7_substitution_post

Sec. 8 - SN2/SN1 Forsey

1

Ionic ReactionsSubstitution Reactions and Transition State Theory

A great number of substitution reactions are known, and all involve a competition between a pair of Lewis bases for a Lewis acid.

In substitution reactions both the nucleophile (Greek for nucleus loving) and the leaving group are Lewis bases. Note that all substitution reactions are at equilibrium and are reversible. Although for a given reaction it may not be reversible in a practical sense. For example in the reaction below HO¯ is a much stronger nucleophile/base than I¯, which causes the equilibrium to be shifted far to the right.


2

Nucleophilic Substitution Reactions

Nucleophile Species with an unshared electron pairLoves or seeks a positive center.

Leaving Group A good leaving group must be able to leave as a relatively stable weakly basic molecule or ion; stable meaning non reactive.

Nu + R – X R – Nu + X

Nucleophile Alkyl halide(substrate)

Product Leaving GroupIn this example a halide

heterolysis or carbon halogen bond breaks


3

SN2 Reactions

CH3Cl + ¯OH CH3OH + Cl¯

Double [methyl chloride], keep [OH¯ ] constant rate of reaction doubles

Double [OH¯ ], keep [methyl chloride] constant rate of reaction doubles

Double both reactants and the rate increases four times.

Rate [CH3Cl][OH¯ ] or rate = k[CH3Cl][OH¯ ] overall second order.

Conclusion The reaction is bimolecular – two species are involved in the rate determining step

SN2

substitution nucleophilic bimolecular

Nu C

H

L

HH

-C

H

Nu

HH

L

Inversion


4

How are the electrons being transferred from one atom or molecule to the other?

FilledHOMO

Empty orbital*

LUMO

Basedonates

electrons

Acidaccepts electrons

Nu

Nu C L

C L

Filled C-L bonding orbital

Net destabilizing interaction between two filled orbitals or weaker interaction

Nu

Electrons are transferred from the nonbonding electrons in a filled molecular orbital of the nucleophile to the empty antibonding molecular orbital of the Lewis acid.

Does the nucleophile donate its electrons to the C – L bond?

LC

Transition Statehigh energy the carbon is surrounded by three

coplanar groups with 10e¯ around it

What is the hybridization of the transition sate. It trigonal planar.


5

Overview of SN2 Mechanism


6

Reaction Coordinate

Transition SateF

ree

Ene

rgy

Products

The transition state involves both the nucleophile and the Lewis acid, which accounts for the bimolecular reaction

Nu C L

Nu LC

Nu C L

G‡

G°

Reactants

G‡ = Free energy of activation

G° = Free energy change

A transition state is the high energy state of the reaction or the rate determining step.It is an unstable entity with a very brief existence (10-12 s)

In the transition state bonds are partially formed and broken. This called a concerted reaction.

*


7

Transition State Theory: Free Energy Diagrams Why does a reaction occur?

G° = H° - TS°

Remember from 1st year

For a spontaneous energy to occur the Gibbs free energy (G°) must be negative. Thus you can have different combinations of H° and S° values for a reaction to occur.

A process that gives off heat (H° is negative, exothermic) and becomes more disordered (S° is positive) a spontaneous reaction will occur and G° will be negative for all temperatures

A process that has an increase in enthalpy (H° is positive, endothermic) and a decrease increase in order (S° is negative) a reaction will be nonspontaneous at all temperatures.

The questionable cases are those in which the entropy and enthalpy factors work in opposition. In general the H factor dominates al low temperature and the TS term at high temperatures.

total orfree energy

change

change inbonding energy

change in disorder


8

Lets look at the reaction:

CH3–Cl + OH¯ CH3–OH + Cl ¯

G° = H° - TS°

G° = -100 kJ mol-1

H° = - 100 kJ mol-1

T = 333 K S° = 0.075 kJ mol-1 K-1

at 60°C

Therefore a spontaneous reaction will occurThe reaction is said to be exergonic, G° is negative

So what does this mean to the equilibrium

This is a very large Keq which means the reaction will go to completionRemember the equilibrium always favors the species lower in energy or less reactive


9

But don’t forget there is an energy barrier to overcome before the reaction will occur

Energy Barrier that the reactants must surmount

in order to reach the lower energy level

Exergonic: The free energy level of the products is lower than that of the reactants


10

Endergonic Reaction Positive free-energy change

Energy Barrier (G‡) that the reactants must

surmount is much larger for a endergonic reaction

than for an exergonic reaction


11

Relationship between Temperature and Rate of Reaction

For many reactions taking place at room temperature, a 10°C increase in temperature will the cause the reaction rate to double.

