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1 Chapter 11 The Unsaturated Hydrocarbons: Alkenes, Alkynes, and Aromatics The Unsaturated Hydrocarbons: Alkenes, Alkynes, and Aromatics.

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Presentation on theme: "1 Chapter 11 The Unsaturated Hydrocarbons: Alkenes, Alkynes, and Aromatics The Unsaturated Hydrocarbons: Alkenes, Alkynes, and Aromatics."— Presentation transcript:

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2 1 Chapter 11 The Unsaturated Hydrocarbons: Alkenes, Alkynes, and Aromatics The Unsaturated Hydrocarbons: Alkenes, Alkynes, and Aromatics

3 2 1. Structure Alkenes are hydrocarbons with a double bond. C n H 2n Alkynes are hydrocarbons with a triple bond. C n H 2n-2 Alkenes and alkynes are unsaturated (don’t have the maximum number of hydrogens bonded to each carbon). Alkenes are hydrocarbons with a double bond. C n H 2n Alkynes are hydrocarbons with a triple bond. C n H 2n-2 Alkenes and alkynes are unsaturated (don’t have the maximum number of hydrogens bonded to each carbon).

4 3 1. Comparison

5 4 1. Geometry

6 5 1. Geometry [3.4 Lewis structures] Four groups of electrons Ethane tetrahedral extend toward the corners of a regular tetrahedron bond angle = o Four groups of electrons Ethane tetrahedral extend toward the corners of a regular tetrahedron bond angle = o

7 6 1. Geometry [3.4 Lewis structures] Three groups of electrons Ethene All in the same plane Trigonal planar Bond angle = 120 o Three groups of electrons Ethene All in the same plane Trigonal planar Bond angle = 120 o

8 7 1. Geometry [3.4 Lewis structures] Two groups of electrons Ethyne Linear Bond angle = 180 o Two groups of electrons Ethyne Linear Bond angle = 180 o

9 8 1. Physical properties NameMelting pointBoiling point ethene o C o C propene o C-47.6 o C 1-butene o C-6.1 o C methylpropene o C-6.6 o C ethyne-81.8 o C-84.0 o C propyne o C-23.2 o C 1-butyne o C8.1 o C 2-butyne-32.3 o C27.0 o C

10 9 1. Physical properties In each case, the alkyne has a higher boiling point than the alkene. Its structure is more linear. The molecules pack together more efficiently. Intermolecular forces are stronger. In each case, the alkyne has a higher boiling point than the alkene. Its structure is more linear. The molecules pack together more efficiently. Intermolecular forces are stronger.

11 10 The root name is based on the longest chain that includes both carbons of the multiple bond. The –ane ending is changed to –ene for double bonds and –yne for triple bonds. The root name is based on the longest chain that includes both carbons of the multiple bond. The –ane ending is changed to –ene for double bonds and –yne for triple bonds. 2. Nomenclature ethene ethyne propene propyne

12 11 2. Nomenclature The chain is numbered from the end nearest the multiple bond. The position of the multiple bond is indicated with the lower-numbered carbon in the bond. The chain is numbered from the end nearest the multiple bond. The position of the multiple bond is indicated with the lower-numbered carbon in the bond. 1-butene [not 3-butene] 2-pentyne [not 3-pentyne]

13 12 2. Nomenclature Determine the name and number of each substituent and add in front of the name of the parent compound. 5-chloro-4-methyl-2-hexene 2,6-dimethyl-3-octene 5-bromo-4-ethyl-2-heptene

14 13 2. Nomenclature Alkenes with more than one double bond are called alkadienes (2 double bonds) alkatrienes (3 double bonds) etc… Each double bond is designated by its lower-numbered carbon. Alkenes with more than one double bond are called alkadienes (2 double bonds) alkatrienes (3 double bonds) etc… Each double bond is designated by its lower-numbered carbon. 2,4-hexadiene

15 14 2. Nomenclature Cycloalkenes must be numbered so the double bond is between carbons one and two. 3-chloro-cyclopentene 4-ethyl-5-methylcyclooctene

16 15 2. Nomenclature Name the following compounds. CH 3 CH=C(CH 2 CH 3 ) 2 H 2 C=C-CH 2 -CH=CH 2 Name the following compounds. CH 3 CH=C(CH 2 CH 3 ) 2 H 2 C=C-CH 2 -CH=CH 2

17 16 2. Nomenclature Name the following compounds.

