SN1 SN2 E1 E2 Reaction PHR-122.ppt

1

SUBSTITUTION AND ELIMINATION

SN1, SN2, E1 & E2 REACTIONS

Dr Md Ashraful Alam

2

When a nucleophile (electron donor, e.g., OH-) reacts with an alkyl halide, the halogen leaves as a halide

There are two competing reactions of alkyl halides with nucleophiles….

1) substitution

2) elimination

C C

H

X

Nu:- + C C

H

Nu

+ X-

+ C C

H

X

Nu:- C C + X- + Nu H

The Nu:- replaces the halogen on the -carbon.

The Nu:- removes an H+ from a -carbon & the halogen leaves forming an alkene.

BrR.... :

..

.. ::BrNu:

3

There are two kinds of substitution reactions, called SN1 and SN2. As well as two kinds of elimination reactions, called E1 and E2.Let’s study SN2 reactions first. SN2 stands for Substitution, Nucleophilic,

bimolecular. Another word for bimolecular is ‘2nd order’. Bimolecular (or 2nd order) means that the rate of an SN2 reaction is

directly proportional to the molar concentration of two reacting molecules, the alkyl halide ‘substrate’ and the nucleophile:

Rate = k [RX] [Nu:-] (This is a rate equation and k is a constant). The mechanism of an SN2 reaction is the one shown on slide #2:

2nd Order Nucleophilic Substitution Reactions, i.e., SN2 reactions

C C

H

X

Nu:- + C C

H

Nu

+ X-

Note that the nucleophile must hit the back side of the -carbon. The nucleophile to C bond forms as the C to X bond breaks. No C+ intermediate forms. An example is shown on the next slide.

4

2nd Order Nucleophilic Substitution Reactions, i.e., SN2 reactions

The rate of an SN2 reaction depends upon 4 factors: 1. The nature of the substrate (the alkyl halide)2. The power of the nucleophile3. The ability of the leaving group to leave4. The nature of the solvent1. Consider the nature of the substrate:

Unhindered alkyl halides, those in which the back side of the -carbon is not blocked, will react fastest in SN2 reactions, that is:

Me° >> 1° >> 2° >> 3° While a methyl halides reacts quickly in SN2 reactions, a 3° does not react.

The back side of an -carbon in a 3° alkyl halide is completely blocked.

O H....: C Br

..

.. :

H

H

H+

transition state

C Br.... :

H H

H

OH.... +

..

.. :Br:C

H

HH

OH....

5

Me°>> 1° >> 2° >> 3°

Effect of nature of substrate on rate of SN2 reactions:

CH3 Br CH3 CH2 Br CH Br

CH3

CH3

C Br

CH3

CH3

CH3

t-butyl bromidemethyl bromide ethyl bromide isopropyl bromide

Back side of -C of a methyl halide

is unhindered.

Back side of -C of a 1° alkyl halide is slightly hindered.

Back side of -C of a 2° alkyl halide is mostly hindered.

Back side of -C of a 3° alkyl halide is

completely blocked.

decreasing rate of SN2 reactions

SPACE FILLING MODELS SHOW ACTUAL SHAPES AND RELATIVE SIZES

6

The -carbon in vinyl and aryl halides, as in 3° carbocations, is completely hindered and these alkyl halides do not undergo SN2 reactions.

Effect of the nucleophile on rate of SN2 reactions:

CH2 CH Br

H2C CH Br

Br

Br

vinyl bromide bromobenzene

The overlapping p-orbitals that form the -bonds in vinyl and aryl halides completely block the access of a nucleophile to the back side of the -carbon.

Nu:- Nu:-

7

2. Consider the power of the nucleophile: The better the nucleophile, the faster the rate of SN2 reactions. The table below show the relative power or various nucleophiles. The best nucleophiles are the best electron donors.

Effect of nature substrate on rate of SN2 reactions:

Reactivity Nu:- Relative Reactivity

very weak HSO4-, H2PO4

-, RCOOH < 0.01 weak ROH 1

HOH, NO3- 100

fair F- 500 Cl-, RCOO- 20 103 NH3, CH3SCH3 300 103

good N3-, Br- 600 103

OH-, CH3O- 2 106 very good CN-, HS-, RS-, (CH3)3P:, NH2

- ,RMgX, I-, H- > 100 106

incr

easi

ng

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3. Consider the nature of the leaving group:

The leaving group usually has a negative charge Groups which best stabilize a negative charge are the best leaving groups,

i.e., the weakest bases are stable as anions and are the best leaving groups. Weak bases are readily identified. They have high pKb values.

