alkyl halides

ALKYL HALIDES & ARYL HALIDES CHEMISTRY

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NOMENCLATUREThe general formula of saturated mono substituted alkyl halide is CnH2n+1X, where X is ahalogen atom. Alkyl halides are usually represented by R – X where R is an alkyl group.For example

CH3Br

CH3

CH3

CH3

I1-Bromopropane 1-Iodo-2,2-dimethylpropane

Some examples of substituted alkyl and aryl halides are:

Br

OH

Cl CH2Br

(Bromomethyl)benzene 2-(chloromethyl)phenol Bromoethylene

CH3Cl

Cl

Cl1,3-Dichlorocyclohexane1-Chloro-1-methylcyclopentane

Br

H

Br

Hcis-1, 3-dibromo cyclobutane

H

H

CH3

CH3Br

H

3-Bromo-2-methyl but-1-ene

CH3

H

CH3

HBr

H

1-Bromo-2-methyl but-2-ene

Classification: Alkyl halides can be classified as methyl halide, primary alkyl halide,secondary alkyl halide (2°) and tertiary alkyl halide (3°), according to the number of othercarbon atoms attached to the carbon bearing the halogen atom.

Methyl halide

XH

HH X

H

HR X

R'

HR X

R'

R"R

Primary alkyl halide (1°) Secondary alkyl

halide (2°) Tertiary alkyl halide (3°)

ALKYL HALIDE & ARYL HALIDE

CHEMISTRY ALKYL HALIDES & ARYL HALIDES

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R, R and R may be same or different.As we know that the alkyl halides contains halogen atom as the functional group, whichis responsible for the characteristic reactions viz. nucleophilic substitution and eliminationreactions, of alkyl halides. These reactions are highly influenced by nature of solvents..

Solvents and its typesa) Non polarb) Polar (These solvents are of two type – polar protic and polar aprotic)

POLAR PROTIC SOLVENTSExamples :1. H2O2. CH3OH3. CH3CH2OH4. H—COOH5. CH3—COOH6. NH3

APROTIC SOLVENTSExamples :

7. OCH3CH3

(acetone)

8.SO

CH3CH3

(DMSO)

9.CO

NH CH3

CH3

(DMF)

10.CO

NCH3 CH3

CH3

(DMA)



11. ´ ´METHODS OF PREPARATION

1. From halogenation of Alkanes: Alkanes can be halogenated to give alkyl halide.h

4 2 3or hightempCH X CH X HX (Where X = Cl, Br and I)On further halogenation, di, tri – or tetrahalomethane can be obtained. This can be achievedby increasing the concentration of halogen (X2).

h3 2 2 2or hightemp

CH X X CH X HX h

2 2 2 3or hightempCH X X CH X HX

h3 2 4or hightemp

CHX X CX HX

Illustration 1 :1. What would be the major product in each of the following reaction?

i) What would be the major product in each of the following reaction? 2 , Heat, Light1. Br

5 6 2 5 2. NaCNH C C H

sol. h5 6 2 3 2H C CH CH Br 5 6H C

Br|CH – 3CH NaCN 5 6H C CN|CH – CH3

ii) What happens when excess chlorine is passed through boiling toluene in thepresence of sunlight?

sol.

CH3

h

CCl3

3HClExcess

Cl2Benzo Chloride

2. From addition of halo acids on alkenes and alkynes: Electrophilic addition of HX onalkenes and alkynes give the alkyl halides.

CH3CH3

HX+ CH3CH3

X



(Where HX = HCl, HBr and HI)

2HX+CH3CH CH3

CH3

X

X

Example

HCl+CH3CH2

CH3Cl

CH3

+CH3CH

CH3CH3

Cl

Cl2HCl

Note:— Free radical addition of HBr on alkenes will give the anti-Markovnikov product.

3. From addition of X2 on alkenes and alkynes: Addition of X2 on alkenes give dihalo andthat on alkyne give tetrahalo alkane.

X2+RCH2 H

R

X

Xdihaloalkane

2X2+ TetrahaloalkaneCH3

X

X

X

XCH3CH3

CH3

Example

+CH3CH2

Br2 CH3Br

Br

1,2-dibromopropane



+CH3CH

2Cl2 CH3

Cl

Cl

Cl

ClH

1,1,2,2-tetrachloropropane

4. From alcohols: When alcohols are treated with PX3 or HX they give alkyl halides.3PX

2 2or HXRCH OH RCH XExample

CH3OH 42SOH ,NaBr

orHBr .conc CH3Br

3PCl

OH

CH3

Cl

CH3

For the preparation of iodoalkane, we take a mixture of red phosphorus and I2.2Red P I

3 2 3 2CH CH OH CH CH IAlkyl chlorides can also be prepared from alcohols by treating it with PCl5 or SOCl2.CH3CH2OH + PCl5 CH3CH2Cl + POCl3 + HClCH3CH2OH + SOCl2 Py CH3CH2Cl + SO2- + HClPreparation of alkyl chloride from alcohols by treating it with SOCl2 is the best method asit gives almost pure alkyl chloride since the by products of the reaction i.e. SO2 and HClare in gaseous phase.

5. From free radical substitution of alkyl benzenes

h ,Cl2CH3 Cl

CH3

NBS

CH3Br

6. Halide Exchange: Alkyl iodides can be prepared from the corresponding alkyl chloridesor alkyl bromides by reacting it with a solution of sodium iodide in acetone.R – Cl + I– acetone R – I + Cl–



As NaCl and NaBr are less soluble in acetone, they precipitate and can be removed fromsolution by filtration.

Physical Properties of Alkyl Halidesa) Dipole moment: The physical properties of alkyl halides are influenced by the polarity of

C – X bond. The bond length of C – X bond in alkyl halides follow the order MeI > MeBr >MeCl > MeF. Vapour phase dipole moments in Me – X varies as MeCl > MeF > MeBr >MeI. This is because dipole moment depends on electronegativity of halogen as well ason the bond length.

b) Boiling Point: The order of boiling points of alkyl halides is MeI > MeBr > MeCl > MeF,which is influenced by the decreasing van der Waals’ forces of attraction between them.Among polyhalogen compounds, boiling point decreases as CCl4 > CHCl3 > CH2Cl2 >CH3Cl. This is due to accumulation of Cl’s on CH4, increases the molecular mass andsize, thus van der Waals’ forces increases and boiling point also increases.

c) Density: The densities of alkyl iodides and bromides are more than that of H2O and thedensities of alkyl chlorides and fluorides are less than that of H2O. The order of densitiesof alkyl halides and H2O is RI > RBr > H2O > RCl > RF and the density of polychloromethane varies as CCl4 > CHCl3 > CH2Cl2 > H2O > CH3Cl. The alkyl halides are in generalinsoluble in water.HSAB (HARD and Soft Acid-Base) Principle: According to hard and soft acid-baseprinciple of ‘Pearson’, Hard acids are those species, which have more tendency toaccept an electron pair (like H+, Li+, Mg2+, Cr3+, Al3+ etc.) and hard bases are those species,which have more tendency to donate electron pair (like F–, O2– etc.) A hard base prefers ahard acid whereas a soft base prefers a soft acid.Basicity And Nucleophilicity:A negatively charged species or the species containing electron pair can function asnucleophile as well as like base but its nucleophlicity and basicity are different.Nucleophilicity of the species is the ability of the species to attack an electrophilic carbonwhile basicity is the ability of the species to remove H+ from an acid. Let us have aspecies. B–. Its function as a nucleophile is shown asB A A B And its role as base is indicated asB H A B H A The nucleophilicity is determined by the kinetics of the reaction, which is reflected by itsrate constant (k) while basicity is determined by the equilibrium constant, which is reflectedby its Kb.The order of nucleophilicity of different species depends on the nature of solvent used. Forinstance, let us take F–, Cl–, Br– and I– with their counter cation as Na+ and see theirnucleophilicity order in different solvents. There are four categories of solvents, namelynon-polar (CCl4), polar protic (H2O), polar aprotic (CH3SOCH3) and weakly polar aprotic(CH3COCH3).



