Haloalkanes and Haloarenes

Introduction

Haloalkanes are the derivatives of hydrocarbons formed by the replacement of hydrogen

atom/atoms by the halogen atoms (X = F, Cl, Br, I).

Haloarenes are derived from aromatic hydrocarbons by the replacement of hydrogen

atom/atoms by halogen atoms.

Classification

Haloalkanes and Haloarenes can be classified as

Perhalo compounds

When all the hydrogen atoms of a hydrocarbon are replaced by halogen atom, the term
per is used e.g.

Classification of Monohalo compounds

The classification is based on the hybridization of carbon atoms to which halogen atom is
bonded.
1. Alkyl halides or haloalkanes
 They represent a homologus series represented by CnH2n+1X
This type of Haloalkanes are classified as primary (1o), secondary (2o), tertiary
(3o) Haloalkanes.

2. Allylic halides

3. Benzylic halides

4. Vinylic halides

5. Aryl halides: These are the compounds in which the halogen atom is bonded nto
sp2 hybridized carbon atom of an aromatic ring.
Nomenclature

Alkyl halides are named in two ways

(1) Common system: In this system the alkyl group is attached to the halogen atom
is named first. This is then followed by an appropriate word chloride, bromide,
iodide. Common name of alkyl halide are two word names.

(2) The IUPAC names of alkyl halides are one word names

Haloarenes

Naming dihaloalkanes
 Dihaloalkanes having the same type of halogen atoms are named as ‘
ALKYLIDENE’ or ‘ALKYLENE DIHALIDE’
  Halogen atoms present at the same carbon atom are named as geminal
halides.
 When halogen atoms are present at adjacent carbon atoms, they are named
as vicinal halides
 In common system, geminal halides are named as alkylidene halides. And
vicinal halides are named as alkylene halides.

Isomerism
Haloalkanes (aliphatic) exhibit the following isomerism
1. Position isomerism

In Haloarenes

2. Chain isomerism

Practice question:

1. Write the structures of the following compounds:

1. 2-Chloro-3-methylpentane
2. 1-Chloro-4-ethylcyclohexane
 3. 4-tert-Butyl-3-iodoheptane
4. 1,4-Dibromobut-2-ene
5. 1-Bromo-4-sec-butyl-2-methylbenzene

Solution:

Methods of preparation

1. Action of phosphorus halides on alcohols.

ROH + PX5 → RX + POCl3 + H2O

e.g. 2CH3CH2OH + PCl5 → 2CH3CH2Cl + POCl3 + H2O
ROH + PX3 → RX + H3PO3
e.g. 2C2H5OH + PCl3 → 2C2H5Cl + H3PO3
 PBr3 and PI3 are obtained in the reaction mixture (in situ) by the reaction
of red P + Br2/I2 respectively.
 The reaction is not applicable for aryl halides.

2. Action of halogen acids on alcohols

ROH + HX → RX + H2O
e.g.

Anhydrous ZnCl2 acts as a catalyst (only for 1o and 2o alcohols), 3o alcohols

simply reacts with conc HCl at room temperature.
 The above reaction is is called lucas recation.
 The mixture of HCl and anhydrous ZnCl2 is called lucas reagent.
 Lucas reagent is useful in identification of primary, secondary and tertiary
alcohols.
 RBr is obtained by constant boiling of alcohol with HBr (48%).
  RI obtained by heating alcohol with NaI/KI in 95% H3PO4.
 Reactivity of alcohol with HX is 3o>2o>1o.
 The method is not applicable for aryl halides.

3. Action of thionyl chloride on alcohols

Significance of the reaction: this recation is the best method of preparing pure
alkyl halides because by products escapes as gases.
ROH + SOCl2 → RCl + SO2 + HCl
e.g. C2H5OH + SOCl2 → C2H5Cl + SO2 + HCl
The reaction is called DARZEN’s REACTION

 These methods are not applicable for aryl halides.

i.e., unlike alcohols phenol cannot give haloarene by any of the above
methods because of sp2 hybridised C and partial double bonded character
of C-O bond .

4. Halogenation of alkanes.

Alkanes react with halogens in presence of UV light or at high temperature.

