Alkenes and Alkynes Chapter #3. Alkene Introduction Hydrocarbon with carbon-carbon double bonds...

Alkenes and AlkynesChapter #3

Alkene Introduction• Hydrocarbon with carbon-carbon double bonds• Sometimes called olefins, “oil-forming gas”• General formula CnH2n n≥2• Examples

n=2 C2H4

Common NamesUsually used for small molecules.Examples:

Vinyl carbons are the carbons sharing a double bond in blue

Vinyl hydrogens are the hydrogens bonded to vinyl carbons in red

CH2=CH2 CH2=CH-CH3 CH2=C-CH3

ethylene propylene isobutylene

CH3

IUPAC Nomenclature

• Parent is longest chain containing the double or triple bond.• -ane changes to –ene (or -diene, -triene) for double bonds, or –yne (or –diyne, -triyne).• Number the chain so that the double bond, or triple bond has the lowest possible number.• In a ring, the double bond is assumed to be between carbon 1 and carbon 2.

Name These Alkenes

CH2 CH CH2 CH3

CH3 C

CH3

CH CH3

CH3

CHCH2CH3H3C

Name These Alkenes

CH2 CH CH2 CH3

CH3 C

CH3

CH CH3

CH3

CHCH2CH3H3C1-butene

Name These Alkenes

CH2 CH CH2 CH3

CH3 C

CH3

CH CH3

CH3

CHCH2CH3H3C

1-butene

2-methyl-2-butene

Name These Alkenes

CH2 CH CH2 CH3

CH3 C

CH3

CH CH3

CH3

CHCH2CH3H3C

1-butene

2-methyl-2-butene

3-methylcyclopentene

Name These Alkenes

CH2 CH CH2 CH3

CH3 C

CH3

CH CH3

CH3

CHCH2CH3H3C

1-butene

2-methyl-2-butene


2-sec-butyl-1,3-cyclohexadiene

Name These Alkenes

CH2 CH CH2 CH3

CH3 C

CH3

CH CH3

CH3

CHCH2CH3H3C

1-butene

2-methyl-2-butene


2-sec-butyl-1,3-cyclohexadiene

3-n-propyl-1-heptene

Alkene Substituents= CH2

methylene

- CH = CH2

vinyl

- CH2 - CH = CH2

allyl

- CH2 - CH = CH2

allyl

Name = ?


methylene

- CH = CH2

vinyl

- CH2 - CH = CH2

allyl

- CH2 - CH = CH2

allyl

Name = Methylenecyclohexane Name =


methylene

- CH = CH2

vinyl

- CH2 - CH = CH2

allyl

Name = Methylenecyclohexane Name = vinylcyclohexane

Alkyne Common Names

• Acetylene is the common name for the two carbon alkyne.

• To give common names to alkynes having more than two carbons, give alkyl names to the carbon groups attached to the vinyl carbons followed by acetylene.

Alkyne Examples

Alkyne Examples

Isopropyl methyl acetylene

Alkyne Examples

Isopropyl methyl acetylene sec-butyl Cyclopropyl acetylene

Cis-trans Isomerism• Similar groups on same side of double bond, alkene is cis.

• Similar groups on opposite sides of double bond, alkene is trans.

• Cycloalkenes are assumed to be cis.

• Trans cycloalkenes are not stable unless the ring has at least 8 carbons.

Name these:

C CCH3

H

H

CH3CH2

Name these:

C CCH3

H

H

CH3CH2

trans-2-pentene

Name these:

C CCH3

H

H

CH3CH2

trans-2-pentene

C CBr

H

Br

H

Name these:

C CCH3

H

H

CH3CH2

trans-2-pentene

C CBr

H

Br

H

cis-1,2-dibromoethene

Which of the following show cis/trans isomers?

a. 1-penteneb. 2-pentenec. 1-chloro-1-pentened. 2-chloro-1-pentenee. 2-chloro-2-pentene

Solution to the Question

H

H

H

Hydrogens cannon be both cis and trans. Reversing the hydrogens on the first carbon produces the same compound.