18 17 2. Nomenclature Write a structural formula for each of the following compounds. 1-hexene 1,3-dicholoro-2-butene 4-methyl-2-hexyne 1,4-cyclohexadiene Write a structural formula for each of the following compounds. 1-hexene 1,3-dicholoro-2-butene 4-methyl-2-hexyne 1,4-cyclohexadiene

19 18 2. Nomenclature Draw a structural formula for each of the following compounds: 1-bromo-3-hexyne 2-butyne dichloroethyne 9-iodo-1-nonyne Draw a structural formula for each of the following compounds: 1-bromo-3-hexyne 2-butyne dichloroethyne 9-iodo-1-nonyne

20 19 3. Geometric isomers Rotation around a double bond is restricted, in much the same was as rotation is restricted for the cycloalkanes. In the alkenes, geometric isomers occur when there are two different groups on each of the double-bonded carbon atoms. Rotation around a double bond is restricted, in much the same was as rotation is restricted for the cycloalkanes. In the alkenes, geometric isomers occur when there are two different groups on each of the double-bonded carbon atoms. 1,2-dichloroethene

21 20 3. Geometric isomers Time for the first Chapter 11 Journal question! [Use tag “difference”] In your own words, explain how constitutional isomers and geometric isomers are different. Be sure to consider BOTH their differences and their similarities! You might want to use examples of actual molecules. Time for the first Chapter 11 Journal question! [Use tag “difference”] In your own words, explain how constitutional isomers and geometric isomers are different. Be sure to consider BOTH their differences and their similarities! You might want to use examples of actual molecules.

22 21 3. Cis-trans isomers If both constituents are on the same side of the double bond, the isomer is cis-. If the constituents are on opposite sides of the double bond, the isomer is trans-. If both constituents are on the same side of the double bond, the isomer is cis-. If the constituents are on opposite sides of the double bond, the isomer is trans-. cis-1,2-dichloroethene trans-1,2-dichloroethene

23 22 3. Cis-trans isomers Alkenes without substituents also may exhibit cis-trans isomerism. cis-4-octene trans-4-octene

24 23 3. Cis-trans isomers In order for cis and trans isomers to exist, neither double- bonded carbon may have two identical substituents. 2-methyl-2-butene no cis/trans isomerism 1-butene no cis/trans isomerism

25 24 3. Cis-trans isomers Which of the following compounds can exist as geometric isomers? 1-bromo-1-chloro-2-methylpropene 1,1-dichloroethene 1,2-dibromoethene 3-ethyl-2-methyl-2-hexene Which of the following compounds can exist as geometric isomers? 1-bromo-1-chloro-2-methylpropene 1,1-dichloroethene 1,2-dibromoethene 3-ethyl-2-methyl-2-hexene

26 25 4. Alkenes in nature Ethene (ethylene) and ripening Ripening agents Ripening bowl Ethene (ethylene) and ripening Ripening agents Ripening bowl

27 26 5. Reactions of alkenes and alkynes The most common reactions of alkenes and alkynes are addition reactions. Hydrogenation: addition of H 2 Halogenation: addition of X 2 Hydration: addition of H 2 O Hydrohalogenation: addition of HX The most common reactions of alkenes and alkynes are addition reactions. Hydrogenation: addition of H 2 Halogenation: addition of X 2 Hydration: addition of H 2 O Hydrohalogenation: addition of HX

28 27 A double bond consists of a sigma bond: two electrons concentrated on a line between the two connected atoms; a pi bond: two electrons concentrated in planes above and below the sigma bond. A double bond consists of a sigma bond: two electrons concentrated on a line between the two connected atoms; a pi bond: two electrons concentrated in planes above and below the sigma bond. 5. General addition reaction

