HOFMANN REACTION

1
HOFMANN REACTION
E2 gone astray
2
HOFMANN RULE
When you have a bulky leaving group
like -N(CH3)3 the least-substituted alkene will
be the major product.
Big is not the same as bulky.
BULKY
Branched at the first atom attached to the chain
chain
chain
trimethyl ammonium
dimethyl sulfonium
OTHER GROUPS FOLLOW THE ZAITSEV RULE
3
HOFMANN ELIMINATION
Hofmann found that when the leaving group was
-N(CH3)3 E2 elimination reactions gave the
least-substituted alkene.
Hofmann
95
5
Zaitsev
31
69
4
EFFECT OF INCREASING SUBSTITUENT BULK
E2
(cis trans)
HOFMANN
ZAITSEV
31
69
30
70
48
52
Big is not the same as bulky.
87
13
98
2
5
ANALYSIS OF 2-SUBSTITUTED PENTANE ELIMINATIONS
ZAITSEV
less crowding in this area of the molecule
cis
H
trans
C
C
C
C
C
H
3
X
H
bulky groups cause crowding and give
Hofmann products
HOFMANN
( all H equivalent )
6
H
most steric crowding
CH3
H
less steric crowding
H
CH2CH3
H
no steric crowding
CH3
H
N(CH3)3
H
CH3CH2
H
CH2CH2CH3
H
N(CH3)3
H
H
N(CH3)3
WOULD MAKE
WOULD MAKE
cis
trans
ZAITSEV PRODUCTS
HOFMANN PRODUCT
FORMED
NOT FORMED
7
BULKY BASES ALSO INCREASE HOFMANN PRODUCT
ZAITSEV
HOFMANN
81 19
methoxide
47 53
t -butoxide
bulky base
8
BULKY b-SUBSTITUENTS
What constitutes bulky?
a methyl group is not bulky
ZAITSEV
HOFMANN
b
a
NO
NaOEt
80
20
even two or three are not bulky
b
NO
NaOEt
79
21
YES
NaOEt
14
86
b
t-butyl is bulky !
9
THE ELIMINATION MOVES TO A LESS CROWDED REGION
less crowded
crowded
crowding
14
86
REACTIVE CONFORMATIONS
crowding
spacer
10
EXAMPLES
11
HOW THE VARIOUS FACTORS AFFECT THE OUTCOME
CH3
CH3
NORMAL
NaOEt
Zaitsev
Br
EtOH / D
90
Bromine is big, not bulky
BULKY LEAVING GROUP
CH3
CH2
Hofmann
KOH

90
N(CH3)3
EtOH / D
-
I
Prototypical Hofmann elimination
CH3
BULKY BASE
CH3
CH2
NaOtBu

Br
tBuOH / D
Bulky base alone not as effective as bulky
leaving group
60/40
12
HOW THE VARIOUS FACTORS AFFECT THE OUTCOME
( CONTINUED )
BULKY BASE LEAVING GROUP
CH3
CH2
NaOtBu
Hofmann

N(CH3)3
tBuOH / D
100
-
I
Double Whammy !
Bulky base bulky leaving group
BULKY b-SUBSTITUENT
CH3
CH3
CH2
NaOEt
Br

EtOH / D
tBu
tBu
tBu
H
no double bond here
either cis or trans to Br - same result
Favors Hofmann products
Use a bulky base here and ...
13
E2 REACTIONS DEVIATE FROM THE ZAITSEV RULE
1. If the favored b-hydrogen cant achieve
anti-coplanar geometry.
2. If the double bond would form at a bridgehead.
(see next slides)
3. If there is a bulky leaving group.
4. If there is a bulky base.
5. If there is a bulky b-substituent.
14
BREDTS RULE
A double bond cannot form at a bridgehead.
15
BREDTS RULE
A double bond cannot form at a bridgehead in
a bicyclic system with small rings.
no way !
z
p orbitals cannot become coplanar
y
y
Try a model !
z
16
SYN ELIMINATION
More difficult than anti-coplanar
elimination, but does occur in some circumstances.
( Usually requires forced conditions
- heat and
pressure. )
17
COPLANAR ARRANGEMENTS
E2






