Ch #12 Alkenes and Alkynes

Alkene Introduction Hydrocarbon with carbon-carbon double bonds Sometimes called olefins, “oil-forming gas” General formula C n H 2n n≥2 Examples n=2 C 2 H 4

Common Names Usually used for small molecules. Examples: Vinyl carbons are the carbons sharing a double bond Vinyl hydrogens are the hydrogens bonded to vinyl carbons

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

1-butene

Name These Alkenes 1-butene 2-methyl-2-butene

Name These Alkenes 1-butene 2-methyl-2-butene 3-methylcyclopentene

Name These Alkenes 1-butene 2-methyl-2-butene 3-methylcyclopentene 2-sec-butyl-1,3-cyclohexadiene

Name These Alkenes 1-butene 2-methyl-2-butene 3-methylcyclopentene 2-sec-butyl-1,3-cyclohexadiene 3-n-propyl-1-heptene

Alkene Substituents = CH 2 methylene - CH = CH 2 vinyl - CH 2 - CH = CH 2 allyl - CH 2 - CH = CH 2 allyl Name = ?

Alkene Substituents = CH 2 methylene - CH = CH 2 vinyl - CH 2 - CH = CH 2 allyl - CH 2 - CH = CH 2 allyl Name = MethylenecyclohexaneName =

Alkene Substituents = CH 2 methylene - CH = CH 2 vinyl - CH 2 - CH = CH 2 allyl Name = MethylenecyclohexaneName = 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

Isopropyl methyl acetylene

Alkyne Examples Isopropyl methyl acetylenesec-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:

trans-2-pentene

Name these: trans-2-pentene

Name these: trans-2-pentenecis-1,2-dibromoethene

Which of the following show cis/trans isomers? a. 1-pentene b. 2-pentene c. 1-chloro-1-pentene d. 2-chloro-1-pentene e. 2-chloro-2-pentene

E-Z Nomenclature Use the Cahn-Ingold-Prelog rules to assign priorities to groups attached to each carbon in the double bond. 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 Z2Z 5E5E

Z2Z 5E5E 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 (-H 2 O) Examples: Zaitsev’s rule: The major product contains the most substituted double bond Elimination Reactions:

Alkene Reactions I. Addition Reactions C=C a. Hydration C-C + H-O-H C=C HO-H b. Hydrogenation C-C + H-H HH c. Halogenation + X-X Catalyst H+H+ Catalyst = Ni, Pt, Pd C-C XX Alcohol Alkane X = Cl, Br, I Dihalide Follows Markovnikov’s Rule

Regiospecificity Markovnikov’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: CH 3 H H HH C=C H CH 3 H + H-O-H H+H+ + H-Cl H C-C H HCl H H C-C H HO-H H CH 3 Major Products

Alkene Reactions (2) I. Addition Reactions (cont.) d. Hydrohalogenation C=C C-C + H-X C=C HX e. Glycol Formation + H-O-O-H C-C H-OH-OO-HO-H Alkyl halide Glycol Follows Markovnikov’s Rule

Alkene Reactions Step 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: Anise oil Bay leaves

Terpenes

head tail head tail head Geraniol (roses) Head to tail link of two isoprenes Called diterpene 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, from bay leaves =>

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.

ALKENE REVIEW

Describe the geometry around the carbon–carbon double bond. a.Tetrahedral b.Trigonal pyramidal c.Trigonal planar d.Bent e.Linear

Answer a.Tetrahedral b.Trigonal pyramidal c.Trigonal planar d.Bent e.Linear

Give the formula for an alkene. a.C n H 2n-4 b.C n H 2n-2 c.C n H 2n d.C n H 2n+2 e.C n H 2n+4

Answer a.C n H 2n-4 b.C n H 2n-2 c.C n H 2n d.C n H 2n+2 e.C n H 2n+4

Name CH 3 CH=CHCH=CH 2. a.2,4-butadiene b.1,3-butadiene c.2,4-pentadiene d.1,3-pentadiene e.1,4-pentadiene

Answer a.2,4-butadiene b.1,3-butadiene c.2,4-pentadiene d.1,3-pentadiene e.1,4-pentadiene

Calculate the unsaturation number for C 6 H 10 BrCl. a.0 b.1 c.2 d.3

Answer a.0 b.1 c.2 d.3 U = 0.5 [2(6) + 2 – (12)] = 1

Name. a. Trans-2-pentene b. Cis-2-pentene c. Trans-3-methyl-2-pentene d. Cis-3-methyl-2-pentene