This method is not used in the laboratory because of difficulty of separating the
products. This reaction involves free radical mechanism. Free radicals are the
species formed by hemolytic fission. Covalent bond breaks in such a way that is
each bonded atom gets one electron.
Species with odd electrons is called free radical.

Free radicals cannot be isolated since they are highly reactive and short lived.
This reaction is a chain reaction and does not depend on the concentration of
halogen or alkane but only depends on the intensity of light. Hence, it is a
classical example of zero order reaction.

Method of preparation of haloarenes

1. By electrophilic substitution
  Ortho and para isomers are formed, because methyl group in benzene is
ortho-para directing.
 Benzene prefers electrophilic substitution because of resonance.
 The reaction occurs in the presence of lewis acids (FeCl3 is lewis acid).
 The separation of ortho and paraisomers is easy since they differs largely
in their boiling points.
 Chlorides and bromides are readily prepared.
 For iodides, the presence of oxidizing agents (HNO 3, HIO4) is required
because without this the reaction is reversible.
 Fluorides are not prepared by this method since fluorine is highly reactive.

2. Sandmeyer’s reaction

It is a two stage preparation.

1st step: preparation of diazonium salt (chloride or bromide).

2nd step: diazonium salt reacts with cuprous chloride or cuprous bromide to form
aryl chloride or aryl bromide respectively.

Replacement of the diazonium group by iodine does not require the presence of
cuprous halide.

3. From alkenes
a) Addition of halogen acids to alkenes

The mode of addition of HCl, HBr, HI follows MARKONIKOV’s RULE

except for the addition of HBr in presence of organic peroxides ROOR.
MARKONIKOV’s RULE: when a reagent adds to an unsymmetrical alkene,
the positive part of the reagent becomes attached to the double bonded carbon
which has more number of hydrogen atoms and the negative part is added to
the double bonded carbon having less number of hydrogen atoms.

Mechanism: the addition of hydrogen halides to alkenes is an electrophilic

addition reaction. Thus during the addition of HBr to propene, the first step
involves the addition of proton. This addition, in principle, can occur in two
ways. If the proton adds on the terminal carbon atom of the double bond, a 2 o
carbocation is formed and if the addition occurs to the middle carbon atom, a
1o carbocation is formed.

Since, a 2o carbocation is more stable than 1 o carbocation, therefore, 2o

carbocation is formed predominantly. This carbocation then rapidly undergoes
nucleophilic attack by the Br- ion forming 2-bromopropane as the major
product. Thus markonikov’s addition occurs through the more stable
carbocation intermediate.

Carbonium ion:
 Carbon atom bearing a positive charge is called carbonium ion. Carbonium
ion formed by hetrolytic fission.
Remember: Tertiary(3o) carbonium ion is more stable than secondary(2 o)
carbonium ion. Secondary(2o) carbonium ion is more stable than primary(1o)
carbonium ion due to resonance.

PEROXIDE EFFECT/KHARASCH EFFECT/ANTI MARKONIKOV

RULE: in the presence of organic peroxide (ROOR), the addition of HBr to
unsymmetrical alkene is reverse to that of markonikov’s rule, that is the
negative part of the reagent, i.e., Br- is added to that double bonded carbon
atom which has more number of hydrogen atoms and H+ is added to that
double bonded carbon atom which has less number of hydrogen atoms.

 Antimarkonikov rule is exhibited only by HBr but not HCl and HI.
This is because,
HCl bond energy (103kcal/mol) is more than the bond energy of HBr
(87kcal/mol). Hence, HCl is not broken by the alkoxy free radical
(RO.) obtained from the peroxide.
HI bond energy (71kcal/mol). It is less than HBr bond energy, yet it
does not show antimarkonikov rule because iodine atoms readily
combine to form iodine.
2HI → 2H + 2I
I + I → I2
b) Addition of halogens to alkenes:

Bromine in CCl4 undergoes addition to alkenes. During the reaction the

reddish brown colour of bromine decolourises. This reaction is used as a test
to detect unsaturation.

