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Alkenes C n H 2n “unsaturated” hydrocarbons C 2 H 4 ethylene Functional group = carbon-carbon double bond PowerPoint Presentation
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Alkenes C n H 2n “unsaturated” hydrocarbons C 2 H 4 ethylene Functional group = carbon-carbon double bond - PowerPoint PPT Presentation


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Alkenes C n H 2n “unsaturated” hydrocarbons C 2 H 4 ethylene Functional group = carbon-carbon double bond sp 2 hybridization => flat, 120 o bond angles σ bond & π bond => H 2 C=CH 2 No rotation about double bond!. C 3 H 6 propylene CH 3 CH=CH 2

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slide1

Alkenes CnH2n “unsaturated” hydrocarbons

C2H4 ethylene

Functional group = carbon-carbon double bond

sp2 hybridization => flat, 120o bond angles

σ bond & π bond => H2C=CH2

No rotation about double bond!

slide2

C3H6 propylene CH3CH=CH2

C4H8 butylenes CH3CH2CH=CH2

α-butylene

1-butene

CH3

CH3CH=CHCH3 CH3C=CH2

β-butylene isobutylene

2-butene 2-methylpropene

slide3

there are two 2-butenes:

cis-2-butene trans-2-butene

“geometric isomers” (diastereomers)

slide4

C=C are called “vinyl” carbons

If eithervinyl carbon is bonded to two equivalent groups, then no geometric isomerism exists.

CH3CH=CHCH3 CH3CH2CH=CH2

yes no

CH3

(CH3)2C=CHCH3 CH3CH=CCH2CH3

no yes

slide5

Confusion about the use of cis- and trans-. According to IUPAC rules it refers to the parent chain.

“cis-”

????????

slide6

E/Z system is now recommended by IUPAC for the designation of geometric isomerism.

  • Use the sequence rules to assign the higher priority * to the two groups attached to each vinyl carbon.
  • 2. * * *
  • *
  • (Z)- “zusammen” (E)- “entgegen”
  • together opposite
slide7

*

*

(Z)-

(E)-

*

*

slide8

Nomenclature, alkenes:

  • Parent chain = longest continuous carbon chain that contains the C=C.
  • alkane => change –ane to –ene
  • prefix a locant for the carbon-carbon double bond using
  • the principle of lower number.
  • Etc.
  • If a geometric isomer, use E/Z (or cis/trans) to indicate which isomer it is.
slide9

*

*

(Z)-3-methyl-2-pentene

(3-methyl-cis-2-pentene)

*

(E)-1-bromo-1-chloropropene

*

slide10

CH3

CH3CH2 CHCH2CH3

\ /

C = C 3-ethyl-5-methyl-3-heptene

/ \

CH3CH2 H (not a geometric isomer)

slide11

-ol takes precedence over –ene

CH2=CHCH2-OH 2-propen-1-ol

CH3CHCH=CH2 3-buten-2-ol

OH

slide12

Physical properties:

non-polar or weakly polar

no hydrogen bonding

relatively low mp/bp ~ alkanes

water insoluble

Importance:

common group in biological molecules

starting material for synthesis of many plastics

slide13

Syntheses, alkenes:

  • dehydrohalogenation of alkyl halides
  • 2. dehydration of alcohols
  • dehalogenation of vicinal dihalide
  • 4. (later)
slide14

dehalogenation of vicinal dihalides

  • | | | |
  • — C — C — + Zn  — C = C — + ZnX2
  • ||
  • X X
  • eg.
  • CH3CH2CHCH2 + Zn  CH3CH2CH=CH2 + ZnBr2
  • Br Br
  • Not generally useful as vicinal dihalides are usually made from alkenes. May be used to “protect” a carbon-carbon double bond.
slide15

dehydrohalogenation of alkyl halides

  • | | | |
  • — C — C — + KOH(alc.)  — C = C — + KX + H2O
  • | |
  • H X
  • RX: 3o > 2o > 1o
  • no rearragement 
  • may yield mixtures 
  • Saytzeff orientation
  • element effect
  • isotope effect
  • rate = k [RX] [KOH]
  • Mechanism = E2
slide16

rate = k [RX] [KOH] => both RX & KOH in RDS

R-I > R-Br > R-Cl “element effect”

=> C—X broken in RDS

R-H > R-D “isotope effect”

=> C—H broken in RDS

Concerted reaction: both the C—X and C—H bonds are broken in the rate determining step.

slide17

Mechanism = elimination, bimolecular E2

One step! “Concerted” reaction.

slide18

CH3CHCH3 + KOH(alc)  CH3CH=CH2

Br

isopropyl bromide propylene

CH3CH2CH2CH2-Br + KOH(alc)  CH3CH2CH=CH2

n-butyl bromide 1-butene

CH3CH2CHCH3 + KOH(alc)  CH3CH2CH=CH2

Br 1-butene19%

sec-butyl bromide +

CH3CH=CHCH3

2-butene81%

slide19

Problem 8.6. What akyl halide (if any) would yield each of the following purealkenes upon dehydrohalogenation by strong base?

CH3 CH3

isobutylene  KOH(alc) + CH3CCH3 or CH3CHCH2-X

X

1-pentene  KOH(alc) + CH3CH2CH2CH2CH2-X

note: CH3CH2CH2CHCH3 would yield a mixture! 

