Alkenes-prep-II-2012-ques

Report
Dehydrohalogenation of
Alkyl Halides
-Elimination Reactions
dehydrohalogenation of alkyl halides:
X = H; Y = Br, etc.
X
C

C Y
C
C
+
X
Y
-Elimination Reactions
dehydrohalogenation of alkyl halides:
X = H; Y = Br, etc.
X
C

C Y
C
requires base
C
+
X
Y
Dehydrohalogenation
Cl
NaOCH2CH3
ethanol, 55°C
(100 %)
likewise, NaOCH3 in methanol, or KOH in ethanol
Dehydrohalogenation
When the alkyl halide is primary, potassium
tert-butoxide in dimethyl sulfoxide (DMSO), a
strong non-protic polar solvent is the
base/solvent system that is normally used.
CH3(CH2)15CH2CH2Cl
KOC(CH3)3
dimethyl sulfoxide
CH3(CH2)15CH
(86%)
CH2
Regioselectivity
KOCH2CH3
Br
+
ethanol, 70°C
29 %
71 %
follows Zaitsev's rule
more highly substituted double bond predominates
Stereoselectivity
KOCH2CH3
ethanol
Br
+
(23%)
(77%)
more stable configuration
of double bond predominates
E2 Energy Diagram
Question
How many alkenes would you expect to be
formed from the E2 elimination of
3-bromo-2-methylpentane?
A) 2
B) 3
C) 4
D) 5
Stereoselectivity
Br
KOCH2CH3
ethanol
+
(85%)
(15%)
more stable configuration
of double bond predominates
The E2 Mechanism of
Dehydrohalogenation of Alkyl Halides
Empirical Data
(1) Dehydrohalogenation of alkyl halides
exhibits second-order kinetics
first order in alkyl halide
first order in base
rate = k[alkyl halide][base]
implies that rate-determining step
involves both base and alkyl halide;
i.e., it is bimolecular
Question
The reaction of 2-bromobutane with
KOCH2CH3 in ethanol produces trans-2butene. If the concentration of both
reactants is doubled, what would be the
effect on the rate of the reaction?
A) halve the rate
B) double the rate
C) quadruple the rate
D) no effect on the rate
Empirircal Data
(2) Rate of elimination depends on halogen
weaker C—X bond; faster rate
rate: RI > RBr > RCl > RF
implies that carbon-halogen bond breaks in
the rate-determining step
The E2 Mechanism
concerted (one-step) bimolecular process
single transition state
C—H bond breaks
 component of double bond forms
C—X bond breaks
The E2 Mechanism
R
.. –
O
.. :
H
C
C
: X:
..
Reactants
The E2 Mechanism
R
–
..
O
..
H
Transition state
C
C
–
: X:
..
E2 Mechanism / Transition State
CH3CH2
•• –
O ••
••
Br
The E2 Mechanism
R
..
O
..
H
C
C
.. –
: X:
..
Products
Question
Which one of the following best describes a mechanistic
feature of the reaction of 3-bromopentane with sodium
ethoxide?
A) The reaction occurs in a single step which is
bimolecular.
B) The reaction occurs in two steps, both of which are
unimolecular.
C) The rate-determining step involves the formation
of the carbocation (CH3CH2)2CH+.
D) The carbon-halogen bond breaks in a rapid
step
that follows the rate-determining step.
Stereochemistry:
Anti Elimination in E2 Reactions
Stereoelectronic Effects
E2 – Stereoelectronic Effect
Consider dehydrohalogenation of chlorocyclohexane.
An anti-periplanar T.S. is required and only the chair
conformation on the left alllows for the elimination to
occur.
Stereoelectronic Effect
An effect on reactivity that has its origin in
the spatial arrangement of orbitals or bonds is
called a stereoelectronic effect.
The preference for an anti coplanar
arrangement of H and Br in the transition
state for E2 dehydrohalogenation is an
example of a stereoelectronic effect.
Stereoelectronic Effect
Br
KOC(CH3)3
(CH3)3COH
(CH3)3C
cis-1-Bromo-4-tertbutylcyclohexane
(CH3)3C
Stereoelectronic Effect
(CH3)3C
trans-1-Bromo-4-tertbutylcyclohexane
Br
(CH3)3C
KOC(CH3)3
(CH3)3COH
Stereoelectronic Effect
cis
Br
KOC(CH3)3
(CH3)3COH
(CH3)3C
Rate constant for
dehydrohalogenation
of 1,4- cis is >500
times than that of 1,4trans
Br
(CH3)3C
trans
(CH3)3C
KOC(CH3)3
(CH3)3COH
Stereoelectronic Effect
cis
Br
KOC(CH3)3
(CH3)3COH
(CH3)3C
H H
(CH3)3C
H that is removed by base must be anti
coplanar to Br
Two anti coplanar H atoms in cis
stereoisomer
Stereoelectronic Effect
trans
H
Br
H
(CH3)3C
KOC(CH3)3
(CH3)3COH
H H
(CH3)3C
H that is removed by base must be anti
coplanar to Br
No anti coplanar H atoms in trans stereoisomer;
all vicinal H atoms are gauche to Br; therefore
infinitesimal or no product is formed
Question
Which of the two molecules below will NOT be able to
undergo an E2 elimination reaction?
A)
B)
Stereoelectronic Effect
1,4- cis
more reactive
1,4- trans
much less reactive
E2 – Regioselectivity
Sterically unhindered bases favor the Zaitsev
product.
Sterically hindered bases favor the Hofmann product.
See: SKILLBUILDER 8.5.
Question
Which would react with KOC(CH3)3/(CH3)3COH
faster?
A) cis-3-tert-butylcyclohexyl bromide
B) trans-3-tert-butylcyclohexyl bromide
Question
Which would react with KOCH2CH3 in ethanol
faster?
A) cis-2-tert-butylcyclohexyl bromide
B) trans-2-tert-butylcyclohexyl bromide
Question
What is the major product of the following reaction?
NaOMe/MeOH
NaBr + MeOH + ??????