Polar solvents are able to dissociate the salts i.e., ion-pairs can be separated. On theother hand, non-polar and weakly polar solvents are unable to dissociate salts, so theyexist as ion-pairs. The ion-pairing is strong when ions are small and have high chargedensity.In non-polar and weakly polar aprotic solvents, all the salts will exist as ion-pairs. The ion-pairing will be strongest with the smallest anion (F–) and weakest with the largest anion (I–

), thus the reactivity of X– decreases with decreasing size. Thus, the nucleophilicity orderof X– in such solvents would be

I Br Cl F In polar protic solvents, hydrogen bonding or ion-dipole interaction diminishes the reactivityof the anion. Stronger the interaction, lesser is the reactivity of anion. F– ion will formstrong H-bonding with polar protic solvent while weakest ion-dipole interaction will be theI– ion. Thus, the nucleophilicity order of X– in polar protic solvent would be I– > Br– > Cl– >F–

Polar aprotic solvents have the ability to solvate only cations, thus anions are left free.The reactivity of anions is then governed by their negative charge density (i.e. their basiccharacter). Thus, the order of nucleophilcity of X– in polar aprotic solvents would beF– > Cl– > Br– > I–

NUCLEOPHILIC SUBSTITUTION REACTIONSA halogen is more negative than a carbon consequently, the two atoms do not share

their bonding electron pairs equally. Because the more electronegative halogen has alarger share of electrons, it has a partial –ve charge and the carbon to which it is bondedhas a partial +ve charge.

2R CH X X = Cl, Br, I, F

This unequal sharing of electrons causes alkyl halides to undergo substitution andelimination reactions. There are two important mechanisms of substitution reaction.

1. A nucleophile is attracted to the partially positively charged atom of carbon. As thenucleophile approaches the carbon, it causes the C – X bond to break heterolytically.

Nu XH

H H

Nu

H H H



2. The C – X bond breaks heterolytically without any assistance from the nucleophile, forminga carbocation. The carbocation then reacts with nucleophile to form the substitution product.

+X X-

+ NuNu

Regardless of the mechanism by which such a substitution reaction occurs, it is called anucleophilic substitution reaction because a nucleophile is substituted for the halogen.The mechanism that predominates depends on– the structure of the alkyl halide.– the reactivity and the structure of the nucleophile.– the concentration of the nucleophile.– the solvent in which the reaction is carried out.MECHANISM OF SN2 REACTIONS

The rate of nucleophilic substitution reaction such as the reaction of methyl bromidewith hydroxide ion depends on concentration of both the reagents. If the concentration ofthe methyl bromide in the reaction mixture is doubled, the rate of SN reaction doubles. Ifthe concentration of OH– is doubled the rate of reaction is also doubled. If the concentrationof both the reactants are doubled, the rate of reaction quadruples.

CH3Br + HO– CH3OH + Br –Rate [Alkyl halide] [Nucleophile]Thus it is the second order reaction, first order with respect to substrate and 1st order

with respect to nucleophile. The rate law tells us which molecules are involved in thetransition state of the rate determining step of the reaction. From the rate law for thereaction of methyl bromide with hydroxide ion, we know that both methyl bromide andhydroxide ion, are involved in the rate determining transition state. The transition state isBIMOLECULAR; i.e. it involves two molecules. The rate constant k describes how difficultit is to overcome the energy barrier of the reaction (how hard it is to reach the transitionstate). The larger the rate constant, the easier it is to reach the transition state.



The mechanism proposed by Hughes and Ingold for an SN2 reaction has one step. Thenucleophile attacks the carbon bearing the leaving group and displaces the leaving group.The nucleophile hits the carbon bearing the leaving group and displaces the leaving group.Because the nucleophile hits the carbon on the side opposite to the side bonded to theleaving group, the carbon is said to undergo back side attack. Backside attack occursbecause the orbital of the nucleophile that contains its non-bonding electrons interactswith the empty anti bonding molecular orbital associated with the C – Br bond. This orbitalhas its larger lobe on the side of the carbon directed away from the C – Br bond.Consequently, the best overlap of the interacting orbitals is achieved through backsideattack. An SN2 reaction is also called direct displacement reaction because the nucleophiledisplaces the leaving group in a single step. A reaction co-ordinate diagram is shown infigure.

+HO CH3 Br CH3 OH+ Br-

Relative Rates of SN2 reactions for several alkyl halides with OH–

A lk yl B ro m id e C lass R e la tive R a te CH 3 Br M e th yl 1200 CH 3

B r P rim ary 40 CH 3

B r P rim ary 16

CH 3C H 3

B r

S eco ndary 1

CH 3Br

C H 3

CH 3 T e rtia ry T oo s low to

m easure

OH C

HB r

H H

G

B rO HC H 3

T .S .

P rogress of r eaction

G

Since the nucleophile attacks from the backside of the carbon that is bonded to the

halogen, bulky substituents attached to this carbon will make it harder for the nucleophileto get to the back side and therefore will decrease the rate of reaction. Thus the mechanismexplains why substituting methyl groups for the hydrogens in methyl bromide progressivelyslows the rate of the substitution reaction.

Steric effect and SN2 reaction: Effects due to groups occupying a certain volume ofspace are called steric effect. A steric effect that decreases the reactivity is called sterichindrance. Steric hindrance results from groups getting in the way at the reaction site. Asa consequence of steric hindrance, alkyl halides have the following, relative reactivities inan SN2 reaction. The steric crowding in tertiary alkyl halide, is too great that they areunable to undergo SN2 reactions.



Relative reactivities of alkyl halides in an SN2 reaction:

Methyl halide > 1° alkyl halide > 2° alkyl halide > 3° alkyl halide.It is not just the number of alkyl groups attached to the carbon undergoing nucleophilic

attack that determines the rate of an SN2 reaction; the size of the alkyl group is alsoimportant. For example, ethyl bromide and n-propyl bromide are both primary alkyl halide,but ethyl bromide is more than twice as reactive in an SN2 reaction because the methylgroup of the ethyl bromide provides less steric hindrance to back side attack than doesthe ethyl group of n-propyl bromide.

As the nucleophile approaches the backside of the carbon of methyl bromide the C – Hbond begins to move away from the nucleophile and its attacking electrons. By the timetransition state is reached the C – H bonds are all in the same plane and the carbon ispentacoordinate (fully bonded to three atoms and partially bonded to two) rather thantetrahedral. As the nucleophile gets closer to the carbon and the bromine moves furtheraway from it, the C – H bonds continue to move in the same direction. Evidently the bondbetween the carbon and the nucleophile is fully formed and the bond between the carbonand the bromine is completely broken, and thus the carbon is once again tetrahedral.

+HO BrBr OH C Br OH +

The best way to visualize the movement of groups bonded to the carbon at whichsubstitution occurs is to picture on umbrella that turns inside out. This is called inversionof configuration. The carbon at which substitution occurs have inverted its configurationduring the course of the reaction just as an umbrella has a tendency to invert in a windstorm.The inversion is known as WALDEN INVERSION, since Paul Walden was the first todiscover that compounds could invert their configuration as a result of substitution reactions.

Because an SN2 reaction takes place with inversion of configuration, only one product isformed when an alkyl halide that has the halogen leaving group bonded to a chiral centreundergoes an SN2 reaction. The configuration of that product is inverted compared withthe configuration of the alkyl halide. Therefore, proposed mechanism accounts for theobserved configuration of the product.

+ HO Br+R'

H HBr

R'

RH OHR'

RH

BrR'

RH

OH



Leaving Group: If an alkyl iodide, alkyl bromide, an alkyl chloride, and an alkyl fluoride(all having the same alkyl group), were allowed to react with the same nucleophile underthe same conditions, we would find that the alkyl iodide is the most reactive and the alkylfluoride is the least reactive.

