlorine or bromine atom
Nitration nitro group (—NO2)
Sulfonation —SO3H group

72. Halogenation In a halogenation
An H atom of benzene is replaced by a chlorine or bromine atom.
A catalyst such as FeCl3 is needed in chlorination.
A catalyst such as FeBr3 is needed in bromination.

73. Nitration In the nitration of benzene
An H atom of benzene is replaced by a nitro (-NO2) group from HNO3.
An acid catalyst such as H2SO4 is needed.

74. Sulfonation In a sulfonation
An H atom on benzene is replaced by a —SO3H group from SO3.
An acid catalyst such as H2SO4 is needed.

75. Learning Check
Write the equation for the bromination of benzene including catalyst.

76. Solution
Write the equation for the bromination of benzene including catalyst.