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.
=>
D and L Assignments
CHO
H
*
CHO
OH
H
CH2OH
D-(+)-glyceraldehyde
HO
H
COOH
H2N
*
H
=>
CH2CH2COOH
L-(+)-glutamic
acid
OH
H
OH
H *
OH
CH2OH
D-(+)-glucose
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.
=>
Resolution of Enantiomers
React a racemic mixture with a chiral compound to form
diastereomers, which can be separated.
=>
Chromatographic
Resolution of Enantiomers
=>
Organic Chemistry, 5th Edition
L. G. Wade, Jr.
Chapter 6
Alkyl Halides: Nucleophilic
Substitution and Elimination
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:
H H
H C C Br
H H
alkyl halide
I
H
H
=>
C C
H
Cl
vinyl halide
aryl halide
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.
=>
H + H C Br
H
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.
=>
Classify These:
CH3
CH CH3
CH3CH2F
Cl
(CH3)3CBr
CH3I
=>
Dihalides
• Geminal dihalide: two halogen atoms are
bonded to the same carbon
•Vicinal dihalide: two halogen atoms are bonded
to adjacent carbons.
H
H
H
C
C
H H
Br
H Br
geminal dihalide
H C C Br
Br H
vicinal dihalide
=>
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
2-chlorobutane
CH2CH2Br
CH3(CH2)2CH(CH2)2CH3
4-(2-bromoethyl)heptane
=>
Systematic Common Names
• Name as alkyl halide.
• Useful only for small alkyl groups.
• Name these:
(CH3)3CBr
CH3
CH CH2CH3
Cl
CH3
CH3
CH CH2F
=>
“Trivial” Names
• CH2X2 called methylene halide.
. •CHX is a haloform.
3
•CX4 is carbon tetrahalide.
•Examples:
–CH2Cl2 is methylene chloride
–CHCl3 is chloroform -CHI3 is iodoform
–CCl4 is carbon tetrachloride
Preparation of RX
• Free radical halogenation (Chapter 4) REVIEW
• Free radical allylic halogenation
– produces alkyl halide with double bond on the
neighboring carbon. LATER
=>
Substitution Reactions
C C
H X
+
Nuc:
-
C C
+
X:
-
H Nuc
• 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.
=>
Elimination Reactions
C C
+
-
B:
C C
+
X:
-
+ HB
H X
• The alkyl halide loses halogen as a halide ion, and
also loses H+ on the adjacent carbon to a base.
•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).
Ingold
Ingold
Sir Christopher
Father of Physical Organic Chemistry
Coined such names and symbols as:
SN1, SN2, E1, E2, nucleophile, electrophile
resonance effect, inductive effect/
In print, he often attacked enemies
vigorously and sometimes in vitrolic
manner.
SN2 Mechanism
H
H
H O
H
C Br
H
HO C Br
H H
H
HO C
• Rate is first order in each reactant
•Both reactants are involved in RDS
Note: one-step reaction with no
intermediate
•Bimolecular nuleophilic substitution.
•Concerted reaction: new bond forming
and old bond breaking at same time
INVERSION OF CONFIGURATION
H
-
+ Br
H
SN2 Energy Diagram
• One-step reaction.
• Transition state is highest in energy. =>
Uses for SN2 Reactions
• Synthesis of other classes of compounds.
• Halogen exchange reaction.
Nucleophile

Product
R-I
Class of Product
akyl halide
-

R-OH
alcohol
R-X + OR'
-

R-OR'
ether
-

R-SH
thiol
-

R-SR'
thioether

amine salt

R-NH3+X
R- N3
R-X + CC-R'
-

R-CC-R'
alkyne
-

R-CN
R-COO-R'
nitrile
R-X + I
-
R-X + OH
R-X + SH
R-X + SR'
R-X + NH3
-
R-X + N3
R-X + CN
R-X + R-COO
-

-
azide
ester
=>
SN2: Nucleophilic Strength
• Stronger nucleophiles react faster.
• Strong bases are strong nucleophiles, but not all
strong nucleophiles are basic.
=>
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-
Polarizability Effect
=>
Bulky Nucleophiles
Sterically hindered for attack on carbon, so
weaker nucleophiles.
CH3 CH2 O
ethoxide (unhindered)
weaker base, but stronger nucleophile
CH3
H3C
C
O
CH3
=>
t-butoxide (hindered)
stronger base, but weaker nucleophile
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.
=>
Solvent Effects (2)
• Polar aprotic solvents (no O-H or N-H) do not form
hydrogen bonds with nucleophile
• Examples:
O
CH3 C N
acetonitrile
H
C
O
N
CH3
CH3
dimethylformamide
(DMF)
C
H3C
CH3
acetone
=>
Crown Ethers
• Solvate the cation, so
nucleophilic strength
of the anion increases.
Fluoride anion becomes a
good nucleophile
O
O
O
K+
O
O
O
18-crown-6
CH2Cl
CH2F
KF, (18-crown-6)
CH3CN
=>
SN2: Reactivity of Substrate
 Carbon must be partially positive.
Must have a good leaving group
Carbon must not be sterically hindered.
=>
Leaving Group Ability
• Electron-withdrawing
•Stable once it has left (not a strong base)
•Polarizable to stabilize the transition state.
=>
Structure of Substrate
• Relative rates for SN2:
CH3X > 1° > 2° >> 3°
• Tertiary halides do not react via the SN2
mechanism, due to steric hindrance.
=>
Effect of Beta Branching