H

H

H

1-Pentene

H

2-Pentene does show cis/trans isomerism

H

H

2-Pentene

H

H

1-Chloro-1-pentene does show cis/trans isomerism

H

1-chloro-1-pentene

Cl

H

H

Cl cis trans

Solution to the QuestionCl

2-Chloro-1-pentene does not show cis/trans isomerism, since flipping the hydrogen atoms in C-1 produces the same compound

Cl

2-chloro-1-pentene

H

H

H

H

Cl

2-Chloro-2-pentene does show cis/trans isomerism, since since the colored atoms on the vinyl carbons are on the same side in cis and opposite sides on the trans isomer.

Cl

2-chloro-2-pentene

H

Htrans cis

Which of the following show cis/trans isomers?

a. 1-pentene-Nob. 2-pentene- Yesc. 1-chloro-1-pentene- Yesd. 2-chloro-1-pentene- Noe. 2-chloro-2-pentene- yes

E-Z Nomenclature• Use the Cahn-Ingold-Prelog rules to assign priorities to groups attached to each carbon in the double bond. Highest priority is #1 and is the element with the largest atomic number.• If high priority groups are on the same side, the name is Z (for zusammen).• If high priority groups are on opposite sides, the name is E (for entgegen).

Example, E-Z

C C

H3C

H

Cl

CH2C C

H

H

CH CH3

Cl1

2

1

2

2

1

1

2

2Z 5E

Example, E-Z

C C

H3C

H

Cl

CH2C C

H

H

CH CH3

Cl1

2

1

2

2

1

1

2

2Z 5E

3,7-dichloro-(2Z, 5E)-2,5-octadiene

Physical Properties• Low boiling points, increasing with mass.

• Branched alkenes have lower boiling points.

• Less dense than water.

•Nonpolar (Hydrophobic)

Alkene Synthesis

• Dehydrohalogenation (-HX)• Dehydration of alcohols (-H2O)

OHH + H2O

minor major

Examples:Cl

NaOHminor major

+ + + NaCl + HOH

Zaitsev’s rule: The major product contains the most substituted double bond

Elimination Reactions:

Alkene ReactionsI. Addition Reactions

C=C

a. Hydration

C-C+ H-O-H

C=C

C=C

H O-H

b. Hydrogenation

C-C+ H-H

HH

c. Halogenation

+ X-X

Catalyst

H+

Catalyst = Ni, Pt, Pd

C-C

X X

Alcohol

Alkane

X = Cl, Br, IDihalide

Follows Markovnikov’s Rule

RegiospecificityMarkovnikov’s Rule: The proton (H+) of an acid adds to the carbon in the double bond that already has the most H’s. “Rich get richer.”

C=C

Examples:

CH3H

H

HH

C=CH CH3

H

+ H-O-HH+

+ H-Cl

H

C-CH

H Cl

H

H

C-CH

H O-H

H

CH3

CH3Major Products

Alkene Reactions (2)I. Addition Reactions (cont.)

d. Hydrohalogenation

C=C C-C+ H-X

C=C

H X

e. Glycol Formation

+ H-O-O-H C-C

H-O O-H

Alkyl halide

Glycol

Follows Markovnikov’s Rule

Alkene ReactionsStep 1: Pi electrons attack the electrophile.

Step 2: Nucleophile attacks the carbocation

Terpenes• Composed of 5-carbon isopentyl groups.• Isolated from plants’ essential oils.• C:H ratio of 5:8, or close to that.• Pleasant taste or fragrant aroma.• Examples:

Myrcene (From bay or myrcia plants)α-Pinene (From pine trees)Β-Selinene (From celery)Menthol (From peppermint oil)Camphor (From evergreen trees)R-Carvone (From spearmint)

Classification

• Terpenes are classified by the number of carbons they contain, in groups of 10.• A monoterpene has 10 C’s, 2 isoprenes. • A diterpene has 20 C’s, 4 isoprenes.• A sesquiterpene has 15 C’s, 3 isoprenes.

