Ch #12 Alkenes and Alkynes

Alkene Introduction Hydrocarbon with carbon-carbon double bonds Sometimes called olefins, “oil-forming gas” General formula CnH2n n≥2 Examples n=2 C2H4

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

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-sec-butyl-1,3-cyclohexadiene 2-methyl-2-butene 3-methylcyclopentene

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

Alkene Substituents = CH2 methylene - CH = CH2 vinyl - CH2 - CH = CH2 allyl - CH2 - CH = CH2 allyl Name = ?

Alkene Substituents = CH2 methylene - CH = CH2 vinyl - CH2 - CH = CH2 allyl - CH2 - CH = CH2 allyl Name = Methylenecyclohexane Name =

Alkene Substituents = CH2 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:

Name these: trans-2-pentene

Name these: trans-2-pentene

Name these: trans-2-pentene cis-1,2-dibromoethene

Which of the following show cis/trans isomers. a. 1-pentene. b 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 1 2 2 1 1 2 1 2 2Z 5E

Example, E-Z 1 2 2 1 1 2 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 Elimination Reactions: Dehydrohalogenation (-HX) Dehydration of alcohols (-H2O) Examples: Zaitsev’s rule: The major product contains the most substituted double bond

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

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.” H O-H Examples: H H H+ C-C H H C=C + H-O-H H CH3 H CH3 Major Products H Cl H H C-C C=C + H-Cl H H H CH3 H CH3

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

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

Terpenes

Terpenes head tail head head tail head tail tail head Geraniol (roses) Head to tail link of two isoprenes Called diterpene 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 40

Answer a. Tetrahedral b. Trigonal pyramidal c. Trigonal planar d. Bent e. Linear 41

Give the formula for an alkene. a. CnH2n-4 b. CnH2n-2 c. CnH2n d. CnH2n+2 e. CnH2n+4 42

Answer a. CnH2n-4 b. CnH2n-2 c. CnH2n d. CnH2n+2 e. CnH2n+4 43

Name CH3CH=CHCH=CH2. a. 2,4-butadiene b. 1,3-butadiene c. 2,4-pentadiene d. 1,3-pentadiene e. 1,4-pentadiene 44

Answer a. 2,4-butadiene b. 1,3-butadiene c. 2,4-pentadiene d. 1,3-pentadiene e. 1,4-pentadiene 45

Calculate the unsaturation number for C6H10BrCl. d. 3 46

Answer a. 0 b. 1 c. 2 d. 3 U = 0.5 [2(6) + 2 – (12)] = 1 47

Name . a. Trans-2-pentene b. Cis-2-pentene c. Trans-3-methyl-2-pentene d. Cis-3-methyl-2-pentene 48

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 49

Answer a. CH3COOH b. CH3CHO c. CH3CH2OH d. HOCH2CH2OH e. CH3CH(OH)2 Ethylene oxide is formed first, followed by a ring opening to form ethylene glycol. 50

a. ClCH2CH2Cl b. ClCH=CHCl c. CH2=CH2 d. CH2=CHCl 51

Answer a. ClCH2CH2Cl b. ClCH=CHCl c. CH2=CH2 d. CH2=CHCl Chlorine is added across the double bond, then HCl is lost. 52

a. (CH3)2CHOH b. CH3CH2CH2OH c. HOCH2CH2CH2OH d. CH3CH(OH)CH2OH 53

Answer a. (CH3)2CHOH b. CH3CH2CH2OH c. HOCH2CH2CH2OH d. CH3CH(OH)CH2OH Water adds by Markovnikov’s orientation across the double bond. 54

a. [CH2CH(CH3)]n b. [CH2CH2]n c. [CH2=CH(CH3)]n d. [CH2=CH2]n 55

Answer a. [CH2CH(CH3)]n b. [CH2CH2]n c. [CH2=CH(CH3)]n d. [CH2=CH2]n 56

Identify the product formed from the polymerization of tetrafluoroethylene. a. Polypropylene b. Poly(vinyl chloride), (PVC) c. Polyethylene d. Poly(tetrafluoroethylene), Teflon 57

Answer a. Polypropylene b. Poly(vinyl chloride), (PVC) c. Polyethylene d. Poly(tetrafluoroethylene), Teflon Teflon is formed from the polymerization of tetrafluoroethylene. 58

a. CH3CCCH3 b. CH2=CHCH=CH2 c. CH3CH=CHCH3 d. CH3CH2CH2CH3 59

Answer a. CH3CCCH3 b. CH2=CHCH=CH2 c. CH3CH=CHCH3 d. CH3CH2CH2CH3 Hydrogen adds across the double bond to form an alkane. 60

7.15 a. (CH3)2CHOSO3H b. CH3CH=CH2 c. (CH3)2C=O d. CH3CH2COOH 61

7.15 Answer a. (CH3)2CHOSO3H b. CH3CH=CH2 c. (CH3)2C=O d. CH3CH2COOH Acid dehydrates alcohols to form alkenes. 62

7.16 Dehydration of alcohols occurs by what mechanism? a. SN1 b. SN2 c. E1 d. E2 63

7.16 Answer a. SN1 b. SN2 c. E1 d. E2 The dehydration of alcohols occurs by an E1 mechanism. 64

7.17 Give the products from the catalytic cracking of alkanes. a. Alkanes b. Alkenes c. Alkynes d. Alkanes + alkenes e. Alkanes + alkynes 65

7.17 Answer a. Alkanes b. Alkenes c. Alkynes d. Alkanes + alkenes e. Alkanes + alkynes 66

7.18 Give the products from the dehydrogenation of alkanes. a. Alkanes b. Alkenes c. Alkynes d. Alkanes + alkenes e. Alkanes + alkynes 67

7.18 Answer a. Alkanes b. Alkenes c. Alkynes d. Alkanes + alkenes e. Alkanes + alkynes 68

7.19 a. (CH3)3CO-, (CH3)3COH b. CH3CH2O-, CH3CH2OH c. NaI, acetone d. H2, Pd 69

7.19 Answer a. (CH3)3CO-, (CH3)3COH b. CH3CH2O-, CH3CH2OH c. NaI, acetone d. H2, Pd The Hofmann product (least substituted) is favored with a bulky base. 70

7.20 a. Pt, 500o C b. H2, Pt c. H2SO4, 150o C d. NaI, acetone e. NaOH 71

7.20 Answer a. Pt, 500o C b. H2, Pt c. H2SO4, 150o C d. NaI, acetone e. NaOH Dehydrogenation occurs with a metal catalyst and heat. 72

End Chapter #3