Lecture%21:%November%6,%2012 -...

CHM 223 • Organic Chemistry IFall 2012, Des Plaines • Prof. Chad Landrie

More Addi)on Reac)ons to Alkenes:Sec)ons 8.2-‐8.5; 287

Oxida)on of Alkenes:Sec)ons 8.7-‐8.8

Lecture 21: November 6, 2012


Compe))on Between Subs)tu)on and Elimina)on:

Sec)on 11.12

We Almost Forgot. . .

© 2012, Dr. Chad L. LandrieOrganic Chemistry 1 (CHM 223)

Slide Lecture 21: November 6

Elimina'on-‐Subs'tu'on Compe''on

3

When the nucleophile is anionic, elimination competes with substitution




4


BrHCH3CH2O

BrHCH3CH2O

E2

SN2

OCH2CH3

Anion must be more basic than hydroxide (pKa of conjugate acid > 15.7) for E1 or E2 to be a competing mechanism




5

Br

Br

E1

SN1

H

OCH2CH3

OCH2CH3

OCH2CH3


Anion must be more basic than hydroxide (pKa of conjugate acid > 15.7) for E1 or E2 to be a competing mechanism



1.Low steric hinderance: a. small nucleophilesb.least substituted alkyl halide possible

2.Neutral nucleophiles (like solvolysis)3.Or anionic nucleophiles less basic than OH–

Condi'ons Favoring Subs'tu'on

6






7






8

KOC(CH3)3 is so large that it prefers E2 even when the alkyl halide is primary.






9

Neutral nucleophiles are not basic enough to deprotonate a β-hydrogen in an E2 mechanism; they also undergo addition to carbocations faster in

SN1 than deprotonation in an E1 mechanism.

CH3Cl+

methanol

CH3OmethanolysisH3COH

H3Ckrel = 4






10

pKa (H2O) = 15.7 pKa (HCN) = 9.1stonger acid = weaker conjugate base



1.Large steric interactions: a. large nucleophilesb.most substituted alkyl halide possible

2. Anionic nucleophiles more basic than OH–

Condi'ons Favoring Elimina'on

11

CH3Br

O CCH3

CH3CH3K++

CH3

DMSO

pKa (H2O) = 15.7 pKa (tert-butanol) = 18stonger acid = weaker conjugate base



i>Clicker Ques'on

12

H3CO

H3CO

CH2

H3CO

H3CO

CH3

H3CO

H3CO

CH3

NC

S

H3CO

H3CO

CH3

SCN

H3CO

H3CO

CH3

ClS C N+ S C N

DMF

DMF = N,N-dimethylformamide(common polar aprotic solvent) N

O

HCH3

H3C

What is the major product of the reaction below?

Chauncey says. . .

A

B

C

D



i>Clicker Ques'on

13

Although the Markovnikov rule generally applies to the addition of hydrogen halides across unsymmetrically substituted alkenes, this terminology is often expanded to include other additions to alkenes. Which reaction, then, would provide the anti-Markovnikoff product?

Markovnikoff rule: H is added to least substituted C of alkene; X is added to most substituted carbon of alkene

A

B

C

D

E. A & D

Markovnikov

anti-Markovnikov

Markovnikov

neither


Sec)on 8.5You are responsible for sec0on 8.4 and 8.9!

Hydrobora)on–Oxida)on



Pa=ern: Hydrobora'on

15

C C X Y+ addition C CX Y

C C H BH2+ hydroboration C CH BH2alkene borane alkyl borane

alcohol

oxidation

C CH OH

oxidation is a separate step and requires a separate set of conditions and reagents



Condi'ons: Hydrobora'on

16

B HHH

• boron is an exception to octet rule• sp2-hybridized (only 3 valence e-s)• contains an empty p-orbital• strong Lewis acid (electron pair acceptor)• forms 3-bonds (neutral) & 4-bonds (-/ve)

2 BH3dimerization B2H6

“Free” borane (BH3) only exists in gas phase, otherwise undergoes dimerization to diborane (B2H6)

borane diborane




17


borane diborane

B BH

H

H HH

HB

HB

HHH

HH+ dimerization

B HHH





18


borane diborane

BH

HH B

H HH dimerization

B HHH





19

OO

OCH3H3C

diglyme: diethylene glycol dimethyl ether

• diglyme is a common solvent (reaction liquid; does not participate in the reaction)

• hydroborations typically at room temperature (rt) or below• generally very fast reactions




20

Other hydroborating reagents with different reactivities may be prepared by adding Lewis bases to diborane (B2H6)

B2H6O

tetrahydrofuran (THF)

H3CS

CH3

dimethyl sulfide (DMS)

N

triethylamine (TEA)

CH3

SCH3

OH3B

H3B•THF

H3B

H3B•DMS

N(CH2CH3)3H3B

H3B•TEA

diborane+

+

+

B2H6diborane

B2H6diborane

}Le

wis

Aci

d-Ba

se C

ompl

exes



Condi'ons: Oxida'on

21

HBH2

H2O2, NaOH (aq)oxidation

HOH

• boron is replaced by more electronegative atom (O); therefore: oxidation

• only requires hydrogen peroxide (H2O2) and hydroxide (OH–); usually added immediately following hydroboration