not coplanar
( very difficult )
other angles
syn- coplanar
(difficult)
0o
anti- coplanar
(easiest)
Difficulty Order
180o
18
SYN-COPLANAR ELIMINATIONS REQUIRE FORCED
CONDITIONS
H
X
syn-coplanar (0o)
C
C
C
C
This often requires heating above the boiling
point of the solvent in a sealed tube (next
slide).
Temperatures above 100 oC are common.
19
SEALED TUBE
Glass tube sealed at both ends.
glass tube
alkyl halide NaOEt / EtOH (bp 78 oC) inside tube
1.25 D
REACTANTS
0.25 wall
Carried out in a hood behind a glass shield.
Allows reactants to be heated to a high
temperature (above bp) without boiling away.
heated oil (bp gt 250 oC)
hot plate
20
A CASE OF SYN ELIMINATION
SYN ELIMINATION OCCURS BECAUSE THERE ARE NO
ANTI-COPLANAR b-H
Bredts Rule
NaOEt EtOH

110 oC
0
NaOEt EtOH
same products
NO DEUTERIUM
110 oC
proves the syn hydrogen is the one removed
not this one
21
SYN ELIMINATION IS SEEN WHEN ANTI-COPLANAR DOES
NOT EXIST
not coplanar
hydrogens are not anti-coplanar to chlorines
NaOH 110o
very slow
difficult reaction
coplanar
NaOH 110o
faster
syn- elimination
22
a-ELIMINATION
occurs when the substrate has NO
b-HYDROGENS
on these carbon atoms
23
a -ELIMINATION
Cl
Cl
-
KOH

C
C
H
Cl
Cl
CHCl3
Cl
Cl
no b -hydrogens
a -elimination
very reactive
Cl
..



C
Cl
Cl

..
a carbene
24
CARBENES ARE ELECTROPHILES !
Because of an incomplete octet carbenes
are electrophilic (need electrons to complete
their valence shell).
missing a pair of electrons
Carbenes will react with an alkene (electron pair
donor).

nucleophile
electrophile
25
CARBENES ADD TO DOUBLE BONDS
alkene carbene cyclopropane ring
H
2
H
1
syn addition
Probably concerted
steps 1,2 for visualization only
26
ANALYSIS OF THE ADDITION
2p
H
C

.
SP2 hybrid
.
H
stepwise analysis of the concerted process
syn addition

-
..

the intermediate does not exist
27
SYN ADDITION
CHCl3
BOTH NEW RING BONDS FORM ON THE SAME SIDE OF
THE DOUBLE BOND
KOH
EtOH
H
Stereospecific
H
28
STEREOSPECIFICITY PROVES THE REACTION IS
CONCERTED
SUBSTITUENTS ON THE DOUBLE BOND RETAIN THEIR
ORIGINAL CIS OR TRANS RELATIONSHIPS IN THE NEW
RING
CHCl3
KOH
cis
cis
EtOH
CHCl3
KOH
trans
trans
EtOH
29
STEREOSPECIFIC CONCERTED
YES !
Concerted.
Both bonds form on the same side.
Stereospecific!
H
NO !
An intermediate would allow rotation.
H

CH3
This does not happen.
..
CH3
Would not be stereospecific.
30
COMPOUNDS WITHOUT b-HYDROGENS
Not a common type of compound !
31
ELIMINATIONS SUMMARY
32
K.I.S.S.
THE MOST BASIC STUFF
alkyl halide strong base heat E2
alkyl halide solvent heat (solvolysis)
E1
alcohol strong acid heat E1
(acid assisted)
typical situation for E1cb
H next to CO (easy to remove)
X strong base (difficult to break bond)
Only E1 reactions have rearrangements
(carbocations)
Only E2 reactions require anti-coplanar
b-hydrogens
33
THE BIG PICTURE
alkyl halides
alcohols
special
E1cb E2
E1 E1 E1 acid


assisted

strong strong weak
acidic
neutral

base base base
stepwise -
concerted
stepwise - carbocation
carbanion
stereospecific
solvolysis
anti-coplanar
special case -
not common
Zaitsev if
not stereospecific
stereochem
requires
allows
Zaitsev Zaitsev Zaitsev
acidic H and
Hofmann if
poor leaving
bulky groups
group
a-elim. if no b-H
carbocation rearrangements