Subst.Addit.Elim._2015_

P r o f . C AR L O S M E L É N D E Z G Ó M E Z Q c o , P h . D .

P R O G R AM A D E Q U Í M I C A

FAC U LTAD D E C I E N C I AS B ÁS I C AS

U N I V E R S I D AD D E L ATL ÁN TI C O

2 0 1 3

Nucleophilic Substitution Reactions

at the Saturated C Atom

• The electronegative halogen atom in alkyl halides creates a

polar C—X bond, making the carbon atom electron deficient.

Electrostatic potential maps of four simple alkyl halides

illustrate this point.

The Polar Carbon-Halogen Bond

Electrostatic potential maps of four halomethanes (CH3X)

Alkyl Halides and Nucleophilic Substitution


The Polar Carbon-Halogen Bond

4

• Three components are necessary in any substitution reaction.

General Features of Nucleophilic Substitution:


• Negatively charged nucleophiles like HO¯ and HS¯ are used as

salts with Li+, Na+, or K+ counterions to balance the charge. Since

the identity of the counterion is usually inconsequential, it is often

omitted from the chemical equation.

• When a neutral nucleophile is used, the substitution product bears

a positive charge.



• Furthermore, when the substitution product bears a positive charge

and also contains a proton bonded to O or N, the initially formed

substitution product readily loses a proton in a BrØnsted-Lowry acid-

base reaction, forming a neutral product.

• To draw any nucleophilic substitution product:

Find the sp3 hybridized carbon with the leaving group.

Identify the nucleophile, the species with a lone pair or bond.

Substitute the nucleophile for the leaving group and assign

charges (if necessary) to any atom that is involved in bond

breaking or bond formation.



Nucleophiles and Electrophiles; Leaving Groups



Stated with some exaggeration, organic chemistry is comparatively simple to learn

because most organic chemical reactions follow a single pattern. This pattern is

Contain an electron pair that is easily available because it is

nonbonding or they must contain a bonding electron pair that can be

donated from the bond involved and thus be made vailable to the

reaction partner.

Are electron pair acceptors. They therefore contain either a deficiency

in the valence electron shell of one of the atoms they consist of or they

are indeed valence-saturated but contain an atom from which a

bonding electron pair can be removed as part of a leaving group

Nucleophile

Electrophile

For most organic chemical reactions, the pattern just specified can thus be written

more briefly as follows:



Alkyl groups are transferred to the nucleophiles. Organic electrophiles of this type

are referred to as alkylating agents. The group X is displaced by the nucleophile

agent. Consequently, both the bound group X and the departing entity X are called

leaving groups.

Some uncharged and a few positively charged three-membered heterocycles also

react as alkylating agents. The most important heterocyclic alkylating agents of

this type are the epoxides.When there are no Brønsted or Lewis acids present,

epoxides act as hydroxy alkylating agents with respect to nucleophiles.

This reaction can also be considered to be an addition reaction. An intermolecular

addition reaction is one that involves the combination of two molecules to form one new

molecule. …


at the Saturated C Atom.

As emerges from these definitions, good and poor nucleophiles, high and low

nucleophilicity, good and poor alkylating agents, good and poor electrophiles, and

high and low electrophilicity are kinetically determined concepts.

Good and Poor Nucleophiles



Answers to all these questions are obtained via pairs of SN reactions, which are carried out as

competition experiments. In a competition experiment two reagents react simultaneously with

one substrate…. The nucleophile that reacts to form the main product is then the “better”

nucleophile.

With this as the basic idea, the experimentally observable nucleophilicity gradations

can be interpreted as follows……

Within a group of nucleophiles that attack at the electrophile with the same atom,

the nucleophilicity decreases with decreasing basicity of the nucleophile

Decreasing basicity is equivalent to decreasing affinity of an electron pair for a proton,

which to a certain extent, is a model electrophile for the electrophiles of SN reactions...





This parallel between nucleophilicity and basicity can be reversed by steric

effects. Less basic but sterically unhindered nucleophiles therefore have a higher

nucleophilicity than strongly basic but sterically hindered nucleophiles.

This is most noticeable in reactions with sterically demanding alkylating agents or

sterically demanding epoxides…..





Nucleophilicity decreases with increasing electronegativity of the attacking atom.

This is true both in comparisons of atomic centers that belong to the same period

of the periodic table of the elements.

and in comparisons of atomic centers from the same group of the periodic table:

The nucleophilicity of a given nucleophilic center is increased by attached heteroatoms

that possess free electron pairs (-effect):

The reason for this is the unavoidable overlap of the orbitals that accommodate the free

electron pairs at the nucleophilic center and its neighboring atom.

Leaving Groups and the Quality of Leaving Groups



Good leaving groups: RHal and

epoxides can be further activated

with Lewis acids.

The Hammond postulate implies that a good leaving group is a stabilized species, not a

high-energy species. Therefore good leaving groups are usually weak bases, not strong

bases.

Very basic leaving groups are produced relatively slowly according to

Hammond. In other words, strong Brønsted bases are poor leaving

groups; weak Brønsted bases are good leaving groups!!!!!!

HOW CAN WE EXPLAIN?

A 1:1 mixture of a strong base with protons would be high in energy relative to the

corresponding conjugate acid. From this we can conclude that a mixture of a strongly basic

leaving group with the product of an SN reaction is also relatively high in energy.. !!!!



Alcohols, ethers, and carboxylic acid esters do not enter into any SN reactions with

nucleophiles.The reason for this is that poor leaving groups would have to be released (OH,

OR, O2CR).

These compounds can enter into SN reactions with nucleophiles when they are activated as

oxonium ions, for example via a reversible protonation, via bonding of a Lewis acid (LA), or via

a phosphorylation. Thus, upon attack by the nucleophile, better leaving groups (e.g., HOH,

HOR, HO2CR, O―PPh3) can be released.

in situ activation of the leaving group

necessary!!!!



Very poor leaving group or not a

leaving group..!!!

Poor leaving group. The effect of nucleophiles on substrates of the latter type does

result in a reaction, but it is not an SN reaction.

The nucleophile abstracts an acidic proton in

the alpha- position to the functional group

instead of replacing the functional group.

Another reaction competing with the substitution of a functional

group by a nucleophile is an attack on the functional group by

the nucleophile.

Hammond’s Postulate

“If two states, for example, a transition state and an unstable intermediate, occur consecutively during a reaction process and have nearly the same energy content, their interconversion will involve only a small reorganization of the molecular structures”

George S. Hammond, J. Am. Chem. Soc. 1955, 77, 334-338.


“The structure of the transition state for an exothermic reaction is reached early in the reaction, so it resembles reactants more than products. Conversely, the structure of the transition state for an endothermic reaction step is reached relatively late, so it resembles products more then reactants.”


In reactions where the starting material is higher in energy (A), the transition state more closely resembles the starting material

In reactions where the product is higher in energy (B), the transition state more closely resembles the product

George S. Hammond, J. Am. Chem. Soc. 1955, 77, 334-338.

SN2 Reactions: Kinetic and Stereochemical Analysis—Substituent

Effects on Reactivity



Energy Profile and Rate Law for SN2

Reactions: Reaction Order.

An SN2 reaction refers to an SN reaction. in

which the nucleophile and the alkylating

agent are converted into the substitution

product in one step, that is, via one transition

state.



Rate laws establish a relationship between…..

• The change in the concentration of a product, an intermediate, or a starting material

as a function of time.

• The concentrations of the starting material(s) and possibly the catalyst

• The rate constants of the elementary reactions that are involved in the overall

reaction.

With the help of the rate laws that describe the elementary reactions involved, it is

possible to derive…

If the right-hand side of Equation does not contain any sums or

differences, the sum of the powers of the concentration(s) of the starting

material(s) in this expression is called the order m of the reaction….