Br
Br

Propyl
bromide
>

Isobutyl
bromide
Br
>

Neopentyl
bromide
NOTE: ALL ABOVE ARE PRIMARY
Miscellaneous Substrate
X
X
X
All have sterically hindered
backsides - No SN2 reactivity
Stereochemistry of SN2
Walden inversion
=>
EXAMPLES
NC-
Br
Me
Me
DMSO
CN
trans
cis
Br
Me
CN
CH3
CH3
NC-
H
NC
Br
H
DMSO
R
D
D
S
EXAMPLES 2
CH3
CH3
F
H
H
Cl
N3DMSO
F
H
N3
H
CH3
CH3
CH3
CH3
Br
H
H
CN
CH3
CNDMSO
H
CN
H
CN
CH3
MESO
EXAMPLE 3
Br
CH3SH
H
Acetone
I
SCH3
I
O
O
I
Cl
-O
CH3
I
O
CH3CN
CH3
EXAMPLE 4
How would you prepare the
following from an alkyl halide?
N3
N3DMSO
TARGET
X
retrosynthetic
analysis
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.
SN1 Mechanism (1)
Formation of carbocation (slow)
(CH3)3C Br
+
(CH3)3C
-
+ Br
=>
SN1 Mechanism (2)
• Nucleophilic attack
+
(CH3)3C
+ H O H
(CH3)3C O H
H
• Loss of H+ (if needed)
(CH3)3C O H + H O H
H
(CH3)3C O H
+
+ H3O
=>
SN1 Energy Diagram
• Forming the
carbocation is
endothermic
• Carbocation
intermediate is in
an energy well.
=>
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
Stereochemistry of SN1
Racemization:
inversion and retention
=>
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.
Hydride Shift
H
Br H
CH3
CH3
C C CH3
C C CH3
H CH3
H CH3
H
H
CH3
CH3
C C CH3
C C CH3
H CH3
H CH3
H
CH3
C C CH3
H CH3
H Nuc
Nuc
CH3
C C CH3
H CH3
=>
Methyl Shift
CH3
Br CH3
CH3
CH3
C C CH3
H CH3
H CH3
CH3
CH3
CH3
C C CH3
CH3
C C CH3
C C CH3
H CH3
H CH3
CH3
CH3
C C CH3
H CH3
Nuc
CH3
CH3 Nuc
C C CH3
H CH3
=>
Interesting Rearrangement =
PUSH PULL
H3C
Ag+
CH3
H3C
CH3
CH3
C
Br
Br

MeOH
push
MeO
H3C

pull
Ag
H3C
-AgBr
CH2
CH3
MeO
H
+
CH3
MeO
CH3
+
-H+
H
CH3
CH3
H3C
CH2
SN2
or
SN1?
• Primary or methyl
• Tertiary
•Strong nucleophile
•Weak nucleophile (may also
be solvent)
•Polar protic solvent, Ag+
•Polar aprotic solvent
•Rate = k[halide][Nuc]
•Rate = k[halide]
•Inversion at chiral carbon •Racemization of optically
active compound
•No rearrangements
Rearranged products
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.
E1 EXAMPLES
CH2-H
CH2-H
MeOH
CH2
+
OMe
Br
heat
SN-1
E-1
CH3
H
H3C
CH2
CH3
CH3CH2OH
Br
CH3
CH3
H
H3C
CH3
CH3
+
CH3
H3C
CH3
E1 Mechanism
H
H Br
H C C CH3
H C C CH3
H CH3
H CH3
H
H O
H
H C C CH3
H CH3
CH3
H
+
C C
H
CH3
• Halide ion leaves, forming carbocation.
•Base removes H+ from adjacent carbon.
Pi bond forms.
+
H3O
A Closer Look
H
H O
H
H C C CH3
H CH3
CH3
H
+
C C
H
+
H3O
CH3
=>
E1 Energy Diagram
• Note: first step is same as SN1
=>
E2 Reaction
• Bimolecular elimination
•Requires a strong base
•Halide leaving and proton abstraction happens
simultaneously - no intermediate
E2 Examples
E2 Mechanism
H Br
CH3
H
C C
H C C CH3
H
O
H CH3
H
-
+ H2O + Br
CH3
=>
Saytzeff’s Rule
• If more than one elimination product is possible, the mostsubstituted alkene is the major product (most stable).
• R2C=CR2>R2C=CHR>RHC=CHR>H2C=CHR
tetra >
tri
>
di > mono
H Br CH3
H C C C CH3
H H H
-
OH
H
CH3
H
CH3
C C C CH3 + H C C C
H
CH3
H H
H H
minor
major
=>
E2 Stereochemistry
=>
E1 or
• Tertiary > Secondary
• Weak base
• Good ionizing solvent
• Rate = k[halide]
• Saytzeff product
• No required geometry
• Rearranged products
E2?
•
•
•
•
•
•
Tertiary > Secondary
Strong base required
Solvent polarity not important
Rate = k[halide][base]
Saytzeff product
Coplanar leaving groups
(usually anti)
• No rearrangements
=>
End of Chapter 6