29 28 In an addition reaction, the pi bond is lost and its electrons become part of the single bonds to A and B. 5. General addition reaction

30 29 5. General addition reaction For hydrogenation, halogenation, hydration, and hydrohalogenation, identify the A and B portions of what is being added to the double bond. hydrogenation, H 2 halogenation, X 2 (where X = F, Cl, Br, or I) hydration, H 2 O hydrohalogenation, HX (where X = F, Cl, Br, or I) For hydrogenation, halogenation, hydration, and hydrohalogenation, identify the A and B portions of what is being added to the double bond. hydrogenation, H 2 halogenation, X 2 (where X = F, Cl, Br, or I) hydration, H 2 O hydrohalogenation, HX (where X = F, Cl, Br, or I)

31 30 5. Hydrogenation In hydrogenation of an alkene, one molecule of hydrogen (H 2 ) adds to one mole of double bonds. Reaction conditions: platinum, palladium, or nickel catalyst [sometimes] heat and/or pressure In hydrogenation of an alkene, one molecule of hydrogen (H 2 ) adds to one mole of double bonds. Reaction conditions: platinum, palladium, or nickel catalyst [sometimes] heat and/or pressure

32 31 5. Hydrogenation In hydrogenation of an alkyne, two molecules of hydrogen (H 2 ) add to one mole of triple bonds. Reaction conditions: same as for alkenes. In hydrogenation of an alkyne, two molecules of hydrogen (H 2 ) add to one mole of triple bonds. Reaction conditions: same as for alkenes.

33 32 5. Hydrogenation Compare the products resulting from the hydrogenation of trans-2-pentene and cis-2-pentene.

34 33 5. Hydrogenation Compare the products resulting from the hydrogenation of 1-butene and cis-2-butene.

35 34 5. Vegetable oil and margarine Why does hydrogenation make oils more solid? MP = o C MP = 69.6 o C MP = 62.9 o C

36 35 5. Halogenation In halogenation of an alkene, one mole of a halogen (Cl 2, Br 2, I 2 ) adds to one mole of double bonds. Since halogens are more reactive than hydrogen, no catalyst is needed. In halogenation of an alkene, one mole of a halogen (Cl 2, Br 2, I 2 ) adds to one mole of double bonds. Since halogens are more reactive than hydrogen, no catalyst is needed.

37 36 5. Halogenation In halogenation of an alkyne, two moles of a halogen (Cl 2, Br 2, I 2 ) add to one mole of double bonds.

38 37 5. Halogenation Draw the structure and write a balanced equation for the halogenation of each of the following compounds. 3-methyl-1,4-hexadiene 4-bromo-1,3-pentadiene 3-chloro-2,4-hexadiene Draw the structure and write a balanced equation for the halogenation of each of the following compounds. 3-methyl-1,4-hexadiene 4-bromo-1,3-pentadiene 3-chloro-2,4-hexadiene

39 38 5. Halogenation A solution of bromine in water has a reddish-orange color. A simple test for the presence of an alkene or alkane is to add bromine water. If a double or triple bond is present, the bromine will be used up in a halogenation reaction and the color will disappear. A solution of bromine in water has a reddish-orange color. A simple test for the presence of an alkene or alkane is to add bromine water. If a double or triple bond is present, the bromine will be used up in a halogenation reaction and the color will disappear.

40 39 5. Hydration In hydration, one mole of water (H 2 O) is added to one mole of double bonds. A trace of acid is required as a catalyst. In hydration, one mole of water (H 2 O) is added to one mole of double bonds. A trace of acid is required as a catalyst.

41 40 5. Hydration Unlike hydrogenation and halogenation, hydration is not a symmetric addition to a double bond. If the double bond is not symmetrically located in the molecule, there are two possible hydration products. Unlike hydrogenation and halogenation, hydration is not a symmetric addition to a double bond. If the double bond is not symmetrically located in the molecule, there are two possible hydration products.