The rate increases because more molecules have sufficient kinetic energy to surmount the energy barrier (G‡).

T1 < T2

At T2 more molecules have sufficient energy to surmount the energy barrier or the free energy of activation (G‡).


12

C L-

+

Nu

Nu

Nu

So how does this all Relate to an SN2 Reaction

An infinite number of possible pathways may lead from reactants to products. The most probable pathway requires the least amount of energy and the most likely transition state lies at the top of the pathway that requires the least amount of energy. A SN2 reaction has a narrow pass through the energy valley which means that the orientation of the reactants has to be very specific before a reaction will occur.

Nu


13

CNu L-

+the only way in

The only way for an SN2 reaction to occur is through a back side attack

CNu L CNu L

+

+-

Transition stateHigh energy

10 electrons around carbonInversion

What happens at a stereogenic carbon?


14

HH

H3C BrCH3CH2S Na

acetone

BrH3C

cis-1-bromo-3-cyclopentane

Stereochemistry of SN2 Reactions

H

H3C

H

Transition State

Na Br+

electrons from bond

lone pair of electrons from

sulfur makes new bond

+

H

H3CSCH2CH3

flip -1-(ethylthio)-3-methylcyclopentane

What is the stereochemistry at the stereogenic centers?

solvent CH3C

O

CH3


15

SN2 Reactions

NaOH + CH3 Br

1. Put charges and partial charges on atoms and molecules2. Identify the electrophile (acid) and nucleophile (base)3. Negative goes to positive. Show the flow of electrons with arrows

Arrows are your thought process

NH3 + CH3 Br CH3NH3 + Br


16

Br

SN2 Reactions

+ CH3 OHNa

CH3OHH2SO4 Na Br


17

C

C6H13

BrH3CH

SN2 Reactions

2-bromooctane-(–)-

[] = - 34.25°

Start with enantiomeric purity = 100%

NaOH

transition state

inversion of configuration

[] = - 9.90°

-(–)-2-octanol

enantiomeric purity = 100%

C

C C

C


18

SN1 Reactions

(CH3)3C-Cl + OH¯ (CH3)3C-OH + Cl¯ H2O

acetone

Double [tert-butyl chloride], keep [OH¯ ] constant rate of reaction doubles

Double [OH], keep [tert-butyl chloride] constant rate stays the same

Rate [(CH3)3C-Cl] or rate = k[(CH3)3C-Cl] overall first order.

Conclusion the reaction is unimolecular

SN1

substitution nucleophilic unimolecular


19

SN1 Reactions

An SN1 occurs through a series or steps and forms intermediate species. The rate of the reaction is determined by the slowest step.

The slow step is called the rate-limiting step or the rate determining step.


20

SN1 Mechanism

(CH3)3CCl + 2 H2O (CH3)3OH + H3O+ + Cl¯

C

CH3

Cl

CH3

H3C

Overall reaction

- +

C

CH3

Cl

CH3

H3C

transition state

CCH3

CH3

H3C

What is the hybridizationor the carbon?

+ Cl

Step 1

1st intermediate


21

+C

CH3

CH3

H3C + Cl

Step 2

1st intermediate

SN1 Mechanism

electrophile(acid)

nucleophile(base)

OH

H C

CH3H3C

H3C

OH2

+

C

CH3H3C

H3C

OH

C

CH3H3C

H3C

OH H+

OH

H

Step 3

acid

base

C

H3CCH3H3C

OH H

+

2nd intermediateTransition State

C

H3C

H3C CH3

OH H O

H

H

+

+ H3O+ + Cl¯


22

Stability of Carbocations

Alkyl groups are electron donating. Hyperconjugation donates electron density via a partial overlap the adjacent sigma bonds. Any time a charge is dispersed or delocalized the system will be stabilized. The more interactions of this kind that are possible, the more stable the ion. Thus tertiary cations will be more stable than secondary and secondary cations are more stable than primary cations

H

C

C

C

HH C

H

H

HH

HH

sp3

sp3

sp2sp3

What is the hybridization of the positively charged

carbon?