The only difference among these four reactions is the nature of the leaving group. ApparentlyI– is the good leaving group and the F– ion is the poor. This brings us to an important rule inorganic chemistry that will keep reappearing. The weaker the base, the better is the leavinggroup.The leaving ability depends on basicity because a weak base does not share its electronsas that as a strong base does. Consequently, a weak base is not bonded as strongly tothe carbon as a strong base would be. Thus a bond between a carbon and a weak base ismore easily broken than a bond between a carbon and a strong base.We know that the hydrogen halides have the following relative acidities.

HI > HBr > HCl > HFBecause we know that stronger the acid weaker is its conjugate base, the order of basicitiesof halide ions is

I– < Br– < Cl– < F–

Further, because weaker bases are better leaving groups, the halide ions have the followingrelative leaving abilities.

I– > Br– > Cl– > F–

As a consequence of relative leaving abilities of the halides, alkyl halides have the followingrelative reactivities in an SN2 reactions.

RI > RBr > RCl > RFNucleophile: When we talk about atoms or molecules that have lone pair of electrons,sometimes we call them bases and sometimes we call them nucleophiles. What is thedifference between a base and a nucleophile?A base shares its lone pair with a proton. Basicity is a measure of how strongly the baseshares those electrons with a proton. The stronger the base the better it shares its electrons.Basicity is measured by the acid dissociation constant (Ka), which indicates the tendencyof the conjugate acid of the base to lose a proton. A nucleophile uses its lone pair ofelectrons to attack on electron deficient atom, other than a proton. NUCLEOPHILICITY is ameasure of how readily the nucleophile is able to attack such an atom. It is measured by rateconstant (k). In the case of an SN2 reaction, nucleophilicity is a measure of how readily thenucleophile attacks an sp3 hybridised carbon bonded to a leaving group.



In comparing molecules with the same attacking atom, there is generally a directrelationship between basicity and nucleophilicity. Stronger bases are better nucleophiles.For example, a compound with negatively charged oxygen is a stronger base and a betternucleophile than a compound with a neutral oxygen.

In comparing molecules with attacking atom of approximately the same size, the strongerbases are again the better nucleophiles. The atoms across the second row of the periodictable have approximately the same size. If hydrogens are attached to the second rowelements the resulting compounds have the following relative acidities.CH4 < NH3 < H2O < HF

Consequently, the conjugate bases have the following relative basicities andnucleophilicites. For example, the methyl anion is the strongest base as well as the bestnucleophile.CH3

– > NH2– > OH– > F–

In comparing molecules with attacking atoms that are very different in sizes, the directrelationship between basicity and nucleophilicity is retained if the reaction occurs in thegas phase. If, however, reaction occurs in solvent the relationship between basicity andnucleophilicity depends upon the solvent.If the solvent is aprotic, the direct relationship between nucleophilicity and basicity holds.For example, both the nucleophilicities and the basicities of halide decreases with increasingsize in the aprotic solvent such as N.N-dimethyl formamide (DMF), Dimethylsulphoxide(DMSO), N, N-dimethylacetamide (DMA).If the solvent is protic, the relationship between basicity and nucleophilcity becomesinverted, as basicity increases, nucleophilicity decreases. Thus I– ion which is the weakestbase of the halogen family, is the poorest nucleophile of the family in an aprotic solventand best nucleophile in a protic solvent.



F-Cl-Br-I-

Size increases Basicity increases increases nucleophilicity in an aprotic solvent increases nucleophilicity

in a protic solvent

How does a solvent ability to be hydrogen bonded does not affect the relationship betweennucleophilicity and basicity? When a –vely charged species are placed in a protic solvent,the solvent molecules arrange themselves so that their partially positively charged hydrogenpoints towards the –vely charged species. An aprotic solvent does not have a partiallypositively charged hydrogen.

OH H

O

H

H

OH

H

O

H

H

Y

Ion-diopole interaction between a nucleophile and water (protic solvent)The interaction between the ion and the dipole of the protic solvent is called an ion-

dipole interaction. The change from a direct relationship between basicity and nucleophilicityin an aprotic solvent to an inverse relationship in a protic solvent results from the ion-dipole interactions between the nucleophile and the protic solvent. This occurs becauseat least one of the ion dipole interactions must be broken before the nucleophile canparticipate in an SN2 reaction. Weak bases interact before the nucleophile can participatein an SN2 reaction weak bases interact weakly with protic solvents, strong bases interactmore strongly because they are better at sharing their electrons. It is, therefore, easier tobreak the ion-dipole interactions between an iodide ion and the solvent than between themore basic fluoride ion and the solvent, because the latter is a stronger base. As a result,iodide ion is a better nucleophile in a protic solvent.Relative Nucleophilicity towards CH3I in Methanol

increasing nuleophilicity

RS-I-CN-CH3O-Br-NH3Cl-F-CH3OH



Because there are many different kinds of nucleophiles, a wide variety of organiccompounds can be synthesized by means of SN2 reactions. The following reactions showjust a few of the many kinds of organic compounds that can be synthesized in this way.

CH3CH2Cl + HO– CH3CH2OH + Cl–CH3CH2Br + HO– CH3CH2OH + Br–

CH3CH2I + RO– CH3CH2OR + I–

CH3CH2Br + RS– CH3CH2SR + Br–

CH3CH2F + NH2– CH3CH2NH2 + F–

CH3CH2Br + C– CR CH3CH2C CR + Br–

CH3CH2I + CN– CH3CH2CN + I–

STEREOCHEMISTRY OF SN2 REACTIONAs we have seen that an SN2 reaction the nucleophilie attacks from the backside, that

is, from the side directly opposite to the leaving group. This mode of attack causes achange in configuration of the carbon atom that is the target of the nucleophile. As thedisplacement takes place, the configuration of the carbon atom under attack inverts it isturned inside out in such a way that an umbrella is turned inside out in a strong gale.

HO BrR"

RR' OH C

RBr

R' R"

OHR"

RR'

Inversion of Configuration

Br+

A compound that contains one stereocentre, and therefore, exists as a pair of enantiomers,is 2-bromooctane. These enantiomers have been obtained separately and are known tohave the configuration and rotation shown here.

CH3

H BrH13C6

R(-)-2-bromooctane CH3

Br HH13C6

S(+)-2-bromooctane 25.34][ 25

D 25D[ ] 34.25

The S-2-octanol is also chiral. The configuration and rotation of the 2-octanol enantiomershave also been determined.



CH3

H OHH13C6

CH3

OH HH13C6

90.9][ 25D 90.9][ 25

DS-(+)-2-octanolR-(-)-2-octanol

When R-(–)-2-bromooctane reacts with NaOH, the only substitution product is obtainedfrom the reaction is S-(+)-2-octanol.

HO BrH13C6

CH3H OH C

CH3Br

H C6H13

OHCH3

C6H13H

25.34][ 25D

R-(-)-2-bromooctaneEnantiomeric purity = 100%

90.9][ 25D

S-(+)-2-octanolEnantiomeric purity = 100%

Br

There still remains a problem. We need to relate the configuration of the reactant and theproduct. We know that the direction of rotation (optical) and the configuration of twodifferent compounds are not necessarily related, for they may have the same sign ofrotation but may have different configurations. The experimental proof of Walden inversionwas obtained by a series of experiments in which optically active alcohol was convertedinto its enantiomer having a rotation opposite in sign but nearly equal in magnitude to thatof the original alcohol.Thus in the series of reactions on an optically active (+) alcohol formation of an ester with4-methylbenzenesulphonyl chloride (Tosyl chloride) is known not to break the C – O bondof the alcohol (that such as the case may be shown by using an alcohol labeled with O18

in its OH group, and demonstrating that this atom is not eliminated on forming the tosylate;it is, however, eliminated when the tosylate; is reacted with MeCOO–) hence the tosylatemust have the same configuration as the original alcohol.