• I will not ask you to learn the mechanism for this step; for the curious, see textbook page 250



• usually written as a two-step process

• works with diborane (B2H6) or a borane complex with another Lewis base

• diglyme is a common solvent

• affords alcohol products; similar to hydration (H2O/H3O+)

Pa=ern & Condi'ons: Hydrobora'on-‐Oxida'on

22

1. B2H6, diglyme

2. H2O2, NaOHOH

CH3 1. H3B•S(CH3)2

2. H2O2, NaOH

CH3

OH

1. H3B•S(CH3)2

2. H2O2, NaOHOH

Which carbon of the alkene becomes bonded to the -OH group? Do you see a pattern here?



Hydrobora'on is Regioselec've

23

O O O OHBH2H3B•THF H2O2/OH–

• boron atom is added to least substituted carbon atom, hydrogen atom is added to most substituted

• after oxidation, gives the least substituted product• opposite regioselectivity of Markovnikov addition

compare to . . .

OO OH2SO4 H2O/heat

OSO3H OH



Hydrobora'on Mechanism

24

• concerted processes = B and H add to same side (face) of double bond

• hydride (H:) adds to most substituted carbon• boron adds to least substituted carbon• this is NOT a PROTONATION

CH3

H2B H

CH3

HH2B

Why this regioselectivity?




25

Regioselectivity Argument One: Boron is larger than hydrogen; it prefers to add to the least sterically hindered side of the double bond, which is the least substituted side.

C

B H

CH3

HH2BH

HH

HH

CH3

BH HH

CH3

BH2H

X




26

Regioselectivity Argument Two: both carbons in alkene develop partial positive charge in transition state; more substituted = more partially positively charged = hydride (H:) transfer prefered (faster) to that carbon atom.

π-complex

Lewis acid

Lewis base

π-complex rearrangedπ-complex

most substituted = most partially positively charged =most electrophilic carbon atom =

want electrons from a nucleophile most



Stereochemistry of Hydrobora'on

27

Hydroboration is a syn addition since the mechanism is concerted. Bonds are being formed to the boron atom and the hydrogen atom at the same time from one face of the π-bond.

syn



i>Clicker Ques'on

28

1. Br2, CH2Cl2, hv2. NaOC(CH3)2, DMSO

3. B2H6, diglyme4. H2O2, NaOH (aq)

HO

HO

HO

Br

HO

What is the major organic product after the following sequence of reactions?

A.

B.

C.

D.

E.


Sec)on 8.2-‐8.3

Halogen Addi)on to Alkenes



Pa=ern: Halogen Addi'on

30

C C X Y+ addition C CX Y

halogen must either be bromine (Br2) or chlorine (Cl2); no other halogens work

vicinal dibromide

vicinal dichloride

C C Br Br+ halide addition C CBr Br

C C Cl Cl+ halide addition C CCl Cl



Condi'ons: Halogen Addi'on

31

• only bromine (Br2) and chlorine (Cl2) undergo this reaction• typically use a halogenated solvent like chloroform (CHCl3)• low temperature (0 ºC)• no light (avoid radical halogenation at saturated carbons)



Halogen Addi'on is Stereoselec've

32

• anti addition of Br2 or Cl2 across double bonds• two halogen atoms are trans to each other in rings



Halogen Addi'on Mechanism? No!

33

X

• concerted process would require syn addition of to two halogen atoms

• therefore mechanism is not conerted.




34

CH2BrHH

• carbocation intermediate are planar (flat)• addition of second halide could approach either lobe of empty p-

orbital on carbocation• therefore this mechanism is not possible because it would not be

stereoselective

X




35

X

Since only trans vicinal dihalides are obtained, a carbocation intermediate is not possible since it would give a mixture of trans and cis products.

+



Correct Halogen Addi'on Mechanism: Halonium Ion Intermediates

36

H3C CH3

Br Br

H3C CH3

Br

Br

δ+ δ+

δ−

H3C CH3

Br

Br H3C

CH3Br

Br

• electron pair in π-bond form new bond to bromine atom as Br-Br bond breaks

• lone pair of electrons on bromide form new bond to alkene carbon losing a bond at the same time (concerted)

• halonium ions (three-membered rings with one halogen) are intermediates; no carbocation intermediates

halonium ion



An# Addi'on: Steric Perspec've

37

BrBr....

....

BrBr

::....

....::::

....

––BrBr:: ::....

....

attack of bromide (Br–) from opposite side C-Br bond that is

breaking gives anti product Br



An# Addi'on: MO Perspec've

38

BrBr....

....

BrBr

::....

....::::

....

––BrBr:: ::....

....

bonding (σ) orbital of Br– overlaps best with the antibonding (σ*) orbital of the C-Br bond from

the back Br




39

BrBr....

....