Iodine (-I) is a good leaving group because iodide (I-) is non basic. The hydroxyl group (-OH) is a poor leaving group because hydroxide (OH-)

is a strong base.

Effect of nature of the leaving group on rate of SN2 reactions:

pKb = 23 pKb = 22 pKb = 21 pKb = 11 pKb = -1.7 pKb = -2 pKb = -21

I- Br - Cl- F- HO- RO- H2N-

30,000 10,000 200 1 0 0 0

Increasing leaving ability

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4. Consider the nature of the solvent:There are 3 classes of organic solvents: Protic solvents, which contain –OH or –NH2 groups. Protic solvents

slow down SN2 reactions. Polar aprotic solvents like acetone, which contain strong dipoles but no

–OH or –NH2 groups. Polar aprotic solvents speed up SN2 reactions. Non polar solvents, e.g., hydrocarbons. SN2 reactions are relatively slow

in non polar solvents.

Effect of the solvent on rate of SN2 reactions:

Protic solvents (e.g., H2O, MeOH, EtOH, CH3COOH, etc.) cluster around the Nu:- (solvate it) and lower its energy (stabilize it) and reduce its reactivity via H-bonding.

X:-

H

H

HH OR

OR

OR

RO

+

+

+

+

-

-

-

-

A solvated anion (Nu:-) has reduced nucleophilicity,reduced reactivity and increased stability

A solvated nucleophile has difficulty hitting the -carbon.

10

Polar Aprotic Solvents solvate the cation counterion of the nucleophile but not the nucleophile.

Examples include acetonitrile (CH3CN), acetone (CH3COCH3), dimethylformamide (DMF) [(CH3)2NC=OH], dimethyl sulfoxide, DMSO [(CH3)2SO], hexamethylphosphoramide, HMPA {[(CH3)2N]3PO} and dimethylacetamide (DMA).


DMF

C

O

H NCH3

CH3

C

O

NCH3

CH3

DMSO

S

O

CH3H3C

HMPA

[(CH3)2N]3P O H3C

DMA

: : : : : :.. .. ..

....

..

CH3 C N :

acetonitrileC

O

CH3H3C

: :

acetone

Polar aprotic solvents solvate metal cationsleaving the anion counterion (Nu: -) bare andthus more reactive

CH3C O

O: :.... :

_Na+

Na+

N C CH3

N C CH3

N C CH3NCH3C

-

-

-

-

+

+

+

+

+ CH3C O

O: :.... :

_

CH3CN::

..

..

:

11

Non polar solvents (benzene, carbon tetrachloride, hexane, etc.) do not solvate or stabilize nucleophiles.

SN2 reactions are relatively slow in non polar solvents similar to that in protic solvents.


benzene

C

Cl

ClCl Cl

carbontetrachloride

CH3CH2CH2CH2CH2CH3

n-hexane

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1st Order Nucleophilic Substitution Reactions, i.e., SN1 reactions

C

CH3

H3C

CH3

Br + Na+ I- C

CH3

H3C

CH3

I + Na+ Br-3°rapid

3 alkyl halides are essentially inert to substitution by the SN2 mechanism because of steric hindrance at the back side of the -carbon.

Despite this, 3 alkyl halides do undergo nucleophilic substitution reactions quite rapidly , but by a different mechanism, i.e., the SN1 mechanism.

SN1 = Substitution, Nucleophilic, 1st order (unimolecular).

SN1 reactions obey 1st order kinetics, i.e., Rate = k[RX].

The rate depends upon the concentration of only 1 reactant, the alkyl halide-not the nucleophile

The order of reactivity of substrates for SN1 reactions is the reverse of SN2

3 > 2 > 1 > vinyl > phenyl > Me°

R3C-Br R2HC-Br RH2C-Br CH2=CH-Br -Br H3C-Br

increasing rate of SN1 reactions

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The mechanism of an SN1 reaction occurs in 2 steps:

Reaction Steps … 1. the slower, rate-limiting dissociation of the alkyl halide forming a C+

intermediate2. a rapid nucleophilic attack on the C+

Mechanism of SN1 reactions

C

CH3

H3C

CH3

3° Br.... : + Na+ Br-C

CH3

H3C

CH3

I....

:1.

