Alkyl halides

ALKYLHALIDES &ARYLHALIDES CHEMISTRY
gdvms www.chemadda.net
NOMENCLATURE
The general formula of saturated mono substituted alkyl halide is Cn
H2n+1
X, where X is a
halogen atom. Alkyl halides are usually represented by R – X where R is an alkyl group.
For example
CH3
Br
CH3
CH3
CH3
I
1-Bromopropane 1-Iodo-2,2-dimethylpropane
Some examples of substituted alkyl and aryl halides are:
Br
OH
Cl CH2
Br
(Bromomethyl)benzene 2-(chloromethyl)phenol Bromoethylene
CH3
Cl
Cl
Cl
1,3-Dichlorocyclohexane
1-Chloro-1-methylcyclopentane
Br
H
Br
H
cis-1, 3-dibromo cyclobutane
H
H
CH3
CH3
Br
H
3-Bromo-2-methyl but-1-ene
CH3
H
CH3
H
Br
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 other
carbon atoms attached to the carbon bearing the halogen atom.
Methyl halide
X
H
H
H X
H
H
R X
R'
H
R X
R'
R"
R
Primary alkyl
halide (1°)
Secondary alkyl
halide (2°)
Tertiary alkyl
halide (3°)
ALKYL HALIDE & ARYL HALIDE

CHEMISTRY ALKYLHALIDES &ARYLHALIDES
www.chemadda.net gdvms
R, R and R may be same or different.
As we know that the alkyl halides contains halogen atom as the functional group, which
is responsible for the characteristic reactions viz. nucleophilic substitution and elimination
reactions, of alkyl halides. These reactions are highly influenced by nature of solvents..
Solvents and its types
a) Non polar
b) Polar (These solvents are of two type – polar protic and polar aprotic)
POLAR PROTIC SOLVENTS
Examples :
1. H2
O
2. CH3
OH
3. CH3
CH2
OH
4. H—COOH
5. CH3
—COOH
6. NH3
APROTIC SOLVENTS
Examples :
7. O
CH3CH3
(acetone)
8.
S
O
CH3CH3
(DMSO)
9.
C
O
NH
CH3
CH3
(DMF)
10.
C
O
NCH3
CH3
CH3
(DMA)

11. ´ ´
METHODS OF PREPARATION
1. From halogenation of Alkanes: Alkanes can be halogenated to give alkyl halide.
h
4 2 3
or high
temp
CH X CH X HX
  
(Where X = Cl, Br and I)
On further halogenation, di, tri – or tetrahalomethane can be obtained. This can be achieved
by increasing the concentration of halogen (X2
).
h
3 2 2 2
or high
temp
CH X X CH X HX
  
h
2 2 2 3
or high
temp
CH X X CH X HX
  
h
3 2 4
or high
temp
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. NaCN
H C C H
sol. h
5 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 the
presence of sunlight?
sol.
CH3
h

CCl3
3HCl
Excess
Cl2
Benzo
Chloride
2. From addition of halo acids on alkenes and alkynes: Electrophilic addition of HX on
alkenes and alkynes give the alkyl halides.
CH3
CH3
HX+ CH3
CH3
X

(Where HX = HCl, HBr and HI)
2HX+
CH3
CH
CH3
CH3
X
X
Example
HCl+CH3
CH2
CH3
Cl
CH3
+CH3
CH
CH3
CH3
Cl
Cl
2HCl
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 and
that on alkyne give tetrahalo alkane.
X2+R
CH2 H
R
X
X
dihaloalkane
2X2
+ TetrahaloalkaneCH3
X
X
X
X
CH3
CH3
CH3
Example
+CH3
CH2
Br2
CH3
Br
Br
1,2-dibromopropane

+CH3
CH
2Cl2 CH3
Cl
Cl
Cl
Cl
H
1,1,2,2-tetrachloropropane
4. From alcohols: When alcohols are treated with PX3
or HX they give alkyl halides.
3PX
2 2
or HX
RCH OH RCH X
Example
CH3
OH
42SOH,NaBr
orHBr.conc
  CH3
Br
  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 I

Alkyl chlorides can also be prepared from alcohols by treating it with PCl5
or SOCl2
.
CH3
CH2
OH + PCl5
 CH3
CH2
Cl + POCl3
+ HCl
CH3
CH2
OH + SOCl2
Py
 CH3
CH2
Cl + SO2
- + HCl
Preparation of alkyl chloride from alcohols by treating it with SOCl2
is the best method as
it gives almost pure alkyl chloride since the by products of the reaction i.e. SO2
and HCl
are in gaseous phase.
5. From free radical substitution of alkyl benzenes
  h,Cl2
CH3 Cl
CH3
NBS
CH3
Br
6. Halide Exchange: Alkyl iodides can be prepared from the corresponding alkyl chlorides
or 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 from
solution by filtration.
Physical Properties of Alkyl Halides
a) 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 as
on 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
> CH2
Cl2
>
CH3
Cl. This is due to accumulation of Cl’s on CH4
, increases the molecular mass and
size, 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 H2
O and the
densities of alkyl chlorides and fluorides are less than that of H2
O. The order of densities
of alkyl halides and H2
O is RI > RBr > H2
O > RCl > RF and the density of polychloro
methane varies as CCl4
> CHCl3
> CH2
Cl2
> H2
O > CH3
Cl. The alkyl halides are in general
insoluble in water.
HSAB (HARD and Soft Acid-Base) Principle: According to hard and soft acid-base
principle of ‘Pearson’, Hard acids are those species, which have more tendency to
accept 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.) Ahard base prefers a
hard acid whereas a soft base prefers a soft acid.
Basicity And Nucleophilicity:
A negatively charged species or the species containing electron pair can function as
nucleophile 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 carbon
while basicity is the ability of the species to remove H+
from an acid. Let us have a
species. B–
. Its function as a nucleophile is shown as
B A A B 
  
And its role as base is indicated as
B H A B H A 
   
The nucleophilicity is determined by the kinetics of the reaction, which is reflected by its
rate constant (k) while basicity is determined by the equilibrium constant, which is reflected
by its Kb
.
The order of nucleophilicity of different species depends on the nature of solvent used. For
instance, let us take F–
, Cl–
, Br–
and I–
with their counter cation as Na+
and see their
nucleophilicity order in different solvents. There are four categories of solvents, namely
non-polar (CCl4
), polar protic (H2
O), polar aprotic (CH3
SOCH3
) and weakly polar aprotic
(CH3
COCH3
).