OHPhH2C

HMe OSO2C6H4CH3

H2CPh

HMe

OH

OCH3

O

OCH2Ph

HMe

CH3O

6 4 2p MeC H SO Cl

OHCH2Ph

HMeCH3

OH

O



Reaction of this ester with CH3COO– is known to be a displacement in which ArSO3– (Ar =

p – MeC6H4) is expelled and MeCO2– is introduced, hence the C – O bond is broken in this

step and inversion of configuration can thus take place in forming the acetate. Alkalinehydrolysis of the acetate can be shown not to involve fission of the alkyl oxygen (C – O)linkage, so the alcohol must have the same configuration as the acetate. (Hydrolysis ofan acetate in which the alcohol oxygen atom is O18 labelled fails to result in the latterreplacement, thus showing that alkyl oxygen bond of the acetate is not broken during itshydrolysis). As it is found to be the mirror image of the starting material – opposite directionof the optical rotation – an inversion of configuration must have taken place during theseries of reactions and can have occurred only during reaction of MeCO2

– with the tosylate.Reaction of this tosylate with a number of nucleophiles showed that inversion ofconfiguration occurred in each case; it may thus be concluded with some confidence thatit occurs on reaction with Br– to yield the bromide, i.e., that the bromide like acetate hasthe opposite configuration to the original alcohol.

The general principle that bimolecular (SN2) displacement reactions are attended byinversion of configuration has been established in an elegant and highly ingeniousexperiment, in which an optically active alkyl halide undergoes displacement by the samethough isotopically labeled – halide ion as a nucleophile e.g. 128I– on (+) – 2-iodooctane.

IH

H13C6CH3

I CC6H13

IH3C H

IH

C6H13

CH 3

I128 + 128

128 + I

( )(+)-2-iodooctane

The displacement was monitored by observing the changing distribution of I128 betweenthe inorganic (sodium) iodide and 2-iodooctane and it was found under these conditions,to be second order overall.If the inversion takes place, as SN2 requires. The optically activity of the solution willdecline zero i.e., racemization will occur. This will happen because inversion of theconfiguration of a molecule of (+) 2-iodooctane results in the formation of a molecule of itsmirror image (-)-2-iodooctane. Which pair off with a second molecule of (+)-2-iodooctaneto form a racemate (±); thus, the observed rate of racemisation will be twice the rate ofinversion.

Illustration 2: Explain why an SN2 solvolysis (where solvent is the nucleophile) appearsto follow a first order rate law, rather than a second order one.Solution:Because an SN2 solvolysis reaction is bimolecular, it would follow the followingrate law Rate = k[RX] [solvent]. However, the “concentration” of solvent is very large, andgreatly exceeds the amount of RX. For this reason, its concentration is effectively constant,and therefore the reaction will follow the following pseudo-first-order rate law: Rate = k[RX],where k = k[solvent]



Illustration 3: Write the structure of the nucleophilic substitution products in each ofthe following in aprotic solvent.

i) CN

CH2Br

ii) PhSCH2 ClCH3

CH3

Solution: i)

CN

ii) PhS CH3

CH3

SN1 REACTIONSGiven our understanding about SN2 reactions, if were to measure the rate of reaction oftert-butyl bromide with water, we would expect a relatively slow substitution reaction sincewater is a poor nucleophile and tert-butyl bromide is sterically hindered to attack by anucleophile.. However, we actually discover that the reaction is surprisingly fast. It is, infact over one million times faster than the reaction of methyl bromide – a compound withno sterric hindrance with water. Clearly, the reaction must be taking place by a mechanismdifferent from that of the SN2 reaction.As we have seen, a study of the kinetics of a reaction is one of the first step undertakenwhen investigating the mechanism of a reaction. If we were to investigate the kinetics ofthe reaction of tert-butyl bromide with water, we would find that doubling the concentrationof the alkyl halide doubles the rate of reaction. We would also find that changing theconcentration of nucleophile has no effect on the rate of the reaction.

OH 2CH3

Br

CH3

CH3

CH3OH

CH3

CH3 Br H

Rate = K [Alkyl Halide]Because the rate of reaction depends upon the concentration of only one reactant, it is afirst order reaction. An SN1 reaction has two steps, the carbon-halogen bond breaksheterolytically with the halogen retaining the previously shared electron pair. In the secondstep, the nucleophile reacts rapidly with the carbocation formed in the first step.

-rBBrCH3

CH3

CH3 CH3CH3

CH3

slow

fast2 OH OH2

CH3

CH3

CH3CH3CH3

CH3+ OH

CH3

CH3

CH3 H



From the observation that the rate of an SN1 reaction depends only upon the concentrationof the RX, we know that the first step is the slow and rate determining step. Because thenucleophile is not involved in the r.d.s., its concentration has no effect on the rate of thereaction. If you look at the reaction – co-ordinate diagram, you will be able to see whyincreasing the rate of the second step will not make an SN1 reaction go any faster.

BrCH3

CH3CH3

OHCH3

CH3CH3

CH3

CH3CH3 OH2

CH3

CH3CH3

0G 0G

H

Br-

H2O

Progress of Reaction

Free E

nergy

Second, a carbocation is formed in the slow step of an SN1 reaction. Because a 3°carbocation is more stable and therefore, easier to form than a secondary carbocationwhich in turn is more stable to form than a 1° carbocation. 3° alkyl halides are morereactive than a 2° alkyl halide which in turn more reactive than a 1° alkyl halide in an SN1reaction.

Relative rates of SN1 for several alkyl halides. (Solvent and Nucleophile = H2O)

Alkyl Halide Class Relative Rate Br

CH3

CH3

CH3

3° 1,200,000

Br

CH3

CH3

2° 11.6

CH3Br 1° 1.00

CH3 – Br Methyl 1.05

Relative Rates of Alkyl Halides in an SN1 reaction3°RX > 2°RX > 1°RX >CH3XActually 1° carbocation and CH3

+ are so unstable that primary RX and CH3X do notundergo SN1 reactions. (The very slow reaction reported for ethyl bromide and methylbromide in table are SN2 reaction).



The positively charged carbon of the carbocation is sp2 hybridised, and the three bondconnected to this carbon are in the same plane. In the 2nd step of the SN1 reaction, thenucleophile can approach the carbocation from either side of the plane.

Br

H2O H2O

OH O+H

H

H+

H O+H

H-H+

H+ OH

(a) (b)

Inverted configuration

Same configuration (retention)

(a) (b)

The SN1 reaction of an alkyl halide in which the leaving group is attached to the chiralitycentre leads to the formation of two stereoisomers; attack of the nucleophile on one sideof the planar cabrocation forms one stereoisomer and attack on the other side producesthe other isomer.

R'

R HBr OH2

R'

RH OH

R'

R HOHHBr

In the reaction of (S) – 2-bromobutane with water two substitution products are formedone product has the same relative configuration. In an SN1 reaction the leaving groupleaves before the nucleophile attacks. This means that the nulceophile is free to attackfrom either side of the planar carbocation.

R HBr

CH3

conditions 1S2 NOH RH OH

CH3

R HOH

CH3

(R)-2-butanol inverted (S)-2-butanol retended



Although one might expect that equal amounts of two isomers would form in an SN1reaction but a greater amount of the inverted product is obtained in most cases. Typically,50 – 70% of the product of an SN1 reaction is the inverted product. If the reaction leads tothe formation of equal amounts of the two products, the reaction is said to take place withcomplete racemization, when more of the inverted product is formed, the reaction is saidto take place with partial racemization.Paul Winstein first postulated that dissociation of alkyl halide initially results in the formationof an intermate ion pair. In an intimate ion pair, the bond between the carbon and theleaving group has broken but the cation and anion remains next to each other. This speciesthen forms an ion pair in which one or more solvent molecules have come between thecation and the anion. This is called the solvent separated ion pair. Further separationbetween the two results in the dissociated ions.

R Br R Br R BrR Br

Intimate ion pair Solvent separated ionThe nucleophile can attack any of these species. If the nucleophile attack on only the

completely dissociated carbocation, the product will be completely racemised. If thenucleophile attacks either the intimate ion-pair or the solvent separated ion-pair, the leavinggroup will be in a position to partially block the approach of the nucleophile to that side ofthe carbocation. As a result, more of the product with the inverted configuration will beobtained. If the nucleophile attacks the undissociated molecule, it will be an SN2 reaction,and all the product will have the inverted configuration).

