Chapter 6:Alkyl Halides Nucleophilic Substitution and

Elimination

Nomenclature (6-2)Alkyl Halides are also known as

HaloalkanesÚMolecules containing: F, Cl, Br, IÚNomenclature

RCHX

RRCX

R

RCH2X

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RCHX RCX

R2o halide 3o halide

RCH2X

1o halide

Dihalides:

R

X

X

R

C

X

C

X

geminal vicinal

Structure of Alkyl Halides (6-4)ÚBecause X is more electronegative than C,

the C-X bond is strongly polarized

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Synthesis of Alkyl Halides (6-6)We have already seen that alkyl halides are available via the corresponding alkanes by radical substitution. The reaction normally favours the alkyl halides formed from

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the most stable radical, however the yields are usually low.

Ú However, free radical haloganation of alkenes is a high yield process since it takes place via a very stable allylic radical (assuming that allylic hydrogen are present). The allylic position is the carbon directly attached to the alkene.

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ÚThe mechanism of the reaction follows the same 3 steps (initiation, propagation and termination) that we have seen before, but the high yield in these reactions is due to the very stable allylic radical produced in the process.

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ÚLarge quantities of Br2 must be avoided as this reagent will also add to the double bonds of alkene (chapter 8). A good way to make large amounts of allylic bromide is by using NBS (N-bromosuccinimide), as this reagent keeps the concentration of Br low.

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Br2 low.

Practice QuestionÚComplete the following reactions.

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ÚThere are many other ways to prepare alkyl halides from other functional groups. These will be covered later in other chapters:

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Reactions of Alkyl Halides (6-7)Alkyl halides can give 2 major types of reactionswhich often compete with each other

– Nucleophilic Substitutions (SN2 or SN1) and Eliminations (E2 and E1)

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ÚBecause of this competition that is always present, it is often difficult to identify the major product in their reactions.

ÚAnalysis of reaction conditions is very

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ÚAnalysis of reaction conditions is very important to decide on the product formed

Second Order Nucleophilic Substitutions (SN2) (6-8)

General Mechanism: SN2

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Ú In SN2 the mechanism of this reaction is “concerted” (i.e. bonds are formed and broken at the same time). In other words, no reactive intermediates are formed.

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ÚThis concerted mechanism is the reason for the name, because the nucleophile and the halide are both involved in the rate determining step of the reaction. Another name often used is: bimolecular nucleophilic substitution (SN2).

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– Nucleophile attacks the back side of the halide

– Angle of attack of the nucleophile is 180o

from the C-X bond– A transition state exists where partial

C-Nu and C-X bonds exist– The halide departs (this is called the

leaving group).

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leaving group).– Reactivity of halides follow the trend:

CH3X > 1o > 2o (3o give no reaction)

Examples of SN2 Reactions (6-9)

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ÚFactors Affecting SN2: Strength of the Nucleophile (6-10)

– Increasing the negative charge on an element, increases the nucleophilicity

– The more nucleophilic is the nucleophile, the better the rate of SN2

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the better the rate of SN2

ÚFactors affecting nucleophilicity- A species with a negative charge is a stronger nucleophile than one without a chargeHO- > H2O HS- > H2S H2N- > NH3

- Nucleophilicity decreases from left to right in the periodic table (because the atoms are getting smaller…therefore electrons are more tightly held)

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held)HO- > F- :NH3 > H2O: (Me)3P: > (Me)3S:

- Nucleophilicity increases down the periodic table (also based on the radius of atoms)I- > Br- > Cl- > F-

HSe- > HS- > HO-

R3P: > R3N:

ÚSteric effect of the nucleophile

– The more sterically hindered is the nucleophile, the poorer it is as a reagent….slower rate. This effect is called: steric hindrance.

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ÚSolvent effectsolvent can also affect the nucleophilicity of a nucleophile.– Polar protic (with acidic proton in the structure,

OH, NH) H-bond with the nucleophile and reduce its reactivity. Slower reactions.

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ÚPolar aprotic solvents (no OH or NH) cannot H-bond and keep the nucleophile more reactive. These are better solvents for SN2 reactions.

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Ú Reactivity of the Substrate (6-11)- Leaving group

The leaving group serves two purposes in a SN2 reaction:(1) polarizes the C-X bond (makes C δ+)(2) leaves the molecules taking 2 electrons originally shared with the carbon electrophile.

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A good leaving group has 3 characteristics:(1) electron withdrawing (polarize C-X bond)(2) stable (weak base) once it has left(3) polarizable to stabilize the transition state.

The better the leaving group, the better will be the rate of SN2 reactions

Ú The ability of a group to leave a molecule is a measure of its ease of displacement.

Ú Since the leaving group takes along the electron pair, its ability to leave can be correlated with its capacity to accommodate a negative charge.