1s2

2s2

2p2

po

ten

tial

en

erg

y

1s2

2sp2

2p

C+

empty

C+


23


No hyperconjugationWill not form in a solution


24

CH3+

(CH3)3C +


Remember: Hyperconjugation delocalizes the positive charge to adjacent bonds but the charge does not resonate onto other atoms thus a carbocation that is resonance stabilized would be more stable than a tertiary carbocation.

+ + more stable than H3C C

CH3

CH3

+

++ ++


25

SN1 MechanismWhat happens at a stereogenic carbon?

C

H3CH2C

H3C CH2CH2CH3

Br

H2O

acetoneC

+ O

H

H

A

B

C

A

-H+

C

B

-H+

C

(R)-3-bromo-3-methylhexane

C

C

use instead of last stepoptically active

flip ( )-3-methyl-3-hexanol

( )-3-methyl-3-hexanol


26

Solvolysis Reactions

The solvent is the nucleophile Solvent + Lysis: cleavage by the solvent

(CH3)3C-Br + H2O (CH3)3C-OH + HBr

(CH3)3C-Br + CH3CH2OH

H2O hydrolysis

CH3CH2OH ethanolysis(CH3)3C-OCH2CH3 + HBr

(CH3)3C-Cl + HCO2H + HCl(CH3)3C-OCH

O

What is the nucleophile?


27

Factors Affecting Rates of Substitution Reactions

Relative Rate If a given alkyl halide and nucleophile reacts rapidly by a SN2 and slowly by a SN1 mechanism then most of the molecules

will follow a SN2 pathway or visa versa. So how can you tell if a reaction will follow a SN2 or a SN1 reaction?

Factors: 1) Structure of Substrate2) The concentration and reactivity of nucleophile (SN2 only)3) Effect of Solvent4) Nature of Leaving Group

The effect of the substrate structure has on the rate reaction for SN2 reactions. The reactions where conducted in a environment that promotes an SN2 reaction.

(we will talk about this more later)

Substituent Compound Relative Rate

Methyl CH3-X 30.0

1 CH3CH2-X 1.0

2 (CH3)2CH-X 0.025

Neopentyl (CH3)3CH2-X 0.0001

3 (CH3)3C-X ≈ 0

Why do wehave these

results?


28

Effect of Substrate on the Rates of SN2 Reactions

Steric Effects: A SN2 reaction requires the nucleophile to attack from the back of the carbon bearing the leaving group to form and break the bonds. If there are bulky groups that interfere or sterically hinder the attack by the nucleophile the rate of reaction will decrease.

G‡ =

Fre

e E

ne

rgy

Energy barrier too largefor a reaction to occur


29

G‡ =

Fre

e E

ne

rgy

Energy barrier too largefor a reaction to occur

Effect of Substrate on the Rates of SN2 Reactions


30

What about benzylic or allylic alkylhalides (not in text book)

Hyperconjugation – adjacent bond helps stabilize the transition state by sharing some the negative charge

Substituent Compound Relative Rate

Methyl CH3-X 30.0

1 CH3CH2-X 1.0

2 (CH3)2CH-X 0.025

Allylic CH2 =CHCH2-X 40.0

Benzylic C6H5CH2-X 120.0

CC

H

H

C HH

Nu

L


31

Relative rates of reaction for SN2 reactions

Effect of Structure on the Rates of SN1 Reactions

The primary factor that determines the reactivity of organic substrates in a SN1 reaction is the relative stability of the carbocation that is

formed (stability of the intermediate).

G‡ =+

H3C C

CH3

CH3+

+

H3C C

CH3

H

+

+

+

H3C C

H

H

H C

H

H+

+

+

+

Fre

e E

ne

rgy


32

Why does the leaving group leave in a SN1 reaction?

G‡ G°

Fre

e E

ne

rgy

Reaction coordinate

Cl +

C

CH3H3C

CH3 The formation of the ions in the 1st step in a SN1 reaction is both endergonic and endothermic and the transition state for this process should show a strong resemblance to the product of that step. Thus when you think about what stabilizes the intermediate the same processes (delocalization of carbocation) will help stabilize the transition state and will help the reaction proceed.

bond breakingthe carbocation must be

somewhat stable or the bondwill not break.


