1 chapter 11 the unsaturated hydrocarbons: alkenes, alkynes, and aromatics the unsaturated...

1

Chapter 11Chapter 11

The Unsaturated Hydrocarbons:Alkenes, Alkynes, and Aromatics

The Unsaturated Hydrocarbons:Alkenes, Alkynes, and Aromatics

2

1. Structure1. Structure

Alkenes are hydrocarbons with a double bond.

CnH2n

Alkynes are hydrocarbons with a triple bond.

CnH2n-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.

CnH2n

Alkynes are hydrocarbons with a triple bond.

CnH2n-2

• Alkenes and alkynes are unsaturated (don’t have the maximum number of hydrogens bonded to each carbon).

3

1. Comparison1. Comparison

4

1. Geometry1. Geometry

5

1. Geometry [3.4 Lewis structures]1. Geometry [3.4 Lewis structures]

Four groups of electronsEthanetetrahedral

extend toward the corners of a regular tetrahedron

bond angle = 109.5o

Four groups of electronsEthanetetrahedral

extend toward the corners of a regular tetrahedron

bond angle = 109.5o

6


Three groups of electronsEtheneAll in the same planeTrigonal planar

Bond angle = 120o

Three groups of electronsEtheneAll in the same planeTrigonal planar

Bond angle = 120o

7


Two groups of electronsEthyneLinear

Bond angle = 180o

Two groups of electronsEthyneLinear

Bond angle = 180o

8

1. Physical properties1. Physical properties

Name Melting point Boiling pointethene -160.1oC -103.7oC

propene -185.0oC -47.6oC1-butene -185.0oC -6.1oC

methylpropene -140.0oC -6.6oCethyne -81.8oC -84.0oC

propyne -101.5oC -23.2oC1-butyne -125.9oC 8.1oC2-butyne -32.3oC 27.0oC

9

1. Physical properties1. 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.

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. Nomenclature2. Nomenclature

ethene

ethyne

propenepropyne

11


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]

12


Determine the name and number of each substituent and add in front of the name of the parent compound.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

13


Alkenes with more than one double bond are calledalkadienes (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 calledalkadienes (2 double bonds)alkatrienes (3 double bonds)etc…

Each double bond is designated by its lower-numbered carbon.

2,4-hexadiene

14


Cycloalkenes must be numbered so the double bond is between carbons one and two.Cycloalkenes must be numbered so the double bond is between carbons one and two.

3-chloro-cyclopentene

4-ethyl-5-methylcyclooctene

15


Name the following compounds.

CH3CH=C(CH2CH3)2

H2C=C-CH2-CH=CH2

Name the following compounds.

CH3CH=C(CH2CH3)2

H2C=C-CH2-CH=CH2

16


Name the following compounds. Name the following compounds.

17


Write a structural formula for each of the following compounds.

1-hexene1,3-dicholoro-2-butene

4-methyl-2-hexyne1,4-cyclohexadiene

Write a structural formula for each of the following compounds.

1-hexene1,3-dicholoro-2-butene

4-methyl-2-hexyne1,4-cyclohexadiene

18


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

19

3. Geometric isomers3. 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

20

3. Geometric isomers3. 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.

21

3. Cis-trans isomers3. 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

22


• Alkenes without substituents also may exhibit cis-trans isomerism.

• Alkenes without substituents also may exhibit cis-trans isomerism.

cis-4-octenetrans-4-octene

23


• In order for cis and trans isomers to exist, neither double-bonded carbon may have two identical substituents.

• In order for cis and trans isomers to exist, neither double-bonded carbon may have two identical substituents.

2-methyl-2-buteneno cis/trans isomerism

1-buteneno cis/trans isomerism

24


• 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

25

4. Alkenes in nature4. Alkenes in nature

• Ethene (ethylene) and ripening

– Ripening agents

– Ripening bowl

• Ethene (ethylene) and ripening

– Ripening agents

– Ripening bowl

26

5. Reactions of alkenes and alkynes 5. Reactions of alkenes and alkynes

• The most common reactions of alkenes and alkynes are addition reactions.

– Hydrogenation: addition of H2

– Halogenation: addition of X2

– Hydration: addition of H2O

– Hydrohalogenation: addition of HX

• The most common reactions of alkenes and alkynes are addition reactions.

– Hydrogenation: addition of H2

– Halogenation: addition of X2

– Hydration: addition of H2O

– Hydrohalogenation: addition of HX

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 reaction5. General addition reaction

28

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


29


• For hydrogenation, halogenation, hydration, and hydrohalogenation, identify the A and B portions of what is being added to the double bond.

– hydrogenation, H2

– halogenation, X2 (where X = F, Cl, Br, or I)– hydration, H2O– 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, H2

– halogenation, X2 (where X = F, Cl, Br, or I)– hydration, H2O– hydrohalogenation, HX (where X = F, Cl, Br, or I)

30

5. Hydrogenation5. Hydrogenation

• In hydrogenation of an alkene, one molecule of hydrogen (H2) 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 (H2) adds to one mole of double bonds.

