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![Page 1: Fischer-Rosanoff Convention Before 1951, only relative configurations could be known. Sugars and amino acids with same relative configuration as (+)-glyceraldehyde.](https://reader035.fdocuments.net/reader035/viewer/2022062515/56649c7d5503460f94932f16/html5/thumbnails/1.jpg)
Fischer-Rosanoff Convention• Before 1951, only relative configurations could be known.
• Sugars and amino acids with same relative configuration as (+)-glyceraldehyde were assigned D and same as (-)-glyceraldehyde were assigned L.
• With X-ray crystallography, now know absolute configurations: D is (R) and L is (S).
• No relationship to dextro- or levorotatory. =>
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D and L Assignments
CHO
H OH
CH2OH
D-(+)-glyceraldehyde
*CHO
H OH
HO H
H OH
H OH
CH2OHD-(+)-glucose
*
COOH
H2N H
CH2CH2COOH
L-(+)-glutamic acid
*=>
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Properties of Diastereomers
• Diastereomers have different physical properties: m.p., b.p.
• They can be separated easily.• Enantiomers differ only in reaction with
other chiral molecules and the direction in which polarized light is rotated.
• Enantiomers are difficult to separate. =>
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Resolution of Enantiomers
React a racemic mixture with a chiral compound to form diastereomers, which can be separated.
=>
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ChromatographicResolution of Enantiomers
=>
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Chapter 6Alkyl Halides: Nucleophilic Substitution and Elimination
Organic Chemistry, 5th EditionL. G. Wade, Jr.
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Classes of Halides• Alkyl: Halogen, X, is directly bonded to sp3
carbon.
C
H
H
H
C
H
H
Br
alkyl halide
C CH
H
H
Cl
vinyl halide
I
aryl halide
=>
•Vinyl: X is bonded to sp2 carbon of alkene.
•Aryl: X is bonded to sp2 carbon on benzene ring.
• Examples:
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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
+ -
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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. =>
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Classify These:
CH3 CH CH3
Cl
CH3CH2F
(CH3)3CBr CH3I =>
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Dihalides• Geminal dihalide: two halogen atoms are
bonded to the same carbon
C
H
H
H
C
H
Br
Br
geminal dihalide
C
H
H
Br
C
H
H
Br
vicinal dihalide
=>
•Vicinal dihalide: two halogen atoms are bonded
to adjacent carbons.
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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=>
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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 =>
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“Trivial” Names• CH2X2 called methylene halide..
•CHX3 is a haloform.
•CX4 is carbon tetrahalide.
•Examples: –CH2Cl2 is methylene chloride
–CHCl3 is chloroform -CHI3 is iodoform
–CCl4 is carbon tetrachloride
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Preparation of RX
• Free radical halogenation (Chapter 4) - REVIEW
• Free radical allylic halogenation– produces alkyl halide with double bond on the
neighboring carbon. LATER =>
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Substitution Reactions
• The halogen atom on the alkyl halide is replaced with another group.
C C
H X
+ Nuc:-C C
H Nuc
+ X:-
•Since the halogen is more electronegative than
carbon, the C-X bond breaks heterolytically and
X- leaves.
The group replacing X- is a nucleophile. =>
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Elimination Reactions
• The alkyl halide loses halogen as a halide ion, and also loses H+ on the adjacent carbon to a base.
C C
H X
+ B:- + X:- + HB C C
•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).
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Ingold
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Sir Christopher
Father of Physical Organic Chemistry
Coined such names and symbols as:
SN1, SN2, E1, E2, nucleophile, electrophileresonance effect, inductive effect/
In print, he often attacked enemies vigorously and sometimes in vitrolic manner.
Ingold
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SN2 Mechanism
• Rate is first order in each reactant
CH
Br
HH
H O CHO Br
H
HH
CHO
H
HH
+ Br-
•Both reactants are involved in RDSNote: one-step reaction with no intermediate
•Bimolecular nuleophilic substitution.•Concerted reaction: new bond forming
and old bond breaking at same time
INVERSION OF CONFIGURATION
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SN2 Energy Diagram
• One-step reaction.
• Transition state is highest in energy. =>
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Uses for SN2 Reactions• Synthesis of other classes of compounds.
• Halogen exchange reaction.
=>
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SN2: Nucleophilic Strength• Stronger nucleophiles react faster.• Strong bases are strong nucleophiles, but not all strong nucleophiles are basic.
=>
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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. Moreelectronegative atoms less likely to form new bond:
OH- > F-, NH3 > H2O
Increases down Periodic Table, as size and polarizability increase: I- > Br- > Cl-
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Polarizability Effect
=>
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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
=>
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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.
=>
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Solvent Effects (2)
• Polar aprotic solvents (no O-H or N-H) do not form hydrogen bonds with nucleophile
• Examples:
CH3 C Nacetonitrile
dimethylformamide (DMF)
CH
O
NCH3
CH3
C
O
H3C CH3
acetone =>
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Crown Ethers• Solvate the cation, so
nucleophilic strength of the anion increases.