Practice questions:
2. Write the products of the following reactions:
3. Why is sulphuric acid not used during the reaction of alcohols with KI?
Solution: H2SO4 is an oxidizing agent. It oxidizes HI produced during the reaction
to I2 and thus prevents the reaction between an alcohol and HI to form an alkyl
iodide.
2KI + H2SO4 → 2KHSO4 + 2HI; 2HI +H2SO4 → H2O + I2 + SO2
To overcome this difficulty, a non-oxidising acid such as H 3PO4 is used instead of
H2SO4.
CH3CH2OH + KI + H3PO4 → CH3CH2I + KH2PO4 + H2O

4. Write the structures of different dihalogen derivatives of propane.

Solution: Four isomeric dihalogen derivativesof propane are formed for each
halogen. For example;

Similarly, we can write four isomeric dihalogen derivatives of other halogens.

5. Draw the structures of major monohalo products in each of the following

reactions.
c) Halogen exchange

FILKENSTEIN REACTION: Cl displaces Br, Cl and Br displaces I from

the respective compounds. Cl2 is more electronegative than Br2 and I2. Br2 is
more electronegative than I2.
 SWART’s REACTION: alkyl fluorides are prepared by this reaction. Alkyl
fluorides are prepared by displacement reaction. This is because, fluorine
being more electronegative displaces chlorine or bromine or iodine from their
compounds.

Physical properties

In general, the physical properties like, M.pt, B.pt, density, physical state, colour,
solubility etc. depend on the structure of the molecules.

Physical state
 Lower alkyl halides are gases at room temperature.
 Higher alkyl halides are both solids and liquids.

Melting point and boiling point

Melting point and boiling point depends on
 Intermolecular forces of attraction
 Vander waal’s forces
 Polarity
 Dipole moment
 Dipole-dipole interactions; all factors are related to structure of the molecule.

1. Boiling points of haloalkanes (RCl, RBr, RI) is greater than the corresponding
hydrocarbon.
e.g. B.pt. of RI>RBr>RCl>RI
Magnitude of vanderwaals forces increases with increase of size (I>Br>Cl>F).
2. Boiling point is also influenced by carbon structure. Straight chain carbon atoms
have more boiling point than branched chain carbon structure.
3. Melting point: Melting point in aromatic compounds, the para isomer has more
melting points than ortho and para isomers.
p- Dichlorobenzene has higher melting point and lower solubility than o- and
m- isomers.
Reason: The p- isomer being more symmetrical fits closely in the crystal lattice
and thus has stronger intermolecular forces of attraction than those of o- and m-
isomers. Since during melting or dissolution, the crystal lattice breaks, therefore
larger amount of energy is needed to melt or dissolve the p- isomer than the
corresponding o- and m- isomers. Thus the melting point of p- isomer is higher
and its solubility is lower than corresponding o- and m- isomers.
4. Boiling point of Haloalkanes is slightly higher than hydrocarbons..
Reason: Due to greater polarity as well as high molecular mass as compared to
hydrocarbons, the intermolecular forces of attraction, i.e., dipole-dipole and Van
 der waal’s forces, are stronger in halogen derivatives. That is why; boiling point
of haloalkanes is slightly higher than hydrocarbons.

Density
Bromo, iodo and polychloro derivatives of hydrocarbons are heavier than water.

Solubility
Haloalkanes are less soluble in water.
Reason: if a haloalkane/haloarene is to dissolve in water, energy is required to overcome
the forces of attraction already existing between haloalkane/haloarene molecules and to
break the hydrogen bonds already existing between water molecules. Less energy is
released when new attractions are set up between the haloalkane/haloarene and water
molecules because these are not as strong as the hydrogen bonds already existing
between water molecules. As a result, Haloalkanes/Haloarenes are less soluble in water.

Dipole moment
Dipole moment in Haloalkanes follows the order:
CH3Cl (1.860D) >CH3F (1.847D) >CH3Br (1.830D) >CH3I (1.636D)

Dipole moment of CH3F is slightly lower than CH3Cl.

Reason: This is because although the magnitude of negative charge on the F atom is
much higher than that of chlorine atom but due to small size of F atom as compared to Cl,
the C-F bond distance is so small that the product of charge and distance, i.e. dipole
moment of CH3F is smaller than CH3Cl.

