alkyl halide

Post on 15-Jul-2015

184 Views

Category:

Education

0 Downloads

Preview:

Click to see full reader

TRANSCRIPT

Chapter 6Alkyl Halides: Nucleophilic Substitution and Elimination

Organic Chemistry

Chapter 6 2

Classes of Halides Alkyl: Halogen, X, is directly bonded to sp3

carbon. Vinyl: X is bonded to sp2 carbon of alkene. Aryl: X is bonded to sp2 carbon on benzene

ring. Examples:

C

H

H

H

C

H

H

Br

alkyl halide

C CH

H

H

Cl

vinyl halide

I

aryl halide

=>

Chapter 6 3

Polarity and Reactivity

Halogens are more electronegative than C. Carbon-halogen bond is polar, so carbon has

partial positive charge. Carbon can be attacked by a nucleophile. Halogen can leave with the electron pair.

=>

CH

HH

Br

δ+ δ-

Chapter 6 4

Classes of Alkyl Halides

Methyl halides: only one C, CH3X

Primary: C to which X is bonded has only one C-C bond.

Secondary: C to which X is bonded has two C-C bonds.

Tertiary: C to which X is bonded has three C-C bonds. =>

Chapter 6 5

Classify These:

CH3 CH CH3

Cl

CH3CH2F

(CH3)3CBr CH3I =>

Chapter 6 6

Dihalides

Geminal dihalide: two halogen atoms are bonded to the same carbon

Vicinal dihalide: two halogen atoms are bonded to adjacent carbons.

C

H

H

H

C

H

Br

Br

geminal dihalide

C

H

H

Br

C

H

H

Br

vicinal dihalide

=>

Chapter 6 7

IUPAC Nomenclature

Name as haloalkane. Choose the longest carbon chain, even if the

halogen is not bonded to any of those C’s. Use lowest possible numbers for position.

CH3 CH CH2CH3

Cl CH3(CH2)2CH(CH2)2CH3

CH2CH2Br

2-chlorobutane 4-(2-bromoethyl)heptane=>

Chapter 6 8

Systematic Common Names

Name as alkyl halide. Useful only for small alkyl groups. Name these:

CH3 CH CH2CH3

Cl

(CH3)3CBr

CH3 CH

CH3

CH2F =>

Chapter 6 9

“Trivial” Names

CH2X2 called methylene halide. CHX3 is a haloform. CX4 is carbon tetrahalide. Examples:

CH2Cl2 is methylene chloride

CHCl3 is chloroform

CCl4 is carbon tetrachloride.

=>

Chapter 6 10

Uses of Alkyl Halides Solvents - degreasers and dry cleaning fluid Reagents for synthesis of other compounds Anesthetic: Halothane is CF3CHClBr

CHCl3 used originally (toxic and carcinogenic)

Freons, chlorofluorocarbons or CFC’s Freon 12, CF2Cl2, now replaced with Freon 22,

CF2CHCl, not as harmful to ozone layer.

Pesticides - DDT banned in U.S. =>

Chapter 6 11

Dipole Moments µ = 4.8 x δ x d, where δ is the charge

(proportional to ∆EN) and d is the distance (bond length) in Angstroms.

Electronegativities: F > Cl > Br > I Bond lengths: C-F < C-Cl < C-Br < C-I Bond dipoles: C-Cl > C-F > C-Br > C-I

1.56 D 1.51 D 1.48 D 1.29 D

Molecular dipoles depend on shape, too!

=>

Chapter 6 12

Boiling Points

Greater intermolecular forces, higher b.p. dipole-dipole attractions not significantly different for

different halides London forces greater for larger atoms

Greater mass, higher b.p. Spherical shape decreases b.p.

(CH3)3CBr CH3(CH2)3Br 73°C 102°C =>

Chapter 6 13

Densities

Alkyl fluorides and chlorides less dense than water.

Alkyl dichlorides, bromides, and iodides more dense than water. =>

Chapter 6 14

Preparation of RX

Free radical halogenation (Chapter 4)produces mixtures, not good lab synthesisunless: all H’s are equivalent, orhalogenation is highly selective.

Free radical allylic halogenationproduces alkyl halide with double bond on the

neighboring carbon. =>

Chapter 6 15

Halogenation of Alkanes All H’s equivalent. Restrict amount of

halogen to prevent di- or trihalide formation

Highly selective: bromination of 3° C =>

+ HBr

H

BrhνBr2+

H

H

90%

+ HBrCH3 C

CH3

CH3

Brhν

Br2+CH3 C

CH3

CH3

H

Chapter 6 16

Allylic Halogenation Allylic radical is resonance stabilized. Bromination occurs with good yield at the

allylic position (sp3 C next to C=C). Avoid a large excess of Br2 by using

N-bromosuccinimide (NBS) to generate Br2 as product HBr is formed.

