CHM 2211C

6th edition Notes


Chapter 11

Reactions of Alkyl Halides:  Nucleophilic Substitutions and Eliminations



Dr. Andrea Wallace

Coastal Georgia Community College


Edited by

John T. Taylor

Florida Community College at Jacksonville
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.









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.





R-Y       Nu:-   à    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:






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 -____________________

Double the concentration of both reactants - _____________________

Half the concentration of one reactant - _________________________


11.4 The SN2 Reaction





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 ®-2-bromobutane? Show the stereochemistry of both reactant and product.







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




------------------Reactivity _____________ -----------------------------------à


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












------------------Nucleophilic Reactivity _____________ -----------------------------------à


(Based on Bromomethane in ethanol.)




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.












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-, OR-            F-         Cl-        Br-                    I-                      TosO-


Relative            <<1                  1          200      10,000             30,000             60,000



-------------------------____________________ Reactivity ----------------------à




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.



CH3OH            H2O     DMSO             DMF                CH3CN            HMPA


Relative    1                  7          1300                2800                5000                200,000



-------------------------------------____________________ Reactivity----------------à


Summary of SN2


1)      Rate = k [RX] [Nu]

2)      Mechanism is a single step (umbrella)

3)      Inversion of Configuration

4)      Faster reaction if the substrate is a methyl or primary

5)      Strong bases work best as nucleophiles – further down a group, negative

6)      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  à  R-OH  + HBr






Relative                        <1                    1                      12                    1,200,000


----------------------------Increasing Reactivity ----------------------------à


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.




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           à                   40% R             +     60% S

                                                (retention)                (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





11.9 Characteristics of SN1


Substrate Effects:


The more stable the carbocation, the faster the SN1 reaction.









-------------------------Increasing Carbocation Stability---------------------------à



Why Allylic and Benzylic? _____________________













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.



SN1 reactions usually occur under neutral or acidic conditions, thus H2O can act as a leaving group.








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 ------------------à


Summary of SN1


1)      Rate = k [RX]

2)      Mechanism is a two step process

3)      Racemic Mixture (Almost)

4)      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.



Problem 11.14, p. 367

Predict whether each of the following substitution reactions is likely to be SN1 or SN2:

















11.10    Elimination Reactions of Alkyl Halides:  Zaitsev’s Rule








Alkyl Halide                 Nucleophile/

                                    Lewis Base








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.




Zaitsev’s Rule Example:







2-bromobutane                                     2-butene                       1-butene

                                                                        (81%)                          (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?











11.11   The E2 Reaction


E2 Reaction


-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.











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







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-tert-butylcyclohexane 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.

















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


-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.


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.
















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.













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   à   1-azidobutane






b) 3-chloropentane  + KOH   à   2-pentene










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 à  Alkyl Halide, SN1 reaction (Ch. 10)


3)      1o Alcohol + PBr3 or SOCl2 à Alkyl Halide, SN2 reaction (Ch. 10)