Br
A.
D.
OMe
B.
C.
E.
Question
What is the major product of the following reaction?
O
NaBr + MeOH + ??????

Br
A.
D.
O
B.
C.
E.
The E1 Mechanism of
Dehydrohalogenation of Alkyl
Halides
Example
CH3
CH3
CH2CH3
C
Br
Ethanol, heat
H3C
CH3
H2C
+
C
CH2CH3
(25%)
H
C
C
CH3
H3C
(75%)
The E1 Mechanism
1. Alkyl halides can undergo elimination in protic
solvents in the absence of base.
2. Carbocation is intermediate.
3. Rate-determining step is unimolecular
ionization of alkyl halide.
CH3
Step 1
CH3
CH2CH3
C
: Br:
..
slow, unimolecular
CH3
C
CH3 + CH2CH3
.. –
: Br :
..
CH3
Step 2
CH3
C
+
CH2CH3
– H+
CH3
CH2
+
C
CH3
CH2CH3
C
CH3
CHCH3
Question
Which reaction would be most likely to proceed
by an E1 mechanism?
A) 2-chloro-2-methylbutane + NaOCH2CH3 in
ethanol (heat)
B) 1-bromo-2-methylbutane + KOC(CH3)3 in
DMSO
C) 2-bromo-2-methylbutane in ethanol
(heat)
D) 2-methyl-2-butanol + KOH
Predicting Substitution vs. Elimination
1. Analyze the function of the reagent (nucleophile
and/ or base).
2. Analyze the substrate (1°, 2°, or 3°).
Predicting Substitution vs. Elimination
1. Analyze the function of the reagent (nucleophile
and/ or base).
2. Analyze the substrate (1°, 2°, or 3°).
Predicting Substitution vs. Elimination
1. Analyze the function of the reagent (nucleophile
and/ or base).
2. Analyze the substrate (1°, 2°, or 3°).
Predicting Substitution vs. Elimination
1. Analyze the function of the reagent (nucleophile
and/ or base).
2. Analyze the substrate (1°, 2°, or 3°).
See SKILLBUILDER 8.11.
Predicting Products
1. Analyze the function of the reagent (nucleophile
and/ or base).
2. Analyze the substrate (1°, 2°, or 3°).
3. Consider regiochemistry and stereochemistry.
REGIOCHEMICAL OUTCOME
STEREOCHEMICAL OUTCOME
SN2 The nucleophile attacks the α
position, where the leaving
group is connected.
The nucleophile replaces the
leaving group with inversion of
configuration.
SN1 The nucleophile attacks the
carbocation, which is where the
leaving group was originally
connected, unless a carbocation
rearrangement took place.
The nucleophile replaces the
leaving group with racemization.
Predicting Products
REGIOCHEMICAL OUTCOME
STEREOCHEMICAL OUTCOME
E2
The Zaitsev product is generally
favored over the Hofmann product,
unless a sterically hindered base is
used, in which case the Hofmann
product will be favored
This process is both stereoselective
and stereospecific.
When applicable, a trans disubstituted
alkene will be favored over a cis
disubstituted alkene. When the β
position of the substrate has only one
proton, the stereoisomeric alkene
resulting from anti-periplanar
elimination will be obtained
(exclusively, in most cases).
E1
The Zaitsev product is always
The process is stereoselective. When
favored over the Hofmann product. applicable, a trans disubstituted
alkene will be favored over a cis
disubstituted alkene.
See: SKILLBUILDER 8.12.
Question
For each reagent, predict which product will predominate.
Br
SCH3
A.
NaOMe
D.
1.
2.
O-
B. Cl-/H2O
E. NaH
H
3.
4.
Cl
C. CH3S-/DMF F.
Cl-/DMF
5.
SH
6.
a. A = 3; B = 1; C = 2; D = 1; E = 1; F = 5
b. A = 4; B = 4; C = 2; D = 4; E = 5; F = 2
c. A = 2; B = 4; C = 2; D = 4; E = 5; F = 2
d. A = 4; B = 4; C = 1; D = 4; E = 3; F = 1
e. A = 3; B = 5; C = 2; D = 1; E = 3; F = 5

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