chapter 6 alkyl halides: nucleophilic substitution and elimination
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DESCRIPTIONOrganic Chemistry , 6 h Edition L. G. Wade, Jr. Chapter 6 Alkyl Halides: Nucleophilic Substitution and Elimination. =>. Classes of Halides. Alkyl: Halogen, X, is directly bonded to sp 3 carbon. Vinyl: X is bonded to sp 2 carbon of alkene. - PowerPoint PPT Presentation
Chapter 6Alkyl Halides: Nucleophilic Substitution and Elimination
Organic Chemistry, 6h EditionL. G. Wade, Jr.
Chapter 6 2
Classes of Halides• Alkyl: Halogen, X, is directly bonded to sp3
• Vinyl: X is bonded to sp2 carbon of alkene.
• Aryl: X is bonded to sp2 carbon on benzene ring. Examples:
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.
Chapter 6 4
6-2 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
Chapter 6 5
Systematic Common Names
• Name as alkyl halide.
• Useful only for small alkyl groups.
• Name these:
CH3 CH CH2CH3
Chapter 6 6
• 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 7
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 8
CH3 CH CH3
(CH3)3CBr CH3I =>
Chapter 6 9
Dihalides• Geminal dihalide: two halogen atoms
are bonded to the same carbon
• Vicinal dihalide: two halogen atoms are bonded to adjacent carbons.
Chapter 6 10
6-3 Uses of Alkyl Halides• Solvents - degreasers and dry cleaning fluid• Reagents for synthesis of other compounds• Anesthetics: Halothane is CF3CHClBr
CHCl3 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
6-5 Physical Properties of RX
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 73C 102C =>
Chapter 6 13
• Alkyl fluorides and chlorides less dense than water.
• Alkyl dichlorides, bromides, and iodides more dense than water. =>
Chapter 6 14
6-6 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, etc formation
• Highly selective: bromination of 3C
CH3CH3 Cl2+ CH3CH2Cl + CH3CHCl2 + CH3CCl3 +ClCH2CCl3 + Cl2CHCCl3 + Cl3CCCl3
Chapter 6 16
• 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.
Chapter 6 17
Reaction MechanismFree radical chain reaction
initiation, propagation, termination.
(1) 2 BrBr2h
Br Br C Br
. + HBr
. + Br.
R R2R .(4)
Chapter 6 18
• 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. =>
+ Nu + X--
Chapter 6 19
• 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).
+ Base + X-
Chapter 6 20
• Bimolecular nucleophilic substitution.
• Concerted reaction: new bond forming and old bond breaking at same time.
• Rate is first order in each reactant.
• Inversion of configuration. =>
Chapter 6 21
SN2 Energy Diagram
• One-step reaction.
• Transition state is highest in energy. =>
Chapter 6 22
6-9 Generality of SN2 RXN• Synthesis of other classes of compounds.• Halogen exchange reaction.
Chapter 6 23
6-10 Factors Affecting SN2: Nucleophilic Strength
• Stronger nucleophiles react faster.• Strong bases are strong nucleophiles, but
not all strong nucleophiles are basic.
Chapter 6 24
Trends in Nucleophilic 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
Chapter 6 26
Steric EffectBulky Nucleophiles
Sterically hindered for attack on carbon, so weaker nucleophiles.
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
Chapter 6 29
Crown Ethers• Solvate the cation, so nucleophilic strength
of the anion increases.• Fluoride becomes a good nucleophile.
Chapter 6 30
Leaving Group Ability
• Stable once it has left (not a strong base)
• Polarizable to stabilize the transition state.
Chapter 6 31
Structure of Substrate
• Relative rates for SN2: CH3X > 1° > 2° >> 3°
• Tertiary halides do not react via the SN2 mechanism, due to steric hindrance.
Structure of Substrate
Chapter 6 32
Steric Hindrance• Nucleophile approaches from the back side.• It must overlap the back lobe of the C-X sp3
Chapter 6 33
Stereochemistry of SN2
Walden inversion or Inversion of Configuration
Chapter 6 34
• 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
Chapter 6 36
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
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
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
• Rearranged products
Chapter 6 41
• Unimolecular elimination
• Two groups lost (usually X- and H+)
• Nucleophile acts as base
• Also have SN1 products (mixture)
Chapter 6 42
E1 Mechanism• Halide ion leaves, forming carbocation.• Base removes H+ from adjacent carbon.• Pi bond forms. =>
Chapter 6 43
A Closer Look
Chapter 6 44
Chapter 6 45
E1 Energy Diagram
Note: first step is same as SN1=>
Chapter 6 46
• 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
Chapter 6 47
Chapter 6 48
• Bimolecular elimination
• Requires a strong base
• Halide leaving and proton abstraction happens simultaneously - no intermediate. =>
Chapter 6 49
• Order of reactivity: 3° > 2 ° > 1°• Mixture may form, but Zaitsev product
Chapter 6 50
Chapter 6 51
Chapter 6 52
Chapter 6 53
Chapter 6 54
E1 or E2?• Tertiary > Secondary• Weak base• Good ionizing solvent
• Rate = k[halide]• Zaitsev product• No required geometry
• Rearranged products
• Tertiary > Secondary• Strong base required• Solvent polarity not
important• Rate = k[halide][base]• Zaitsev product• Coplanar leaving
groups (usually anti)• No rearrangements
Chapter 6 55
Substitution or Elimination?
• Strength of the nucleophile determines order: Strong nucleophile, bimolecular, 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 56
Mixtures of products are common.
Chapter 6 57
End of Chapter 6