CH3 H

CH3

CH3 H

CH3

H2O OH2 H2O OH2Br-

Br- has diffused away giving H2O equal access

to both sides of the carbocationBr-

has not diffused away, so it blocks the approach of H2O to one side of the carbocation

The difference between the products obtained from an SN1 reaction and from an SN2reaction is a little easier to visualize in the case of cyclic compounds. When cis-1-bromo-4-methylcyclohexane undergoes an SN2 reaction, only the trans product is obtainedbecause the carbon bonded to the leaving group is attacked by nucleophile only on itsback side.

CH3H

BrH

1-bromo-4-methylcyclohexane

-Br CH3H

HOH

4-methylcyclohexanol2S

HON



When however, cis-1-bromo-4-methylcyclohexane undergoes SN1 reaction, both the cisand the trans products are formed because the nucleophile can approach the carbocationintermediate from either side.

CH3H

BrH CH3

HHOH

4-methylcyclohexanol1S

HON

CH3H

OHH HBr

cis-isomerThe Leaving Group: Because the r.d.s. of an SN1 reaction is dissociation of the alkylhalide to form a carbocation, two factors affects the rate of SN1 reaction, the case withwhich the leaving group departs from the carbon and the stability of the carbocation that isformed. In the preceding section, we saw that the tertiary alkyl halides are more reactivethan 2° and which in turn more reactive than 1° alkyl halide. This is because moresubstituted carbocation is, more easily it is formed. But about a series of alkyl halideswith different leaving groups that dissociate to form the same carbocation? The answer issame for the SN1 reaction as for as for the SN2 reaction. The weaker the bases, the lesslightly it is bonded to the carbon and the easier it is to break the C – halogen bond. So analkyl iodide is the most reactive of the alkyl halides in both SN1 and SN2 reactions.Relative reactivities of alkyl halides in an SN1 reaction.

RI > RBr > RCl > RFThe Nucleophile: The nucleophile traps the carbocation that is formed in the r.d.s. of theSN1 reaction. Because the nucleophile comes into play after the r.d.s., the reactivity ofthe nucleophile has no effect on the rate of the SN1 reaction.In some SN1 reactions, the solvent is the nucleophile. For example, relative rates given intable are for the reaction of alkyl halides with water serves as both the nucleophie and asthe solvent. When the solvent is the nucleophile the reaction is called the solvolysisreaction.Illustration 4: What happens when (–)–2– bromooctane is allowed to react withsodium hydroxide under SN2 conditions.

Solution: NNaOHS 2Br

CH3

HC6H13

OHCH3

HC6H13

(-)-2-Bromooctane (+)-2-OctanolIn Fisher projection the above reaction can be represented as follows:

NNaOHS 2H

CH3

C6H13Br OH

CH3

C6H13Br



We see that OH group has not taken the position previously occupied by Br–, the alcoholobtained has a configuration opposite to the bromide. A reaction that yields a productwhose configuration is opposite to that of the reactant is said to proceed with inversion ofconfiguration.

NNaOHS 2Br

CH3

HC6H13

OHCH3

HC6H13

(-)-2-Bromooctane (+)-2-Octanol

Illustration 5: Alkyl halides are hydrolysed to alcohol very slowly by water, but rapidlyby silver oxide suspended in boiling water.

Solution:R––X + Ag+ [R––X––Ag]+

[R––X––Ag]+ Slow R+ + AgXR+ + OH– ROHHeavy metal ions, particularly silver ions, catalyse SN1 reaction and mechanism involvesa fast pre- equilibrium step (the silver ions has an empty orbital).

SNI MECHANISM: RETENTION OF CONFIGURATIONDespite what has been said above about displacement reactions leading to inversion ofconfiguration, to racemisation, or to a mixture of both, a number of cases are known ofreactions that proceed with actual retention of configuration, i.e., in which the startingmaterial and product have the same configuration. One reaction in which this has beenshown to occur is in the replacement of OH by Cl through the use of thionyl chloride,SOCl2:

(I)

OHMe

HPh2SOCl Cl

Me

HPh 2 SO HCl

(IIa)The reaction has been shown to follow a second order rate equation, rate = k2[ROH][SOCl2],but clearly cannot proceed by the simple SN2 mode for this would lead to inversion ofconfiguration in the product, which is not observed.



Carrying out the reaction under milder conditions allows of the isolation of an alkylchlorosulphite, ROSOCl (III), and this can be shown to be a true intermediate. Thechlorosulphite is formed with retention of configuration, the R – O bond not being brokenduring the reaction. The rate at which the alkyl chlorosulphite intermediate (III) break downto the product, RCl (II), is found to increase with increasing polarity of the solvent, andalso with increasing stability of the carbocation R+: an ion pair, R OSOCl (IV), isalmost certainly involved. Provided collapse of the ion pair to produce then occurs rapidly,i.e., in the intimate ion pair (V) within a solvent cage, then attack by Cl– is likely to occuron the same side of R+ from which –OSOCl departed, i.e., with retention of configuration:

(III)

OMe

HPh S OCl

slowMe

Ph HO S

Cl

O Me

Ph HCl Cl

Me

HPh

(IV) (V) (IIa)

Whether the breaking of the C – O and the S – Cl bonds occurs simultaneously, orwhether the former occurs first, is still a matter of debate.It is interesting that if the SOCl2 reaction on ROH (I) is carried out in the presence ofpyridine, the product RCl is found now to have undergoing inversion of configuration (IIb).This occurs because the HCl produced during the formation of (III) from ROH and SOCl2 isconverted by pyridine into C5H5NH+Cl– and Cl–, being an effective nucleophile, attacks (III)‘from the back’ in a normal SN2 reaction with inversion of configuration.

OMe

HPh S OCl

+2 SO +ClCMe

Ph HCl O S

O

ClCl

Me

HPh Cl +

(III) (IIb)

NEIGHBOURING GROUP PARTICIPATIONSubstitution with Retention of Configuration (Outlines of Neighbouring GroupParticipation) and Anchimeric Assistance

Often the rate of a reaction is greater than expected and retention of configuration at achiral carbon may be observed and not inverted or racemized. This usually happens whenthere is a group (Z) in the substrate with an unshared pair of electrons in a position b to theleaving group which can play a transient part in the reaction. These assisted reactions bya more-or-less remote functional group is termed the neighbouring group mechanism, andinvolves two successive inversions of configuration. In other words, the neighbouring groupeffect involves essentially two SN2 substitutions, each causing inversion, the net result



being retention of configuration. In the first step the neighbouring group Z acting as anucleophile pushes out the leaving group but still retains attachment to the molecule. Inthe subsequent step the external nucleophile pushes out the neighbouring group.

Step 1RR

ZR L

R

RRR

R

Z+ L

RRR

R

Z+ Nu Step 2 RR

ZR Nu

R

Often when the neighbouring group effect is operative, one may not get the substitutionproduct but a rearrangement. In these situations the nucleophile does not attack thecarbon from which the leaving group had left but instead attacks the carbon to which theneighbouring group was originally linked. Thus the alkaline hydrolysis of (I) affords therearranged product (III). The cyclic intermediate (II) formed after the neighbouring groupparticipation by nitrogen undergoes attack at the methylene (–CH2–) group rather than (–CHEt) because of less steric crowding on the former carbon atom.

CHCH3

ZCCH3 LCH3

CHCH3

C CH3CH3

Z+ NuCCH3

HCZ CH3CH3

Nu

Et2N CHEt Cl Clslow

Et2N CHEt OH Et2NOH

Et

(I) (II) (III)In such cases both substitution and rearrangement occur.

ELIMINATION REACTIONS OF ALKYL HALIDESThe E2 Reaction: There are two important mechanism for elimination. The reaction ofethyl bromide with hydroxide ion is an example of an E2 reaction. It is a second-orderreaction because the rate of reaction depends on the concentration of both ethyl bromideand hydroxide ion.