Energy Profile and Rate Law for SN2

Reactions: Reaction Order.

This reaction is a bimolecular substitutions. And

referred to a SN2 mechanism .The

bimolecularity makes it possible to distinguish

between this type of substitution and SN1

reactions.





The attack by potassium acetate

on the trans-tosylate A gives

exclusively the cyclohexyl

acetate cis-B. No trans-isomer

is formed…..!!!

In the substitution product cis-B the acetate is axial and the adjacent H atom is

equatorial. Thus a 100% inversion of the configuration has taken place in this SN2

reaction.

The reason for the inversion of configuration is that SN2

reactions take place with a backside attack by the

nucleophile on the bond between the C atom and the leaving

group.





The SN2 reactions take place with a backside attack

by the nucleophile on the bond between the C atom

and the leaving group.

The SN2 mechanism is casually also referred to as an ―umbrella mechanism.‖The

nucleophile enters in the direction of the umbrella handle and displaces the leaving

group, which was originally lying above the tip of the umbrella.

The geometry of the transition state corresponds to the geometry of the umbrella, which

is just flipping over.


at the Saturated C Atom. A Refined Transition State Model for the SN2 Reaction;

Crossover Experiment and Endocyclic Restriction Test

Determination of the mechanism of one-step SN

reactions on methyl (arenesulfonates): intra- or

intermolecularity.

Experiment 1. The perprotio-sulfonyl anion [H6]-A reacts to

form the methylated perprotio-sulfone [H6]-B.

Experiment 2. The sulfonyl anion [D6]-A, (perdeuterated in

both methyl groups), reacts to form the hexadeuterated

methylsulfone [D6]-B.

The intramolecular methylation of the substrate, (not

observed), would had to take place through a six-

membered cyclic transition state. Why a cyclic

transition state not able to explain the SN reactions in

Figure?

In other cases cyclic six-membered transition states of intramolecular reactions are so favored that

intermolecular reactions usually do not occur.

INTRAMOLECULAR O INTERMOLECULAR REACTION?



The conformational degrees of freedom of cyclic transition states are considerably limited or

―restricted‖ relative to the conformational degrees of freedom of noncyclic transition states

(cf. the fewer conformational degrees of freedom of cyclohexane vs n-hexane).

Mechanistic investigations of this type are

therefore also referred to as endocyclic

restriction tests. They prove or refute in a

very simple way certain transition state

geometries because of this conformational

restriction.

The endocyclic restriction imposes limitations of the conceivable geometries on cyclic transition states.

These geometries do not comprise all the possibilities that could be realized for intermolecular

reactions proceeding through acyclic transition states.



In the SN reactions of Figure, the endocyclic restriction would therefore impose a geometry in an

intramolecular substitution that is energetically disfavored relative to the geometry that can be obtained in

an intermolecular substitution.

In an intramolecular substitution the

nucleophile, the attacked C atom, and the

leaving group cannot lie on a common axis.

The colinear geometry can be obtained

in an intermolecular way!!!!!



In the SN reactions of Figure, the endocyclic restriction would therefore impose a geometry in an

intramolecular substitution that is energetically disfavored relative to the geometry that can be obtained in

an intermolecular substitution



This approach path is preferred in order to achieve a transition state with optimum bonding interactions.

Let us assume that in the transition state, the distance between the nucleophile and the attacked C

atom and between the leaving group and this C atom are exactly the same.

• The geometry of the transition state would

then correspond precisely to the geometry of

an umbrella that is just flipping over.

• The attacked C atom would be—as shown

in Figure (left)—sp2-hybridized and at the

center of a trigonal bipyramid.

• The nucleophile and the leaving group

would be bound to this C atom via sigma

bonds, both suffer a overlap with one lobe of

the 2pz AO. A linear arrangement of the

nucleophile, the attacked C atom, and the

leaving group is preferred in the transition

state of SN reactions.

The figure shows that in a bent transition state of the SN reaction neither the nucleophile nor the leaving

group can form similarly stable bonds by overlap with the 2pz AO of the attacked C atom. Because the

orbital lobes under consideration are not parallel Bent bonds are weaker than linear bonds because

of the smaller orbital overlap



Substituent Effects on SN2 Reactivity

How substituents in the alkylating agent influence the rate constants of SN2 reactions can be

explained by means of the transition state model developed previously. This model makes it

possible to understand both the steric and the electronic substituent effects.

• The bond angle between these

substituents and the leaving group

decreases from approximately the

tetrahedral angle 109 to approximately 90.

• The attacking nucleophile approaches the

inert substituents until the bond angle that

separates it from them is also

approximately 90.

The resulting increased steric interactions destabilize the transition state. It should be noted that this

destabilization is not compensated for by the simultaneous increase in the bond angle between the inert

substituents from approximately the tetrahedral angle to approximately 120.




The effect of the substituent in the transition state has various consecuences…..

• The SN2 react alkylating agent decreases with an increasing number of the alkyl substituents at

the attacked C ativity of an aom. In other words, alpha branching at the C atom of the alkylating

agent reduces its SN2 reactivity. This reduces the reactivity so much that tertiary C atoms can no

longer be attacked according to an SN2 mechanism at all:

• The SN2 reactivity of an alkylating agent decreases with an increase in size of the alkyl

substituents at the attacked C atom. In other words, beta branching in the alkylating agent reduces

its SN2 reactivity. This reduces the reactivity so much that a C atom with a tertiary C atom in the

position can no longer be attacked at all according to an SN2 mechanism:



Besides these rate-reducing steric substituent effects, in SN2 reactions there is a rate increasing

electronic substituent effect. It is due to facilitation of the rehybridization of the attacked C atom from sp3

to sp2.

Electronic effects on SN2 reactivity:

conjugative stabilization of the transition

state by suitably aligned unsaturated

substituents.

This effect is exerted by unsaturated substituents bound to the attacked C atom!!!

These include substituents, such as alkenyl,

aryl, or the C=O double bond of ketones or

esters.

When it is not prevented by the occurrence of strain, the p-electron system of

these substituents can line up in the transition state parallel to the 2pz AO at

the attacked C atom!!!

SN1 Reactions: Kinetic and Stereochemical Analysis;

Substituent Effects on Reactivity



Energy Profile and Rate Law of SN1 Reactions;

Steady State Approximation

Take place in two steps. In the first and slower step, heterolysis of the bond between the C atom

and the leaving group takes place. A carbenium ion is produced as a high-energy, and

consequently short-lived, intermediate. In a considerably faster second step, it combines with

the nucleophile to form the substitution product Nu¬R.

Mechanism and energy profile of SN1

reactions: Ea,het designates the activation

energy of the heterolysis, khet and kattack

designate the rate constants for the

heterolysis and the nucleophilic attack on the

carbenium ion, respectively.

In a substitution according to the SN1 mechanism, the nucleophile does not activel attack

the alkylating agent.The reaction mechanism consists of the alkylating agent dissociating

by itself into a carbenium ion and the leaving group

The energy profile shows that here—as everywhere else—the rate-determining step of a multistep

sequence is the step in which the highest activation barrier must be overcome.



Rate and selectivity of SN1

reactions.vBoth reactions 1 and

2vtake place via the benzhydryl

cation. Both reactions (with pyridine

and triethylamine) take place with the

same rate constant. The higher

nucleophilicity of pyridine

does not become noticeable until

experiment 3, the ―competition

experiment‖: The pyridinium salt is by

far the major product.

• Both take place via the

benzhydryl cation Ph2CH. In

the first experiment, this cation

is intercepted by pyridine as

the nucleophile to form a

pyridinium salt. In the second

experiment, it is intercepted by

triethylamine to form a

tetraalkylammonium salt.