Haloarenes have lower dipole moment than haloalkanes.

Reason: (i) Due to a greater s- character, a sp 2 hybrid carbon is more electronegative than
a sp3 hybrid carbon. Therefore, the sp2 hybrid carbon of C-Cl bond in chlorobenzene has
less tendency to release e-s to chlorine than a sp3 hybrid carbon of haloalkanes. As a result
the C-Cl bond in haloarenes is less polar than a haloalkanes.
 (ii) Due to delocalization of lone pair of electrons of chlorine atom over the
benzene ring, C-Cl bond in haloarenes acquires some double bond character, while C-
Cl bond in haloalkanes is a pure single bond, i.e. C-Cl bond in chlorobenzene is
shorter than in haloalkanes.
Since dipole moment is a product of charge and distance, therefore
chlorobenzene has lower dipole moment than haloalkanes due lower magnitude of
negative charge on chlorine atom and shorter C-Cl distance.

Chemical properties

The reaction of Haloalkanes may be divided as:

1. Nucleophilic substitution reactions
2. Elimination reaction
3. Reaction with metal

Nucleophilic substitution reaction

Nucleophile: A reagent which can donate an electronpair in a reaction is called

nucleophile. It attacks regions of low electron density (i.e. positive centres). Nucleophile
are electron rich. It is represented as
e.g.; Cl-, Br-, I-, CN-, OH-, RCH2-, NH3, RNH2, H2O, ROH.

Nu- + RX → R-Nu + X-
S.No Reagent Nucleophile Substitution Class of the main
(Nu ) -
product (R-Nu) product
1. NaOH/KOH/Moist Ag2O HO- R-OH Alcohol
2. H2O H2O R-OH Alcohol
3. R’ONa R’O- R-O-R’ Ether
4. NaI I- R-I Alkyl iodide
5. NH3 NH3 R-NH2 Primary amine
6. R’NH2 R’NH2 R-NH-R’ Secondary amine
7. R’NHR’’ R’NHR’’ RNR’R’’ Tertiary amine
8. KCN -
C≡N R-CN Nitrile (cyanide)
9. AgCN Ag-C ≡ N: R-N ≡ C Isonitrile (Isocyanide)
10. KNO2 O = N –O- R-O-N = O Alkyl nitrile
11. AgNO2 Ag-O-N = O R-NO2 Nitroalkane
12. R’COOAg R’COO -
R’COOR Ester
13. NaHS HS- R-SH Thioalcohol
14. R’SNa R’S -
R-S-R’ Thioether
15. LiAlH4 H- R-H Hydrocarbon
16. R’ M
- +
R’ -
R-R’ Alkane

1. RX + NaOH → ROH + KX
2. RX + HOH → ROH + HX
 3. RX + NaOR’ → ROR’ + NaX
4. RX + NH3 → RNH2 + HX
5. RX + R’NH2 → RNHR’ + HX
6. RX + R’’NHR’ → RN(R’’)R’ + HX
7. RX + KCN → RCN + KX
RX + AgCN → RNC + AgCN

Reaction of alkyl halides with KCN gives cyanide and with AgCN gives
isocyanide.
Reason: Cyanide ion is a ambident nucleophile, i.e., either carbon or nitrogen can act
as a electron donor to an alkyl halide.
KCN is predominantly ionic in nature; hence, reaction can occur either
through carbon or nitrogen. But C-C bonds are stronger than C-N bonds and hence
alkyl cyanides are main products.
AgCN is covalent in nature, therefore lone pair of electrons over nitrogen are
available for the bond formation, hence isocyanides are main products.

8. RX + KNO2 → RONO + KX
RX + AgNO2 → RNO2 + AgX

Reaction of alkyl halide with potassium nitrite gives alkyl nitrite whereas with
AgONO gives nitroalkane.
Reason: Nitrite ion O=N-O- is a ambident nucleophile, i.e., either oxygen or nitrogen
can act as a electron donor to an alkyl halide.
KONO is predominantly ionic in nature, hence, reaction can occur through
oxygen as the electron are delocalized over the oxygen atoms and hence is more
available. Hence, alkyl nitrites are main products.
AgONO is covalent in nature, therefore lone pair of electrons over nitrogen
are available for the bond formation, hence nitroalkanes are the main products.