CH3CH2Br + HO– CH2 = CH2 + H2O + Br–

Rate = K [Alkyl halide] [Base]The rate law tells us that alkyl bromide and hydroxide ion both are involved in the transitionstate of the rate determining step of the reaction. The following mechanism agrees withthe observed second order kinetic.The E2 reaction is a concerted one step reaction. The proton and the bromide ion areremoved in the same step.

H O H

Br

2 H O Br

In an E2 reaction, a base removes a proton from a carbon adjacent to the carbon bondedto the halogen. As the proton is removed, the electrons, the hydrogen shared with carbonmoves forward to the carbon bonded to the halogen. As these electrons move in thehalogen leaves taking its bonding electrons with it. The removal of the proton and thehalide ion from alkyl halide is known as dehydrogenation.The carbon to which the halogen is attached is called the -carbon. An adjacent is calleda -carbon. Because the reaction is initiated by the removal of proton from the -carbon,on E2 reaction is sometimes called a -elimination reaction. It is also called a 1,2-elimination reaction because the atoms being removed from adjacent carbon.

_Br BH H

R Br

RR

R

B carbon

carbon2-bromopropane has two -carbon atoms from which a proton can be removed in an E2reaction. Because the two carbons are identical, the proton can be removed from eitherone. The product is propene.

3 CH O CH3Br

CH3CH3

CH23 CH OH Br

How one will known which product will be formed in greater yield? You must determinewhich of the product is formed more easily – that is, which product is formed faster.One must know that alkenes stability depends upon the number of alkyl substituentsbonded to the sp2 carbons, the greater that number, the more stable the alkene. In thetransition state leading to an alkene, the C – H and C – Br bonds are partially, broken andthe double bond is partially formed giving the transition state an alkene like structure.



Because the transition state has an alkene like structure, the transition state leading to 2-butene is more stable than the T.S. leading to 1-butene. The more stable transition stateallows the formation of 2-butene faster than 1-butene.

CH3 CHBr

CH CH3

HOCH3

H2C CH

H

BrCH2 CH3

OCH3

T.S. leading to 2-butene more stable T.S. leading to 1-butene less stableThe difference in rate of formation of two alkenes is not very great, consequently, bothproducts are formed but the more stable alkene is the major product of the eliminationreaction. Thus an E2 reaction is regio-selective; i.e., more of one isomer is formed thanthe other.The reaction of 2-bromo-2-methyl butane with hydroxide ion produces both 2-methyl-2-butene and 2-methyl-1-butene. Because-2-methyl-2-butene has greater number of alkylsubstitutents attached to sp2 carbon. It is more stable alkene and is the major product ofthe elimination reaction.

G°

Progress of reaction

1-butene

2-butene

2-bromobutane

2H O HO

(70%)

CH3CH3

BrCH3

CH3CH3

CH3

CH2CH3

CH3(30%)

Saytzeff Elimination vs Hoffmann Elimination: Alkyl halides have the following relativereactivities in an E2 reaction, because elimination from a 3° alkyl halide typically leads toa more highly substituted alkene. But one must exercise some care in using Saytzeff’srule to predict the major product of the elimination reaction because the most substitutedalkene is not always the one that is easiest to form.



For example, if the base in an E2 reaction is sterically hindered, it will preferentially removethe most accessible -hydrogen. In the following, it is easier for the bulky tert-butoxideion to remove one of the more exposed terminal hydrogens which leads to the lesssubstituted alkene. Because the less substituted alkene is the one that is the moreeasily formed, it is the major product of the reaction.

OHC)CH( 33

(28%)

CH3CH3

BrCH3

CH3CH3

CH3O CH3

CH3

CH3

CH2CH3

CH3(72%)

sterically hindered baseSaytzeff product Hofmann product

The data in table below show that when an alkyl halide undergoes an E2 reaction with avariety of alkoxide ion, as the size of the base increases, the % of the less substitutedalkene (i.e. Hofmann product) increases.

Effect of steric properties of the base on the distribution of products in an E2 reaction.

-RO CH3CH3

CH3CH3Br

CH3

CH3 CH2

CH3CH3

CH3 CH3

CH3

Base Saytzeff Product More

substituted product Hoffmann product Less

substituted product CH3

O 79% 21%

OCH3

CH3

CH3

27% 73%

CH3O

CH3

CH3 19% 81%

CH3

OCH3CH3

8% 92%

Although the major product of the E2 dehydrohalogenation of the alkyl chlorides, alkylbromides and alkyl iodides is normally the most substituted alkene, the major product ofthe E2 dehydrohalogenation of alkyl of fluorides is the least substituted alkene (i.e. Hofmannelimination).



(30%) OHCH3 3

OCH CH3 CH3

FCH2 CH3 CH3 CH3

(70%)

When a hydrogen and a chlorine, bromine or iodine are eliminated from an alkyl halide,you have seen that the halogen starts to leave as soon as the base begins to remove theproton. Departure of the halogen and its bonding electrons prevents the build up of –vecharge on the carbon that is loosing the proton, giving the T.S. alkene character. Of thehalogens ions, fluoride ion is the strongest base and therefore, the poorest leaving group.So when a base begins to remove a proton from an alkyl fluoride, the F– ion has lesstendency to leave than do the other halide ions. As a result, -ve charge develops on thecarbon that is losing a proton, giving the T.S. a carbanion character rather than an alkenelike T.S. Normally, carbanion T.S. are unstable, but in this case the carbanion T.S. isstabilized by the strongly electron withdrawing fluorine.

CarbanonicT.S. leading to 2-pentene (less stable)

H2C CH3

F

HOCH3

H2C CH3

F

HOCH3

Carbanonic T.S. leading to 1-pentene (more stable)

The T.S. leading to 1-pentene has the developing –ve charge on a primary carbon. This ismore stable than the T.S. leading to 2-pentene which has the developing –ve charge onthe 2° carbon. Because the T.S. leading to 1-pentene is more stable, 1-pentene is formedmore rapidly and is therefore major product of the reaction.In the following reaction, Saytzeff rule does not lead to the more stable alkene, becausethe rule does not take into account the fact that the conjugated double bonds are morestable than the isolated double bonds. Since the conjugated alkene is the more stablealkene, it is the one that is most easily formed and is therefore, the major product of thereaction. So, if there is a double bond or a benzene ring in the alkyl halide, do not useSaytzeff rule to predict the major product of the elimination reaction.



Minor product

HO CH2CH3

CH3

CH2CH3

CH3Major product

CH2CH3

CH3Br

Minor product

HO

Major productBr CH3

CH3

CH3

CH3

CH3

CH3

We can summarise by saying that the major product of the E2 reaction is the mostsubstituted alkene unless one of the following applies.

· The base is sterically hindered.· The alkyl halide is alkyl fluoride.· The alkyl halide contains one or more double bonds.Illustration 6: What are the major products of the following reactions?

a) 3 OCH CH3CH3

CH3Cl b)

3 CH X CH3CH3

CH3O

Solution: a) 3 OCH CH3CH3

CH3Cl CH3

CH2

CH3

3OCH (nucleophile) can’t attack 3° carbon having high electron - density hence eliminationtakes place giving alkene.

b) 3 CH X CH3CH3

CH3O CH3

CH3

CH3O

CH3Nucleophilic attack on methyl carbon is possible giving ether (Williamson synthesis).



Illustration 7: Explain the fact that a small amount of NaI catalyzes the general reactionRCl + RO–Na+ ROR + NaCl

Solution: With I– the overall reaction occurs in two steps, each of which is faster than theuncatalysed reaction.Step 1: RCl + I– RI + Cl–This step is faster because I–, a soft base has more nucleophilicity than OR–, a hard baseStep 2: RI + RO:– ROR + I–

This step is faster because I– is a better leaving group than Cl–

THE E1 REACTIONThe reaction of tert-butyl bromide with water is the first order elimination reaction because

the rate of the reaction depends only on the concentration of the alkyl halide. It is calledan E1 reaction.