• In experiment 3, both SN1

reactions are carried out as

competitive reactions:

benzhydryl chloride heterolyzes

(just as fast as before) in the

presence of equal amounts of

both amines. The benzhydryl

cation is now intercepted faster

by the more nucleophilic

pyridine. The major product is

the pyridinium salt


at the Saturated C Atom. Stereochemistry of SN1 Reactions; Ion Pairs

In the carbenium ion intermediates R1R2R3C+, the valence-unsaturated C atom has a

trigonal planar geometry. These intermediates are therefore achiral (if the substituents Ri

themselves do not contain chiral centers).

The carbenium ions react with achiral

nucleophiles to form chiral substitution products

R1R2R3C-Nu. These must be produced as a

1:1 mixture of both enantiomers (i.e., as a

racemic mixture).

Stereochemistry of an SN1 reaction

that takes place via a contact ion pair.

The reaction proceeds with 66%

inversion of configuration and 34%

racemization.

The SN1 reaction with optically pure R-2-bromooctane carried out as a solvolysis. By

solvolysis one means an SN1 reaction performed in a polar solvent that also functions as the

nucleophile



The solvolysis reaction takes place in a water/ethanol mixture. In the rate-

determining step a secondary carbenium ion is produced.

The reacting carbenium ion is still almost in contact with this bromide ion. Thus it exists

as part of a so-called contact ion pair (the emphasis is on ion pair) R1R2HC. . . Br.

• In this ion pair, the bromide ion adjacent to the

carbenium ion center partially protects one side of the

carbenium ion from the attack of the nucleophile.

• The nucleophile preferentially attacks from the side that

lies opposite the bromide ion. Thus the solvolysis

product in which the onfiguration at the attacked C

atom has been inverted is the major product.

When in an SN1 reaction the nucleophile attacks

the contact ion pair, the reaction proceeds with

partial inversion of configuration (or

alternatively, with partial but not complete

racemization).





The solvolysis reaction takes place in a water/ethanol mixture. In the rate-

determining step a secondary carbenium ion is produced.

The reacting carbenium ion is still almost in contact with this bromide ion. Thus it exists

as part of a so-called contact ion pair (the emphasis is on ion pair) R1R2HC. . . Br.

When in an SN1 reaction the nucleophile

attacks the carbenium ion after it has

separated from the leaving group, the

reaction takes place with complete

racemization.This is the case with more

stable and consequently longer-lived

carbenium ions.

The bromide ion moves so far away from the alpha-

methylbenzyl cation intermediate that it allows the

solvent to attack both sides of the carbenium ion

with equal probability!!!!!



Solvent Effects on SN1 Reactivity

Heterolyses of alkylating agents, and consequently SN1 reactions, therefore,

succeed only in highly solvating media. These include the polar protic solvents

such as methanol, ethanol, acetic acid, and aqueous acetone as well as the polar

aprotic solvents acetone, acetonitrile, DMF, NMP, DMSO, and DMPU.

It is important to note that the use of polar solvents is only a necessary but not a sufficient

condition for the feasibility of SN1 substitutions or for the preference of the SN1 over the

SN2 mechanism.




The effects of group substitution in the SN1 reactivity are compared with

the carbocation stability as well.

Stability: Stabilization via alkyl substituents (hyperconjugation)

R

R

R

H

R

R

H

H

R

H

H

H

Order of carbocation stability: 3Þ>2Þ>1Þ

>> > Due to increasing number of substituents capable of hyperconjugation

C C+H

314

276

249

231

287

386

239

Hydride ion affinities

The relative stabilities of various carbocations can be measured in the gas phase by theiraffinity for hydride ion.

J. Beauchamp, J. Am. Chem. Soc. 1984, 106, 3917.

+ H

Note: As S-character increases, cation stability decreases due to more electronegative carbon.

+ HI

HI increases C(+) stability decreases

Hydride Affinity = –G°

C C C C

CH3+

CH3CH2+

(CH3)2CH+

(CH3)3C+

H2C=CH+

PhCH2+

R R–H

Carbocation StabilityCarbocation Stability



Hydride abstraction from neutral precursors

R3C H + Lewis-Acid R3C H =

HH

H

RS

RS

H

H

R2N

R2N

H

Hetc.

Lewis-Acid: Ph3C BF4, BF3, PCl5

Removal of an energy-poor anion from a neutral precursor via Lewis Acids

R3C X + LA LA–X LA: Ag , AlCl3, SnCl4, SbCl5, SbF5, BF3, FeCl3, ZnCl2, PCl3, PCl5, POCl3 ...

X: F, Cl, Br, I, OR

Acidic dehydratization of secondary and tertiary alcohols

R3C OH- H2O R: Aryl + other charge stabilizing substituents

X: SO42-, ClO4

-, FSO3-, CF3SO3

-

From neutral precursors via heterolytic dissociation (solvolysis) - First step in SN1 or E1 reactions

solventAbility of X to function as a leaving group:

-N2+ > -OSO2R' > -OPO(OR')2 > -I • -Br > Cl > OH2

+ ...

Carbocation Generation

R3C

R3C +

+ R3C +H–X X

R3C X R3C + X

Addition of electrophiles to š-systems

R

R

R

R

H R

R

R

R

H R RH R

H

R

H

Carbocation Generation



Vinyl & Phenyl Cations: Highly Unstable

Phenyl Cations

H3C CH2

276

H2C CH

287

+21HC C

386

+81

Hydride ion affinities (HI)

H2C CH

287

+11

298

Allyl & Benzyl Carbocations

Carbocation Stabilization via -delocalization

Stabilization by Phenyl-groups

The Benzyl cation is as stable as a t-Butylcation. This is shown in the subsequent isodesmic equations:

Ph CH2

239

Hydride ion affinities (HI)

231

Me3 C

Carbocation Stability

Br

Carbocation Stability



Solvolysis rates represent the extend of that cyclopropyl orbital overlap contributing to the stabiliziation of the carbenium ion which is involved as a

reactive intermediate:

Me

Me

OTs

OTs

Cl

Cl

krel = 1 krel = 1

krel = 106 krel = 10-3

OTs

OTs

krel = 1

krel = 108

Me

Me

Me

OTs

TsO TsO TsO

Bridgehead Carbocations

1 10-7 10-13 104

Bridgehead carbocations are highly disfavored due to a strain increase in achieving planarity. Systems with the greatest strain increase upon passing from ground state to transition state react slowest.

why so reactive?

Cyclopropyl Carbocations

Carbocations in Bridged SystemsCarbocations In Bridged Systems



Carbocation [1,2] Sigmatropic Rearrangements

B

AC

D

B

AC

D

1,2 Sigmatropic shifts are the most commonly encountered cationic rearrangements. When either an alkyl substituent or a hydride is involved, the term Wagner-Meerwein shift is employed to identify this class of rearrangments.

Stereoelectronic requirement for migration....

B

AC

D

bridging T.S.

retention of stereochemistry





Me

MeMe

Me

HMe

Me

H

HMe OH

Me

Me

H

H

Me

Me

Me

HOH

-caryophyllene alcoholE. J. Corey J. Am. Chem. Soc. 1964, 86, 1652.

Demjanov-rearrangement (Driving force: relief of ring strain)

OH

H

Me

Me

Me

Meequiv to

H2SO4



Factors that influence carbocation formation

N ATU R E O F TH E L E AV I N G G R O U P

The leaving group may be a stable, neutral atom or molecule

i) CH 3 T -decay

R Zheterolytic

cleavageR

++ Z

CH 3 He3

( )+1

+ 1e-1

CH 3+ + He

3 0

gas phase

ii)KNO 2

HCl+ Cl

-NH 2C6H 5 N 2+

C6H 5 + Cl-

C6H 5+

N 2+ C6H 5 Cl

iii) ROR'HX

ROR'

H

++ R

+ + R'OHX-

X- + RX

iv) R+

++

NR 3NR 3R




AN I O N I C L E AV I N G G R O U P S R Z R

++ Z

-

The greater the ability of the departing anionic species to

stabilize negative charge, the better a leaving group it is.