Br --C

CH3

H3C

CH3

+

3° C+

rapid

Na+ I -....: :

2.

Note that the nucleophile is not involved in the slower, rate-limiting step.

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The rate of an SN1 reaction depends upon 3 factors: 1. The nature of the substrate (the alkyl halide)2. The ability of the leaving group to leave3. The nature of the solventThe rate is independent of the power of the nucleophile.

1. Consider the nature of the substrate:

Highly substituted alkyl halides (substrates) form a more stable C+.

The Rate of SN1 reactions

C

H

H

H +C

CH3

H

H +C

CH3

H

H3C +C

CH3

CH3

H3C +

tertiary 3º

secondary 2º

primary 1º

methyl

morestable

lessstable

> > >

increasing rate of SN1 reactions

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Alkyl groups are weak electron donors. They stabilize carbocations by donating electron density by induction

(through bonds)

They stabilize carbocations by hyperconjugation (by partial overlap of the alkyl C-to-H bonds with the empty p-orbital of the carbocation).

Stability of Carbocations

C

CH3

CH3

H3C +

Inductive effects: Alkyl groups donate (shift) electrondensity through sigma bonds toelectron deficient atoms.This stabilizes the carbocation.

vacant p orbitalof a carbocation

sp2hybridizedcarbocation

Csp3-Hssigma bondorbital

overlap (hyperconjugation)

HYPERCONJUGATION

+C C

..H

H

H

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Allyl and benzyl halides also react quickly by SN1 reactions because their carbocations are unusually stable due to their resonance forms which delocalize charge over an extended system

Stability of Carbocations

H2C CH +CH2 CH2HCH2C+

1º allyl carbocation

H2C CH +CHR CHRHCH2C+

2º allyl carbocation

2º benzylic1º benzylic

C

H

R

+C

H

H

+C

H

HC

H

HC

H

H

+ +

+

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Relative Stability of All Types of Carbocations

2º allylic

> >

3º allylic

> > >

3º C+

CCH3

CH3

CH3

+

CH2 CH CHR+CH2 CH CR2

+

C R2+

3º benzylic

C HR+

2º benzylic

1º allylic

CCH3

CH3

H+

2º C+

CH2 CH CH2+

C H2+

1º benzylic

1º C+

CCH3

H

H+

+

+

CH

H

Hm ethyl C

+

phenyl

> CH2 CH+

+vinyl C

Increasing C+ stability and rate of SN1 reaction

Note that 1° allylic and 1° benzylic C+’s are about as stable as 2°alkyl C+’s.Note that 2° allylic and 2° benzylic C+’s are about as stable as 3° alkyl C+’s.Note that 3° allylic and 3° benzlic C+’s are more stable than 3° alkyl C+’sNote that phenyl and vinyl C+’s are unstable. Phenyl and vinyl halides do not usually react by SN1 or SN2 reactions

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2. Consider the nature of the leaving group:The nature of the leaving group has the same effect on both SN1 and SN2

reactions.The better the leaving group, the faster a C+ can form and hence the faster will

be the SN1 reaction.The leaving group usually has a negative charge Groups which best stabilize a negative charge are the best leaving groups,

i.e., the weakest bases are stable as anions and are the best leaving groups. Weak bases are readily identified. They have high pKb values.

Effect of nature of the leaving group on rate of SN1 reactions:

pKb = 23 pKb = 22 pKb = 21 pKb = 11 pKb = -1.7 pKb = -2 pKb = -21

I- Br - Cl- F- HO- RO- H2N-

30,000 10,000 200 1 0 0 0

Increasing leaving ability Iodine (-I) is a good leaving group because iodide (I-) is non basic. The hydroxyl group (-OH) is a poor leaving group because hydroxide (OH-) is

a strong base.

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3. Consider the nature of the solvent: For SN1 reactions, the solvent affects the rate only if it influences the

stability of the charged transition state, i.e., the C+. The Nu:- is not involved in the rate determining step so solvent effects on the Nu:- do not affect the rate of SN1 reactions.

Polar solvents, both protic and aprotic, will solvate and stabilize the charged transition state (C+ intermediate), lowering the activation energy and accelerating SN1 reactions.

Nonpolar solvents do not lower the activation energy and thus make SN1 reactions relatively slower


reaction rate increases with polarity of solvent

The relative rates of an SN1 reaction due to solvent effects are given

(CH3)3C-Cl + ROH (CH3)3C-OR + HCl

H2O 20% EtOH (aq) 40% EtOH (aq) EtOH100,000 14,000 100 1

20

Solvent polarity is usually expressed by the “dielectric constant”, , which is a measure of the ability of a solvent to act as an electric insulator.