42 41 5. Hydration The predominant product is determined by Markovnikov’s rule: The rich get richer. OR: The carbon that already has more hydrogens will get the hydrogen from the water. Hydration of propene: The predominant product is determined by Markovnikov’s rule: The rich get richer. OR: The carbon that already has more hydrogens will get the hydrogen from the water. Hydration of propene: + H 2 O 

43 42 5. Hydration Write a balanced equation for the hydration of each of the following compounds: 2-butene 2-ethyl-3-hexene 2,3-dimethylcyclohexene Alkynes undergo a much more complicated hydration that you don’t need to remember at this time! Write a balanced equation for the hydration of each of the following compounds: 2-butene 2-ethyl-3-hexene 2,3-dimethylcyclohexene Alkynes undergo a much more complicated hydration that you don’t need to remember at this time!

44 43 5. Hydrohalogenation Like hydration, hydrohalogenation is an asymmetric addition to a double bond. Hydrohalogenation also follows Markovnikov’s rule. Like hydration, hydrohalogenation is an asymmetric addition to a double bond. Hydrohalogenation also follows Markovnikov’s rule.

45 44 5. Hydrohalogenation 2-butene + HBr  ? 3-methyl-2-hexene + HCl  ? cyclopentene + HI  ? 2-butene + HBr  ? 3-methyl-2-hexene + HCl  ? cyclopentene + HI  ?

46 45 5. Hydrohalogenation Here’s your second Journal question! [Use tag “addition”] Explain how hydrogenation and halogenation are different from hydration and hydrohalogenation as addition reactions. [Hint: There’s a rule involved!] Here’s your second Journal question! [Use tag “addition”] Explain how hydrogenation and halogenation are different from hydration and hydrohalogenation as addition reactions. [Hint: There’s a rule involved!]

47 46 6. Aromatic compounds Consider the following molecular formulas for unsaturated hydrocarbons: Hexane (all single bonds): C 6 H 14 Cyclohexane (one ring): C 6 H 12 Hexene (one double bond): C 6 H 12 Hexadiene (two double bonds): C 6 H 10 Cyclohexene (one ring, one double bond): C 6 H 10 Hexatriene (three double bonds): C 6 H 8 Cyclohexadiene (one ring, two double bonds): C 6 H 8 Consider the following molecular formulas for unsaturated hydrocarbons: Hexane (all single bonds): C 6 H 14 Cyclohexane (one ring): C 6 H 12 Hexene (one double bond): C 6 H 12 Hexadiene (two double bonds): C 6 H 10 Cyclohexene (one ring, one double bond): C 6 H 10 Hexatriene (three double bonds): C 6 H 8 Cyclohexadiene (one ring, two double bonds): C 6 H 8

48 47 6. Aromatic compounds The molecular formula for benzene is C 6 H 6. The structure must be highly unsaturated. One ring, three double bonds? Reactions of benzene: Benzene does not decolorize bromine solutions. Benzene does not undergo typical addition reactions. Benzene reacts mainly by substitution. The first three items are opposite from what is expected from unsaturated compounds. The last item is identical to what is expected for alkanes. The molecular formula for benzene is C 6 H 6. The structure must be highly unsaturated. One ring, three double bonds? Reactions of benzene: Benzene does not decolorize bromine solutions. Benzene does not undergo typical addition reactions. Benzene reacts mainly by substitution. The first three items are opposite from what is expected from unsaturated compounds. The last item is identical to what is expected for alkanes.

49 48 6. Benzene structure The benzene ring consists of: six carbon atoms joined in a planar hexagonal arrangement with each carbon bonded to one hydrogen atom. Two equivalent structures proposed by Kekulé are recognized today as resonance structures. The real benzene molecule is a hybrid with each resonance structure contributing equally to the true structure. The benzene ring consists of: six carbon atoms joined in a planar hexagonal arrangement with each carbon bonded to one hydrogen atom. Two equivalent structures proposed by Kekulé are recognized today as resonance structures. The real benzene molecule is a hybrid with each resonance structure contributing equally to the true structure.