Ease of displacement of common anionic leaving groups:

O2N S

O

O

O- S

O

O

O-

Br S

O

O

O-

H 3C> >

nosylate brosylate tosylate

S

O

O

O-

Z I-

Br-

Cl-

F-

> > > > = CH 3COO-

> OH-

= OR-

> NR 2-~ ~

Other particularly good anionic leaving groups:

CF3 S

O

O

O-

triflate

O2N

NO 2

NO 2

O-

picrate

CF3 CO

O-

trifluoroacetate




Structural factors Alkyl substitution - The greater the degree of alkyl substitution at the cationic center,

the greater will be the stability of the carbocation produced.

3 > 2 > 1 > CH 3+oo o

Aryl substitution - The greater the degree of conjugation of the cationic center with

delocalizing groups, the greater will be the stability of the carbo-

cation produced.

(C6H 5)3C+

> (C6H 5)2CH+

> C6H 5CH 2+

Conjugation with heteroatoms:

CH 3C

H 3COR

+

CH 3C

H 3COR+

R = H or alkyl

CH 3C

H 3CNR 2

+

CH 3C

H 3CNR 2 R = H or alkyl+

Steric factors - Carbocations prefer a planar geometry: Any structural feature that

interferes or prevents the attainment of 120

(CH 3)3C+

o

interbond angleswill hinder (retard) carbocation formation.

> CH 3

+>

+

120 105 60oo o

~ ~

(all are 3 o)




Solvent effects: Any property of a solvent

system that can lower the energy of

activation for heterolytic bond cleavage will

favor carbocation formation.



The role of solvent in carbocation formation

Dielectric constant - a rough measure of the

ability of the solvent to separate oppositely

charged ions.

Hydrogen-bonding ability

Acid-base properties

Nucleophilicity: As the nucleophilicity of a

solvent decreases, the likelihood of discrete

carbocation formation increases.



Preparatively Useful SN2 Reactions: Alkylations



synthetically important reactions, shows the various nucleophiles that can be

alkylated according to the SN2 mechanism.

Enolate Synthesis




C-nucleophiles

N-nucleophiles




S-nucleophiles

Hal-nucleophiles



P-nucleophiles


O-nucleophiles

Preparation of methyl

esters from carboxylic

acids and

diazomethane.




With its reactions is possible to invert the configuration of a stereocenter equipped with an OH group.

The enantiomers of optically pure alcohols, which contain a single stereocenter that bears an OH group,

are easily accessible through a Mitsunobu inversion process.

Mitsunobu inversion




Mitsunobu inversion

Mechanism of

the Mitsunobu

inversion

Additions to the Olefinic C=C

Double Bond

Chapter 8

Reactivity of C=C

Electrophiles are attracted to the pi electrons.

Carbocation intermediate forms.

Nucleophile adds to the carbocation.

Net result is addition to the double bond.



Step 1: Pi electrons attack the electrophile.

C C+ E

+C

E

C +

C

E

C + + Nuc:

_

C

E

C

Nuc

=>

• Step 2: Nucleophile attacks the carbocation.

Electrophilic Addition

Chapter 8

Addition of HX (1)

Protonation of double bond yields the most stable carbocation.

Positive charge goes to the carbon that was not protonated.

X

+ Br_

+

+CH3 C

CH3

CH CH3

H

CH3 C

CH3

CH CH3

H

H Br

CH3 C

CH3

CH CH3



WWU -- Chemistry

MECHANISM STEP-BY-STEP ACCOUNT OF WHAT HAPPENS

C C

E +

C C

E +

C C

E

X step 2 step 1

intermediates are

formed during a

reaction but are

not products

:X-

Intermediate



WWU -- Chemistry

product

H

intermediate

TS2

TS1

ENERGY PROFILE two step reaction

E

N

E

R

G

Y

step 1 step 2 C C

E +

X -

+ C C

C C

E

X

E



WWU -- Chemistry

ACTIVATED COMPLEXES

correspond to transition states for each step

ACTIVATED

COMPLEXES

C C

E

X X -

+ C C

E

E +

C C

C C

E

C C

E

X

intermediate

show bonds in the process of breaking or forming

(bonds are half formed or half broken)

+

+

-

+



Chapter 8

Regiospecificity

Markovnikov‘s Rule: The proton of an acid adds to the carbon in the double bond that already has the most H‘s. ―Rich get richer.‖

More general Markovnikov‘s Rule: In an electrophilic addition to an alkene, the electrophile adds in such a way as to form the most stable intermediate.

HCl, HBr, and HI add to alkenes to form Markovnikov products.



WWU -- Chemistry

MARKOVNIKOV RULE

C H 2

+ HCl

C H 3

C l

When adding HX to a double bond the

hydrogen of HX goes to the carbon

which already has the most hydrogens

..... conversely, the anion X adds to the most

highly substituted carbon ( the carbon with

most alkyl groups attached).



WWU -- Chemistry

REGIOSELECTIVE

REACTION

C C H 3

C H 3

C H 2 C C H 3

C H 3

C H 3

C l

+ C H C H 3 C H2

C l

C H 3 HCl

major minor

one of the possible products is formed

in larger amounts than the other one

Compare

REGIOSPECIFIC only one of the possible products is

formed (100%).



WWU -- Chemistry

Mechanism (Markovnikov)

R CH CH2 R CH-CH2

R CH CH2 H

Br

1)+

+

slow

2)_ fast

+

+

H

Br

H

R CH-CH2

+H

Electrophile

Nucleophile

Secondary C+

Major product



WWU -- Chemistry

Mechanism (anti-Markovnikov)

R CH CH2

R CH CH2

H

R CH CH2

H

R CH CH2

Br

H

1)+ slow

+2)

_ fast

+

+

H

Br

+

Minor!

Primary carbocation



WWU -- Chemistry

COMPETING PATHWAYS

Lower energy

intermediate

Higher energy

intermediate

rate-determininng

(slow) step

rate-determining step

faster

slower

1 o

2 o



WWU -- Chemistry

SOME ADDITIONAL EXAMPLES

C H 3

+ HCl

C H 3

C l

C H 2

+ HCl

C H 3

C l

C H C H 2 C H C H 3

C l + HCl

only major product is shown



Chapter 8

Free-Radical Addition of HBr

In the presence of peroxides, HBr adds to an

alkene to form the ―anti-Markovnikov‖ product.

Only HBr has the right bond energy.

HCl bond is too strong.

HI bond tends to break heterolytically to form

ions.



Chapter 8

Free Radical Initiation

Peroxide O-O bond breaks easily to form free

radicals.

+R O H Br R O H + Br

O OR R +R O O Rheat

• Hydrogen is abstracted from HBr.



Chapter 8

Propagation Steps

Bromine adds to the double bond.

+C

Br

C H Br+ C

Br

C

H

Br

C

Br

CC CBr +

• Hydrogen is abstracted from HBr.



Chapter 8

Anti-Markovnikov ??

Tertiary radical is more stable, so that

intermediate forms faster.

CH3 C

CH3

CH CH3 Br+

CH3 C

CH3

CH CH3

Br

CH3 C

CH3

CH CH3

Br

X



β-Eliminations



Concepts of β-Elimination Reactions



The Concepts of α,β- and 1,n-Elimination

Reactions in which two atoms or atom groups X and Y are removed from a

compound are referred to as eliminations

In many eliminations X and Y are removed in such a way that they do not

become constituents of one and the same molecule. In other eliminations they

become attached to one another such that they leave as a molecule

of type X-Y or X=Y.