BrBr

::....

....::::

....

––BrBr:: ::....

....


the back Br

attack from the side involves equal amounts of constructive and destructive interference = no change




40

BrBr....

....

BrBr

::....

....::::

....


the back Br

attack at Br involves electrostatic repulsion as well as weaker orbital overlap; notice in the C-Br bond, the orbital on Br is smaller than the orbital on carbon. Don’t worry about why.

––BrBr:: ::....

....



Rela've Rates of Bromina'on

41

C C CHH

H

• more alkyl groups attached to double bond =• more σ➞π orbital donation/overlap (hyperconjugation) • double bond has more electron density =• double bond is a stronger nucleophile (Lewis base) =• reacts faster with electrophiles like Br-Br

H

H H

H

H3C

H H

H

H3C

H3C H

H

H3C

H3C CH3

CH3

alkene relative rate (krel)

1

61

5400

920,000

σ➞∏donation



Halogen Addi'on Modifica'on: Aqueous Solu'ons Give Halohydrins

42

• if the solvent for the reaction is changed to water, halohydrins are formed

• since water is much more concentrated, it will add to halonium ion intermediate faster than the halide

bromohydrin



Forma'on of Halohydrins is Regioselec've

43

H3C H

Br Br

H3C H

Br

OH2HH3C

H3CH3C

HH

Br

OH

HH3C H

Br H3CH3C

HH

Br

HOOH2

• product is most substituted alcohol; least substituted halide

• most substituted carbon is most partially positively charged in transition state = strongest electrophile = water most attracted to that carbon



i>Clicker Ques'on

44

What is the major organic product obtained after the following sequence of reactions?

A

B

C

D

E. none of the above

Nucleophile adds to the most substituted carbon of the halonium ion.


Sec)ons 8.7-‐8.8

Oxida)on of Alkenes: Epoxida)on, Ozonolysis and

Dihydroxyla)on



Expoxides

46

• three-membered rings containing an oxygen atom• common intermediates in organic synthesis



Epoxida'on: Pa=ern

47

H3C

O

OHacetic acid

H3C

O

Operoxyacetic acid

OH

• peroxyacids are source of electrophilic oxygen atoms• addition of a single oxygen atom across double bond• similar to formation of bromonium ion intermediate• don’t worry about mechanism; for curious see textbook page 259



Epoxida'on: Examples

48

• any peroxyacid will work• more substituted double bonds react faster• because only adding one atom across double bond, must be syn addition



Ozonolysis: Pa=ern

49

ozone

• ozone adds across double bonds to give ozonides• ozonides react further with reducing agents to provide

carbonyl compounds• we will not discuss mechanism for this reaction

C O CO



Ozonolyis: Condi'ons

50

reducing agents;reduce ozonide intermediate

pattern = alkene is cut in half and replaced by an oxygen atom at each end

carbonyl product depends on substitution of alkene



Prepara'on of Diols: Dihydroxyla'on

51

We will not learn the mechanism, just the pattern.



Prepara'on of Diols: Dihydroxyla'on

52

Dihydroxylation is syn addition of vicinal hydroxyl (-OH) groups. This reaction can be stereospecific then.


Synthesis



Retrosynthe'c Analysis

54

How could you prepare 1-methylcyclohexene from methylcyclohexane?

CH3 ? CH3

Step One: Devise a retrosynthesis. Work backwards from the target molecule. For each step backward, you should ask, “How can I prepare the target functional group?”

CH3 CH3Br

CH3

dehydrohalogenation radical bromination



Synthe'c Route

55

How could you prepare 1-methylcyclohexene from methylcyclohexane?

CH3 ? CH3

Step Two: Write the synthesis out in the forward direction including all necessary reagents and conditions.

CH3 Br2, CHCl3

light, 60 ºC

CH3Br

NaOCH2CH3

DMSO, 55 ºC

CH3



Complimentary Retrosyntheses

56

?CH3 H3C OH

?CH3 CH3

OH

H3C OH

CH3OH

CH3 H3C Br CH3

1. H2SO42. H2O/heat

1. B2H6

2. NaOH

/H2O2



i>Clicker Ques'on

57

OHBr

Br

OHBr

OH

OHBr

OH

OHBr Br

OH

B.

C.

D.

Br2, CHCl3

light, 60 ºCBr

NaOCH2CH3

DMSO, 55 ºC

Br2

H2O/THF

OHBr

Which retrosynthetic analysis provides the best route for the synthesis of 2-bromocyclopentanol?

A.



i>Clicker Ques'on

58

Devise a retrosynthesis for the synthetic target below.

?

Cl

Cl

Cl

Cl OH

Cl O

A.

B.

C.

D.

1. B2H6, diglyme2. H2O2, NaOH

OH

SOCl2K2CO3



i>Clicker Ques'on

59

HO

HO

H

OH

OBr

A.

B.

C.

D.

E.

Which starting material could be used to prepare the molecule below? Write out your synthesis. Don’t guess.


Next Lecture. . .9.1 -‐ 9.9

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