Polar solvents are able to dissociate the salts i.e., ion-pairs can be separated. On the
other hand, non-polar and weakly polar solvents are unable to dissociate salts, so they
exist as ion-pairs. The ion-pairing is strong when ions are small and have high charge
density.
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 order
of X–
in such solvents would be
I Br Cl F   
  
In polar protic solvents, hydrogen bonding or ion-dipole interaction diminishes the reactivity
of the anion. Stronger the interaction, lesser is the reactivity of anion. F–
ion will form
strong H-bonding with polar protic solvent while weakest ion-dipole interaction will be the
I–
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 basic
character). Thus, the order of nucleophilcity of X–
in polar aprotic solvents would be
F–
> Cl–
> Br–
> I–
NUCLEOPHILIC SUBSTITUTION REACTIONS
A 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 a
larger share of electrons, it has a partial –ve charge and the carbon to which it is bonded
has a partial +ve charge.
2R CH X
 
  X = Cl, Br, I, F
This unequal sharing of electrons causes alkyl halides to undergo substitution and
eliminationreactions. There are two important mechanisms of substitution reaction.
1. A nucleophile is attracted to the partially positively charged atom of carbon. As the
nucleophile approaches the carbon, it causes the C – X bond to break heterolytically.
Nu
X
H
H
H
Nu
H
H
H

2. The C – X bond breaks heterolytically without any assistance from the nucleophile, forming
a 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 a
nucleophilic 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 SN
2 REACTIONS
The rate of nucleophilic substitution reaction such as the reaction of methyl bromide
with hydroxide ion depends on concentration of both the reagents. If the concentration of
the methyl bromide in the reaction mixture is doubled, the rate of SN
reaction doubles. If
the concentration of OH–
is doubled the rate of reaction is also doubled. If the concentration
of both the reactants are doubled, the rate of reaction quadruples.
CH3
Br + HO–
 CH3
OH + 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 the
transition state of the rate determining step of the reaction. From the rate law for the
reaction of methyl bromide with hydroxide ion, we know that both methyl bromide and
hydroxide ion, are involved in the rate determining transition state. The transition state is
BIMOLECULAR; i.e. it involves two molecules. The rate constant k describes how difficult
it is to overcome the energy barrier of the reaction (how hard it is to reach the transition
state). The larger the rate constant, the easier it is to reach the transition state.

The mechanism proposed by Hughes and Ingold for an SN
2 reaction has one step. The
nucleophile 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 the
leaving group, the carbon is said to undergo back side attack. Backside attack occurs
because the orbital of the nucleophile that contains its non-bonding electrons interacts
with the empty anti bonding molecular orbital associated with the C – Br bond. This orbital
has 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 backside
attack.An SN
2 reaction is also called direct displacement reaction because the nucleophile
displaces the leaving group in a single step. A reaction co-ordinate diagram is shown in
figure.
+HO CH3 Br CH3 OH+ Br
-
Relative Rates of SN
2 reactions for several alkyl halides with OH–
Alkyl B rom ide C lass R elative R ate
CH3 Br M ethyl 1200
CH 3
Br
Prim ary 40
CH 3
B r
Prim ary 16
CH3
CH 3
Br
S econdary 1
CH3
Br
CH 3
CH 3
Tertiary
Too slow to
m easure
OH C
H
Br
H H



G

 B rO HC H 3
T.S.
Progress of reaction
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 nucleophile
to get to the back side and therefore will decrease the rate of reaction. Thus the mechanism
explains why substituting methyl groups for the hydrogens in methyl bromide progressively
slows the rate of the substitution reaction.
Steric effect and SN
2 reaction: Effects due to groups occupying a certain volume of
space are called steric effect. A steric effect that decreases the reactivity is called steric
hindrance. Steric hindrance results from groups getting in the way at the reaction site. As
a consequence of steric hindrance, alkyl halides have the following, relative reactivities in
an SN
2 reaction. The steric crowding in tertiary alkyl halide, is too great that they are
unable to undergo SN
2 reactions.

Relative reactivities of alkyl halides in an SN
2 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 SN
2 reaction; the size of the alkyl group is also
important. For example, ethyl bromide and n-propyl bromide are both primary alkyl halide,
but ethyl bromide is more than twice as reactive in an SN
2 reaction because the methyl
group of the ethyl bromide provides less steric hindrance to back side attack than does
the ethyl group of n-propyl bromide.
As the nucleophile approaches the backside of the carbon of methyl bromide the C – H
bond begins to move away from the nucleophile and its attacking electrons. By the time
transition state is reached the C – H bonds are all in the same plane and the carbon is
pentacoordinate (fully bonded to three atoms and partially bonded to two) rather than
tetrahedral. As the nucleophile gets closer to the carbon and the bromine moves further
away from it, the C – H bonds continue to move in the same direction. Evidently the bond
between the carbon and the nucleophile is fully formed and the bond between the carbon
and 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 which
substitution occurs is to picture on umbrella that turns inside out. This is called inversion
of configuration. The carbon at which substitution occurs have inverted its configuration
during the course of the reaction just as an umbrella has a tendencyto invert in a windstorm.
The inversion is known as WALDEN INVERSION, since Paul Walden was the first to
discover that compounds could invert their configuration as a result of substitution reactions.
Because an SN
2 reaction takes place with inversion of configuration, only one product is
formed when an alkyl halide that has the halogen leaving group bonded to a chiral centre
undergoes an SN
2 reaction. The configuration of that product is inverted compared with
the configuration of the alkyl halide. Therefore, proposed mechanism accounts for the
observed configuration of the product.
+ HO Br+
R'
H H
Br
R'
RH
OH
R'
R
H
Br
R'
R
H
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 under
the same conditions, we would find that the alkyl iodide is the most reactive and the alkyl
fluoride is the least reactive.
The onlydifference among these four reactions is the nature of the leaving group.Apparently
I–
is the good leaving group and the F–
ion is the poor. This brings us to an important rule in
organic chemistry that will keep reappearing. The weaker the base, the better is the leaving
group.
The leaving ability depends on basicity because a weak base does not share its electrons
as that as a strong base does. Consequently, a weak base is not bonded as strongly to
the carbon as a strong base would be. Thus a bond between a carbon and a weak base is
more 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 > HF
Because we know that stronger the acid weaker is its conjugate base, the order of basicities
of halide ions is
I–
< Br–
< Cl–
< F–
Further, because weaker bases are better leaving groups, the halide ions have the following
relative leaving abilities.
I–
> Br–
> Cl–
> F–
As a consequence of relative leaving abilities of the halides, alkyl halides have the following
relative reactivities in an SN
2 reactions.
RI > RBr > RCl > RF
Nucleophile: 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 the
difference between a base and a nucleophile?
A base shares its lone pair with a proton. Basicity is a measure of how strongly the base
shares 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 tendency
of the conjugate acid of the base to lose a proton. A nucleophile uses its lone pair of
electrons to attack on electron deficient atom, other than a proton. NUCLEOPHILICITY is a
measure of how readilythe nucleophile is able to attack such an atom. It is measured byrate
constant (k). In the case of an SN
2 reaction, nucleophilicity is a measure of how readily the
nucleophile attacks an sp3
hybridised carbon bonded to a leaving group.