Name. a. E-2-pentene b. Z-2-pentene c. E-3-methyl-2-pentene d. Z-3-methyl-2-pentene e. Z-2-methyl-2-pentene

Answer a.CH 3 COOH b.CH 3 CHO c.CH 3 CH 2 OH d.HOCH 2 CH 2 OH e.CH 3 CH(OH) 2 Ethylene oxide is formed first, followed by a ring opening to form ethylene glycol.

a.ClCH 2 CH 2 Cl b.ClCH=CHCl c.CH 2 =CH 2 d.CH 2 =CHCl

Answer a.ClCH 2 CH 2 Cl b.ClCH=CHCl c.CH 2 =CH 2 d.CH 2 =CHCl Chlorine is added across the double bond, then HCl is lost.

a.(CH 3 ) 2 CHOH b.CH 3 CH 2 CH 2 OH c.HOCH 2 CH 2 CH 2 OH d.CH 3 CH(OH)CH 2 OH

Answer a.(CH 3 ) 2 CHOH b.CH 3 CH 2 CH 2 OH c.HOCH 2 CH 2 CH 2 OH d.CH 3 CH(OH)CH 2 OH Water adds by Markovnikov’s orientation across the double bond.

a.[CH 2 CH(CH 3 )] n b.[CH 2 CH 2 ] n c.[CH 2 =CH(CH 3 )] n d.[CH 2 =CH 2 ] n

Answer a.[CH 2 CH(CH 3 )] n b.[CH 2 CH 2 ] n c.[CH 2 =CH(CH 3 )] n d.[CH 2 =CH 2 ] n

Identify the product formed from the polymerization of tetrafluoroethylene. a.Polypropylene b.Poly(vinyl chloride), (PVC) c.Polyethylene d.Poly(tetrafluoroethylene), Teflon

Answer a.Polypropylene b.Poly(vinyl chloride), (PVC) c.Polyethylene d.Poly(tetrafluoroethylene), Teflon Teflon is formed from the polymerization of tetrafluoroethylene.

a.CH 3 C  CCH 3 b.CH 2 =CHCH=CH 2 c.CH 3 CH=CHCH 3 d.CH 3 CH 2 CH 2 CH 3

Answer a.CH 3 C  CCH 3 b.CH 2 =CHCH=CH 2 c.CH 3 CH=CHCH 3 d.CH 3 CH 2 CH 2 CH 3 Hydrogen adds across the double bond to form an alkane.

7.15 a.(CH 3 ) 2 CHOSO 3 H b.CH 3 CH=CH 2 c.(CH 3 ) 2 C=O d.CH 3 CH 2 COOH

7.15 Answer a.(CH 3 ) 2 CHOSO 3 H b.CH 3 CH=CH 2 c.(CH 3 ) 2 C=O d.CH 3 CH 2 COOH Acid dehydrates alcohols to form alkenes.

7.16 Dehydration of alcohols occurs by what mechanism? a.S N 1 b.S N 2 c.E1 d.E2

7.16 Answer a.S N 1 b.S N 2 c.E1 d.E2 The dehydration of alcohols occurs by an E1 mechanism.

7.17 Give the products from the catalytic cracking of alkanes. a.Alkanes b.Alkenes c.Alkynes d.Alkanes + alkenes e.Alkanes + alkynes

7.17 Answer a.Alkanes b.Alkenes c.Alkynes d.Alkanes + alkenes e.Alkanes + alkynes

7.18 Give the products from the dehydrogenation of alkanes. a.Alkanes b.Alkenes c.Alkynes d.Alkanes + alkenes e.Alkanes + alkynes

7.18 Answer a.Alkanes b.Alkenes c.Alkynes d.Alkanes + alkenes e.Alkanes + alkynes

7.19 a.(CH 3 ) 3 CO -, (CH 3 ) 3 COH b.CH 3 CH 2 O -, CH 3 CH 2 OH c.NaI, acetone d.H 2, Pd

7.19 Answer a.(CH 3 ) 3 CO -, (CH 3 ) 3 COH b.CH 3 CH 2 O -, CH 3 CH 2 OH c.NaI, acetone d.H 2, Pd The Hofmann product (least substituted) is favored with a bulky base.

7.20 a.Pt, 500 o C b.H 2, Pt c.H 2 SO 4, 150 o C d.NaI, acetone e.NaOH

7.20 Answer a.Pt, 500 o C b.H 2, Pt c.H 2 SO 4, 150 o C d.NaI, acetone e.NaOH Dehydrogenation occurs with a metal catalyst and heat.

End Chapter #3