Polar solvents are good electric insulators because their dipoles surround and associate with charged species.

Dielectric constants of some common solvents are given in the following table


name dielectric constant name dielectric constant aprotic solvents protic solvents

hexane 1.9 acetic acid 6.2

benzene 2.3 acetone 20.7

diethyl ether 4.3 ethanol 24.3

chloroform 4.8 methanol 33.6

HMPA 30 formic acid 58.0

DMF 38 water 80.4

DMSO 48

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Consider the nature of the Nucleophile: Recall again that the nature of the nucleophile has no effect on the rate of

SN1 reactions because the slowest (rate-determining) step of an SN1 reaction is the dissociation of the leaving group and formation of the carbocation.

All carbocations are very good electrophiles (electron acceptors) and even weak nucleophiles, like H2O and methanol, will react quickly with them.

The two SN1 reactions will proceed at essentially the same rate since the only difference is the nucleophile.

Effect of the nucleophile on rate of SN1 reactions:

C

CH3

H3C

CH3

Br + Na+ I- C

CH3

H3C

CH3

I + Na+ Br-3°

C

CH3

H3C

CH3

Br + C

CH3

H3C

CH3

F + K+ Br-3° K+ F-

22

We have seen that alkyl halides may react with basic nucleophiles such as NaOH via substitution reactions.

Also recall our study of the preparation of alkenes. When a 2° or 3° alkyl halide is treated with a strong base such as NaOH, dehydrohalogenation occurs producing an alkene – an elimination (E2) reaction.

bromocyclohexane + KOH cyclohexene (80 % yield)

Substitution and elimination reactions are often in competition. We shall consider the determining factors after studying the mechanisms of elimination.

Elimination Reactions, E1 and E2:

BrKOH in ethanol + KBr + H2O

-HBr

O H....: C Br

..

.. :

H

H

H+

transition state

C Br.... :

H H

H

OH.... +

..

.. :Br:C

H

HH

OH....

23

+ C C

H

Br

OH- C C + Br- + HO H

There are 2 kinds of elimination reactions, E1 and E2.E2 = Elimination, Bimolecular (2nd order). Rate = k [RX] [Nu:-] E2 reactions occur when a 2° or 3° alkyl halide is treated with a strong

base such as OH-, OR-, NH2-, H-, etc.

E2 Reaction Mechanism

The Nu:- removes an H+ from a -carbon & the halogen leaves forming an alkene.

All strong bases, like OH-, are good nucleophiles. In 2° and 3° alkyl halides the -carbon in the alkyl halide is hindered. In such cases, a strong base will ‘abstract’ (remove) a hydrogen ion (H+) from a -carbon, before it hits the -carbon. Thus strong bases cause elimination (E2) in 2° and 3° alkyl halides and cause substitution (SN2) in unhindered methyl° and 1° alkyl halides.

24

In E2 reactions, the Base to H bond formation, the C to H bond breaking, the C to C bond formation, and the C to Br bond breaking all occur simultaneously. No carbocation intermediate forms.

Reactions in which several steps occur simultaneously are called ‘concerted’ reactions. Zaitsev’s Rule:Recall that in elimination of HX from alkenes, the more highly substituted (more stable) alkene product predominates.

E2 Reaction Mechanism

B:-

C

H

CR

R

X

B

RR

H

C CRR

R

R X

+

-

C C

R R

RR

+ B H + X-

CH3CH2CHCH3

Br CH3CH2O-Na+

EtOHCH3CH CHCH3 + CH3CH2CH CH2

2-butene 1-butene

major product( > 80%)

minor product( < 20%)

25

E2 reactions, do not always follow Zaitsev’s rule.

E2 eliminations occur with anti-periplanar geometry, i.e., periplanar means that all 4 reacting atoms - H, C, C, & X - all lie in the same plane. Anti means that H and X (the eliminated atoms) are on opposite sides of the molecules.

Look at the mechanism again and note the opposite side & same plane orientation of the mechanism:

E2 Reactions are ‘antiperiplanar’

B:-

C

H

CR

R

X

B

RR

H

C CRR

R

R X

+

-

C C

R R

RR

+ B H + X-

26

Antiperiplanar E2 Reactions in Cyclic Alkyl Halides

When E2 reactions occur in open chain alkyl halides, the Zaitsev product is usually the major product. Single bonds can rotate to the proper alignment to allow the antiperiplanar elimination.