In comparing molecules with the same attacking atom, there is generally a direct
relationship between basicity and nucleophilicity. Stronger bases are better nucleophiles.
For example, a compound with negatively charged oxygen is a stronger base and a better
nucleophile than a compound with a neutral oxygen.
In comparing molecules with attacking atom of approximately the same size, the stronger
bases are again the better nucleophiles. The atoms across the second row of the periodic
table have approximately the same size. If hydrogens are attached to the second row
elements the resulting compounds have the following relative acidities.
CH4
< NH3
< H2
O < HF
Consequently, the conjugate bases have the following relative basicities and
nucleophilicites. For example, the methyl anion is the strongest base as well as the best
nucleophile.
CH3
–
> NH2
–
> OH–
> F–
In comparing molecules with attacking atoms that are very different in sizes, the direct
relationship between basicity and nucleophilicity is retained if the reaction occurs in the
gas phase. If, however, reaction occurs in solvent the relationship between basicity and
nucleophilicity 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 increasing
size 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 becomes
inverted, as basicity increases, nucleophilicity decreases. Thus I–
ion which is the weakest
base of the halogen family, is the poorest nucleophile of the family in an aprotic solvent
and 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 between
nucleophilicity and basicity? When a –vely charged species are placed in a protic solvent,
the solvent molecules arrange themselves so that their partiallypositively charged hydrogen
points towards the –vely charged species. An aprotic solvent does not have a partially
positively charged hydrogen.
O
H H





 O
H
H






O
H
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 nucleophilicity
in 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 because
at least one of the ion dipole interactions must be broken before the nucleophile can
participate in an SN
2 reaction. Weak bases interact before the nucleophile can participate
in an SN
2 reaction weak bases interact weakly with protic solvents, strong bases interact
more strongly because they are better at sharing their electrons. It is, therefore, easier to
break the ion-dipole interactions between an iodide ion and the solvent than between the
more 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 CH3
I in Methanol
increasing nuleophilicity
RS
-
I
-
CN
-
CH3O
-
Br
-
NH3
Cl
-
F
-
CH3OH

Because there are many different kinds of nucleophiles, a wide variety of organic
compounds can be synthesized by means of SN
2 reactions. The following reactions show
just a few of the many kinds of organic compounds that can be synthesized in this way.
CH3
CH2
Cl + HO–
 CH3
CH2
OH + Cl–
CH3
CH2
Br + HO–
 CH3
CH2
OH + Br–
CH3
CH2
I + RO–
 CH3
CH2
OR + I–
CH3
CH2
Br + RS–
 CH3
CH2
SR + Br–
CH3
CH2
F + NH2
–
 CH3
CH2
NH2
+ F–
CH3
CH2
Br + C–
 CR  CH3
CH2
C  CR + Br–
CH3
CH2
I + CN–
 CH3
CH2
CN + I–
STEREOCHEMISTRY OF SN
2 REACTION
As we have seen that an SN
2 reaction the nucleophilie attacks from the backside, that
is, from the side directly opposite to the leaving group. This mode of attack causes a
change in configuration of the carbon atom that is the target of the nucleophile. As the
displacement takes place, the configuration of the carbon atom under attack inverts it is
turned inside out in such a way that an umbrella is turned inside out in a strong gale.
HO Br
R"
R
R'
OH C
R
Br
R' R"


OH
R"
R
R'


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 to
have the configuration and rotation shown here.
CH3
H Br
H13C6
R(-)-2-bromooctane
CH3
Br H
H13C6
S(+)-2-bromooctane
 
25.34][ 25
D
25
D[ ] 34.25
   
The S-2-octanol is also chiral. The configuration and rotation of the 2-octanol enantiomers
have also been determined.

CH3
H OH
H13C6
CH3
OH H
H13C6
 
90.9][ 25
D  
90.9][ 25
D
S-(+)-2-octanolR-(-)-2-octanol
When R-(–)-2-bromooctane reacts with NaOH, the only substitution product is obtained
from the reaction is S-(+)-2-octanol.
HO Br
H13C6
CH3
H
OH C
CH3
Br
H C6H13


OH
CH3
C6H13
H


 
25.34][ 25
D
R-(-)-2-bromooctane
Enantiomeric purity = 100%
 
90.9][ 25
D
S-(+)-2-octanol
Enantiomeric purity = 100%
Br
There still remains a problem. We need to relate the configuration of the reactant and the
product. We know that the direction of rotation (optical) and the configuration of two
different compounds are not necessarily related, for they may have the same sign of
rotation but may have different configurations. The experimental proof of Walden inversion
was obtained by a series of experiments in which optically active alcohol was converted
into its enantiomer having a rotation opposite in sign but nearly equal in magnitude to that
of the original alcohol.
Thus in the series of reactions on an optically active (+) alcohol formation of an ester with
4-methylbenzenesulphonyl chloride (Tosyl chloride) is known not to break the C – O bond
of 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 tosylate
must have the same configuration as the original alcohol.
OH
PhH2C
H
Me
OSO2C6H4CH3
H2CPh
H
Me
OH
O
CH3
O
O
CH2Ph
H
Me
CH3
O
6 4 2p MeC H SO Cl

OH
CH2Ph
H
Me
CH3
OH
O

Reaction of this ester with CH3
COO–
is known to be a displacement in which ArSO3
–
(Ar =
p – MeC6
H4
) 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. Alkaline
hydrolysis 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 of
an acetate in which the alcohol oxygen atom is O18
labelled fails to result in the latter
replacement, thus showing that alkyl oxygen bond of the acetate is not broken during its
hydrolysis).As it is found to be the mirror image of the starting material – opposite direction
of the optical rotation – an inversion of configuration must have taken place during the
series of reactions and can have occurred onlyduring reaction of MeCO2
–
with the tosylate.
Reaction of this tosylate with a number of nucleophiles showed that inversion of
configuration occurred in each case; it may thus be concluded with some confidence that
it occurs on reaction with Br–
to yield the bromide, i.e., that the bromide like acetate has
the opposite configuration to the original alcohol.
The general principle that bimolecular (SN
2) displacement reactions are attended by
inversion of configuration has been established in an elegant and highly ingenious
experiment, in which an optically active alkyl halide undergoes displacement by the same
though isotopically labeled – halide ion as a nucleophile e.g. 128
I–
on (+) – 2-iodooctane.
I
H
H13C6
CH3
I C
C6H13
I
H3C H