2H OCH3CH3

CH3Br CH3

CH2

CH3

Rate = K[Alkyl Halide]We know that only the alkyl halide is involved in the T.S. of the r.d.s. of the reactiontherefore, there must be at least two steps in the reaction.An E1 reaction is the two step reaction. In the first step, alkyl halide dissociatesheterolytically. This is the r.d.s. of the reaction. In the second step of the reaction, thebase forms an elimination product by removing a proton from a carbon adjacent to the+vely charged carbon. Because the first step is the r.d.s., increasing the concentration ofthe base which comes into play in the second step of the reaction, has no effect on therate of the reaction.

Mechanism

CH3CH3

CH3Br CH3

CH3

CH3Br-slow

CH3CH3

H

H3O+fast CH3CH2

CH3

OH2



When two elimination products can be formed, the major product is generally the oneobtained by following Saytzeff rule.

Minor product

Major product

CH3 CH3Cl

CH3OH

CH3CH3

CH3

CH3

CH2CH3

Because the first step is the r.d.s. of an E1 reaction depends on both the ease with whichthe leaving group leaves, and the stability of the carbocation that is formed. Thus therelative reactivities of the series of alkyl halides with the same leaving group parallel therelative stabilities of carbocations. 3° benzylic alkyl halide is the most reactive alkyl halidebecause the 3° benzylic carbocation, is the most stable carbocation, is the easiest toform.Relative Reactivities of Alkyl Halides in an E1 Reaction = Relative Stabillities ofthe carbocations.

3° benzylic 3° allylic > 2° benzylic @ 2° allylic 3° > 1° benzylic 1° allylic 2°> 1° vinyl.Because the E1 reaction involves the formation of carbocation intermediate, therearrangement of the carbon skeleton can occur before the proton is lost. For example,the 2° carbocation that is formed when a chloride ion dissociates from 3-chloro-2-methyl-2-phenylbutane undergoes 1,2-methanide shift to form a more stable 3° benzyliccarbocation.

OHCH 3

CH3

CH3

Cl

CH3

CH3

CH3

CH3 shift Methanide2,1

CH3

CH3

CH3

CH3

CH3

CH3

-H +

Substitution Versus Elimination ReactionsWe know that an alkyl halide can undergo four types of reactions; SN1, SN2, E1 and E2. Agiven alkyl halide under the given conditions will follow which pathway, can be decided infollowing manner.The first thing you must look at is the alkyl halide, is it 1°, 2° or 3°. If the reactant were aprimary alkyl halide, it would undergo E2/SN2 reactions (as their their carbocations are notstable). If the reactant is a secondary or a tertiary alkyl halide, then it can undergo E1/SN1or E2/SN2 reactions depending upon reaction conditions. E2/SN2 reactions are favoured bya high concentration of a good nucleophile/strong base, whereas a poor nucleophile/weakbase favour E1/SN1 reactions.



Once you have decided whether the conditions will favour E2/SN2 reactions or E1/SN1reactions, then you should decide how much of the product will be substitution and howmuch will be the elimination product. The relative amount of substitution and eliminationproduct can be decided again on the basis of structure of alkyl halide (i.e. 1°, 2° or 3°) andon the nature of the nucleophile/base. Relative reactivities of alkyl halides in various reactionsare:In an SN2 reaction: 1° > 2° > 3°In an E2 reaction: 3° > 2° > 1°In an SN1 reaction: 3° > 2° > 1°In an E1 reaction: 3° > 2° > 1°For instance, propyl bromide when treated with methoxide ion in methanol can undergoeither substitution reaction give methyl propyl ether or elimination reaction to give propene.The major product of the reaction would be substitution product.

CH3CH2CH2 – Br + CH3O– 3CH OH3 2 2 3 3 2 3(90%) (10%)

CH CH CH OCH CH CH CH CH OH Br But when the primary alkyl halide or the nucleophile/base is sterically hindered, thenucleophile will have difficulty getting to the back of -carbon and thus, elimination productwill predominate. For example,

3CH OH3CH O CH3

BrCH3

CH3CH3

CH 2OCH 3 3CH OH Br 1-bromo-2-methyl propane (40%) (60%)

A secondary alkyl halide can form both substitution and elimination products, whoserelative amount depend on the base strength of the nucleophile. The stronger and bulkierthe base, greater will be the percent of the elimination product.

2 5C H OH3 2CH CH O 3 2 2 5CH CH CH C H OH Cl

(25%) (75%)CH3

ClCH3 CH3

OCH2CH3CH3

CH3OCOCH3

CH3 Cl(100%)

3 2CH CO(in acetic acid)



Increasing the temperature at which the reaction is carried out increases the rates of boththe substitution and elimination reaction but increase in the rate of elimination reactionsis more than that of substitution reaction. Thus, if the substitution product is desired, thereaction should be carried out a low temperature and high temperature promotes eliminationproduct.A tertiary alkyl halide is least reactive towards SN2 reaction but most reactive towards E2reaction. Thus, only elimination product is formed.

ARYL HALIDESBenzene and its homologues react with halogens to produce either addition or substitutionproducts.

i) Addition Compounds: These compounds are obtained by exposing the mixture of aromatichydrocarbon and the halogen to direct sunlight, e.g., benzene hexachloride (BHC), C6H6Cl6;benzene hexabromide, C6H6Br6 etc.

ii) Substitution products: Two types of halogen substituted products are known.a) Nuclear halogen substitution product: In these products, the halogen islinked directly to the carbon of the benzene nucleus. These products are generally calledarylhalides.

Cl Br CH3

Br

Cl

Clchlorobenzene bromobenzene1-bromo-4-methylbenzene 1,4-dichlorobenzene

b) Side chain halogen substitution products: In these products, the halogen islinked to the carbon atom of the side chain.

Cl ClCl

ClClCl

(dichloromethyl)benzene (trichloromethyl)benzene(chloromethyl)benzene

These are also called side chain aryl halides. They have properties of alkyl halides.Aryl Halides: According to IUPAC system, aryl halides are named as Haloarenes. Ifmore than one halogen is present their positions in the ring are indicated by numbers orappropriate prefixes, ortho, meta, para.



Br

Br

CH3Cl

CH3Cl

CH31,3-dibromobenzene or m-dibromobenzene 1-chloro-2-methylbenzene

or o-chlorotoluene 2-chloro-1,4-dimethylbenzene

GENERAL METHODS OF PREPARATIONi) By direct halogenation: This method is used for the preparation of chloro and bromo

derivatives. Halgoens react with aromatic hydrocarbons in presence of catalysts or halogencarriers such as iron, iodine or aluminium chloride at room temperature in absence ofdirect sunlight.

Fe6 6 2 6 5C H Cl C H Cl HCl

For further halogenation, more halogen is usedFe

6 6 2 6 5 2Chlorobenzene o and p-DichlorobenzeneC H Cl Cl C H Cl HCl

Toluene in presence of iron reacts with Cl2 or Br2 form a mixture of o- and p-chloro or

bromotoluenes, respectively.

CH3

Fe2Cl

CH3Cl

CH3

ClMixture of o- and p-chlorotoluene

Iodo derivatives cannot be obtained by direction reaction with iodine as the reaction isreversible.C6H5 + I2 C6H5I + HIIodo-derivative can be obtained if the reaction is carried out in presence of an oxidizingagent, e.g., iodic acid, or nitric acid, etc. The oxidizing agent oxidizes HI to iodine andthus, the reaction moves to proceed to the right

2HI + [O] H2O + I2



ii)From diazonium salts: Aryl halides can be obtained most satisfactorily by thedecomposition of aryl diazonium salts in presence of copper halide solution dissolved inthe corresponding halogen acid, the diazo group is replaced by a halogen atom (Sandmeyerreaction).

heatHCl/CuCl

Cl

chlorobenzeneBr

bromobenzeneI

iodobenzeneF

fluorobenzene

heatHBr/CuBr

heatKI

heatNaBF4

N2+Cl-

Benzene diazonium

chloride

Sandmayer reaction

Balzsehiencn reaction

Iodo compounds may be obtained by boiling the diazonium salt solution with aqueouspotassium iodide.

iii) Hunsdiecker reaction: Aryl bromides are obtained by heating the silver salts of aromaticacid (in CCl4 or xylene) with bromine.