In cyclic structures, however, single bonds cannot rotate. We need to be mindful of the stereochemistry in cyclic alkyl halides undergoing E2 reactions.

See the following example.

Trans –1-chloro-2-methylcyclopentane undergoes E2 elimination with NaOH. Draw and name the major product.

H

Cl

H3C

H

H

H

Na+ OH-

E2

H3C

HH

H

Non Zaitsev productis major product.

H

HH3C

H

Little or no Zaitsev (more stable) product is formed.

HOH

NaCl++

3-methylcyclopentene 1-methylcyclopentene

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Just as SN2 reactions are analogous to E2 reactions, so SN1 reactions have an analog, E1 reaction.

E1 = Elimination, unimolecular (1st order); Rate = k [RX]

E1 eliminations, like SN1 substitutions, begin with unimolecular dissociation, but the dissociation is followed by loss of a proton from the -carbon (attached to the C+) rather than by substitution.

E1 & SN1 normally occur in competition, whenever an alkyl halide is treated in a protic solvent with a nonbasic, poor nucleophile.

Note: The best E1 substrates are also the best SN1 substrates, and mixtures of products are usually obtained.

E1 Reactions

CCH3

CH3

CH3

Brslow

B:-

CCH3

CH3

C H

H

H

+ rapidC C

CH3 H

CH3 H

+ B H + Br-Br--

28

As with E2 reactions, E1 reactions also produce the more highly substituted alkene (Zaitsev’s rule). However, unlike E2 reactions where no C+ is produced, C+ rearrangements can occur in E1 reactions.

e.g., t-butyl chloride + H2O (in EtOH) at 65 C t-butanol + 2-methylpropene

In most unimolecular reactions, SN1 is favored over E1, especially at low temperature. Such reactions with mixed products are not often used in synthetic chemistry.

If the E1 product is desired, it is better to use a strong base and force the E2 reaction.

Note that increasing the strength of the nucleophile favors SN1 over E1. Can you postulate an explanation?

E1 Reactions

CCH3

CH3

CH3

Cl +

H2O, EtOH

65ºCCCH3

CH3

CH3

OH C CCH3 H

CH3 H

64%36%

SN1product

E1product

29

1. Non basic, good nucleophiles, like Br- and I- will cause substitution not elimination. In 3° substrates, only SN1 is possible. In Me° and 1° substrates, SN2 is faster. For 2° substrates, the mechanism of substitution depends upon the solvent.

2. Strong bases, like OH- and OR-, are also good nucleophiles. Substitution and elimination compete. In 3° and 2° alkyl halides, E2 is faster. In 1° and Me° alkyl halides, SN2 occurs.

3. Weakly basic, weak nucleophiles, like H2O, EtOH, CH3COOH, etc., cannot react unless a C+ forms. This only occurs with 2° or 3° substrates. Once the C+ forms, both SN1 and E1 occur in competition. The substitution product is usually predominant.

4. High temperatures increase the yield of elimination product over substitution product. (G = H –TS) Elimination produces more products than substitution, hence creates greater entropy (disorder).

5. Polar solvents, both protic and aprotic, like H2O and CH3CN, respectively, favor unimolecular reactions (SN1 and E1) by stabilizing the C+ intermediate. Polar aprotic solvents enhance bimolecular reactions (SN2 and E2) by activating the nucleophile.

Predicting Reaction Mechanisms

30


alkyl

halide (substrate)

good Nu-

nonbasic e.g., bromide

Br-

good Nu- strong base e.g., ethoxide

C2H5O-

good Nu- strong bulky base

e.g., t-butoxide (CH3)3CO-

very poor Nu- nonbasic

e.g., acetic acid CH3COOH

Me

1°

2

3

SN1, E1

SN2

E2

SN2

SN2

SN1

SN2

SN2

E2

E2

SN2

E2 (SN2)

E2

no reaction

no reaction

SN1, E1

Strong bulky bases like t-butoxide are hindered. They have difficulty hitting the -carbon in a 1° alkyl halide. As a result, they favor E2 over SN2 products.