X

2-pentene  KOH(alc) + CH3CH2CHCH2CH3

X

2-methyl-2-butene  KOH(alc) + NONE!

slide21

Saytzeff orientation:

Ease of formation of alkenes:

R2C=CR2 > R2C=CHR > R2C=CH2, RCH=CHR > RCH=CH2 > CH2=CH2

Stability of alkenes:

R2C=CR2 > R2C=CHR > R2C=CH2, RCH=CHR > RCH=CH2 > CH2=CH2

CH3CH2CHCH3 + KOH(alc)  CH3CH2CH=CH2RCH=CH2

Br 1-butene 19%

sec-butyl bromide +

CH3CH=CHCH3RCH=CHR

2-butene 81%

slide22

KOH (alc)

CH3CH2CH2CHBrCH3 CH3CH2CH=CHCH3 + CH3CH2CH2CH=CH2

71% 29%

CH3 CH3 CH3

CH3CH2CCH3 + KOH(alc)  CH3CH=CCH3 + CH3CH2C=CH2

Br 71% 29%

CH3 CH3 CH3

CH3CHCHCH3 + KOH(alc)  CH2=CHCHCH3 + CH3CH=CCH3

Br major product

slide23

Order of reactivity in E2: 3o > 2o > 1o

CH3CH2-X  CH2=CH2 3 adj. H’s

CH3CHCH3  CH3CH=CH2 6 adj. H’s & more stable

X alkene

CH3 CH3

CH3CCH3  CH=CCH3 9 adj. H’s & most stable

X alkene

slide25

Elimination, unimolecular E1

  • a)RX: 3o > 2o > 1o
  • b) rearragement possible 
  • c) may yield mixtures 
  • d) Saytzeff orientation
  • e) element effect
  • f) no isotope effect
  • g) rate = k [RW]
slide26

E1:

Rate = k [RW] => only RW involved in RDS

R-I > R-Br > R-Cl “element effect” =>

C—X is broken in RDS

R-H  R-D no“isotope effect” =>

C—H is not broken in the RDS

slide27

Elimination, unimolecular E1

  • RX: 3o > 2o > 1o carbocation
  • rearragement possible “
  • c) may yield mixtures
  • d) Saytzeff orientation
  • e) element effect C—W broken in RDS
  • f) no isotope effect C—H not broken in RDS
  • g) rate = k [RW] only R-W in RDS
slide29

R-X + base  ????????

  • 1) If strong, conc. base:
  • CH3 > 1o => SN2 R-Z 
  • 3o > 2o => E2 alkene(s)
  • If weak, dilute base:
  • 3o > 2o > 1o => SN1 and E1 R-Z + alkene(s) 
  • If KOH(alc.)
  • 3o > 2o > 1o => E2 alkene(s) 
slide30

SN2

CH3CH2CH2-Br + NaOCH3 CH3CH2CH2-O-CH3

1o

CH3E2 CH3

CH3CCH3 + NaOCH3  CH3C=CH2 + HOCH3

Br

3o

E2

CH3CH2CH2-Br + KOH(alc)  CH3CH=CH2

slide31

CH3 CH3

CH3CHCHCH3 + dilute OH- CH3CCH2CH3SN1

Br OH

 CH3

+ CH3C=CHCH2E1

CH3

CH3CHCHCH3 CH3

 + CH2=CCH2CH3E1

 [1,2-H]

CH3

CH3CCH2CH3

slide32

dehydration of alcohols:

  • | |||
  • — C — C — acid, heat  — C = C — + H2O
  • | |
  • H OH
  • ROH: 3o > 2o > 1o
  • acid is a catalyst
  • rearrangements are possible 
  • mixtures are possible 
  • Saytzeff
  • mechanism is E1
  • note: reaction #3 for alcohols!
slide34

CH3CH2-OH + 95% H2SO4, 170oC  CH2=CH2

CH3 CH3

CH3CCH3 + 20% H2SO4, 85-90oC  CH3C=CH2

OH

CH3CH2CHCH3 + 60% H2SO4, 100oC  CH3CH=CHCH3

OH

+ CH3CH2CH=CH2

CH3CH2CH2CH2-OH + H+, 140oC  CH3CH2CH=CH2

rearrangement!  + CH3CH=CHCH3

slide35

Synthesis of 1-butene from 1-butanol:

CH3CH2CH2CH2-OH + HBr  CH3CH2CH2CH2-Br

SN2 E2  KOH(alc)

CH3CH2CH=CH2

only!

To avoid the rearrangement in the dehydration of the alcohol the alcohol is first converted into an alkyl halide.

slide36

Syntheses, alkenes:

  • dehydrohalogenation of alkyl halides
  • E2
  • 2. dehydration of alcohols
  • E1
  • dehalogenation of vicinal dihalide
  • 4. (later)
slide37

R-OH

H+

R-X

KOH

Alkene

(alc.)

Zn

vicinal

dihalide

slide38

Alkyl halides:

nomenclature

syntheses:

1. from alcohols

a) HX b) PX3

2. halogenation of certain alkanes

3.

4.

5. halide exchange for iodide

reactions:

1. nucleophilic substitution

2. dehydrohalgenation

3. formation of Grignard reagent

4. reduction

slide39

Alcohols:

nomenclature

syntheses

later

reactions

1. HX

2. PX3

3. dehydration

4. as acids

5. ester formation

6. oxidation