33

The Hammond-Leffler Postulate

SN2 SN11st step

The transition state for an exergonic reaction looks very much like starting the materials. If a bond is forming in the reaction, that bond is less than half formed and if a bond is breaking, it is less than half broken in the transitions state

The transition state for an endergonic reaction looks very much like the products. Any bonds that are forming are more than half formed, and any bonds that are breaking are more than half broken

Generally the transition state looks most like the species it is closest to in energy


34

So what does this mean?

In a endergonic process, like a SN1 reaction, the free energy of the transition state is closely related to the intermediate that is formed. Thus there is parallelism between the formation of the intermediate and the transition state energy and it can be assumed that the more stable the intermediate, the faster it will form. This associates a thermodynamic property to a kinetic rate and for most reactions this statement is true. Thus a tertiary carbocation will form at a faster rate than a secondary.

For a exergonic process the transition state is closely related to the reactants. If the carbon center is statically unhindered and the attacking species is a strong nucleophile the reaction will occur readily.

C

H H

H

BrNu

C

H3C H3C

H3C

BrNu

G‡ =

The steric interaction between the nucleophile and the Lewis acid

generates more strain in the transition state – increases G‡

C

H H

H

Br

C

H3C CH3

H3C

Br

A methyl carbocation is very unstable and

destabilizes the transition state – increases G‡

+

+

+

+

Fre

e E

ne

rgy

Reaction coordinate

Reaction coordinate

Fre

e E

ner

gy


35

Relatively stable

carbocationHighly unstable carbocation

Efficiency of SN1

No SN1 No SN1

3 2 1 methyl

No SN2 Efficiency of SN2

Steric interaction

Little steric interaction

C

CH3

H3C

CH3

L C

H

H3C

CH3

L C

H

H

CH3

L C

H

H

H

L

Relative Reactivity

Will t-butylbromide ((CH3)3CBr) undergo a SN2 reaction?Will n-butylbromide undergo a SN1 reaction?


36

Rearrangements of Carbocations

C

CH3

H3C

CH3

C

H

CH3methanide

migrationCH3C

CH3

C

H

CH3

CH3++

C

C

C

H3CH3C H

H

HHC

H

H

H

Lewisbase

Lewisacid

Why does this happen?

C

CH3

H3C

CH3

CH2

methanide

migration

+

CH3C

CH3

CH2

CH3+

CH3CH2 CH

H

CH2hydride

migration

+

CH3CH2 CH CH2

H+

3°

3°

2°

2°

1°

1°

C

CH3

CH2C

H2C H

C

HH HH

H

+

+


37

C C

CH3

H3C

CH3

Br

CH3

H

CH3OH

What is the product of the following SN1 reaction?

weak basepoor nucleophile


38

What is the mechanism for the following reaction?OH

OH

conc. H2SO4O


39

SN1resonance stabilized

SN1empty p orbital is at

90° w.r.t. bond and will not resonate

The cation will not form under most conditions

SN2hyperconjugation

How about vinyl and aryl compounds?

allylicbenzylic

vinyl

aryl

Br Br

Br

Br

sp3

sp3

sp2

sp2

+

allylic and benzylic very reactive in both SN1 and SN2

sp2 short strong bondUnder most conditions a

nucleophile will not displace the leaving group in a SN2 reaction

+

+ Br

+ Br

Br

Nu

Br

Br+

vinyl and aryl compounds are very nonreactive in both SN1 and SN2


40

Other Factors

What effect does the concentration and strength of the nucleophile have on the rate of reaction?

For an SN2 reaction increasing the concentration of the nucleophile increases the rate of reaction (2nd order).

For an SN1 reaction increasing the concentration of the nucleophile while keeping the temperature and ionic strength constant, the rate of the reaction will not increase (1st order)

NucleophilicityWhat makes a good nucleophile?

+

CH3O + CH3I CH3OCH3 + I

CH3OH + CH3I CH3OCH3 + I

H

very fast

very slow

Why?


41

Trend A negatively charged nucleophile is always a stronger nucleophile than its conjugate acid.

¯ NH2 > NH3 , ¯ SH > SH2 , ¯ OH > H2O , ¯ OR > HOR

Trend In a group of nucleophiles in which the nucleophilic atom is the same, nucleophilicity parallels basicities.