9. RX + AgOOCR’ → RCOOR’ + AgX

10. RX + LiAlH4 → RH + HX
11. RX + R’N+[R+MgI-] → R-R’ + MgX

Terms related to stereochemical aspects:

1. Plane polarized light: light vibrating in many different planes, when passed
through a Nicol prism is found to vibratein one plane and is said to be plane
polarized.
2. Optical activity: olutions of some organic compound have the ability to rotate the
plane of polarized light and the compound is called optically active and the
property is called optical activity. The substance may rotate the plane of polarized
light either to right or left. Optical isomerism is an account of asymmetric carbon
or stereocentre.
If rotated to right, it is called dextrotatory.
If rotated to left, it is called laevorotatory.
 Such compounds are optical isomers.
3. Enantiomers: The optical isomers which like an object and mirror images are
called as enantiomers.
4. Racemic mixture: A mixture containing an equal amount of d(+) and l(-) isomers
is called recemic mixture.

Mechanism of nucleophilic substitution reaction

There are two types of nucleophilic substitution reactions. They are:

1. SN2 (Substitution, nucleophilic, bimolecular)
2. SN1 (Substitution, nucleophilic, unimolecular)

Substitution, nucleophilic, bimolecular, SN2.

In SN2 mechanism, rate of the reaction is dependent on the concentration of haloalkane

and concentration of nucleophile. Hence, it is called bimolecular. In other words, it is a
second order reaction.

Salient features of SN2 mechanism:

1. R-X + OH- → R-OH + X- (leaving group)
Rate α [Substrate][Nucleophile]
2. Mechanism

The hydroxide ion approaches carbon from the side opposite to the bromine atom.
Reason: Both hydroxide ion and bromide ion is electron rich and repel from each
other.
In the transition state, both hydroxide ion and bromide ion are partially bonded to
the substrate carbon, i.e., C-Br bond is not completely cleaved and C-OH bond is
not completely formed. In transition state, the three C-H bond, lie in one plane
whereas C-OH and C-Br bonds are perpendicular to the plane of C-H bonds.
3. In the cause of the reaction, the configuration of the carbon is inverted rather like
umbrella blown inside out. This change in configuration is called WALDEN
INVERSION.
 4. The C-Br bond dissociation energy helps in bond formation. Reactivity of
haloalkanes is RI>RBr>RCl>RF.
5. Overall order of the reactivity in SN2 reaction is: Methyl halide>1o>2o>3o.
6. The reaction is accelerated in the presence of aprotic solvents.

Substitution, Nucleophilic, unimolecular, SN1

In SN1 mechanism, rate of the reaction depends only on the concentration of alkyl halide.

Salient features of SN1 mechanism:

1. R-X + OH- → R-OH + X- (leaving group)
Rate α [Substrate]
2. Mechanism
Step 1: Alkyl halide ionizes to give the carbonium ion. This step is a slow step
and is a rate determining step. Thus, the rate of reaction is controlled by the
concentration of alkyl halide only. Hence the reaction is a first order reaction and
reaction is called SN1 reaction. The carbonium ion is planar. This is because the
central positively charged carbon atom is sp2 hybridized.

Step 2: The nucleophile can attack the planar carbonium ion from either side to
give t-Butyl alcohol. This step is fast and hence does not affect the rate of the
reaction.

3. The reaction leads to racemisation.

 4. The C-Br bond dissociation energy helps in bond formation. Reactivity of
haloalkanes is RI>RBr>RCl>RF.
5. Overall order of the reactivity in SN1 reaction is: Methyl halide<1o<2o<3o.
6. The energy needed for the cleavage of C-Br bond is obtained through the
salvation of the bromide ion with the proton of protic solvents.

Differences between SN2 and SN1 mechanism:

S.No. SN1 Reaction SN2 Reaction

1. A two step mechanism A one step mechanism
2. It is a unimolecular reaction A bimolecular reaction
3. Rate depends on the structure of Rate depends on the structure of the halide,
the halide, i.e., i.e.,
Methyl halide<1 <2 <3
o o o
Methyl halide>1o>2o>3o
4. This reaction leads to This reaction leads to inversion of
racemisation. configuration.

