CHM 2211C
6th edition Notes
Chapter 11
Reactions of Alkyl
Halides: Nucleophilic
Substitutions and
Eliminations
By
Dr. Andrea Wallace
Coastal Georgia Community College
Edited by
John T. Taylor
Florida Community College at Jacksonville
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Chapter 11: Reactions of Alkyl Halides:
Substitutions and Eliminations
Much of the chemistry in this chapter involves polar reactions between an electrophilic
Carbon and a Nucleophile or Base.
Figure:
Substitution Reactions:
(Substitute a Nu for X.)
Elimination Reactions:
(Eliminate HX and form an alkene.)
11.1 The Discovery of the Walden Inversion
See p. 344 – Walden’s Cycle (1896) of Reactions interconverting (+) and (-) malic acids.
Example of a nucleophilic substitution.
See inversion of configuration.
This work was thought to be a great break through in optically active reaction chemistry.
11.2 Stereochemistry of Nucleophilic Substitutions
Kenyon and Philips (1920’s) set out to determine the mechanism of the nucleophilic
substitution reactions.
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Reaction:
R-Y
Nu:- 
R-Nu
+ Y:-
Y = Cl, Br, I, or OTos
OTos is alkyl toluenesulfonate also referred to as a tosylate. Looks really complicated –
see p. 345, but it works just like a simple halogen.
See Figure 11.2, p. 345, Talk through mechanism.
1) Break O-H, replace with a tosylate group. No inversion of configuration.
2) Break C-OTos, replace with an acetate group. See inversion of configuration.
3) Break O-C which is part of the acetate group and not a stereogenic center, add H.
No inversion of configuration.
After many reactions –
Conclusion: Nucleophilic substitutions of 1o and 2o alkyl halides and tosylates always
proceed with inversion of configuration.
11.3 Kinetics of Nucleophilic Substitutions
Kinetics – rate or speed of a chemical reaction. Reactions are sometimes said to be “fast”
or “slow”.
Determining the reaction rate or how the rate depends on the reactant concentrations can
lead to a determination of the mechanism.
Relationship between reaction rate and reactant concentration:
Reaction:
Rate = k [CH3Br] [OH-]
(Rate equation is determined experimentally)
k = rate constant, units vary, in this equation – L / mol sec
[reactant] = reactant concentration in moles/liter or Molarity (M)
This reaction would be first order for each reactant and would be second order overall.
Thus, this is an SN2 reaction.
How would the rate be affected by the following?
Double the concentration of one reactant -____________________
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Double the concentration of both reactants - _____________________
Half the concentration of one reactant - _________________________
11.4 The SN2 Reaction
SN2
S-substitution
N-nuclophilic
2-Bimolecular (2nd order)
1) Occurs with 1o and 2o Alkyl Halide and Tosylates
2) Accompanied by inversion of configuration if the reaction occurs at a stereogenic
center.
3) Reaction shows 2nd order kinetics: Rate = k [RX] [Nu:-]
The mechanism that meets these criteria was proposed by Hughes and Ingold in 1937.
Single step process
Nucleophile attacks from one face (back face) and the leaving group leaves from the
opposite face (front face). It is necessary to invert the configuration.
Mechanism of an SN2 (p. 348):
Nu
Substrate
Transition State
Product
Leaving Group
The mechanism is consistent with the expected results.
See p. 349 for another view.
Problem 11.2, p. 349
What product would you expect to obtain from SN2 reaction of OH- with ®-2bromobutane? Show the stereochemistry of both reactant and product.
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11.5 Characteristics of the SN2 Reaction
Substrate Effects:
Slower reactions occur with bulkier more sterically hindered substrates.
(See p. 350, Figure 11.6)
Tertiary
Neopentyl
Secondary
Primary
Methyl
Relative
Reactivity
------------------Reactivity _____________ -----------------------------------
SN2 are generally on useful with methyl, primary, and a few simple secondary alkyl
halides.
Vinylic and Aryl halides do not undergo SN2 reactions due to steric problems.
(See p. 351)
Nulceophile Effects:
Nucleophiles must have a lone pair.
Negative nucleophiles produce ____________ products.
Neutral nucleophiles produce ________________ products.
(See p. 351)
Table of Nucleophiles on p. 352
Relative
Reactivity
------------------Nucleophilic Reactivity _____________ -----------------------------------
(Based on Bromomethane in ethanol.)
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Trends for Nucleophiles
1) Nucleophilicity appears to follow basicity. (The more basic, the more
nucleophilic.) See Table 11.2, p. 3353.
2) Nuclophilicity usually increases going down a column of the periodic table.
HS- is more nucleophilic than OH-, I – is more nuclophilic than Br- (However, can
change based on solvent.)
3) Negatively charged nucleophiles are usually more reactive than neutral ones.
Problem 11.4, p. 353
What product would you expect from SN2 reaction of 1-bromobutane with each of the
following?
a) NaI
b) KOH
c) H-CtripleC-Li
d) NH3
Problem 11.5, p. 353
Which substance in each of the following pairs is more reactive as a nucleophile?