N

O

O

Br + HBr N

O

O

H + Br2 =>

Chapter 6 17

Reaction Mechanism

Free radical chain reaction initiation, propagation, termination.

H H

BrH

+ HBrBr

Br

H Br

+ Br

=>

2 BrBr2hν

Chapter 6 18

Substitution Reactions

The halogen atom on the alkyl halide is replaced with another group.

Since the halogen is more electronegative than carbon, the C-X bond breaks

heterolytically and X- leaves.

The group replacing X- is a nucleophile. =>

C C

H X

+ Nuc:-C C

H Nuc

+ X:-

Chapter 6 19

Elimination Reactions

The alkyl halide loses halogen as a halide ion, and also loses H+ on the adjacent carbon to a base.

A pi bond is formed. Product is alkene. Also called dehydrohalogenation (-HX).

=>

C C

H X

+ B:- + X:- + HB C C

Chapter 6 20

SN2 Mechanism

Bimolecular nucleophilic substitution. Concerted reaction: new bond forming

and old bond breaking at same time. Rate is first order in each reactant. Walden inversion.

=>

CH

Br

HH

H O CHO Br

H

HH

CHO

H

HH

+ Br-

Chapter 6 21

SN2 Energy Diagram

One-step reaction. Transition state is highest in energy. =>

Chapter 6 22

Uses for SN2 Reactions Synthesis of other classes of compounds. Halogen exchange reaction.

=>

Chapter 6 23

SN2: Nucleophilic Strength Stronger nucleophiles react faster. Strong bases are strong nucleophiles, but

not all strong nucleophiles are basic.

=>

Chapter 6 24

Trends in Nuc. Strength

Of a conjugate acid-base pair, the base is stronger: OH- > H2O, NH2

- > NH3

Decreases left to right on Periodic Table. More electronegative atoms less likely to form new bond: OH- > F-, NH3 > H2O

Increases down Periodic Table, as size and polarizability increase: I- > Br- > Cl-

=>

Chapter 6 25

Polarizability Effect

=>

Chapter 6 26

Bulky Nucleophiles

Sterically hindered for attack on carbon, so weaker nucleophiles.

CH3 CH2 O ethoxide (unhindered)weaker base, but stronger nucleophile

C

CH3

H3C

CH3

O

t-butoxide (hindered)stronger base, but weaker nucleophile

=>

Chapter 6 27

Solvent Effects (1)Polar protic solvents (O-H or N-H) reduce

the strength of the nucleophile. Hydrogen bonds must be broken before nucleophile can attack the carbon.

=>

Chapter 6 28

Solvent Effects (2)

Polar aprotic solvents (no O-H or N-H) do not form hydrogen bonds with nucleophile

Examples:

CH3 C Nacetonitrile C

O

H3C CH3

acetone =>

dimethylformamide (DMF)

CH

O

NCH3

CH3

Chapter 6 29

Crown Ethers

Solvate the cation, so nucleophilic strength of the anion increases.

Fluoride becomes a good nucleophile.

O

O

O

O

OO

K+

18-crown-6

CH2Cl

KF, (18-crown-6)

CH3CN

CH2F

=>

Chapter 6 30

SN2: Reactivity of Substrate

Carbon must be partially positive. Must have a good leaving group Carbon must not be sterically hindered.

=>

Chapter 6 31

Leaving Group Ability Electron-withdrawing Stable once it has left (not a strong base) Polarizable to stabilize the transition state.

=>

Chapter 6 32

Structure of Substrate

Relative rates for SN2: CH3X > 1° > 2° >> 3°

Tertiary halides do not react via the SN2 mechanism, due to steric hindrance. =>

Chapter 6 33

Stereochemistry of SN2

Walden inversion

=>

Chapter 6 34

SN1 Reaction

Unimolecular nucleophilic substitution. Two step reaction with carbocation

intermediate. Rate is first order in the alkyl halide,

zero order in the nucleophile. Racemization occurs.