31

The nucleophiles in the table on slide 30 are extremes. Some nucleophiles have basicity and nucleophilicity in between these extremes. The reaction mechanisms that they will predominate can be interpolated with good success. Predict the predominant reaction mechanisms the following table.


alkyl

halide (substrate)

……….. Nu- ………. .base

e.g., cyanide CN-

pkb = ………

……….. Nu- ……….. base

e.g., alkyl sulfide RS-, also HS-

pkb = ………

……….. Nu- ……….. base e.g., carboxylate

RCOO-

pkb = ………

alcohol

(substrate)

HI HBr HCl

Me Me

1 1

2 2

3 3

4.7

v. gd. moderate

SN2

SN2

SN2

E2

6.0 / 7.0

moderatev. gd.

SN2

SN2

E2

SN2

9

weakfair

SN2

SN2

E2

E2

SN2

SN2

SN1

SN1

HCl, HBr and HI are assumed to be in aqueous solution, a protic solvent.

32

Recall the preparation of long alkynes. 1. A terminal alkyne (pKa = 25) is deprotonated with a very strong base…

R-C C-H + NaNH2 R-C C:- Na+ + NH3

2. An alkynide anion is a good Nu:- which can substitute (replace) halogen atoms in methyl or 1 alkyl halides producing longer terminal alkynes

R-C C: - Na+ + CH3CH2-X R-C C-CH2CH3 + NaX

The reaction is straightforward with Me and 1 alkyl halides and proceeds via an SN2 mechanism

Alkynide anions are also strong bases (pKb = -11) as well as good Nu:-’s, so E2 competes with SN2 for 2 and 3 alkyl halides

CH3(CH2)3CC:- Na+ + CH3-CH(Br)-CH3 CH3(CH2)3CCCH(CH3)2 (7% SN2) + CH3(CH2)3CCH + CH3CH=CH2 (93% E2)

Alkylation of Alkynides

33

Alkyl halides can be prepared from alcohols by reaction with HX, i.e., the substitution of a halide on a protonated alcohol.

Preparation of Alkyl Halides from Alcohols

+ H Cl(CH3)3C OH....

(CH3)3C OH2..+

H2O-(CH3)3C + (CH3)3C

..: H2O+

:-

..

..:Cl

Cl..SN1(Lucas Test)

3º

..1º

CH3CH2 OH..+ H Cl CH3CH2 OH2

+

:-

..

..:Cl

CH3CH2Cl + H2OSN2

Very slow. Protic solvent inhibits the nucleophile.

Rapid. 3° C+ is stabilized by protic sovent (H2O)

Draw the mechanism of the reaction of isopropyl alcohol with HBr.What products form if concentrated H2SO4 is used in place of aq. HCl?

OH- is a poor leaving group, i.e., is not displaced directly by nucleophiles. Reaction in acid media protonates the OH group producing a better leaving group (H2O). 2 and 3 alcohols react by SN1 but Me° and 1 alcohols react by SN2.

34

Alternative to using hydrohalic acids (HCl, HBr, HI), alcohols can be converted to alkyl halides by reaction with PBr3 which transforms OH- into a better leaving group allowing substitution (SN2) to occur without rearrangement.

Preparation of Alkyl Halides from Alcohols

SN2 CH3(CH2)4CH2 OH

PBr

BrBr

:

etherCH3(CH2)4CH2 O

PBr2....

..+

H

Br-

CH3(CH2)4CH2Br1º

35

On Slide 22 we noted that 2° and 3° alkyl halides can be dehydrohalogenated with a strong base such as OH- producing an alkene.

bromocyclohexane + KOH cyclohexene (80 % yield)

Clearly, this is an E2 reaction.

Preparation of Alkenes from Alkyl Halides

BrKOH in ethanol + KBr + H2O

-HBr

Predict the mechanism that occurs with a Me° or 1° alkyl halide.

Predict the products and mechanism that occur with isopentyl chloride and KOH

36

Alkyl Halide Substrate Reactivity:

Summary of SN /Elimination Reactions

unhindered substrates favor SN2do not form a stable C +do not react by S N1 or E1

hindered substrates. SN2 increasingly unfavorable, E2 is OKform increasingly stable C+ favors SN1 and E1. E2 is OK

CH

H

H

Br CCH3

H

H

Br CCH3

CH3

H

Br CCH3

CH3

CH3

Br

E2 reactions possible with strong bases

E2 reactions possible with strong bulky bases (t-butoxide)

methyl 1º 2º 3º

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SN1 SN2 E1 E2 Reaction PHR-122.ppt