Explain.
a)
b)
c)
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Leaving Group Effects:
Most leaving groups carry a negative charge. Leaving groups which can better stabilize
this negative charge are better. The best leaving groups are the weakest conjugate bases
derived from the strongest acids.
OH-, NH2-, ORRelative
Reactivity
<<1
F-
Cl-
Br-
I-
TosO-
1
200
10,000
30,000
60,000
-------------------------____________________ Reactivity ----------------------
Why?
In the transition state the negative charge is delocalized over both the Nucleophile and the
leaving group. If the leaving group can stabilize the charge, then the transition state will
have a lower energy and will proceed faster.
Transition State
R-F, R-OH, R-OR, R-NH2 are poor leaving groups and are generally not displaced by
nucleophiles and thus do not undergo SN2 reactions.
Problem 11.6, p. 354
Rank the following compounds in order of their expected reactivity toward SN2 reaction:
CH3Br,
CH3OTos,
(CH3)3CCl,
(CH3)2CHCl
Solvent Effects:
Protic solvents which contain –OH or –NH groups (like methanol or ethanol) are the
worst for SN2. Solvating the nucleophile makes it less reactive. (See Figure on p. 355)
Polar aprotic solvents with strong dipoles but without -OH or –NH groups are the best.
Such as acetonitrile, CH3CN, or DMF, [(CH3)2NCHO] – dimethylformamide or DMSO,
(CH3)2SO – dimethyl sulfoxide, HMPA, [(CH3)2N]PO – hexamethylphoshoramide.
Highly polar aprotic solvents readily dissolve salts. They prefer to solvate metal cations
such as Na+ in NaOH which causes release of OH-, the nucleophile.
Solvent
7
CH3OH
Relative 1
Reactivity
H2O
DMSO
DMF
CH3CN
HMPA
7
1300
2800
5000
200,000
-------------------------------------____________________ Reactivity----------------
Summary of SN2
1)
2)
3)
4)
5)
6)
Rate = k [RX] [Nu]
Mechanism is a single step (umbrella)
Inversion of Configuration
Faster reaction if the substrate is a methyl or primary
Strong bases work best as nucleophiles – further down a group, negative
The better leaving group is the more stable, weaker conjugate base ion
(OTos-> I-> Br- > Cl-)
7) Polar aprotic solvents are best (DMF, DMSO, acetonitrile). They dissolve the
metal salts without protonating/solvating the nucleophile.
11.6 The SN1 Reaction
Another mechanistic pathway which makes nucleophilic substitutions possible for
tertiary alkyl halides.
In the reaction,
R-Br + H2O  R-OH + HBr
Relative
Reactivity
<1
1
12
1,200,000
----------------------------Increasing Reactivity ----------------------------
11.7 Kinetics of SN1
First order process. The rate is dependent only on the substrate concentration.
Rate = k [RX]
This is the rate determining/limiting step – slowest step. No reaction can proceed faster
than the slowest step.
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This is a two step process and the nucleophile enters at the second step.
SN1 Mechanism:
See p. 359, Figure 11.10 for the Energy Diagram
(First step is rate limiting.)
11.8 Stereochemistry
The carbocation intermediate is sp2, planar, and achiral.
Reaction with a chiral reagent will form an inoptically active _____________________.
However, most reactions actually give a minor excess of the inversion product.
Approximately 80% is racemized and 20% is inverted.
See p. 360
R

40% R
(retention)
+
60% S
(inversion)
See p. 360, Figure 11.12
Suggested reason why product is not completely racemic – Ion pairing
The nucleophile entering from the opposite face is more likely if the leaving group is not
fully disengaged. (Kind of like SN2)
Ion Pair
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11.9 Characteristics of SN1
Substrate Effects:
The more stable the carbocation, the faster the SN1 reaction.
-------------------------Increasing Carbocation Stability---------------------------
Why Allylic and Benzylic? _____________________
Draw:
Note: Allylic and Benzylic Substrates are also reactive in SN2 reactions. There C-X
bonds are weak and more easily broken.
Problem 11.11,p. 363
Rank the following substances in order of their expected SN1 reactivity:
Leaving Group Effects:
Same order as in SN2.
Since the leaving group is directly involved in the rate limiting step. The more stable, the
better.
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SN1 reactions usually occur under neutral or acidic conditions, thus H2O can act as a
leaving group.
Example:
Nucleophile Effects:
The nucelophile does not affect the reaction rate since it is not part of the rate
determining step. Neutral and negatively charged nucleophiles are just as effective.
Most reactions occur under neutral or acidic conditions. Nonbasic is better. It prevents
competitive elimination of HX.
Solvent Effects:
Polar solvents are good. Solvent lone pairs can stabilize the carbocation and act as a
driving force for its formation. Solvent polarity is given by the dielectric constant. See
Table 11.3 on p. 365.
See p. 365
---------------------- Increasing Solvent Reactivity ------------------
Summary of SN1
1)
2)
3)
4)
Rate = k [RX]
Mechanism is a two step process
Racemic Mixture (Almost)
Faster reaction if the substrate is a Tertiary, (Allyic, Benzlic or Secondary –
all same)
5) Nucleophile unimportant
6) The better leaving group is the more stable, weaker conjugate base ion
(OTos-> I-> Br- > Cl-)
7) Polar solvents are best . They stabilize the carbocation.