I
H
C6H13
CH3



I128 +
128
128 + 
I
( )
(+)-2-iodooctane
The displacement was monitored by observing the changing distribution of I128
between
the inorganic (sodium) iodide and 2-iodooctane and it was found under these conditions,
to be second order overall.
If the inversion takes place, as SN
2 requires. The optically activity of the solution will
decline zero i.e., racemization will occur. This will happen because inversion of the
configuration of a molecule of (+) 2-iodooctane results in the formation of a molecule of its
mirror image (-)-2-iodooctane. Which pair off with a second molecule of (+)-2-iodooctane
to form a racemate (±); thus, the observed rate of racemisation will be twice the rate of
inversion.
Illustration 2: Explain why an SN
2solvolysis (where solvent is the nucleophile) appears
to follow a first order rate law, rather than a second order one.
Solution:Because an SN
2 solvolysis reaction is bimolecular, it would follow the following
rate law Rate = k[RX] [solvent]. However, the “concentration” of solvent is very large, and
greatly 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 of
the following in aprotic solvent.
i) CN

CH2
Br
ii)
PhS
CH2
Cl
CH3
CH3
Solution: i)
CN
ii)
Ph
S CH3
CH3
SN
1 REACTIONS
Given our understanding about SN
2 reactions, if were to measure the rate of reaction of
tert-butyl bromide with water, we would expect a relatively slow substitution reaction since
water is a poor nucleophile and tert-butyl bromide is sterically hindered to attack by a
nucleophile.. However, we actually discover that the reaction is surprisingly fast. It is, in
fact over one million times faster than the reaction of methyl bromide – a compound with
no sterric hindrance with water. Clearly, the reaction must be taking place by a mechanism
different from that of the SN
2 reaction.
As we have seen, a study of the kinetics of a reaction is one of the first step undertaken
when investigating the mechanism of a reaction. If we were to investigate the kinetics of
the reaction of tert-butyl bromide with water, we would find that doubling the concentration
of the alkyl halide doubles the rate of reaction. We would also find that changing the
concentration of nucleophile has no effect on the rate of the reaction.
 OH2
CH3
Br
CH3
CH3
CH3
OH
CH3
CH3
Br H
 
Rate = K [Alkyl Halide]
Because the rate of reaction depends upon the concentration of only one reactant, it is a
first order reaction. An SN
1 reaction has two steps, the carbon-halogen bond breaks
heterolytically with the halogen retaining the previously shared electron pair. In the second
step, the nucleophile reacts rapidly with the carbocation formed in the first step.
-
rBBr
CH3
CH3
CH3
CH3
CH3
CH3
slow
fast
2OH OH2
CH3
CH3
CH3CH3
CH3
CH3
+ OH
CH3
CH3
CH3

 H

From the observation that the rate of an SN
1 reaction depends onlyupon the concentration
of the RX, we know that the first step is the slow and rate determining step. Because the
nucleophile is not involved in the r.d.s., its concentration has no effect on the rate of the
reaction. If you look at the reaction – co-ordinate diagram, you will be able to see why
increasing the rate of the second step will not make an SN
1 reaction go any faster.
Br
CH3
CH3
CH3
OH
CH3
CH3
CH3
CH3
CH3
CH3
OH2
CH3
CH3
CH3
0
G
0
G

 H
Br
-
H2O
Progress of Reaction
FreeEnergy
Second, a carbocation is formed in the slow step of an SN
1 reaction. Because a 3°
carbocation is more stable and therefore, easier to form than a secondary carbocation
which in turn is more stable to form than a 1° carbocation. 3° alkyl halides are more
reactive than a 2° alkyl halide which in turn more reactive than a 1° alkyl halide in an SN
1
reaction.
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
CH3
Br
1° 1.00
CH3 – Br Methyl 1.05
Relative Rates of Alkyl Halides in an SN
1 reaction
3°RX > 2°RX > 1°RX >CH3
X
Actually 1° carbocation and CH3
+
are so unstable that primary RX and CH3
X do not
undergo SN
1 reactions. (The very slow reaction reported for ethyl bromide and methyl
bromide in table are SN
2 reaction).

The positively charged carbon of the carbocation is sp2
hybridised, and the three bond
connected to this carbon are in the same plane. In the 2nd
step of the SN
1 reaction, the
nucleophile 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 SN
1 reaction of an alkyl halide in which the leaving group is attached to the chirality
centre leads to the formation of two stereoisomers; attack of the nucleophile on one side
of the planar cabrocation forms one stereoisomer and attack on the other side produces
the other isomer.
R'
R H
Br
 OH2
R'
RH
OH

R'
R H
OH
HBr
In the reaction of (S) – 2-bromobutane with water two substitution products are formed
one product has the same relative configuration. In an SN
1 reaction the leaving group
leaves before the nucleophile attacks. This means that the nulceophile is free to attack
from either side of the planar carbocation.
R H
Br
CH3
  conditions1S
2
N
OH 
RH
OH
CH3
R H
OH
CH3
(R)-2-butanol inverted (S)-2-butanol retended

Although one might expect that equal amounts of two isomers would form in an SN
1
reaction but a greater amount of the inverted product is obtained in most cases. Typically,
50 – 70% of the product of an SN
1 reaction is the inverted product. If the reaction leads to
the formation of equal amounts of the two products, the reaction is said to take place with
complete racemization, when more of the inverted product is formed, the reaction is said
to take place with partial racemization.
Paul Winstein first postulated that dissociation of alkyl halide initially results in the formation
of an intermate ion pair. In an intimate ion pair, the bond between the carbon and the
leaving group has broken but the cation and anion remains next to each other. This species
then forms an ion pair in which one or more solvent molecules have come between the
cation and the anion. This is called the solvent separated ion pair. Further separation
between the two results in the dissociated ions.
R Br

 

R Br R  BrR Br
Intimate ion pair Solvent separated ion
The nucleophile can attack any of these species. If the nucleophile attack on only the
completely dissociated carbocation, the product will be completely racemised. If the
nucleophile attacks either the intimate ion-pair or the solvent separated ion-pair, the leaving
group will be in a position to partially block the approach of the nucleophile to that side of
the carbocation. As a result, more of the product with the inverted configuration will be
obtained. If the nucleophile attacks the undissociated molecule, it will be an SN
2 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 carbocation
Br-
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 SN
1 reaction and from an SN
2
reaction is a little easier to visualize in the case of cyclic compounds. When cis-1-bromo-
4-methylcyclohexane undergoes an SN
2 reaction, only the trans product is obtained
because the carbon bonded to the leaving group is attacked by nucleophile only on its
back side.
CH3
H
Br
H
1-bromo-4-methylcyclohexane
-
Br
CH3
H
H
OH
4-methylcyclohexanol
2S
HO
N
 


When however, cis-1-bromo-4-methylcyclohexane undergoes SN
1 reaction, both the cis
and the trans products are formed because the nucleophile can approach the carbocation
intermediate from either side.
CH3
H
Br
H