Concepts of β-Elimination Reactions



The Concepts of α,β- and 1,n-Elimination

Reactions in which two atoms or atom groups X and Y are removed from a

compound are referred to as eliminations

Depending on the distance between the atoms or groups X and Y removed from the

substrate, their elimination has a distinct designation. If X and Y are geminal, their

removal is an a-elimination. If they are vicinal, it is a b-elimination. If X and Y are

separated from each other by n atoms, their removal is called 1,n-elimination, that

is, 1,3-, 1,4-elimination, and so on

1,n-Eliminations (n 1–4) of

two atoms or groups X and Y,

which are bound to sp3-

hybridized C atoms.

78 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:

Br

KOH 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..

..



+ 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.



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

C

R

R

X

B

R

R

H

C CR

R

R

R X

+

-

C C

R R

RR

+ B H + X-

CH3CH2CHCH3

Br CH3CH2O-Na

+

EtOH

CH3CH CHCH3 + CH3CH2CH CH2

2-butene 1-butene

major product( > 80%)

minor product( < 20%)



81

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

C

R

R

X

B

R

R

H

C CR

R

R

R X

+

-

C C

R R

RR

+ B H + X-



82

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



83 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

Br

slow

B:-

CCH3

CH3

C H

H

H

+ rapid

C C

CH3H

CH3H

+ B H + Br-

Br--



84 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 C

CH3H

CH3H

64%36%

SN1

product

E1product



85 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. (DG = DH –TDS) 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



86


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 a-carbon in a 1° alkyl halide. As a result, they favor E2 over

SN2 products.



87 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

moderate

v. gd.

SN2

SN2

E2

SN2

9

weak

fair

SN2

SN2

E2

E2

SN2

SN2

SN1

SN1

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



88 Recall the preparation of long alkynes.

1. A terminal alkyne (pKa = 25) is deprotonated with a very strong base…

R-C º C-H + NaNH2 ® R-C º 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 ¼

R-C º C: - Na+ + CH3CH2-X ® R-C º 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



89 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.



90 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

P

Br

Br

Br

:

etherCH3(CH2)4CH2 O

PBr2

..

..

..

+

H

Br-

CH3(CH2)4CH2Br

1º



91

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

Br

KOH 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



92 Alkyl Halide Substrate Reactivity:

Summary of SN /Elimination Reactions

unhindered substrates favor SN2

do not form a stable C +

do not react by S N1 or E1

hindered substrates. SN2 increasingly unfavorable, E2 is OK

form 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



Polar Reactions under

Basic Conditions

β-Elimination by the E2 and E1cb Mechanisms

C(sp3)–X electrophiles can undergo β -elimination reactions as well as

substitutions. -Elimination reactions proceed by the E2 or E1cb mechanism

under basic conditions.

The concerted E2 mechanism is

more common. The lone pair of

the base moves to form a bond

to a H atom on a C atom

adjacent to the electrophilic C

atom.

Because the H–C and the C–X bonds break simultaneously in the E2

mechanism, there is a stereoelectronic requirement that the orbitals making up

these two bonds be periplanar, (i.e., parallel to each other), in the transition state

of the elimination.

Acyclic compounds can orient the

two bonds in a periplanar fashion

in two ways, synperiplanar (by

eclipsing them at a 0° dihedral

angle) and antiperiplanar (by

staggering them at a 180° angle).

In cyclic compounds, there are much greater restrictions on conformational flexibility. In six-

membered rings, the antiperiplanar requirement for E2 elimination is satisfied when both the

leaving group and the adjacent H atom are axial.

Moreover, two C–H bonds are antiperiplanar to the C–Cl bond in the reactive conformation

of neomenthyl chloride, so two products are obtained upon E2 elimination, whereas only

one C–H bond is antiperiplanar to the C–Cl bond in the reactive conformation of menthyl

chloride, so only one product is obtained.

Not only can 1° and 2° C(sp3)–X undergo -elimination by the E2 mechanism

under basic conditions, but so can 3° C(sp3)–X systems. Alkenyl halides

also undergo β -elimination readily. When there are H atoms on either side of

an alkenyl halide, either an alkyne or an allene may be obtained. Even alkenyl

ethers (enol ethers) can undergo -elimination to give an alkyne.

Draw a mechanism for the following reaction.

Substitution at C(sp2)–X Bonds

Substitution at Carbonyl C

Many carbonyl compounds, including esters, acyl chlorides, and acid anhydrides,

have leaving groups attached to the carbonyl C, and many reactions proceed with

substitution of this leaving group by a nucleophile

Primary amines react with many esters just upon mixing to give amides. The

amines are sufficiently nucleophilic to add to the ester carbonyl.

After the nucleophilic N is

deprotonated, the alkoxy group is

a much better leaving group, so

collapse of the tetrahedral

intermediate occurs with expulsion

of OR to give the amide as the

product.

Transesterification of esters occurs by a mechanism very much like amide synthesis, but

the reaction requires a catalytic amount of base (usually the Na salt of the alcohol). The

nucleophile is the alkoxide. The reaction is driven in the forward direction by the use of a

large excess of the starting alcohol.

The reaction of alcohols with enolizable acyl chlorides or anhydrides can proceed by two

different mechanisms. One is the addition–elimination mechanism that has already been

discussed.

The other is a two-step, elimination–addition mechanism. In the elimination step,

-elimination occurs by an E2 mechanism to give a ketene, a very reactive compound

that is not usually isolable.

In the addition step, the alkoxide adds to the electrophilic carbonyl C of the ketene to

give the enolate of an ester. Acyl chlorides lacking -hydrogens (t-BuCOCl, ArCOCl), of

course, can react only by the addition–elimination mechanism.

Often a nucleophilic catalyst such as DMAP (4-dimethylaminopyridine) is

added to accelerate the acylation of alcohols ROH with acyl chlorides. The

catalyst is a better nucleophile than RO, so it reacts more quickly with the acyl

chloride than RO to give an acylpyridinium ion by addition–elimination. The

acylpyridinium ion, though, is more reactive than an acyl chloride, so RO adds

more quickly to it than to the acyl chloride.

Draw mechanisms for the following two acylation reactions.

Substitution by the Elimination–Addition Mechanism

A third mechanism for substitution at C(sp3)–X bonds under basic conditions,

elimination– addition, is occasionally seen. The stereochemical outcome of the

substitution reaction shown in the figure tells us that a direct SN2 substitution is not

occurring

An elimination–addition mechanism can be proposed. MeO is both a good nucleophile

and a good base. It can induce -elimination of HBr to give a compound that is now -

electrophilic at the C atom that was formerly -electrophilic. Another equivalent of MeO

can now undergo conjugate addition to the electrophilic C atom; the addition occurs from

the bottom face for steric reasons.

Base-Promoted Rearrangements

A rearrangement is a reaction in which C–C bonds have both broken and

formed in the course of a reaction involving just a single substrate. It is often

more difficult to draw a reasonable mechanism for a rearrangement than for

any other kind of reaction. When you draw a mechanism for a rearrangement

reaction, it is especially important to determine exactly which bonds are being

broken and which are formed.

Migration from C to C

In the benzylic acid rearrangement of 1,2-diketones, HO adds to one ketone.

The tetrahedral intermediate can collapse by expelling HO, giving back

starting material, or it can expel Ph. The Ph group leaves because there is an

adjacent electrophilic C to which it can migrate to give the product.