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Problem 11.14, p. 367
Predict whether each of the following substitution reactions is likely to be SN1 or SN2:
a)
b)
11.10 Elimination Reactions of Alkyl Halides: Zaitsev’s Rule
Substitution
Alkyl Halide
Nucleophile/
Lewis Base
Elimination
Alkyl Halide
Nucleophile/
Lewis Base
Elimination reactions are more complex than substitution reactions.
One reason for this is regiochemistry.
Dehydrohalogenation of an unsymmetrical halide leads to a mixture of alkene products.
Zaitsev’s Rule is used to predict the major product – In a base induced elimination
reaction, the more highly substituted (more stable) alkene product predominates.
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Zaitsev’s Rule Example:
2-bromobutane
2-butene
(81%)
1-butene
(19%)
Problem 11.15, p. 368
What products would you expect from elimination reactions of the following alkyl
halides? Which product will be major in each case?
c)
11.11 The E2 Reaction
E2 Reaction
-Bimolecular
-Rate = k[RX] [Base]
-One Step
-Nucleophile is a strong base such as OH- or OR-Mechanism occurs with periplanar (all four reacting atoms lie in the same plane)
geometry.
Mechanism:
There are two possible periplanar arrangements – syn and anti.
Syn – same side, eclipsed, higher energy.
Anti- opposite side, staggered, lower energy. (As seen in mechanism.) PREFERRED
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Why is periplanar geometry important?
As the alkene forms, a pi bond forms which requires a p-p orbital overlap in the same
plane. Periplanar geometry provides a transition state which is already set up in the
correct plane.
11.12 Elimination Reactions and Cyclohexane Conformations
Anti periplanar occurs when H and the leaving group (halide) are trans diaxial. If the
leaving group or H are equatorial, E2 does not occur.
See Figure 11.19, p. 372.
See Figure 11.20, p. 373.
Reaction a) is 200 times faster than Reaction b).
Problem 11.19, p. 373
Which isomer would you expect to undergo E2 elimination faster, trans-1-bromo-4-tertbutylcyclohexane or cis-1-bromo-4-tert-butylcyclohexane? Draw each molecule in its
more stable chair conformation, and explain your answer.
11.13 The Deuterium Isotope Effect
Proves the existence of E2.
C-H bond is weaker and more easily broken than a C-D bond by about 5 kJ/mole.
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Examples:
Faster
Slower
C-H and C-D bonds are broken in the rate determining step. Results are consistent with
the one step (E2) mechanism that was proposed.
11.14 The E1 Reaction
E1 Reaction
-Unimolecular
-2 steps
-Rate = k[RX]
Mechanism of E1 (starts the same as SN1):
E1 and SN1 are in competition in protic solvents with nonbasic nucelophiles.
Since H and a halogen are lost in separate steps, there is no geometric requirement.
The transition state does not have to be anti periplanar.
Evidence of E1 mechanism:
-There is no deuterium isotope effect.
-The rate determining step is the loss of the leaving group.
-Breaking of C-H or C-D occurs in the fast step after the carbocation is formed.
Thus, there is no difference in reaction rates.
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11.15 Summary of Reactivity: SN1, SN2, E1, E2
See Table 11.4, p. 376
Primary Alkyl Halides
SN2 occurs if a good nucleophile such as RS-, I-, CN-, NH3, or Br- is used.
E2 occurs if a strong, sterically hindered base such as tert-butoxide is used.
Examples:
Secondary Alkyl Halides
SN2 and E2 occur in competition often leading to a mixture of products.
SN2 produces the predominant product in polar aprotic solvents with weakly basic
nucleophiles.
E2 produces the predominant product if a strong base such as CH3CH2O-, OH-, or NH2- is
used.
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SN1 and E1 reactions are common for allylic and benzylic alkyl halides if weakly basic
nucleophiles are used in protic solvents such as ethanol or acetic acid.
Tertiary Alkyl Halides
E2 occurs almost exclusively when a base such as OH- or RO- is used.
SN1 and E1 reactions occur under neutral conditions such as using ethanol as the
nucelophile and the solvent. The SN1 product predominates in hydroxylic solvents.
Problem 11.20, p. 378
Tell whether each of the following reactions is likely to be SN1, SN2, E1, or E2:
a) 1-bromobutane + NaN3  1-azidobutane
b) 3-chloropentane + KOH  2-pentene
c)
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11.16 Substitution Reactions in Synthesis
Very important in synthesis
See p. 379 and 380 – previously seen examples.
1) Alkylation – Reaction of acetylide + 1o alkyl halide, SN2 reaction (2o alkyl
halides give mostly E2 products.) (Ch. 8)
2) 3o Alcohol + HCl  Alkyl Halide, SN1 reaction (Ch. 10)
3) 1o Alcohol + PBr3 or SOCl2  Alkyl Halide, SN2 reaction (Ch. 10)
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