RO¯ > HO¯ >> RCOO¯ > ROH > H2O

Trend Nucleophilicity decreases from left to right in the periodic table, following the increasing in electronegativity from left to right. The more electronegative elements have more tightly held nonbinding electrons that are less reactive to forming new bonds.

HO¯ > F ¯ NH3 > H2O R3P > R3S

Trends in Nucleophilicity


42

Trend Nucleophilicity increases down the periodic tab following the increase in size and polarizability (in solvents that can hydrogen bond).

I ¯ > Br¯ > Cl ¯ > F ¯ HSe¯ > HS ¯ > HO ¯ R3P > R3S

Steric Effects (Hindered nucleophile) The presence of bulky groups on the nucleophile will slow down the rate of reaction of a SN2 reaction.

O

H3CH3C

H3C

t-butoxide is a stronger base

than ethoxide but because it is hinderedt-butoxide is a weaker

nucleophile

CH3CH2O

t-butoxide

ethoxide

Trends in Nucleophilicity

When bulky groups interfere with a reaction because of their size we call this steric hindrance. Steric hindrance has little effect on basicity because basicity usually involves attack on an unhindered proton. Thus by choosing a less hindered or hindered base you can make the attacking species act like a nucleophile and perform a SN2 reaction or react as a Bronsted-Lowry base and abstract a proton. We will talk more about this later

Strong basegood nucleophile

Strong basepoor nucleophile


43

All strong bases are good nucleophiles but not all weak bases are poor nucleophiles. Why is iodide such a strong nucleophile compared to fluoride

even though it is a very weak base?

Polarizability

+

C

H H

H

Br

Iodine’s electrons are far away from the nucleus, are loosely held and are polarizable. Its electrons can move freely towards a positive charge and can engage in partial bonding as it attacks the electrophilic carbon atom. Remember from Hammond-Leffler Postulate that for a exergonic reaction the transition state resembles the reactants and bond formation and breaking is less than half formed.

Nucleophilicity increases down the periodic tab following the increase in size and polarizability (in solvents that can hydrogen bond). Why?

C

H H

H

Br

C

H H

H

Br

C

H H

H

Br

“hard” non polarizable valence

shell

“soft” polarizable

valence shell

Transition State

little bondingF

Imore bonding

+

G‡ =

Reaction coordinate

Fre

e E

ner

gy

Reaction coordinate

Fre

e E

ner

gy


44

Excellent Nucleophile

Relative Rate

Good Nucleophile Relative Rate

Fair Nucleophile Relative Rate

NC¯ Cyanide 126,000 HO¯ Hydroxide 16,000 Cl¯ Chloride 1,000

HS¯ Thiolate 126,000 Br¯ Bromide 10,000 CH3COO¯ Acetate 630

I¯ Iodide 80,000 N3¯ Azide 8,000 F¯ Fluoride 80

NH3 Ammonia 8,000 CH3OH Methanol 1

NO2 Nitrite 5,000 H2O Water 1

Remember bonding occurs between the highest occupied molecular orbital of the nucleophile and the lowest unoccupied molecular orbital of the Lewis acid. The table below shows some nucleophiles segregated into categories. It is not possible to do much better than this as nucleophilicity is a property that depends on the reaction partner. A good nucleophile with respect to carbon may or may not be a good nucleophile with respect to displacement on some other atom. Energy matching is important in the reaction. If we change the reaction partner and thus the energy of the orbitals involved, we will change the stabilization involved as well


45

Solvent Effects

Protic Solvents solvents whose molecules have hydrogen atoms bonded to electronegative atoms such as R-OH, R-NH2, H2O, RCOOH. These solvents can

extensively hydrogen bond with the solute, or reagents

Aprotic Solvents solvents whose molecules do not have hydrogen atoms bonded to electronegative atoms such as chloroform, hexane, acetone, and benzene. These solvents can not extensively hydrogen bond with the solute, or reagents

Solvent Effects on SN2 Reactions: Polar Protic and Aprotic Solvents

Small “hard” ions form stronger hydrogen bonds than larger polarizable “soft” ions. Thus for an ion like F¯ is hindered and must shed or displace the solvent as the ion is attacking the electrophilic carbon