=>

Chapter 6 35

SN1 Mechanism (1)Formation of carbocation (slow)

(CH3)3C Br (CH3)3C+

+ Br-

=>

Chapter 6 36

SN1 Mechanism (2) Nucleophilic attack

(CH3)3C+

+ H O H (CH3)3C O H

H

(CH3)3C O H

H

H O H+ (CH3)3C O H + H3O+

=>

• Loss of H+ (if needed)

Chapter 6 37

SN1 Energy Diagram

Forming the carbocation is endothermic

Carbocation intermediate is in an energy well.

=>

Chapter 6 38

Rates of SN1 Reactions

3° > 2° > 1° >> CH3X Order follows stability of carbocations (opposite to

SN2)

More stable ion requires less energy to form

Better leaving group, faster reaction (like SN2)

Polar protic solvent best: It solvates ions strongly with hydrogen bonding. =>

Chapter 6 39

Stereochemistry of SN1Racemization:

inversion and retention

=>

Chapter 6 40

Rearrangements

Carbocations can rearrange to form a more stable carbocation.

Hydride shift: H- on adjacent carbon bonds with C+.

Methyl shift: CH3- moves from adjacent

carbon if no H’s are available.

=>

Chapter 6 41

Hydride Shift

CH3 C

Br

H

C

H

CH3

CH3CH3 C

H

C

H

CH3

CH3

CH3 C

H

C

H

CH3

CH3CH3 C

H

C

CH3

CH3

H

CH3 C

H

C

CH3

CH3

HNuc

CH3 C

H

C

CH3

CH3

H Nuc

=>

Chapter 6 42

Methyl Shift

CH3 C

Br

H

C

CH3

CH3

CH3CH3 C

H

C

CH3

CH3

CH3

CH3 C

H

C

CH3

CH3

CH3CH3 C

H

C

CH3

CH3

CH3

CH3 C

H

C

CH3

CH3

CH3

NucCH3 C

H

C

CH3

CH3

CH3 Nuc

=>

Chapter 6 43

SN2 or SN1?

Primary or methyl Strong nucleophile

Polar aprotic solvent

Rate = k[halide][Nuc] Inversion at chiral

carbon No rearrangements

Tertiary Weak nucleophile (may

also be solvent)

Polar protic solvent, silver salts

Rate = k[halide] Racemization of optically

active compound

Rearranged products

=>

Chapter 6 44

E1 Reaction

Unimolecular elimination Two groups lost (usually X- and H+) Nucleophile acts as base Also have SN1 products (mixture)

=>

Chapter 6 45

E1 Mechanism

Halide ion leaves, forming carbocation. Base removes H+ from adjacent carbon. Pi bond forms. =>

H C

H

H

C

CH3

CH3

Br

C

H

H

H

C CH3

CH3

O

H

H

C

H

H

H

C CH3

CH3

C C

H

CH3

CH3

H+ H3O+

Chapter 6 46

A Closer LookO

H

H

C

H

H

H

C CH3

CH3

C C

H

CH3

CH3

H+ H3O+

=>

Chapter 6 47

E1 Energy Diagram

Note: first step is same as SN1=>

Chapter 6 48

E2 Reaction

Bimolecular elimination Requires a strong base Halide leaving and proton abstraction

happens simultaneously - no intermediate.

=>

Chapter 6 49

E2 Mechanism

H C

H

H

C

CH3

CH3

Br

C C

H

CH3

CH3

H

O

H+ H2O Br

-+

=>

Chapter 6 50

Saytzeff’s Rule If more than one elimination product is possible,

the most-substituted alkene is the major product (most stable).

R2C=CR2>R2C=CHR>RHC=CHR>H2C=CHR tetra > tri > di > mono

C C

Br

H

C

H

CH3

H

H

H

CH3

OH-

C CH

HC

H H

CH3

CH3

C

H

H

H

C

H

CCH3

CH3

+

=>minor major

Chapter 6 51

E2 Stereochemistry

=>

Chapter 6 52

E1 or E2? Tertiary > Secondary Weak base Good ionizing solvent

Rate = k[halide] Saytzeff product No required geometry

Rearranged products

Tertiary > Secondary Strong base required Solvent polarity not

important Rate = k[halide][base] Saytzeff product Coplanar leaving

groups (usually anti) No rearrangements

=>

Chapter 6 53

Substitution or Elimination? Strength of the nucleophile determines

order: Strong nuc. will go SN2 or E2.

Primary halide usually SN2.

Tertiary halide mixture of SN1, E1 or E2

High temperature favors elimination. Bulky bases favor elimination. Good nucleophiles, but weak bases,

favor substitution. =>

Chapter 6 54

Secondary Halides?

Mixtures of products are common.

=>

Chapter 6 55

End of Chapter 6

top related