dehydrohalogenation of alkyl halides
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Dehydrohalogenation of Alkyl Halides
X Y
dehydrohalogenation of alkyl halides:X = H; Y = Br, etc.
C CC C + X Y
-Elimination Reactions
X Y
dehydrohalogenation of alkyl halides:X = H; Y = Br, etc.
C CC C + X Y
-Elimination Reactions
requires base
(100 %)
likewise, NaOCH3 in methanol, or KOH in ethanol
NaOCH2CH3
ethanol, 55°C
Dehydrohalogenation
Cl
CH3(CH2)15CH2CH2ClKOC(CH3)3
dimethyl sulfoxide
(86%)
CH2CH3(CH2)15CH
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.
Br
29 % 71 %
+
Regioselectivity
follows Zaitsev's rulemore highly substituted double bond predominates
KOCH2CH3
ethanol, 70°C
more stable configurationof double bond predominates
Stereoselectivity
KOCH2CH3
ethanolBr
+
(23%) (77%)
E2 Energy Diagram
Question
How many alkenes would you expect to be formed from the E2 elimination of3-bromo-2-methylpentane?A) 2B) 3C) 4D) 5
more stable configurationof double bond predominates
Stereoselectivity
KOCH2CH3
ethanol
+
(85%) (15%)
Br
The E2 Mechanism of Dehydrohalogenation of Alkyl
Halides
Empirical Data
(1) Dehydrohalogenation of alkyl halides exhibits second-order kineticsfirst order in alkyl halidefirst order in baserate = 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-2-butene. If the concentration of both reactants is doubled, what would be the effect on the rate of the reaction?A) halve the rateB) double the rateC) quadruple the rateD) no effect on the rate
Empirircal Data
(2) Rate of elimination depends on halogenweaker C—X bond; faster raterate: RI > RBr > RCl > RF
implies that carbon-halogen bond breaks in the rate-determining step
concerted (one-step) bimolecular process
single transition state
C—H bond breaks
component of double bond forms
C—X bond breaks
The E2 Mechanism
–OR.... :
C C
H
X..::
Reactants
The E2 Mechanism
C C
–OR.... H
X..:: –
Transition state
The E2 Mechanism
Br
E2 Mechanism / Transition State
CH3CH2 O••
••••–
OR.... H
C C
–X..::..
Products
The E2 Mechanism
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.
Stereoelectronic Effects
Stereochemistry:Anti Elimination in E2 Reactions
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.
E2 – Stereoelectronic Effect
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.
(CH3)3C
(CH3)3C
BrKOC(CH3)3
(CH3)3COH
cis-1-Bromo-4-tert- butylcyclohexane
Stereoelectronic Effect
(CH3)3C
(CH3)3CBr KOC(CH3)3
(CH3)3COH
trans-1-Bromo-4-tert- butylcyclohexane
Stereoelectronic Effect
(CH3)3C
(CH3)3C
Br
(CH3)3CBr
KOC(CH3)3
(CH3)3COH
KOC(CH3)3
(CH3)3COH
cis
trans
Rate constant for dehydrohalogenation of 1,4- cis is >500 times than that of 1,4- trans
Stereoelectronic Effect
(CH3)3C
(CH3)3C
BrKOC(CH3)3
(CH3)3COH
cis
H that is removed by base must be anti coplanar to BrTwo anti coplanar H atoms in cis stereoisomer
HH
Stereoelectronic Effect
(CH3)3C
KOC(CH3)3
(CH3)3COH
trans
H that is removed by base must be anti coplanar to BrNo anti coplanar H atoms in trans stereoisomer; all vicinal H atoms are gauche to Br; therefore infinitesimal or no product is formed
HH
(CH3)3CBr
H
H
Stereoelectronic Effect
Which of the two molecules below will NOT be able to undergo an E2 elimination reaction?
Question
A) B)
1,4- cismore reactive
1,4- trans
much less reactive
Stereoelectronic Effect
Sterically unhindered bases favor the Zaitsev product.
Sterically hindered bases favor the Hofmann product.
See: SKILLBUILDER 8.5.
E2 – Regioselectivity
Question
Which would react with KOC(CH3)3/(CH3)3COH faster?A) cis-3-tert-butylcyclohexyl bromideB) trans-3-tert-butylcyclohexyl bromide
Question
Which would react with KOCH2CH3 in ethanol faster?A) cis-2-tert-butylcyclohexyl bromideB) trans-2-tert-butylcyclohexyl bromide
Question
NaOMe/MeOH
NaBr + MeOH + ??????
OMe
A. D.
B. E.
C.
Br
What is the major product of the following reaction?
Question
What is the major product of the following reaction?
NaBr + MeOH + ??????
O
A. D.
B. E.
C.
Br
O
The E1 Mechanism ofDehydrohalogenation of Alkyl
Halides
CH3 CH2CH3
Br
CH3
C
Ethanol, heat
+
(25%) (75%)
H3C
CH3
C C
H3C
H
CH2CH3
CH3
CH2C
Example
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.
The E1 Mechanism
CH3 CH2CH3
Br
CH3
C
:..:
CCH2CH3CH3
CH3
+
:..: Br.. –
slow, unimolecular
Step 1
CCH2CH3CH3
CH3
+
CCH2CH3CH3
CH2
+ CCHCH3CH3
CH3
– H+
Step 2
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 DMSOC) 2-bromo-2-methylbutane in ethanol
(heat)D) 2-methyl-2-butanol + KOH
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 Substitution vs. Elimination
1. Analyze the function of the reagent (nucleophile and/ or base).
2. Analyze the substrate (1°, 2°, or 3°).3. Consider regiochemistry and stereochemistry.
Predicting Products
REGIOCHEMICAL OUTCOME STEREOCHEMICAL OUTCOMESN2 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.
See: SKILLBUILDER 8.12.
Predicting Products
REGIOCHEMICAL OUTCOME STEREOCHEMICAL OUTCOMEE2 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 favored over the Hofmann product.
The process is stereoselective. When applicable, a trans disubstituted alkene will be favored over a cis disubstituted alkene.
a. A = 3; B = 1; C = 2; D = 1; E = 1; F = 5b. A = 4; B = 4; C = 2; D = 4; E = 5; F = 2c. A = 2; B = 4; C = 2; D = 4; E = 5; F = 2d. A = 4; B = 4; C = 1; D = 4; E = 3; F = 1e. A = 3; B = 5; C = 2; D = 1; E = 3; F = 5
Question
For each reagent, predict which product will predominate.
NaOMe
Br
Cl-/DMF
Cl-/H2O
O-
NaH
CH3S-/DMF
A. D. 1. 2.
B. E. 3. 4.
C. F. 5. 6.Cl
SCH3
H
SH
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