Chapter 6Alkyl Halides: Nucleophilic Substitution and Elimination

Organic Chemistry, 5th EditionL. G. Wade, Jr.

Jo BlackburnRichland College, Dallas, TX

Dallas County Community College District2003,Prentice Hall

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

+ -

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 chlorideCHCl3 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 CF3CHClBrCHCl3 used originally (toxic and carcinogenic)

• Freons, chlorofluorocarbons or CFC’sFreon 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

• = 4.8 x x d, where 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 halidesLondon forces greater for larger atoms

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

(CH3)3CBr CH3(CH2)3Br 73C 102C =>

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 3C =>

+ HBr

H

BrhBr2+

H

H

90%

+ HBrCH3 C

CH3

CH3

Brh

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 reactioninitiation, propagation, termination.

H H

BrH

+ HBrBr

Br

H Br

+ Br

=>

2 BrBr2h

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° >> CH3XOrder 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