Ú Weak bases are good leaving groups

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Ú Leaving group ability is inversely related to the base strength.

Ú To recognize the weak bases, compare them to the conjugate acids, since the weaker the base, the stronger the conjugate acid…pka table in appendix 5 (pages 1261-1262)

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Ú One way to make the previous reaction (in red) work, is by changing the nature of the leaving group chemically. Increasing the stability and decreasing the basicity of the leaving group allows methanol to be transformed into bromomethane.

H3C OHBr- H3C Br + OH-

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H3C OHH Br

H3C O+

HH

Br-

H3C Br + H2O

by protonating the hydroxyl group,the leaving group is now stable anda weak base (H2O) and the reactionworks.

Steric Factors on the Substrate– Halides with less steric factors at the

reacting carbon will react faster in a SN2 reaction. Primary halides (no steric hindrance) will react this way 99% of the time.

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CH3X > 1° > 2° >> 3°

– The same effect is observed at the carbon adjacent to the C-X bond

Since the nucleophile must approach from a 180o angleof the C-X bond, more substitution of the reactingcarbon slows down the reaction until no reaction is observed for tertiary alkyl groups.

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Table 6-5

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Stereochemistry in SN2 Reaction (6-12)

Since the nucleophile attacks the alkyl halide from the opposite side of the C-X bond, for chiral molecule, there will be a change in configuration

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Practice QuestionsÚ For each of the following pairs of SN2 reactions, indicate which

reaction occurs faster:

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ÚWhat is the product of the reaction of ethyl bromide with each of the following nucleophiles?

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ÚFirst-Order Nucleophilic Substitution: SN1 Reaction (6-13)– Reaction is also known as Unimolecular

Nucleophilic Substitution (SN1) since the rate determining step is the halide dissociation (this first step is spontaneous). This results in the formation of a carbocation (by solvolysis)

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ÚThe second step of the reaction is fast since the electrophile (carbocation) is charged and only a small amount of Ea is required.

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ÚTertiary alkyl halides normally react via this mechanism, while primary halides never do. The order is the opposite to the SN2 reaction and follows the order of stability of carbocations.

3° > 2° > 1° >> CH3X

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3° > 2° > 1° >> CH3X

ÚSimilar to the SN2 reaction, many factors influence the SN1 process.

ÚLeaving groupThe easiest it is for the leaving group to depart, the better is the rate of SN1 reaction. This is because breaking the C-X bond is the rate-determining step. Therefore, a better leaving

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determining step. Therefore, a better leaving group will have more tendency to depart than a poor one, increasing the rate of the reaction.

Trend is the same as for SN2: I- > Br- > Cl-

Strength of Nucleophile

Does not affect the rate of SN1 reaction, since the nucleophile is not involved in the rate-determining step below.

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Solvent EffectPolar protic (with HO group) solvents accelerate the SN1 reaction (via the solvation of the resulting intermediate ions)

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Stereochemistry of the SN1 Reaction (6-14)Chiral carbon will be racemized (formation of equal amounts of both enantiomers) in SN1 reactions. This is simply due to the geometry of the carbocation. Since it is now hybridized sp2, nucleophilic attack can take place from two different directions resulting in retention or inversion of configuration.

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Rearrangement in SN1 Reactions (6-15)Ú Since a carbocation is formed, you should be

careful to consider carbocation rearrangement.– Whenever a carbocation is formed, it can

rearrange to a more stable species by hydrogen or alkyl shifts. Smaller species shift more easily.

Ease of hydride oralkyl shift

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– This process is always a possibility whenever a carbocation is formed as a reactive intermediate in any reactions.

H > CH3 > CH3CH2 > Phenyl

better worst

alkyl shift

ÚDriving force: more stable carbocation is produced

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ÚWith primary halide, since carbocation cannot be formed, a concerted rearrangement may take place as seen below in the methyl shift leading to a tertiary carbocation.

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Practice QuestionÚ Arrange the following alkyl halides in order of decreasing

reactivity (most to least reactive) in SN1 reaction.2-bromopentane, 2-chloropentane,1-chloropentane, 3-bromo-3-methylpentane

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Ú Which of the following alkyl halides form a substitution product in a SN1 reaction that is different from the substitution product formed in a SN2 reaction? Draw the products.

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Comparison of SN1 and SN2 Reactions (6-16)

SN2 SN1

CH3X > 1º > 2º 3º > 2º

Strong nucleophile Weak nucleophile (may also be solvent)

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also be solvent)

Polar aprotic solvent Polar protic solvent

Rate = k[alkyl halide][Nuc] Rate = k[alkyl halide]

Inversion at chiral carbon Racemization

No rearrangements Rearranged products

Practice QuestionsÚ Which reaction in each of the following pairs will take

place more rapidly?