CH3
H
H
OH
4-methylcyclohexanol
1S
HO
N
 

CH3
H
OH
H
HBr
cis-isomer
The Leaving Group: Because the r.d.s. of an SN
1 reaction is dissociation of the alkyl
halide to form a carbocation, two factors affects the rate of SN
1 reaction, the case with
which the leaving group departs from the carbon and the stability of the carbocation that is
formed. In the preceding section, we saw that the tertiary alkyl halides are more reactive
than 2° and which in turn more reactive than 1° alkyl halide. This is because more
substituted carbocation is, more easily it is formed. But about a series of alkyl halides
with different leaving groups that dissociate to form the same carbocation? The answer is
same for the SN
1 reaction as for as for the SN
2 reaction. The weaker the bases, the less
lightly it is bonded to the carbon and the easier it is to break the C – halogen bond. So an
alkyl iodide is the most reactive of the alkyl halides in both SN
1 and SN
2 reactions.
Relative reactivities of alkyl halides in an SN
1 reaction.
RI > RBr > RCl > RF
The Nucleophile: The nucleophile traps the carbocation that is formed in the r.d.s. of the
SN
1 reaction. Because the nucleophile comes into play after the r.d.s., the reactivity of
the nucleophile has no effect on the rate of the SN
1 reaction.
In some SN
1 reactions, the solvent is the nucleophile. For example, relative rates given in
table are for the reaction of alkyl halides with water serves as both the nucleophie and as
the solvent. When the solvent is the nucleophile the reaction is called the solvolysis
reaction.
Illustration 4: What happens when (–)–2– bromooctane is allowed to react with
sodium hydroxide under SN
2 conditions.
Solution:
N
NaOH
S 2
Br
CH3
H
C6H13
OH
CH3
H
C6H13
(-)-2-Bromooctane (+)-2-Octanol
In Fisher projection the above reaction can be represented as follows:
N
NaOH
S 2
H
CH3
C6H13
Br OH
CH3
C6H13
Br

We see that OH group has not taken the position previously occupied by Br–
, the alcohol
obtained has a configuration opposite to the bromide. A reaction that yields a product
whose configuration is opposite to that of the reactant is said to proceed with inversion of
configuration.
N
NaOH
S 2
Br
CH3
H
C6H13
OH
CH3
H
C6H13
(-)-2-Bromooctane (+)-2-Octanol
Illustration 5: Alkyl halides are hydrolysed to alcohol very slowly by water, but rapidly
by silver oxide suspended in boiling water.
Solution:R––X + Ag+  [R––X––Ag]+
[R––X––Ag]+ Slow R+
+AgX
R+
+ OH–
 ROH
Heavy metal ions, particularly silver ions, catalyse SN
1 reaction and mechanism involves
a fast pre- equilibrium step (the silver ions has an empty orbital).
SN
I MECHANISM: RETENTION OF CONFIGURATION
Despite what has been said above about displacement reactions leading to inversion of
configuration, to racemisation, or to a mixture of both, a number of cases are known of
reactions that proceed with actual retention of configuration, i.e., in which the starting
material and product have the same configuration. One reaction in which this has been
shown to occur is in the replacement of OH by Cl through the use of thionyl chloride,
SOCl2
:
(I)
OH
Me
H
Ph
2SOCl
 Cl
Me
H
Ph
2SO 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 SN
2 mode for this would lead to inversion of
configuration in the product, which is not observed.

Carrying out the reaction under milder conditions allows of the isolation of an alkyl
chlorosulphite, ROSOCl (III), and this can be shown to be a true intermediate. The
chlorosulphite is formed with retention of configuration, the R – O bond not being broken
during the reaction. The rate at which the alkyl chlorosulphite intermediate (III) break down
to the product, RCl (II), is found to increase with increasing polarity of the solvent, and
also with increasing stability of the carbocation R+
: an ion pair, R OSOCl (IV), is
almost 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 occur
on the same side of R+
from which –
OSOCl departed, i.e., with retention of configuration:
(III)
O
Me
H
Ph
S O
Cl
slow

Me
Ph H
O S
Cl
O Me
Ph H
Cl Cl
Me
H
Ph
(IV) (V) (IIa)
Whether the breaking of the C – O and the S – Cl bonds occurs simultaneously, or
whether 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 of
pyridine, 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
is
converted by pyridine into C5
H5
NH+
Cl–
and Cl–
, being an effective nucleophile, attacks (III)
‘from the back’ in a normal SN
2 reaction with inversion of configuration.
O
Me
H
Ph
S O
Cl
+
2SO +ClC
Me
Ph H
Cl O S
O
Cl
Cl
Me
H
Ph
  
Cl +
(III) (IIb)
NEIGHBOURING GROUP PARTICIPATION
Substitution with Retention of Configuration (Outlines of Neighbouring Group
Participation) and Anchimeric Assistance
Often the rate of a reaction is greater than expected and retention of configuration at a
chiral carbon may be observed and not inverted or racemized. This usually happens when
there is a group (Z) in the substrate with an unshared pair of electrons in a position b to the
leaving group which can play a transient part in the reaction. These assisted reactions by
a more-or-less remote functional group is termed the neighbouring group mechanism, and
involves two successive inversions of configuration. In other words, the neighbouring group
effect involves essentially two SN
2 substitutions, each causing inversion, the net result

being retention of configuration. In the first step the neighbouring group Z acting as a
nucleophile pushes out the leaving group but still retains attachment to the molecule. In
the subsequent step the external nucleophile pushes out the neighbouring group.
Step 1

R
R
Z
R L
R
R
R
R
R
Z
+
L
R
R
R
R
Z
+
Nu Step 2

R
R
Z
R Nu
R
Often when the neighbouring group effect is operative, one may not get the substitution
product but a rearrangement. In these situations the nucleophile does not attack the
carbon from which the leaving group had left but instead attacks the carbon to which the
neighbouring group was originally linked. Thus the alkaline hydrolysis of (I) affords the
rearranged product (III). The cyclic intermediate (II) formed after the neighbouring group
participation by nitrogen undergoes attack at the methylene (–CH2
–) group rather than (–
CHEt) because of less steric crowding on the former carbon atom.
CH
CH3
Z
CCH3 L
CH3
CH
CH3
C CH3
CH3
Z
+ Nu
C
CH3
H
CZ CH3
CH3
Nu
Et2N CHEt Cl Cl
slow


 Et2N CHEt
OH
 Et2N
OH
Et
(I) (II) (III)
In such cases both substitution and rearrangement occur.
ELIMINATION REACTIONS OF ALKYL HALIDES
The E2
Reaction: There are two important mechanism for elimination. The reaction of
ethyl bromide with hydroxide ion is an example of an E2
reaction. It is a second-order
reaction because the rate of reaction depends on the concentration of both ethyl bromide
and hydroxide ion.

CH3
CH2
Br + HO–
 CH2
= CH2
+ H2
O + Br–
Rate = K [Alkyl halide] [Base]
The rate law tells us that alkyl bromide and hydroxide ion both are involved in the transition
state of the rate determining step of the reaction. The following mechanism agrees with
the observed second order kinetic.
The E2
reaction is a concerted one step reaction. The proton and the bromide ion are
removed in the same step.

