17
Chemistry 1102 Charlie Bond MCS Rm 4.16/4.27 [email protected] What is Organic Chemistry? Organic Reactions I II Alkanes (Ch 21) Conformational Analysis (Ch 21) Stereochemistry I II III (Ch 22) Alkyl Halides I II II (Ch 24) Alcohols and Ether I II (Ch 24)

CHEM1102 Lecture Notes 10

Embed Size (px)

DESCRIPTION

greag

Citation preview

Page 1: CHEM1102 Lecture Notes 10

Chemistry 1102Charlie Bond

MCS Rm 4.16/4.27

[email protected]

What is Organic Chemistry?

Organic Reactions I II

Alkanes (Ch 21)

Conformational Analysis (Ch 21)

Stereochemistry I II III (Ch 22)

S

Alkyl Halides I IIII (Ch 24)

Alcohols and Ether I II (Ch 24)

Page 2: CHEM1102 Lecture Notes 10

2

Substitution & Elimination• Alkyl halides are used to demonstrate 2 types of reaction

– nucleophilic substitution

− β -elimination

Br

H

EtO-Na+

OEt

EtOH

Na+Br

-

Na+Br-

ethanol

ethanol

+

+

+

as a nucleophile,ethoxide ion attacksthis carbon

a nucleophile and a base

as a base, ethoxide ionattacks this hydrogen

+

nucleophilicsubstitution

β­elimination

Page 3: CHEM1102 Lecture Notes 10

3

β -Elimination∀ ββ -Elimination-Elimination: removal of atoms or

groups of atoms from adjacent carbons to form a carbon-carbon double bond– we study a type of β -elimination called

dehydrohalogenationdehydrohalogenation (the elimination of HX)

C CH X

CH3CH2O-Na+

C C

CH3CH2OH

CH3CH2OH Na+X -+

+β α

+

An alkyl halide

Base

An alkene

Page 4: CHEM1102 Lecture Notes 10

4

β -Elimination• Zaitsev rule:Zaitsev rule: the major product of a β -

elimination is the more stable (the more highly substituted) alkene

Br CH3CH2O-Na+

CH3CH2OH2­Methyl­2­butene  (major product)

   2­Bromo­2­methylbutane

2­Methyl­1­butene

+

Br CH3O-Na+

CH3OH+

1­Methyl­cyclopentene

(major product)

1­Bromo­1­methyl­cyclopentane

Methylene­cyclopentane

Page 5: CHEM1102 Lecture Notes 10

5

β -Elimination• Zaitsev rule:Zaitsev rule: the major product of a β -

elimination is the more stable alkene

• Note: Trans alkenes are more stable than cis.

Page 6: CHEM1102 Lecture Notes 10

6

β -Elimination

• There are two limiting mechanisms for β -elimination reactions

• E1 mechanism:E1 mechanism: at one extreme, breaking of the C-X bond is complete before reaction with base breaks the C-H bond– only R-X is involved in the rate-determining step

• E2 mechanism:E2 mechanism: at the other extreme, breaking of the C-X and C-H bonds is concerted– both R-X and base are involved in the rate-

determining step

Page 7: CHEM1102 Lecture Notes 10

7

E1 Mechanism

– Step 1: ionization of C-X gives a carbocation intermediate

– Step 2: proton transfer from the carbocation intermediate to a base (in this case, the solvent) gives the alkene

CH2-C-CH3

Br

CH3

CH3-C-CH3

CH3

Br –slow, rate

determining+

(A carbocation intermediate)

+

HO

H3CH-CH2-C-CH3

CH3

HOH

H3CCH2=C-CH3

CH3fast+

+ ++

Page 8: CHEM1102 Lecture Notes 10

8

Page 9: CHEM1102 Lecture Notes 10

9

E2 Mechanism

• A one-step mechanism; all bond-breaking and bond-forming steps are concerted

CH3CH2OCH3

H-CH-CH2-Br

CH3CH2O-H CH3CH=CH2 Br

+

+ +

Page 10: CHEM1102 Lecture Notes 10

10

Page 11: CHEM1102 Lecture Notes 10

11

Elimination Reactions

• Summary of E1 versus E2 Reactions for Haloalkanes

RCH2X

R2CHX

R3CX

Haloalkane E1 E2

Primary

Secondary

Tertiary

E1 does not occur.Primary carbocations areso unstable that they are never observed in solution.

E2 is favored.

Main reaction with strong bases such as OH­ and OR­.

Main reaction with weak bases such as H2O and ROH.

Main reaction with strong bases such as OH­ and OR­.

Main reaction with weak bases such as H2O and ROH.

Page 12: CHEM1102 Lecture Notes 10

12

Substitution vs Elimination

• Many nucleophiles are also strong bases (OH- and RO-) and SN and E reactions often compete– the ratio of SN/E products depends on the

relative rates of the two reactions

nucleophilicsubstitution

β­eliminationC CH X + Nu­

C CH Nu +

C C H­Nu+ +

Page 13: CHEM1102 Lecture Notes 10

13

SN1 versus E1

• Reactions of 2° and 3° haloalkanes in polar protic solvents give mixtures of substitution and elimination products

CH3

CH3

ICCH3 -I-

CH3 C

CH3 SN1

CH3-Cl-H2O

CH3OH

SN1ClCH3

CH3

CH3C

CH2

CH3

CH3

C

C

CH3

CH3

CH3

CCH3

CH3

CH3

OH

OCH3

H+

H+

H+

E1

+

+

+

+

or

Page 14: CHEM1102 Lecture Notes 10

14

SN2 versus E2

• It is considerably easier to predict the ratio of SN2 to E2 products

α

βleaving groupC

C

RR

H

RRAttack of base on a β­hydrogen by E2 is  only slightly affected by branching at the α­carbon; alkene formation is accelerated

SN2 attack of a nucleophile isimpeded by branching at theα­ and β­carbons

Page 15: CHEM1102 Lecture Notes 10

15

Summary of S vs E for Haloalkanes

– for methyl and 1°haloalkanes

RCH2X

CH3X

SN1 and E1 reactions of primary halides are never observed.

SN2

SN1 reactions of methyl halides are never observed.The methyl cation is so unstable that it is never formed in solution.

SN2

E2 The main reaction with strong, bulky bases, such as potassium tert­butoxide.

Primary cations are never formed in solution; therefore,

Methyl

Primary

SN1/E1

SN1

The only substitution reactions observed

The main reaction with strong bases such as OH­  andEtO­. Also, the main reaction with good nucleophiles/weak bases, such as I­ and CH3COO­.

Page 16: CHEM1102 Lecture Notes 10

16

Summary of S vs E for Haloalkanes

– for 2° and 3° haloalkanes

The main reaction with strong bases/good nucleophiles

R3CX

such as I­ and CH3COO­.R2CHX

Main reaction with strong bases, such as HO­ and RO­.

Main reactions with poor nucleophiles/weak bases.

The main reaction with weak bases/good nucleophiles,

E2

SN2

E2

SN 2 reactions of tertiary halides are never observed

SN1/ E1

Secondary

Tertiarybecause of the extreme crowding around the 3° carbon.

SN1/ E1 Common in reactions with weak nucleophiles in polarprotic solvents, such as water, methanol, and ethanol.

such as OH­ and CH3CH2O­.

SN2

Page 17: CHEM1102 Lecture Notes 10

17

Summary of S vs E for Haloalkanes

– Examples: predict the major product and the mechanism for each reaction

ClNaOH 80°C

H2O+1.

Br(C2H5)3N

30°CCH2Cl2

+2.

BrCH3O- Na+

methanol3. +

Cl

Na+

I-4. + acetone