• Reaction conditions:– platinum, palladium, or nickel catalyst– [sometimes] heat and/or pressure

31


• In hydrogenation of an alkyne, two molecules of hydrogen (H2) add to one mole of triple bonds.

• Reaction conditions: same as for alkenes.

• In hydrogenation of an alkyne, two molecules of hydrogen (H2) add to one mole of triple bonds.

• Reaction conditions: same as for alkenes.

32


• Compare the products resulting from the hydrogenation of trans-2-pentene and cis-2-pentene.

• Compare the products resulting from the hydrogenation of trans-2-pentene and cis-2-pentene.

33


• Compare the products resulting from the hydrogenation of 1-butene and cis-2-butene.

• Compare the products resulting from the hydrogenation of 1-butene and cis-2-butene.

34

5. Vegetable oil and margarine5. Vegetable oil and margarine

Why does hydrogenation make oils more solid?Why does hydrogenation make oils more solid?

MP = 13-14oC

MP = 69.6oC

MP = 62.9oC

35

5. Halogenation5. Halogenation

• In halogenation of an alkene, one mole of a halogen (Cl2, Br2, I2) 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 (Cl2, Br2, I2) adds to one mole of double bonds.

• Since halogens are more reactive than hydrogen, no catalyst is needed.

36


• In halogenation of an alkyne, two moles of a halogen (Cl2, Br2, I2) add to one mole of double bonds.

• In halogenation of an alkyne, two moles of a halogen (Cl2, Br2, I2) add to one mole of double bonds.

37


• 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

38


• 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.

39

5. Hydration5. Hydration

• In hydration, one mole of water (H2O) is added to one mole of double bonds.

• A trace of acid is required as a catalyst.

• In hydration, one mole of water (H2O) is added to one mole of double bonds.

• A trace of acid is required as a catalyst.

40


• 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.

41


• 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:

+ H2O

42


• 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!

43

5. Hydrohalogenation5. 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.

44


2-butene + HBr ?

3-methyl-2-hexene + HCl ?

cyclopentene + HI ?

2-butene + HBr ?

3-methyl-2-hexene + HCl ?

cyclopentene + HI ?

45


• 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!]

46

6. Aromatic compounds6. Aromatic compounds

• Consider the following molecular formulas for unsaturated hydrocarbons:

– Hexane (all single bonds): C6H14

– Cyclohexane (one ring): C6H12

– Hexene (one double bond): C6H12

– Hexadiene (two double bonds): C6H10

– Cyclohexene (one ring, one double bond): C6H10

– Hexatriene (three double bonds): C6H8

– Cyclohexadiene (one ring, two double bonds): C6H8

• Consider the following molecular formulas for unsaturated hydrocarbons:

– Hexane (all single bonds): C6H14

– Cyclohexane (one ring): C6H12

– Hexene (one double bond): C6H12

– Hexadiene (two double bonds): C6H10

– Cyclohexene (one ring, one double bond): C6H10

– Hexatriene (three double bonds): C6H8

– Cyclohexadiene (one ring, two double bonds): C6H8

47

6. Aromatic compounds6. Aromatic compounds

• The molecular formula for benzene is C6H6.– 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 C6H6.– 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.

48

6. Benzene structure6. 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.

49

6. Benzene structure6. 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.

50


• Most single-substituent compounds are named as derivatives of benzene.

– Bromobenzene

– Ethylbenzene

• Most single-substituent compounds are named as derivatives of benzene.

– Bromobenzene

– Ethylbenzene

51


• 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)

52


• 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]

53


• 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

54


• The substituent consisting of a –CH2 attached to a benzene ring is called benzyl-.

– Benzyl chloride

• The substituent consisting of a –CH2 attached to a benzene ring is called benzyl-.

– Benzyl chloride

55

6. Polynuclear aromatic hydrocarbons6. 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!

56

6. Reactions of benzene6. 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.

• Cl2 requires FeCl3 catalyst.• Br2 requires FeBr3 catalyst.

– Nitration: substitution of one or more nitro- (-NO2) groups for hydrogen atoms.

• Requires nitric acid and concentration sulfuric acid.

– Sulfonation: substitution of one sulfonic acid (-SO3H) group for a hydrogen atom.

• SO3 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.

• Cl2 requires FeCl3 catalyst.• Br2 requires FeBr3 catalyst.

– Nitration: substitution of one or more nitro- (-NO2) groups for hydrogen atoms.

• Requires nitric acid and concentration sulfuric acid.

– Sulfonation: substitution of one sulfonic acid (-SO3H) group for a hydrogen atom.

• SO3 reactant and concentration sulfuric acid.

57

7. Heterocyclic aromatic compounds7. 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

58

7. Heterocyclic aromatic compounds7. 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?

1 chapter 11 the unsaturated hydrocarbons: alkenes, alkynes, and aromatics the unsaturated...

Documents

o c slide

o slide

o c ethyne

o c propene

o c propyne

o c methylpropene

multiple bond

geometry slide