Diazomethane (CH2N2) reacts with ketones to insert a CH2 unit between the carbonyl

and -carbon atoms. The diazo compound acts as a nucleophile toward the electrophilic

carbonyl C, and migration of R to the diazo C then occurs with expulsion of the great

leaving group N2

In the Wolff rearrangement, an -diazoketone is heated to give a ketene. The mechanism of

the Wolff rearrangement consists of one step: the carbonyl susstituent migrates to the diazo

C and expels N2. When the reaction is executed in H2O or an alcohol, the ketene reacts

with solvent to give a carboxylic acid or an ester as the ultimate product.

Migration from C to O or N

Ketones react with peracids (RCO3H) to give esters in the Baeyer–Villiger

oxidation. The peracid is usually m-chloroperbenzoic acid (m-CPBA);

peracetic acid and trifluoroperacetic acid are also commonly used.

Peracids have an O–OH group attached to the carbonyl C.

The terminal O of the peracid then adds to the carbonyl C. The O–O bond is very weak,

so migration of R from the carbonyl C to O occurs, displacing the good leaving group

carboxylate. (Proton transfer occurs first to transform the carboxylate into an even better

leaving group.) The product ester is obtained after final loss of H from the carbonyl O.




Mitsunobu inversion

Mitsunobu inversion

Mechanism of

the Mitsunobu

inversion

Polar Reactions under

Acidic Conditions

In the rearrangement reaction shown next, a 1,2-shift occurs concerted

with loss of a leaving group. Draw a mechanism for this reaction. FeCl3

is a Lewis acid. What by-products might be obtained if the migration

were not concerted with loss of the leaving group?

Draw a reasonable mechanism for the following substitution reaction.

Which O is protonated first?

Elimination of small molecules like water and MeOH is often carried out

under acidic conditions. The reaction is often driven to completion by

azeotropic distillation with benzene or petroleum ether.

Dihydropyran (DHP) reacts with alcohols under acid catalysis to give

tetrahydropyranyl (THP) ethers. The alcohols can be released again by

treating the THP ether with MeOH and catalytic acid. Thus, the THP group

acts as a protecting group for alcohols. Draw mechanisms for the formation

and cleavage of the THP ether.

Electrophilic Aliphatic Substitution

Alkenes undergo electrophilic substitution reactions by the same

addition–fragmentation mechanism as do arenes.

The Sakurai reaction

TiCl4 is a Lewis acid; it coordinates to the carbonyl O, making the

carbonyl and -carbons into cations. The alkene then attacks the -carbon

in such a way as to put the new carbocation on the C adjacent to the C–

Si bond. Fragmentation of the C–Si bond gives a Ti enolate. Aqueous

workup protonates the enolate to give the observed product.

TiCl4 is a Lewis acid; it coordinates to the carbonyl O, making the

carbonyl and -carbons into cations. The alkene then attacks the -carbon

in such a way as to put the new carbocation on the C adjacent to the C–

Si bond. Fragmentation of the C–Si bond gives a Ti enolate. Aqueous

workup protonates the enolate to give the observed product.

Carbon Nucleophiles

The carbocation that is formed upon protonation of a carbonyl compound

canlose H from the -carbon to give an enol. Enols are good nucleophiles.

Thus,under acidic conditions, carbonyl compounds are electrophilic at the

carbonyl C and nucleophilic at the -carbon and on oxygen, just like they are

under basicconditions. Resonance-stabilized carbonyl compounds such as

amides and esters are much less prone to enolize under acidic conditions

than less stable carbonyl compounds such as ketones, aldehydes, and acyl

chlorides; in fact, esters and amides rarely undergo reactions at the -carbon

under acidic conditions.

Polarización del enlace en el haluro de alquilo

En un haluro de alquilo el átomo de halógeno está enlazado a un

átomo de carbono con hibridación sp3. El halógeno es más

electronegativo que el carbono, y el enlace C-X está polarizado con

una carga parcial positiva en el carbono y una carga parcial negativa

en el halógeno

Cuanto más electronegativo es el átomo de halógeno, mayor es la

polarización entre el enlace carbono-halógeno.

Halogenuros de alquilo: sustitución nucleofílica y eliminación

Representaciones con modelos CPK de los haluros de alquilo

La electronegatividad aumenta

hacia la derecha y hacia arriba

en la tabla periódica.

Los halógenos más pesados son

más voluminosos, con áreas

superficiales mucho más

grandes


El tamaño aumenta hacia la derecha y hacia abajo, por lo que el tamaño

de los halógenos incrementa en el orden de F < Cl < Br < I.

Sustitución nucleofílica y reacciones de eliminación.

Los haluros de alquilo se convierten fácilmente en otros grupos funcionales. El átomo de halógeno puede salir con su par de electrones de enlace para formar un ión haluro estable; se dice que un haluro es un buen grupo saliente.

Cuando otro átomo reemplaza al ión haluro, la reacción es una sustitución.

Si el ión haluro abandona la molécula junto con otro átomo o ión (con frecuencia el H+), la reacción es una eliminación

En una reacción de sustitución nucleofílica, el átomo de haluro se sustituye por un nucleófilo.

En una reacción de eliminación, el haluro se "elimina" de la molécula después de la abstracción de un hidrógeno por medio de una base fuerte. Las reacciones de eliminación producen alquenos.


Sustitución nucleofílica y reacciones de eliminación.


Sustitución nucleofílica del yodometano con ión hidróxido.

Al yodometano se le denomina sustrato, es decir, el compuesto que es atacado por

el reactivo. El átomo de carbono del yodometano es electrofílico, ya que va unido a

un átomo de yodo electronegativo. La densidad electrónica se representa alejada

del carbono debido a la inducción ejercida por el átomo de halógeno, lo que hace

que el átomo de carbono tenga una cierta carga positiva parcial. La carga negativa

del ión hidróxido es atraída hacia esta carga positiva parcial.

El ión hidróxido ataca a la molécula de yodometano desplazando al ión yoduro

y formando un enlace de carbono-oxígeno !!!!


El ión hidróxido es un nucleófilo fuerte porque el átomo de oxígeno tiene

pares de electrones no compartidos y una carga negativa .

Mecanismo SN2

El mecanismo siguiente muestra el ataque por el nucleófilo (ión

hidróxido), el estado de transición y el desprendimiento del grupo

saliente (ión yoduro).

El ión hidróxido ataca al carbono de la molécula de yodometano. En el estado de

transición, existe un enlace que comienza a formarse entre el carbono y el

oxígeno, y el enlace carbono-yoduro se rompe. La reacción produce un alcohol.

Éste es un mecanismo concertado en el que todo pasa en un único paso.


Diagrama de energía de reacción

El diagrama de energía de reacción

muestra sólo un máximo de energía,

(estado de transición); no hay

intermedios

La formación y ruptura de enlaces

tiene lugar al mismo tiempo.

Solamente hay un estado de

transición formado por las dos

moléculas (el ión hidróxido y el

yodometano). No hay intermedios en

esta reacción


El mecanismo SN2 es un ejemplo de reacción concertada!!!

Reacciones de intercambio de halógenos

El yoduro es un buen nucleófilo y muchos cloruros de alquilo reaccionan con yoduro de

sodio para obtener yoduros de alquilo. Los fluoruros de alquilo son difíciles de sintetizar

directamente, por lo que con frecuencia se obtienen tratando cloruros o bromuros de alquilo

con KF; utilizando un éter y un disolvente aprótico que aumente la nucleofilia normalmente

débil del ión fluoruro.

El haluro de un haluro de alquilo se pueden intercambiar por un haluro diferente. Esta

reacción de intercambio a menudo se utiliza para sintetizar los fluoruros de alquilo. El ión

fluoruro no es un buen nucleófilo, pero su nucleofilicidad se puede aumentar llevando a

cabo la reacción en disolventes apróticos y utilizando un éter corona para "apartar" al

catión del fluoruro.