H O H
Br

 

2H O Br
 
In an E2
reaction, a base removes a proton from a carbon adjacent to the carbon bonded
to the halogen.As the proton is removed, the electrons, the hydrogen shared with carbon
moves forward to the carbon bonded to the halogen. As these electrons move in the
halogen leaves taking its bonding electrons with it. The removal of the proton and the
halide ion from alkyl halide is known as dehydrogenation.
The carbon to which the halogen is attached is called the  -carbon.An adjacent is called
a  -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.
_
BrBH 
H
R Br
R
R
R
B
carbon
carbon
2-bromopropane has two  -carbon atoms from which a proton can be removed in an E2
reaction. Because the two carbons are identical, the proton can be removed from either
one. The product is propene.
3CH O CH3
Br
CH3
CH3
CH2
3CH OH Br 
How one will known which product will be formed in greater yield? You must determine
which 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 substituents
bonded to the sp2
carbons, the greater that number, the more stable the alkene. In the
transition state leading to an alkene, the C – H and C – Br bonds are partially, broken and
the 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 state
allows the formation of 2-butene faster than 1-butene.


CH3 CH
Br
CH CH3
H
OCH3


H2C CH
H
Br
CH2 CH3
OCH3




T.S. leading to 2-butene more stable T.S. leading to 1-butene less stable
The difference in rate of formation of two alkenes is not very great, consequently, both
products are formed but the more stable alkene is the major product of the elimination
reaction. Thus an E2
reaction is regio-selective; i.e., more of one isomer is formed than
the 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 alkyl
substitutents attached to sp2
carbon. It is more stable alkene and is the major product of
the elimination reaction.
G°
Progress of reaction
1-butene
2-butene
2-bromobutane
2H O
HO 
(70%)
CH3
CH3
Br
CH3
CH3
CH3
CH3
 CH2
CH3
CH3
(30%)
Saytzeff Elimination vs Hoffmann Elimination: Alkyl halides have the following relative
reactivities in an E2
reaction, because elimination from a 3° alkyl halide typically leads to
a more highly substituted alkene. But one must exercise some care in using Saytzeff’s
rule to predict the major product of the elimination reaction because the most substituted
alkene 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 remove
the most accessible  -hydrogen. In the following, it is easier for the bulky tert-butoxide
ion to remove one of the more exposed terminal hydrogens which leads to the less
substituted alkene. Because the less substituted alkene is the one that is the more
easily formed, it is the major product of the reaction.
  OHC)CH( 33
(28%)
CH3
CH3
Br
CH3
 CH3
CH3
CH3
O CH3
CH3
CH3
CH2
CH3
CH3
(72%)
sterically hindered base
Saytzeff product Hofmann product
The data in table below show that when an alkyl halide undergoes an E2
reaction with a
variety of alkoxide ion, as the size of the base increases, the % of the less substituted
alkene (i.e. Hofmann product) increases.
Effect of steric properties of the base on the distribution of products in an E2 reaction.
 -
ROCH3
CH3
CH3
CH3
Br 
CH3
CH3 CH2
CH3CH3
CH3 CH3
CH3
Base Saytzeff Product More
substituted product
Hoffmann product Less
substituted product
CH3
O
79% 21%
O
CH3
CH3
CH3
27% 73%
CH3
O
CH3
CH3
19% 81%
CH3
O
CH3CH3 8% 92%
Although the major product of the E2
dehydrohalogenation of the alkyl chlorides, alkyl
bromides and alkyl iodides is normally the most substituted alkene, the major product of
the E2
dehydrohalogenation of alkyl of fluorides is the least substituted alkene (i.e. Hofmann
elimination).

(30%)
  
OHCH3 3
OCH
CH3 CH3
F
CH2 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 the
proton. Departure of the halogen and its bonding electrons prevents the build up of –ve
charge on the carbon that is loosing the proton, giving the T.S. alkene character. Of the
halogens 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 less
tendency to leave than do the other halide ions. As a result, -ve charge develops on the
carbon that is losing a proton, giving the T.S. a carbanion character rather than an alkene
like T.S. Normally, carbanion T.S. are unstable, but in this case the carbanion T.S. is
stabilized by the strongly electron withdrawing fluorine.


CarbanonicT.S. leading to 2-pentene
(less stable)
H2C CH3
F
H
OCH3

H2C CH3
F
H
OCH3




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 is
more stable than the T.S. leading to 2-pentene which has the developing –ve charge on
the 2° carbon. Because the T.S. leading to 1-pentene is more stable, 1-pentene is formed
more rapidly and is therefore major product of the reaction.
In the following reaction, Saytzeff rule does not lead to the more stable alkene, because
the rule does not take into account the fact that the conjugated double bonds are more
stable than the isolated double bonds. Since the conjugated alkene is the more stable
alkene, it is the one that is most easily formed and is therefore, the major product of the
reaction. So, if there is a double bond or a benzene ring in the alkyl halide, do not use
Saytzeff rule to predict the major product of the elimination reaction.

Minor product
 

HO
CH2
CH3
CH3
 CH2
CH3
CH3
Major product
CH2
CH3
CH3
Br
Minor product
 

HO

Major product
Br CH3
CH3
CH3
CH3
CH3
CH3
We can summarise by saying that the major product of the E2
reaction is the most
substituted 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)
3OCH CH3
CH3
CH3
Cl
b)
3CH X  CH3
CH3
CH3
O
Solution: a)
3OCH CH3
CH3
CH3
Cl CH3
CH2
CH3
3OCH (nucleophile) can’t attack 3° carbon having high electron - densityhence elimination
takes place giving alkene.
b)
3CH X  CH3
CH3
CH3
O CH3
CH3
CH3
O
CH3
Nucleophilic attack on methyl carbon is possible giving ether (Williamson synthesis).