2Br O-Ag+

OBr

2CO AgBr

iv) Decarboxylation of halogenated acids: Sodium salts of halogenated acid when heatedwith soda lime produce aryl halides.BrC6H5COONa + NaOH C6H5Br + Na2CO3



Properties: Aryl halides are colourless stable liquids with pleasant odour. These areinsoluble in water but readily miscible with organic solvents. Most of them are steamvolatile, heavier than water. Their boiling points are higher than corresponding alkyl halides.The boiling points rise gradually from fluoro to iodo compounds.

NUCLEOPHILIC AROMATIC SUBSTITUTIONNormally, alkenes are unaffected by treatment with nucleophiles, and it is no surprise tofind that benzene also survives treatment with strong nucleophiles. In particular, no reactionoccurs when simple substituted benzenes are treated with base.

Nu No reaction

Cl

No reactionNa+ -OCH3HOCH3

But a few specially substituted benzenes do undergo reaction in base. For example, 2, 4-dinitrochlorobenzene is converted into 2,4-dinitrophenol by treatment with sodium hydroxidein water.

ClNO2

NO2

NaOH/H2Owarming

OHNO2

NO2

(95%)

The nitro group is essential, and it must be substituted in the right place relative to thechlorine. Chlorobenzene and m-nitrochlorobenzene do not react under these conditions,but p-nitrochlorobenzene does, although it substitutes much more slowly than the 2,4-dinitro compound.

Relative Rate

ClNO2

NO2Na+-OCH3

HOCH3, 50°COCH3

NO2

NO2

115,000

Cl

NO2Na+-OCH3

HOCH3, 50°COCH3

NO2

3.4



ClNO2

Na+-OCH3HOCH3, 50°C

OCH3NO2 1.0

Cl

NO2 Na+-OCH3HOCH3, 50°Cor

Cl

No reaction 0

Neither an SN2 nor an SN1 reaction seems likely. Attack from the rear is impossible, asthe ring blocks the path of any entering nucleophile, and ionization would produce a mostunstable carbocation.

C LNu

C

SN1

SN2

An unstable phenyl cation - the SN1 reaction will not be easy.

the path of attack in the SN2 reaction is blocked.

The mechanism of the reaction involves addition to the benzene ring by methoxide togenerate a resonance – stabilized anionic intermediate.

Cl

NO2

NO2

HOCH3

OCH3

H3CO

NO2

NO2Cl H3CO

NO2

NO2Cl H3CO

NO2

NO2Cl H3CO

NO2

NO2Cl

The nitro groups stabilize the negative charge through resonance, and therefore will exerttheir effect only when they are substituted at a carbon that helps to bear the negativecharge. This stabilization by nitro is the reason that o- or p-nitrochlorobenzene undergoesthe reaction, but the meta isomer does not.

Cl

NO2

Na+-OCH3

H3CO

N+

Cl

OO

HOCH3H3CO

N+

Cl

OOPara addition



Cl

NO2

NO2

Na+-OCH3

H3CON+

Cl O

OHOCH3H3CO

N+

Cl O

O

Meta addition – here the nitro groups do not help stabilize the negative charge

Cl

NO2 Na+-OCH3

H3CO Cl

NO2

HOCH3H3CO Cl

NO2

H3CO Cl

NO2

The aromatic system can be regenerated by reversal of the attack of methoxide or by lossof chloride to give 2,4-dinitroanisole.

Cl

NO2

NO2

HOCH3(a)

Na+-OCH3H3CO

NO2

NO2Cl

HOCH3(b)

OCH3

NO2

NO2

(b)(a)There are similarities between nucleophilic aromatic substitution and its more usualelectrophilic counterpart. Each involves the formation of a resonance – stabilizedintermediate, and each involves a temporary loss of aromaticity that is regained in thefinal step of the reaction. But the similarities are only so deep. One reaction is cationic;the other involves anions. Use the differing effects of a nitro group, strongly decelerating inthe electrophilic substitution and strongly accelerating in the nucleophilic reaction, tokeep the two mechanisms distinct in your mind.

BENZYNE MECHANISMWe mentioned the lack of reactivity of halobenzenes with bases. It look the activation of anitro group for substitution to occur for (fig). In truth, if the base is strong enough, evenchlorobenzene will react. In the very strongly basic medium potassium amide in liquidammonia, chlorobenzene is converted into aniline.The mechanism is surely not an SN2 displacement, and there is no reason to thinknucleophilic addition would be possible to an unstabilized benzene ring.



Cl Na+-NH2NH3

NH2+

NH2

Which reactions of halides do you know besides displacement? The E1 and E2 reactionscompete with displacement reactions. If HCl were lost from chlorobenzene, a cyclic alkyne,called dehydrobenzene or benzyne, would be formed. This bent acetylene might bereactive enough to undergo an addition reaction with the amide ion. If this were the case,the labeled material must produce two differently labeled products of addition.

Cl

HNa +- NH 2 NaClNH 3 2NH

(a)

(b)Benzyne(b)(a)

NH2NH2

NH 3H

NH2

NH2

H2NH

Benzyne is surely an unusual species and deserves a close look. Although this intermediateretains the aromatic sextet, the triple bond is badly bent, and there must be very severeangle strain indeed. Remember, the optimal angle in triple bonds is 180°. Moreover, the“third” bond is not composed of 2p-2p overlap like a normal alkyne, but of overlap of twohybrid orbitals.Chlorobenzene with chlorine bonded to 14C gives almost 50% aniline having NH2 bonded to14C and 50% aniline with NH2 bonded to normal 12C atom.

23

NHBr, NH

32 NHNH

CH3Br

CH3 CH3NH2(major product)

23

NHBr, NH

32 NHNH

OCH3

Br

OCH3 OCH3

NH2(major product)



23

NHBr, NH

32 NHNH

CH3

Br

CH3 CH3

NH2(major product)

CH3

NH2(minor product)

23

NHBr, NH

32 NHNH

OCH3

Br

OCH3 OCH3

NH2

OCH3

NH2

(both are in equal in amount)

23

NHBr, NH

32 NHNH

CH3

Br

CH3 CH3

NH2(major product)

One would therefore not be too surprised to find that this molecule is able to add strongbases such as the amide ion. In fact, one would expect very high reactivity for this strainedacetylene.



KEY POINTS

1.R — X Nu

NS 2 reation NS 1reationR — Nu X

Rate = k[RX] [Nu–] Rate = k¢[RX]Inversion of configuration Partial racemizationTransition state less crowded Stable carbocationRate generally favoured by Rate generally favoured by polar solvents.Non-polar solventsCH3X > 1° > 2° > 3° 3° > 2° > 1° > CH3X

2. EliminationDry

EtherR X Mg RMgX HX

3 3 3 2CH CH CH CH CH CH|X

Elimination occurs by E1 and E2 mechanisms. 1° RX reacts by E2 mechanism and 2° & 3°RX by E1 and E2 mechanisms bothE2 reaction is favoured by high concentration of a strong base and polar aprotic solventwhile E1 reaction is favoured by a weak base and polar protic solvent.

3. Ambident Nucleophiles

R — X + KCN R — CN (major) + R—NC (minor)

R — X + AgCN Ag – NC (major) + R – CN (minor)4. Aryl Halides

Less reactive than alkyl halides2o

HNOSn/HCl CuX2 2 20 5 CArNO ArNH ArN Ar X

Nucleophilic substitution by SNAr or Benzyne mechanism.SNAr mechanism operates for activated aryl halides, involves carbanion formation.Benzyne mechanism occurs for activated and deactivated aryl halides.(ii) Chlorination at C-3 produces a chiral carbon marked with star (d and l form).(iii)Chlorination at C-1 also produces a chiral carbon marked with star (d and l form).

alkyl halides

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