CH3BrOH-H2O

CH3OH + Br-CH3OH + HBr

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CH3BrH2ONH3 CH3OH + HBrCH3N+H3 + Br-

Ú Identify the most likely mechanism for the following reactions, SN2 or SN1, and draw the major organic product of each reaction.

EtOHI

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EtO-Br

CH3COOHCl

First-Order Elimination E1 Reaction (6-17)Ú Another type of reaction always compete with

the substitution reaction, the elimination. In this case an alkene is produced. In the case of the E1 reaction it is in competition with the SN1 process. In this case the nucleophile acts as a base.

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base.

Ú Remember that the rate determining step is the dissociation of the halide, so as long as a nucleophile is also basic, both E1 and SN1 will take place…a mixture of products is usually obtained in those reactions. However, in the presence of weak bases, SN1 usually predominates, while small amount of a strong base will give more E1 product.

H3CCH3

OH + H2C CCH3

major

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H3CCH3

CH3

Cl H3C C+CH3

CH3

H2O3

CH3

H2O (1% NaOH)

H3CCH3

CH3

OH

2CH3

+ H2C CCH3

CH3

80% 20%

40% 60%major

Ú Since carbocations are formed, rearrangement is once again a possibility leading to an isomeric product.

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Zaitsev’s Rule (6-18)Ú In elimination reactions, the most

substituted alkene usually predominates (this is due to thermodynamics…more substituted alkenes are more stable)

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Ú Practice QuestionFour alkenes are formed from the E1 reaction of 3-bromo-2,3-dimethylpentane. Give the structures of the alkenes, and rank them according to the amounts that would be formed (most to least).

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Second-Order Elimination: E2 Reaction (6-19)

Also known as bimolecular elimination, the E2 reaction is always a competing reaction when attempting to carry out SN2 reactions. This is especially true when the SN2 reaction site is sterically hindered.

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hindered.

ÚSimilar to the SN2 reaction, the mechanism of the E2 elimination is concerted (bonds break and form at the same time).

HB-

a base deprotonate the carbon, theelectrons of the C-H bond are used toform a new C C bond between the and

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ÚSince the β β β β carbon is the site of attack of the base, E2 eliminations are often called a ββββ-elimination or 1,2-elimination.

Br

form a new C-C bond between the andcarbons while the leaving group departs

Ú Since this reaction is in competition with SN2, the reactivity of the alkyl halides towards the E2 reaction follow the same trend as for the SN2 reaction.

Ú The E2 reaction is also regioselective, i.e. when more than one product can be formed, the

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more than one product can be formed, the most substituted alkene is the major product of the reaction, following Zaitsev’s rule

Br

carbons

CH3O-

20% 80%

Practice QuestionÚ What would be the major product obtained from the

reaction of each of the following alkyl halides with hydroxide ion in an E2 reaction?

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Ú Which of the alkyl halides in each pair is more reactive in an E2 reaction?

BrBr

or

Cl Br

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or

ClCl

or

Competition between E2 and E1 (9.4)ÚCompetition occurs only for 2o and 3o

halides since primary halides cannot react via E1 because the carbocation that would form is too unstable.

ÚFor 2o and 3o halides, the same factors that influenced S 1 or S 2 also affect E /E

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ÚFor 2 and 3 halides, the same factors that influenced SN1 or SN2 also affect E1/E2competition.

– [strong base] + polar aprotic solvent = E2– weak base + protic solvent = E1

Stereochemistry of E2 Reactions (9.5)

Ú The E2 reactions is very specific as to the reactive conformation. An Anti-coplanar arrangement of the ββββ

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arrangement of the ββββproton and the leaving group is preferred rather than the syn-coplanar due to repulsion of between the leaving group and the base.

ÚFor example:

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Comparison of E1 and E2 Eliminations (6-21)

E1 E2

* better with 3o halide* base strenght in noti t t ( ll k

* better with 3o halide* strong base required* l it f l t t

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important (usually weakbase* good ionizing solvent (protic* Zaitsev product formed* Rearrangement products

* polarity of solvent notimportant* Zaitsev products*anticoplanar arrangementnecessary* no rearrangements

Competition between Substitution and Elimination

Since eliminations and substitutions always compete with one another, it is sometimes difficult to predict what product will be obtained. Here are a few guidelines.

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Practice QuestionsWhich reacts faster in an SN2 reaction? Why?

Which reacts faster in an SN1 reaction? Why?

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Which reacts faster in an E1 reaction? Why?

Cl Bror

Study ProblemsÚ 6-46

Show how each of the following compounds might be synthesized by a SN2 displacement of an alkyl halide.

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Ú 6-53List the following carbocations in decreasing order (most to least) of their stability.

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Ú 6-60Predict the major organic products of the following E1 eliminations.

Br

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Br

Ú When 1-bromoethylcyclohexene undergoes solvolysis in ethanol, three major products are formed. Give mechanisms to account for these three products

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