Illustration 7: Explain the fact that a small amount of NaI catalyzes the general reaction
RCl + RO–
Na+
 ROR + NaCl
Solution: With I–
the overall reaction occurs in two steps, each of which is faster than the
uncatalysed reaction.
Step 1: RCl + I–
 RI + Cl–
This step is faster because I–
, a soft base has more nucleophilicity than OR–
, a hard base
Step 2: RI + RO:–
 ROR + I–
This step is faster because I–
is a better leaving group than Cl–
THE E1
REACTION
The 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 called
an E1
reaction.
2H OCH3
CH3
CH3
Br 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 reaction
therefore, there must be at least two steps in the reaction.
An E1
reaction is the two step reaction. In the first step, alkyl halide dissociates
heterolytically. This is the r.d.s. of the reaction. In the second step of the reaction, the
base 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 of
the base which comes into play in the second step of the reaction, has no effect on the
rate of the reaction.
Mechanism
CH3
CH3
CH3
Br CH3
CH3
CH3
Br
-slow
CH3
CH3
H
H3O
+fast CH3
CH2
CH3
OH2

When two elimination products can be formed, the major product is generally the one
obtained by following Saytzeff rule.
Minor product

Major product
CH3
CH3
Cl
CH3
OH
CH3
CH3
CH3

CH3
CH2
CH3
Because the first step is the r.d.s. of an E1
reaction depends on both the ease with which
the leaving group leaves, and the stability of the carbocation that is formed. Thus the
relative reactivities of the series of alkyl halides with the same leaving group parallel the
relative stabilities of carbocations. 3° benzylic alkyl halide is the most reactive alkyl halide
because the 3° benzylic carbocation, is the most stable carbocation, is the easiest to
form.
Relative Reactivities of Alkyl Halides in an E1
Reaction = Relative Stabillities of
the 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, the
rearrangement 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° benzylic
carbocation.
  OHCH 3
CH3
CH3
Cl
CH3
CH3
CH3
CH3
   shiftMethanide2,1
CH3
CH3
CH3
CH3
CH3
CH3
-H
+
Substitution Versus Elimination Reactions
We know that an alkyl halide can undergo four types of reactions; SN
1, SN
2, E1
and E2
. A
given alkyl halide under the given conditions will follow which pathway, can be decided in
following manner.
The first thing you must look at is the alkyl halide, is it 1°, 2° or 3°. If the reactant were a
primary alkyl halide, it would undergo E2
/SN
2 reactions (as their their carbocations are not
stable). If the reactant is a secondary or a tertiary alkyl halide, then it can undergo E1
/SN
1
or E2
/SN
2 reactions depending upon reaction conditions. E2
/SN
2 reactions are favoured by
a high concentration of a good nucleophile/strong base, whereas a poor nucleophile/weak
base favour E1
/SN
1 reactions.

Once you have decided whether the conditions will favour E2
/SN
2 reactions or E1
/SN
1
reactions, then you should decide how much of the product will be substitution and how
much will be the elimination product. The relative amount of substitution and elimination
product can be decided again on the basis of structure of alkyl halide (i.e. 1°, 2° or 3°) and
on the nature of the nucleophile/base. Relative reactivities of alkyl halides in various reactions
are:
In an SN
2 reaction: 1° > 2° > 3°
In an E2 reaction: 3° > 2° > 1°
In an SN
1 reaction: 3° > 2° > 1°
In an E1 reaction: 3° > 2° > 1°
For instance, propyl bromide when treated with methoxide ion in methanol can undergo
either substitution reaction give methyl propyl ether or elimination reaction to give propene.
The major product of the reaction would be substitution product.
CH3
CH2
CH2
– Br + CH3
O–
3CH OH
3 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, the
nucleophile will have difficulty getting to the back of  -carbon and thus, elimination product
will predominate. For example,
3CH OH
3CH O
 
CH3
Br
CH3
CH3
CH3
CH2OCH 3 3CH OH Br
 
1-bromo-2-methyl propane (40%) (60%)
A secondary alkyl halide can form both substitution and elimination products, whose
relative amount depend on the base strength of the nucleophile. The stronger and bulkier
the base, greater will be the percent of the elimination product.
2 5C H OH
3 2CH CH O
  3 2 2 5CH CH CH C H OH Cl
   
(25%) (75%)
CH3
Cl
CH3 CH3
OCH2CH3
CH3
CH3
OCOCH3
CH3 Cl

(100%)
3 2CH CO
(in acetic acid)

Increasing the temperature at which the reaction is carried out increases the rates of both
the substitution and elimination reaction but increase in the rate of elimination reactions
is more than that of substitution reaction. Thus, if the substitution product is desired, the
reaction should be carried out a low temperature and high temperature promotes elimination
product.
A tertiary alkyl halide is least reactive towards SN
2 reaction but most reactive towards E2
reaction. Thus, only elimination product is formed.
ARYL HALIDES
Benzene and its homologues react with halogens to produce either addition or substitution
products.
i) Addition Compounds: These compounds are obtained byexposing the mixture of aromatic
hydrocarbon and the halogen to direct sunlight, e.g., benzene hexachloride (BHC), C6
H6
Cl6
;
benzene hexabromide, C6
H6
Br6
etc.
ii) Substitution products: Two types of halogen substituted products are known.
a) Nuclear halogen substitution product: In these products, the halogen is
linked directly to the carbon of the benzene nucleus. These products are generally called
arylhalides.
Cl Br CH3
Br
Cl
Cl
chlorobenzene bromobenzene
1-bromo-4-methylbenzene 1,4-dichlorobenzene
b) Side chain halogen substitution products: In these products, the halogen is
linked to the carbon atom of the side chain.
Cl Cl
Cl
Cl
Cl
Cl
(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. If
more than one halogen is present their positions in the ring are indicated by numbers or
appropriate prefixes, ortho, meta, para.

Br
Br
CH3
Cl
CH3
Cl
CH3
1,3-dibromobenzene or
m-dibromobenzene
1-chloro-2-methylbenzene
or o-chlorotoluene
2-chloro-1,4-dimethylbenzene
GENERAL METHODS OF PREPARATION
i) 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 halogen
carriers such as iron, iodine or aluminium chloride at room temperature in absence of
direct sunlight.
Fe
6 6 2 6 5C H Cl C H Cl HCl  
For further halogenation, more halogen is used
Fe
6 6 2 6 5 2
Chlorobenzene o and p-
Dichlorobenzene
C 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
 Fe
2Cl
CH3
Cl

CH3
Cl
Mixture of o- and p-chlorotoluene
Iodo derivatives cannot be obtained by direction reaction with iodine as the reaction is
reversible.
C6
H5
+ I2
 C6
H5
I + HI
Iodo-derivative can be obtained if the reaction is carried out in presence of an oxidizing
agent, e.g., iodic acid, or nitric acid, etc. The oxidizing agent oxidizes HI to iodine and
thus, the reaction moves to proceed to the right
2HI + [O]  H2
O + I2

ii)From diazonium salts: Aryl halides can be obtained most satisfactorily by the
decomposition of aryl diazonium salts in presence of copper halide solution dissolved in
the corresponding halogen acid, the diazo group is replaced bya halogen atom (Sandmeyer
reaction).
heat
HCl/CuCl
 
Cl
chlorobenzene
Br
bromobenzene
I
iodobenzene
F
fluorobenzene
heat
HBr/CuBr
 
heat
KI
 
heat
NaBF4
 
N2
+
Cl
-
Benzene
diazonium
chloride
Sandmayer reaction
Balzsehiencn reaction
Iodo compounds may be obtained by boiling the diazonium salt solution with aqueous
potassium iodide.
iii) Hunsdiecker reaction: Aryl bromides are obtained by heating the silver salts of aromatic
acid (in CCl4
or xylene) with bromine.
 2Br
O
-
Ag
+
O
Br
2COAgBr 
iv) Decarboxylation of halogenated acids: Sodium salts of halogenated acid when heated
with soda lime produce aryl halides.
BrC6
H5
COONa + NaOH  C6
H5
Br + Na2
CO3

