CHE 240 Unit IV Stereochemistry, Substitution and Elimination Reactions CHAPTER SIX Terrence P....
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Transcript of CHE 240 Unit IV Stereochemistry, Substitution and Elimination Reactions CHAPTER SIX Terrence P....
CHE 240Unit IV
Stereochemistry, Substitution and Elimination Reactions
CHAPTER SIX
Terrence P. Sherlock
Burlington County College
2004
Chapter 6 2
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 3
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
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 6
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 7
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 8
“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 9
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 10
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 11
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 12
Densities
• Alkyl fluorides and chlorides less dense than water.
• Alkyl dichlorides, bromides, and iodides more dense than water. =>
Chapter 6 13
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 14
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 15
Reaction Mechanism
Free radical chain reactioninitiation, propagation, termination.
H H
BrH
+ HBrBr
Br
H Br
+ Br
=>
2 BrBr2h
Chapter 6 16
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 17
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 18
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 20
Uses for SN2 Reactions• Synthesis of other classes of compounds.• Halogen exchange reaction.
=>
Chapter 6 21
SN2: Nucleophilic Strength• Stronger nucleophiles react faster.• Strong bases are strong nucleophiles, but
not all strong nucleophiles are basic.
=>
Chapter 6 22
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 24
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 25
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 26
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 27
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 28
SN2: Reactivity of Substrate
• Carbon must be partially positive.• Must have a good leaving group• Carbon must not be sterically hindered.
=>
Chapter 6 29
Leaving Group Ability
• Electron-withdrawing
• Stable once it has left (not a strong base)
• Polarizable to stabilize the transition state.
=>
Chapter 6 30
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 32
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 34
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 35
SN1 Energy Diagram
• Forming the carbocation is endothermic
• Carbocation intermediate is in an energy well.
=>
Chapter 6 36
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 38
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 39
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 40
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 41
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 42
E1 Reaction
• Unimolecular elimination
• Two groups lost (usually X- and H+)
• Nucleophile acts as base
• Also have SN1 products (mixture)
=>
Chapter 6 43
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
E2 Reaction
• Bimolecular elimination
• Requires a strong base
• Halide leaving and proton abstraction happens simultaneously - no intermediate. =>
Chapter 6 48
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 50
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 51
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. =>