C

H H

H

BrF

O

H

HOH

H

O

H

H

O

H

H

O H

H

OH

H

OHH

OH

H

O

H

H

O

H

HO H

H O

H

H

O

H

H

O

H

H

OH

H

OH

H

OH

H

O

H

H

O

H

H

OH

H

OH

H

OH

H

O

H

HOH

H

O

H

H

OH

H

O

H

H

Polar Protic

strong Hbonding


46

Polar Protic Solvents

Polar solvents have a hydrogen atom attached to strongly electronegative atoms

They solvate nucleophiles and make them less reactive

Larger nucleophilic atoms are less solvated and therefore more reactive in polar protic solvents

Larger nucleophiles are also more polarizable and can donate more electron density

Relative nucleophilicity in polar solvents:


47

Polar Aprotic Solvents

Polar aprotic solvents do not have a hydrogen attached to an electronegative atom

They solvate cations well but leave anions unsolvated because positive centers in the solvent are sterically hindered

Polar aprotic solvents lead to generation of “naked” and very reactive nucleophiles.

Trends for nucleophilicity are the same as for basicity

They are excellent solvents for SN2 reactions


48

Polar Aprotic Solvents

Polar aprotic solvents not only make the anion less stable or more reactive, polar aprotic solvents help stabilize the slightly polar transition state of an SN2 reaction.

Would the following SN2 reaction react faster in DMF or ethanol?

CH3CH2CH2Br + NaC≡N CH3CH2CH2C≡N + NaBr Aprotic solvent DMF

O

CCH3H3C +

-

Remember: Like Dissolves LikeAcetone


For most SN1 reactions ions are formed in the first step. A polar solvent that stabilizes the ions generated by ionization will favor the ionization process. Also if the transition state is stabilized by the solvent the reaction is more likely to occur because the activation energy is decreased. Thus polar protic solvents will greatly increase the rate of ionization or the rate of reaction

O

+

-

+

Diethylether


49


C CH3H3C

H3C +

+

Cl

Cl

H3C

H3CCH

3

C

Cl

CH3H3C

H3C

+

G‡1

+

Fre

e E

ne

rgy

Reaction coordinate

G‡2

R

+R

Less polar solvent (slower)

more polar solvent (slower)

Remember from Hammond-Leffler Postulate that for a endergonic reaction the transition state resembles the products and bond formation and breaking is more than half formed. Thus the polar transition state in a SN1reaction is stabilized by a polar solvent as well as the intermediate that is formed.

G‡1G‡

2 >


50

Effect of Leaving Group

The best leaving groups are those that are stable ions or molecules after they depart, i.e. weak bases. The stability of the anions (L ¯ ) is related to the ease of dissociation of its conjugate acid, H-L and thus the pka of H-L.

Low pka values (strong acids) good leaving groups, L ¯ weak base

High pka values (weak acids) poor leaving groups. L ¯ strong base

Acid pka Leaving Group

Name

HI -10 I¯ Iodide GoodLeavingGroups

HBr -9 Br¯ Bromide

HCl -7 Cl¯ Chloride

HOSO2R -6.5 ¯ OSO2R Sulfonate

H3O + -1.7 H2O Water

HF 3.2 F¯ Fluoride PoorLeavinggroups

H2S 7 SH¯ Thiolate

HCN 9.2 CN¯ Cyanide

H2O 15.7 OH¯ Hydroxide

HOR 16 to 18 OR¯ Alkoxide


51

Summary SN1 vs. SN2

Stereochemistry

SN2

SN1


52

Br + NaC≡N

Br + CH3OH

Cl + CH3OH

NH2Br

What are the products of the following reactions ?

+

acetone

a lot missing


53

OH

O S

O

O

CH3

O

O

+

+


Na SCH3

NaH + CH3CH2C≡C-Hhexane

Br

acetic acid

a lot missing


54

ClCH2CH2CH2COOH + Na OH


acid-base reaction very fast

CH2CH2CH2C

O

OCl + H2ONa

SN2

1234

a lactone

Br

+ CH3CH2OH

1°Intramolecular ring formation reactions are more favorable than intermolecular reactions (between two different molecules) because the nucleophile is held in close proximity to the electrophile by the carbon chain at all times. Thus attaining the proper orientation required for the SN2 reaction is much more probable than for a intermolecular reaction. For example:

CH3CH2O + CH3CH2Cl CH3CH2OCH2CH3 + ClO

Cl O + Clintermolecular

Intramolecular

is 5000times

slowerthan

a lot missing


55

Organic Synthesis: Functional Group Transformations Using SN2 Reactions

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