Properties: Aryl halides are colourless stable liquids with pleasant odour. These are
insoluble in water but readily miscible with organic solvents. Most of them are steam
volatile, heavier than water. Their boiling points are higher than corresponding alkyl halides.
The boiling points rise gradually from fluoro to iodo compounds.
NUCLEOPHILIC AROMATIC SUBSTITUTION
Normally, alkenes are unaffected by treatment with nucleophiles, and it is no surprise to
find that benzene also survives treatment with strong nucleophiles. In particular, no reaction
occurs when simple substituted benzenes are treated with base.
Nu No reaction
Cl
No reaction
Na
+ -
OCH3
HOCH3
But a few specially substituted benzenes do undergo reaction in base. For example, 2, 4-
dinitrochlorobenzene is converted into 2,4-dinitrophenol bytreatment with sodium hydroxide
in water.
Cl
NO2
NO2
NaOH/H2O
warming
OH
NO2
NO2
(95%)
The nitro group is essential, and it must be substituted in the right place relative to the
chlorine. 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
Cl
NO2
NO2
Na
+-
OCH3
HOCH3, 50°C
OCH3
NO2
NO2
115,000
Cl
NO2
Na
+-
OCH3
HOCH3, 50°C
OCH3
NO2
3.4

Cl
NO2
Na
+-
OCH3
HOCH3, 50°C
OCH3
NO2
1.0
Cl
NO2
Na
+-
OCH3
HOCH3, 50°C
or
Cl
No reaction
0
Neither an SN
2 nor an SN
1 reaction seems likely. Attack from the rear is impossible, as
the ring blocks the path of any entering nucleophile, and ionization would produce a most
unstable carbocation.
C L
Nu
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 to
generate a resonance – stabilized anionic intermediate.
Cl
NO2
NO2
HOCH3
OCH3
H3CO
NO2
NO2
Cl H3CO
NO2
NO2
Cl H3CO
NO2
NO2
Cl H3CO
NO2
NO2
Cl
The nitro groups stabilize the negative charge through resonance, and therefore will exert
their effect only when they are substituted at a carbon that helps to bear the negative
charge. This stabilization by nitro is the reason that o- or p-nitrochlorobenzene undergoes
the reaction, but the meta isomer does not.
Cl
NO2
Na
+-
OCH3
H3CO
N
+
Cl
OO
HOCH3
H3CO
N
+
Cl
OO
Para addition

Cl
NO2
NO2
Na
+-
OCH3
H3CO
N
+
Cl
O
OHOCH3
H3CO
N
+
Cl
O
O
Meta addition – here the nitro groups do not help stabilize the negative charge
Cl
NO2
Na
+-
OCH3
H3CO Cl
NO2
HOCH3
H3CO Cl
NO2
H3CO Cl
NO2
The aromatic system can be regenerated by reversal of the attack of methoxide or by loss
of chloride to give 2,4-dinitroanisole.
Cl
NO2
NO2
HOCH3
(a)
Na
+-
OCH3
H3CO
NO2
NO2
Cl
HOCH3
(b)
OCH3
NO2
NO2
(b)(a)
There are similarities between nucleophilic aromatic substitution and its more usual
electrophilic counterpart. Each involves the formation of a resonance – stabilized
intermediate, and each involves a temporary loss of aromaticity that is regained in the
final 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 in
the electrophilic substitution and strongly accelerating in the nucleophilic reaction, to
keep the two mechanisms distinct in your mind.
BENZYNEMECHANISM
We mentioned the lack of reactivity of halobenzenes with bases. It look the activation of a
nitro group for substitution to occur for (fig). In truth, if the base is strong enough, even
chlorobenzene will react. In the very strongly basic medium potassium amide in liquid
ammonia, chlorobenzene is converted into aniline.
The mechanism is surely not an SN
2 displacement, and there is no reason to think
nucleophilic addition would be possible to an unstabilized benzene ring.

Cl
Na
+-
NH2
NH3
NH2
+
NH2
Which reactions of halides do you know besides displacement? The E1 and E2 reactions
compete with displacement reactions. If HCl were lost from chlorobenzene, a cyclic alkyne,
called dehydrobenzene or benzyne, would be formed. This bent acetylene might be
reactive 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
H
Na
+-
NH 2
 NaClNH 3 2NH
(a)
(b)
Benzyne
(b)(a)
NH2

NH2
NH 3
H
NH2

NH2
H
2NH
Benzyne is surely an unusual species and deserves a close look.Although this intermediate
retains the aromatic sextet, the triple bond is badly bent, and there must be very severe
angle 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 two
hybrid orbitals.
Chlorobenzene with chlorine bonded to 14
C gives almost 50% aniline having NH2
bonded to
14
C and 50% aniline with NH2
bonded to normal 12
C atom.
2
3
NH
Br, NH

 
 32 NHNH
 
CH3
Br
CH3
CH3
NH2
(major product)
2
3
NH
Br, NH

 
 32 NHNH
 
OCH3
Br
OCH3
OCH3
NH2
(major product)

2
3
NH
Br, NH

 
 32 NHNH
 
CH3
Br
CH3
CH3
NH2
(major product)
CH3
NH2
(minor product)
2
3
NH
Br, NH

 
 32 NHNH
 
OCH3
Br
OCH3
OCH3
NH2
OCH3
NH2
(both are in equal in amount)
2
3
NH
Br, 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 strong
bases such as the amide ion. In fact, one would expect very high reactivity for this strained
acetylene.

KEY POINTS
1.
R — X Nu

NS 2 reation NS 1reation
R — Nu X

Rate = k[RX] [Nu–
] Rate = k¢[RX]
Inversion of configuration Partial racemization
Transition state less crowded Stable carbocation
Rate generally favoured by Rate generally favoured by polar solvents.
Non-polar solvents
CH3
X > 1° > 2° > 3° 3° > 2° > 1° > CH3
X
2. Elimination
Dry
Ether
R 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 both
E2 reaction is favoured by high concentration of a strong base and polar aprotic solvent
while 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 halides
2
o
HNOSn/HCl CuX
2 2 20 5 C
ArNO ArNH ArN Ar X

   
Nucleophilic substitution by SN
Ar or Benzyne mechanism.
SN
Ar 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

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Viewers also liked

Viewers also liked (20)

Similar to Alkyl halides

Similar to Alkyl halides (16)

More from suresh gdvm

More from suresh gdvm (20)

Recently uploaded

Recently uploaded (20)

Alkyl halides