Estados de transición para ataques del nucleófilo fuerte y débil.


Basicidad y nucleofilicidad

Podríamos pensar que el metóxido es mucho mejor nucleófilo

porque es mucho más básico. Esto sería un error, ya que la

basicidad y la nucleofilicidad son propiedades diferentes. La

basicidad viene determinada por la constante de equilibrio para

abstraer un protón. La nucleofilicidad se define por la velocidad

de ataque sobre un átomo de carbono electrofílico para dar

sustituciones o adiciones. En ambos casos, el nucleófilo (o base)

forma un nuevo enlace; si el nuevo enlace lo forma un protón, ha

reaccionado como una base, si el nuevo enlace lo forma con el

carbono, ha reaccionado como un nucleófilo.


Basicidad y nucleofilicidad

Predecir de qué forma puede reaccionar una especie

podría ser difícil. La mayoría de los buenos nucleófilos

(pero no todos) también son bases fuertes, y viceversa.

Observando el producto formado podemos decidir si la

base conjugada ha actuado como una base o como un

nucleófilo. Si el nuevo enlace es un protón, ha

reaccionado como una base; si el nuevo enlace lo forma

con el carbono, ha reaccionado como un nucleófilo.


Basicidad y nucleofilia


Nucleófilos comunes

Una base es un nucleófilo más fuerte que su ácido conjugado. En general, cuanto más

electronegativo es el átomo, menos nucleofílico es.


Efecto de la polarizabilidad del nucleófilo en las

reacciones SN2

Comparación de los iones

fluoruro y yoduro como

nucleófilos en las reacciones

SN2. El fluoruro retiene

fuertemente los electrones,

que no pueden formar enlace

C-F hasta que los átomos

están próximos.


El yoduro retiene los electrones externos con menos fuerza, por lo que es más

fácil que se forme un enlace en la reacción !!!

Influencia del disolvente en la nucleofilicidad.

En un disolvente prótico, los aniones pequeños se solvatan

más fuertemente que los grandes, ya que el disolvente se

aproxima más a un ión pequeño y forma enlaces de hidrógeno

más fuertes.

Si el ión es más pequeño, como el fluoruro, se requiere más

energía para poder separar el disolvente de este ión que está

fuertemente solvatado que de otro ión más grande, que está

más débilmente solvatado, como el yoduro.

Los disolventes próticos polares rodearán al nucleófilo y

reducirán su nucleofilia. Cuanto más pequeño sea el átomo,

más solvatado estará y mayor será la reducción de la

reactividad.


Influencia del disolvente en la nucleofilicidad


Grupos salientes comunes

Los sulfonatos, los sulfatos y los fosfatos son buenos grupos salientes porque pueden

deslocalizar la carga negativa sobre los átomos de oxígeno. Las moléculas neutrales son

grupos salientes cuando la reacción se lleva a cabo en medios ácidos.


Influencia estérica del sustrato sobre las reacciones SN2

El ataque SN2 en un haluro de alquilo primario sencillo no está impedido; el ataque en un

haluro de alquilo secundario está impedido, y el ataque en un haluro de alquilo terciario es

imposible.


Para formar un enlace, el nucleófilo tiene que encontrarse dentro de la distancia enlazante

del carbono. Un haluro de alquilo estéricamente impedido evitará que el nucleófilo se

acerque lo suficiente como para reaccionar.

Vista molecular de un ataque SN2

La reacción SN2 tiene lugar a través del ataque del nucleófilo sobre el lóbulo posterior del

OM antienlazante del enlace C-Br. El ataque posterior o dorsal invierte el tetraedro del

átomo de carbono, de forma similar a como el viento invierte un paraguas.

El nucleófilo ataca al carbono desde la parte posterior, opuesto al grupo saliente. Las

reacciones SN2 siempre tendrán como resultado una inversión de la configuración del carbono

que está siendo atacado. Los estereocentros que no están involucrados en la reacción no se

invertirán


Mecanismo de inversión

El estado de transición de la reacción SN2 tiene una geometría

bipiramidal trigonal con el nucleófilo y el grupo saliente a 180º de

cada uno.


Inversión de la configuración

En algunos casos, la inversión de la configuración es muy clara; por ejemplo,

cuando el cis-1-bromo-3-metilciclopentano experimenta un desplazamiento SN2

por el ión hidróxido, la inversión de la configuración da lugar al trans-3-

metilciclopentanol

La inversión de la configuración no sólo puede cambiar la configuración

absoluta, en el caso de los compuestos cíclicos también cambiará la geometría

del cis al trans o viceversa. Las reacciones SN2 son estereoespecíficas


Mecanismo de una reacción SN1

El primer paso del mecanismo es la formación del carbocatión. Es paso es lento y es el paso

limitante de la velocidad. El segundo paso es el ataque nucleofílico del nucleófilo en el

carbocatión. Si el nucleófilo fuera una molécula de agua o alcohol, se tendría que perder un

protón con objeto de obtener un producto neutro

El término unimolecular quiere decir que sólo una molécula

está implicada en el estado de transición del paso limitante

de la velocidad de reacción


Pasos clave del mecanismo SN1

El mecanismo SN1 es un proceso de múltiples pasos. El primer paso es una

ionización lenta para formar un carbocatión. El segundo paso es un ataque rápido

de un nucleófilo al carbocatión. El carbocatión es un electrófilo fuerte y reacciona

rápidamente tanto con un nucleófilo fuerte como con uno débil. En el caso de

ataque de una molécula de alcohol o de agua (como nucleófilos), la pérdida de un

protón da lugar al producto neutro final.

El mecanismo principal está formado por dos pasos: formación de un carbocatión

y ataque nucleofílico. Dependiendo del nucleófilo se puede necesitar un tercer

paso. Si actúa una molécula de alcohol o agua como un nucleófilo, se tendrá que

perder un protón con objeto de obtener un producto neutro.


Diagramas de energía de las reacciones SN1 y SN2

La reacción SN1 tiene un mecanismo que consta de dos pasos, con dos estados de

transición (‡1 y ‡2) y un carbocatión intermedio. La reacción SN2 sólo tiene un

estado de transición y no tiene intermedios.

La reacción SN1 tiene un carbocatión intermedio y proporcionará una mezcla de

enantiómeros. La reacción SN2 tendrá como resultado una inversión de la

configuración.


Efecto inductivo e hiperconjugación El paso limitante de la velocidad de la reacción SN1 es la ionización para formar un

carbocatión, proceso fuertemente endotérmico. El estado de transición para este proceso

endotérmico se asemeja al carbocatión (postulado de Hammond), por lo que las

velocidades de las reacciones SN1 tienen una dependencia de la estabilidad del

carbocatión.


Los cationes alquilo del carbocatión estan estabilizados por la donación de

electrones a través de los enlaces sigma (efecto inductivo), además mediante

el solapamiento de los orbitales llenos (hiperconjugación); por tanto, los

carbocationes altamente sustituidos son más estables.

Estado de transición de la reacción SN1.

En el estado de transición de la ionización SN1, el grupo saliente

forma parte de la carga negativa. El enlace C-X se rompe y un grupo

saliente polarizable todavía puede mantener un solapamiento

sustancial

El estado de transición se asemeja al carbocatión


Disolución de los iones

La reacción SN1 está favorecida en disolventes polares, que estabilizan los iones intermedios.

El paso limitante de la velocidad forma dos iones y la ionización tiene lugar en el estado de

transición.