50 49 6. Benzene structure Sigma and pi bonding in benzene: The sharing of six electrons over the entire ring gives the benzene structure extra stability. Removing any one of the six electrons would destroy that stability. Sigma and pi bonding in benzene: The sharing of six electrons over the entire ring gives the benzene structure extra stability. Removing any one of the six electrons would destroy that stability.

51 50 6. Nomenclature Most single-substituent compounds are named as derivatives of benzene. Bromobenzene Ethylbenzene Most single-substituent compounds are named as derivatives of benzene. Bromobenzene Ethylbenzene

52 51 6. Nomenclature A few “common” names have been adopted as IUPAC nomenclature. toluene phenol aniline xylene (any benzene ring with two methyl groups) A few “common” names have been adopted as IUPAC nomenclature. toluene phenol aniline xylene (any benzene ring with two methyl groups)

53 52 6. Nomenclature There are three ways for the methyl groups on xylene to be arranged. 1,2 [ortho-xylene] 1,3 [meta-xylene] 1,4 [para-xylene] There are three ways for the methyl groups on xylene to be arranged. 1,2 [ortho-xylene] 1,3 [meta-xylene] 1,4 [para-xylene]

54 53 6. Nomenclature The substituent created by removing one hydrogen from the benzene ring is called phenyl-. 2-phenylhexane 3-phenylcyclopentene The substituent created by removing one hydrogen from the benzene ring is called phenyl-. 2-phenylhexane 3-phenylcyclopentene

55 54 6. Nomenclature The substituent consisting of a –CH 2 attached to a benzene ring is called benzyl-. Benzyl chloride The substituent consisting of a –CH 2 attached to a benzene ring is called benzyl-. Benzyl chloride

56 55 6. Polynuclear aromatic hydrocarbons These consist of rings joined along one side. Good news! You don’t have to memorize these names! These consist of rings joined along one side. Good news! You don’t have to memorize these names!

57 56 6. Reactions of benzene Because of the stability of benzene’s ring structure, only substitution reactions are characteristic. Halogenation: substitution of one or more halogen atoms for hydrogen atoms. Cl 2 requires FeCl 3 catalyst. Br 2 requires FeBr 3 catalyst. Nitration: substitution of one or more nitro- (-NO 2 ) groups for hydrogen atoms. Requires nitric acid and concentration sulfuric acid. Sulfonation: substitution of one sulfonic acid (-SO 3 H) group for a hydrogen atom. SO 3 reactant and concentration sulfuric acid. Because of the stability of benzene’s ring structure, only substitution reactions are characteristic. Halogenation: substitution of one or more halogen atoms for hydrogen atoms. Cl 2 requires FeCl 3 catalyst. Br 2 requires FeBr 3 catalyst. Nitration: substitution of one or more nitro- (-NO 2 ) groups for hydrogen atoms. Requires nitric acid and concentration sulfuric acid. Sulfonation: substitution of one sulfonic acid (-SO 3 H) group for a hydrogen atom. SO 3 reactant and concentration sulfuric acid.

58 57 7. Heterocyclic aromatic compounds Heterocyclic aromatic compounds have at least one non- carbon atom incorporated in an aromatic ring or polynuclear aromatic compound. Many of these compounds are biologically important. Components of DNA and RNA Components of hemoglobin and chlorophyll Pharmaceuticals Heterocyclic aromatic compounds have at least one non- carbon atom incorporated in an aromatic ring or polynuclear aromatic compound. Many of these compounds are biologically important. Components of DNA and RNA Components of hemoglobin and chlorophyll Pharmaceuticals pyridine

59 58 7. Heterocyclic aromatic compounds Final Journal question for this unit! [Use tag “common”] What do DNA, RNA, nicotine, hemoglobin, chlorophyll, and a drug used to treat ulcers have in common? Final Journal question for this unit! [Use tag “common”] What do DNA, RNA, nicotine, hemoglobin, chlorophyll, and a drug used to treat ulcers have in common?


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