O
O
O
O
OO
K+
18-crown-6
CH2Cl
KF, (18-crown-6)
CH3CN
CH2F
=>
Fluoride anion becomes a good nucleophile
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SN2: Reactivity of Substrate
Carbon must be partially positive.
=>
Must have a good leaving group
Carbon must not be sterically hindered.
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Leaving Group Ability• Electron-withdrawing
=>
•Stable once it has left (not a strong base)
•Polarizable to stabilize the transition state.
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Structure of Substrate
• Relative rates for SN2: CH3X > 1° > 2° >> 3°
• Tertiary halides do not react via the SN2 mechanism, due to steric hindrance. =>
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Effect of Beta Branching
Br
Br
Br
> >
Propyl Isobutylbromide
Neopentylbromide
bromide
NOTE: ALL ABOVE ARE PRIMARY
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Miscellaneous Substrate
X
X X
All have sterically hinderedbacksides - No SN2 reactivity
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Stereochemistry of SN2
Walden inversion
=>
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EXAMPLES
Me Br
cis
DMSO
NC-
Me
CN
trans
MeBr
CN
Br
CH3
D
H
DMSO
NC-
H
CH3
D
NC
SR
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EXAMPLES 2
N3-
DMSO
CH3
HF
HN3
CH3
CH3
HBr
CNH
CH3
CH3
HF
ClH
CH3
DMSO
CN-
CH3
CNH
CNH
CH3
MESO
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EXAMPLE 3
I
Br
H
CH3S-
Acetone
I
H
SCH3
I
Cl -O CH3
O
CH3CN
I
O
O
CH3
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EXAMPLE 4
How would you prepare the following from an alkyl halide?N3
TARGET
XN3-
DMSOretrosyntheticanalysis
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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.
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SN1 Mechanism (1)
Formation of carbocation (slow)
(CH3)3C Br (CH3)3C+
+ Br-
=>
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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)
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SN1 Energy Diagram
• Forming the carbocation is endothermic
• Carbocation intermediate is in an energy well.
=>
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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
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Stereochemistry of SN1Racemization:
inversion and retention
=>
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Rearrangements
• Carbocations can rearrange to form a more stable carbocation.
•Methyl shift: CH3- moves from adjacent carbon
if no H’s are available.
Hydride shift: H- on adjacent carbon bonds with C+.
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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
=>
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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
=>
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Interesting Rearrangement = PUSH PULL
H3C CH3
Br
Ag
push pull
H3C
C
CH2
CH3
+
CH3
MeO
H
-AgBrBr
H3C CH3 Ag+
MeOH
H3C CH2
CH3MeO
H+
CH3
-H+H3C
CH3MeO
CH3
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SN2 or SN1?• Primary or methyl • Tertiary
•Strong nucleophile•Polar aprotic solvent•Rate = k[halide][Nuc]
•Inversion at chiral carbon•No rearrangements
•Weak nucleophile (may also be solvent)
•Polar protic solvent, Ag+
•Rate = k[halide]
•Racemization of optically active compound
Rearranged products
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E1 Reaction
• Unimolecular elimination
•Two groups lost (usually X- and H+)
•Nucleophile acts as base)
Also have SN1 products (mixture
SN1 and E1 have common first step.
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E1 EXAMPLES
Br
CH2-H
heat
MeOH CH2
E-1
+
OMe
CH2-H
SN-1
H3C
CH3
CH3
CH3
Br
H
CH3CH2OH
H3C
CH3
CH3
H
CH2
+H3C
CH3
CH3
CH3
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E1 Mechanism
• Halide ion leaves, forming carbocation.
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+
•Base removes H+ from adjacent carbon.Pi bond forms.
![Page 55: Fischer-Rosanoff Convention Before 1951, only relative configurations could be known. Sugars and amino acids with same relative configuration as (+)-glyceraldehyde.](https://reader035.fdocuments.net/reader035/viewer/2022062515/56649c7d5503460f94932f16/html5/thumbnails/55.jpg)
A Closer LookO
H
H
C
H
H
H
C CH3
CH3
C C
H
CH3
CH3
H+ H3O+
=>
![Page 56: Fischer-Rosanoff Convention Before 1951, only relative configurations could be known. Sugars and amino acids with same relative configuration as (+)-glyceraldehyde.](https://reader035.fdocuments.net/reader035/viewer/2022062515/56649c7d5503460f94932f16/html5/thumbnails/56.jpg)
E1 Energy Diagram
• Note: first step is same as SN1=>
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E2 Reaction
• Bimolecular elimination
•Requires a strong base
•Halide leaving and proton abstraction happens
simultaneously - no intermediate
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E2 Examples
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E2 Mechanism
H C
H
H
C
CH3
CH3
Br
C C
H
CH3
CH3
H
O
H+ H2O Br
-+
=>
![Page 60: Fischer-Rosanoff Convention Before 1951, only relative configurations could be known. Sugars and amino acids with same relative configuration as (+)-glyceraldehyde.](https://reader035.fdocuments.net/reader035/viewer/2022062515/56649c7d5503460f94932f16/html5/thumbnails/60.jpg)
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
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E2 Stereochemistry
=>
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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
=>
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End of Chapter 6