Los disolventes polares solvatan estos iones debido a la interacción de los

dipolos del disolvente con la carga del ión. Los disolventes próticos como los

alcoholes y el agua son incluso disolventes más efectivos, ya que los aniones

forman enlaces de hidrógeno con el átomo de hidrógeno del grupo -OH y los

cationes, complejos con los electrones no enlazantes del átomo de oxígeno del

grupo -OH.

Racemización en la reacción SN1

Un átomo de carbono asimétrico experimenta racemización cuando

se ioniza y se transforma en un carbocatión aquiral, plano. Un

nucleófilo puede atacar al carbocatión desde cualquier cara, dando

lugar, como producto, a cualquier enantiómero.


Racemización en la reacción SN1

Los carbocationes tienen orbitales híbridos sp2 y un

orbital vacío p. El nucleófilo puede aproximarse al plano

del carbocatión desde arriba o desde abajo. Si el

nucleófilo ataca por el mismo lado del grupo saliente,

habrá una retención de configuración, pero si el nucleófilo

ataca por el lado opuesto del grupo saliente la

configuración se invertirá. Las reacciones SN1 forman

mezclas de enantiómeros (racemización).


Reacción SN1 sobre un anillo

En la reacción SN1 del cis-1-bromo-3-deuteriociclopentano con metanol, el carbocatión puede ser atacado por cualquier cara. Como el grupo saliente (bromuro) bloquea parcialmente la cara frontal cuando se desprende, el ataque posterior (inversión de configuración) está ligeramente favorecido.

El átomo de deuterio se utiliza para distinguir las caras del ciclopentano.


Reacción SN1 sobre un anillo


En la reacción SN1 del cis-1-bromo-3-deuteriociclopentano con metanol, el carbocatión

puede ser atacado por cualquier cara.

Como el grupo saliente (bromuro) bloquea parcialmente la cara frontal

cuando se desprende, el ataque posterior (inversión de configuración) está

ligeramente favorecido!!!

Transposición de hidruro en una reacción SN1

El producto reordenado, 2-etoxi-2-metilbutano, es el resultado de una transposición

de hidruro: movimiento de un átomo de hidrógeno con su par de electrones de

enlace.


Una transposición de hidruro se representa por el símbolo ~H. En este

caso, la transposición de hidruro convierte el carbocatión secundario

inicialmente formado en un carbocatión terciario más estable. El ataque

del disolvente a este último da lugar al producto de reordenamiento.

Transposición del metilo en una reacción SN1

Cuando se calienta el bromuro de neopentilo con etanol, la reacción sólo da un producto de

sustitución reordenado. Este producto es debido a la transposición del metilo (representada

por el símbolo ~CH3), la migración de un grupo metilo junto con su par de electrones.


La formación de un carbocatión primario no es posible, por lo que la transposición

del metilo y la formación del carbocatión se produce en un único paso para formar

un carbocatión terciario. El ataque por el nucleófilo sobre el carbocatión proporciona

el único producto obtenido de esta reacción.

Reacciones de eliminación: E1 y E2.

Una eliminación implica la pérdida de dos átomos o grupos del

sustrato, generalmente con la formación de un enlace pi.

Dependiendo de los reactivos y de las condiciones en las que se

encuentren, una eliminación debería ser un proceso de primer

orden (E1) o de segundo orden (E2). Los siguientes ejemplos

ilustran los tipos de eliminación que se tratarán en este capítulo.

De la misma forma que existe un SN1 y un SN2, existen dos

mecanismos de eliminación, E1 y E2. Qué eliminación se

producirá es una cuestión de las condiciones empleadas durante

la reacción.


Reacciones de eliminación: E1 y E2


Mecanismo E1

En un segundo paso rápido, una base abstrae un protón del átomo de carbono

adyacente al C+. Los electrones que antes formaban el enlace carbono-hidrógeno

ahora forman el enlace pi entre dos átomos de carbono.


Competencia entre las reacciones SN1 y E1

El primer producto (2-metilpropeno) es el resultado de la

deshidrohalogenación, eliminación de hidrógeno y un

átomo de halógeno. Bajo estas condiciones de

primer orden (ausencia de

base fuerte), se produce la

deshidrohalogenación por un

mecanismo E1.

El producto de sustitución es el resultado del ataque nucleofílico al carbocatión. El

etanol sirve como base para la eliminación y como nucleófilo en la sustitución.


Competencia entre las reacciones SN1 y E1

El primer paso de los dos mecanismos es el mismo:

formación del intermedio carbocatiónico. El alcohol

puede actuar como una base o como un nucleófilo. Si

actúa como una base, la abstracción de un protón

producirá un alqueno. Si el alcohol actúa como un

nucleófilo, se obtendrá el producto de sustitución.


Modelo orbital para la eliminación E1


En el segundo paso del mecanismo E1, el átomo de carbono adyacente debe

rehibridarse a sp2 cuando la base ataca al protón y los electrones fluyen hacia

el nuevo enlace pi.

Las bases débiles se pueden utilizar en las reacciones E1, puesto que no se

encuentran implicadas en el paso limitante de la velocidad de la reacción.

Diagrama de energía de reacción para las reacciones E1 Diagrama de energía de reacción. El primer paso es una ionización

limitante de la velocidad de reacción. Compare este perfil de energía con

el de las reacciones SN1.


El mecanismo E1 conlleva dos pasos y un intermedio. La formación del carbocatión

intermedio tiene la energía de activación más elevada, por lo que será el paso

limitante de la velocidad de la reacción.

Reordenamiento en el mecanismo E1.

Como en otras reacciones mediadas por un carbocatión intermedio, en la E1 se

pueden producir reordenamientos. Compare la siguiente reacción E1 (con

reordenamiento) con la reacción SN1 del mismo sustrato

Cuando la reacción conlleva intermedios carbocatiónicos, habrá una posibilidad

de reordenamiento, por lo que se obtendrá la mezcla de productos


Las reacciones E2

La eliminación también puede ser bimolecular en presencia de una base fuerte. A manera de

ejemplo, considérese la reacción del bromuro de terc-butilo con ión metóxido en metanol

E2 muestra la misma preferencia de sustrato como E1, 3o > 2o > 1o . E2 necesita una base

fuerte.


Mecanismo E2

La velocidad de esta eliminación es proporcional a las concentraciones tanto del

haluro de alquilo como de la base, dando lugar a una ecuación de velocidad de

segundo orden. Esto es un proceso bimolecular; ya que participan el haluro de

alquilo y la base en el estado de transición, por lo que este mecanismo se expresa

como E2, forma abreviada de eliminación bimolecular.

La reacción tiene lugar en un único paso, por lo que es una reacción

concertada !!!


Mezcla de productos en las reacciones E2 Las reacciones E2 requieren la abstracción de un protón de un átomo de carbono próximo

al carbono que lleva el halógeno. Si hay dos o más posibilidades, se obtienen mezclas de

productos. Los ejemplos siguientes muestran cómo la abstracción de protones diferentes

puede dar lugar a productos diferentes.

En general, el alqueno más sustituido será el producto principal de la reacción !!!


Estado de transición para las reacciones E2 Halogenuros de alquilo: sustitución nucleofílica y eliminación

Estados de transición concentrados de la reacción E2. Los orbitales del átomo de hidrógeno

y el haluro deben estar alineados para que puedan comenzar a formar un enlace pi en el

estado de transición.

El hidrógeno que se va a abstraer debería ser anti-coplanar al grupo saliente para

minimizar cualquier impedimento estérico entre la base y el grupo saliente. Una

orientación sin-coplanar entre el hidrógeno y el grupo saliente creará una interacción

repulsiva y aumentará la energía del estado de transición !!!







Mitsunobu inversion




Mitsunobu inversion

Mechanism of

the Mitsunobu

inversion

Subst.Addit.Elim._2015_

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