Terpenes

2-methyl-1,3-butadieneIsoprene

headtail

head

tail

head

Geraniol (roses)Head to tail link of two isoprenesCalled diterpene

OH

head

tail

head

tail

Menthol (pepermint)Head to tail link of two isoprenes another diterpene

Structure of Terpenes

Two or more isoprene units, 2-methyl-1,3-butadiene with some modification of the double bonds.

myrcene, frombay leaves

oCamphor (monoterpene)

ALKENE REVIEW

Describe the geometry around the carbon–carbon double bond.

a. Tetrahedralb.Trigonal pyramidalc. Trigonal planard.Bente.Linear

Answer

a. Tetrahedralb.Trigonal pyramidalc. Trigonal planard.Bente.Linear

Give the formula for an alkene.

a. CnH2n-4

b.CnH2n-2

c. CnH2n

d.CnH2n+2

e.CnH2n+4

Answer

a. CnH2n-4

b.CnH2n-2

c. CnH2n

d.CnH2n+2

e.CnH2n+4

Name CH3CH=CHCH=CH2.

a. 2,4-butadieneb.1,3-butadienec. 2,4-pentadiened.1,3-pentadienee.1,4-pentadiene

Answer

a. 2,4-butadieneb.1,3-butadienec. 2,4-pentadiened.1,3-pentadienee.1,4-pentadiene

Calculate the unsaturation number for C6H10BrCl.

a. 0b.1c. 2d.3

Answer

a. 0b.1c. 2d.3

U = 0.5 [2(6) + 2 – (12)] = 1

Name .

a. Trans-2-penteneb. Cis-2-pentenec. Trans-3-methyl-2-pentened. Cis-3-methyl-2-pentene

CC

H

H3C CH3

CH2CH3

Name

a. Trans-2-penteneb. Cis-2-pentenec. Trans-3-methyl-2-pentened. Cis-3-methyl-2-pentene

CC

H

H3C CH3

CH2CH3

Name

a. E-2-penteneb. Z-2-pentenec. E-3-methyl-2-pentened. Z-3-methyl-2-pentenee. Z-2-methyl-2-pentene

CC

H

H3C CH3

CH2CH3

Name

a. E-2-penteneb. Z-2-pentenec. E-3-methyl-2-pentened. Z-3-methyl-2-pentenee. Z-2-methyl-2-pentene

CC

H

H3C CH3

CH2CH3

a. ClCH2CH2Cl

b.ClCH=CHClc. CH2=CH2

d.CH2=CHCl

CC

H

H H

H

Cl2 NaOH

Answer

a. ClCH2CH2Cl

b.ClCH=CHClc. CH2=CH2

d.CH2=CHCl

Chlorine is added across the double bond, then HCl is lost.

a. (CH3)2CHOH

b.CH3CH2CH2OH

c. HOCH2CH2CH2OH

d.CH3CH(OH)CH2OH

CC

H

H CH3

H

H2O

catalyst

Answer

a. (CH3)2CHOH

b.CH3CH2CH2OH

c. HOCH2CH2CH2OH

d.CH3CH(OH)CH2OH

Water adds by Markovnikov’s orientation across the double bond.

Identify the product formed from the polymerization of tetrafluoroethylene.

a. Polypropyleneb.Poly(vinyl chloride), (PVC)c. Polyethylened.Poly(tetrafluoroethylene), Teflon

Answer

a. Polypropyleneb.Poly(vinyl chloride), (PVC)c. Polyethylened.Poly(tetrafluoroethylene), Teflon

Teflon is formed from the polymerization of tetrafluoroethylene.

a. CH3CCCH3

b.CH2=CHCH=CH2

c. CH3CH=CHCH3

d.CH3CH2CH2CH3

CC

H

H3C CH3

H

H2

Pd

Answer

a. CH3CCCH3

b.CH2=CHCH=CH2

c. CH3CH=CHCH3

d.CH3CH2CH2CH3

Hydrogen adds across the double bond to form an alkane.

a. (CH3)2CHOSO3H

b.CH3CH=CH2

c. (CH3)2C=O

d.CH3CH2COOH

HCH3C

OH

CH3

H2SO4

heat

7.15 Answer

a. (CH3)2CHOSO3H

b.CH3CH=CH2

c. (CH3)2C=O

d.CH3CH2COOH

Acid dehydrates alcohols to form alkenes.

Give the products from the catalytic cracking of alkanes.

a. Alkanesb.Alkenesc. Alkynesd.Alkanes + alkenese.Alkanes + alkynes

Answer


Give the products from the dehydrogenation of alkanes.


Give the products from the dehydrogenation of alkanes.


End Chapter #3

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Alkenes and Alkynes Chapter #3. Alkene Introduction Hydrocarbon with carbon-carbon double bonds...

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