Elimination reactions:
Elimination reactions are reverse of addition reactions. Here two or more atoms or groups
attached to the adjacent carbon atoms in substrate molecules are eliminated to form a
multiple bond in presence of alcoholic KOH. E.g.

These reactions like substitution reations are of two types:

1) E2 elimination
2) E1 elimination
E2 elimination:
E2 stands for bimolecular (two molecules). The rate of elimination reactiondepends both
on the concentration of a substrate and the nucleophile. It is a one step reaction like S N2,
i.e., in one step process, the abstraction of protons from the β carbon and the expulsion
of the halide from the α carbon occurs simultaneously.
Mechanism

E1 elimination
E1 stands for the elimination is a unimolecular reaction. The rate depends only on the
concentration of the substrate. It is a two step process, i.e.

Step 1: Formation of a carbonium ion

Step 2: A proton is abstracted by the base from the adjacent β carbon atom to give
alkene.
When alkyl halides are heated with alcoholic KOH, elimination of hydrogen atom from
the β carbon atom and halogen atom from the α carbon atom are eliminated giving an
alkene.

When alkyl halides are heated with alcoholic KOH, elimination of hydrogen atom from β
carbon atom and a hydrogen atom from α carbon atom are eliminated giving an alkene.

In ethanol (alcohol) an equilibrium occurs between the solvent and potassium hydroxide
to produce potassium ethoxide.
Why alcoholic KOH undergoes elimination reaction and aqueous KOH undergoes
substitution reaction?
Reason: In case of alcoholic KOH, base is RO- and in aqueous KOH base is HO-. RO- is
a stronger base than HO- hence RO- abstracts a proton from an alkyl halide leading to
elimination reaction (leading to formation of an alkene) and HO- is a weak base, hence
undergoes substitution reaction. Moreover in aqueous solution HO- is highly solvated
hence reduces its basic character.
Saytzeff rule: If the dehydrohalogenation (β-elimination) can yield more than one
alkene, then according to Saytzeff rule, then the main product is the most highly
substituted alkene.

Factors influencing the rate of reaction

 Nature of the alkyl halide.
 Size/strength of base/nucleophile.
 Reaction conditions.
 Bulkier nucleophile prefer to act as a base rather than approaching carbon atom
and vice versa.
 1o alkyl halide prefers SN2 reaction.
 2o alkyl halide prefers SN2 or β-elimination depending on the strength of base or
nucleophile.
 3o alkyl halide prefers SN1 or elimination reaction depending on the stability of
carbonium ion or more substituted alkene.

Reaction with metals

The compounds of organic compounds with metal atoms are called organomettalic
compounds e.g., CH3MgI; methyl magnesium iodide, (CH3CH2)4Pb; lead tetraethyl,
Na-C≡C-Na; sodium acetylide
Grignard reagents (RMgX):
Alkyl halide reacts with magnesium metal in presence of dry ether to form ethyl
magnesium halide, e.g., CH3MgI methyl magnesium iodide
Reaction
Grignard reagent is a source for preparing organic compounds. Grignard reagents are
highly reactive and react with any source of proton to give hydrocarbons. Even H 2O,
ROH, NH2, are sufficiently acidic to convert RMgX to hydrocarbon
RMgX + H2O → RH + Mg(OH)X

Wurtz reaction
A reaction of RX with metal sodium in presence of dry ether is wurtz reaction.

Higher alkane are prepared from lower alkanes. Symmetrical alkanes are prepared by this
method. When unsymmetrical alkanes are prepared by this method, mixture of alkanes
are obtained which are difficult to separate..
Reaction of Haloarenes

1. Nucleophilic substitution reaction: aryl halides are less reactive towards

nucleophilic substitution reactions than alkyl halides.

Reason: (i) In haloalkanes, the halogen is attached to sp 3 hybridized carbon atom

while in haloarenes; the halogen is attached to sp 2 hybridized carbon atom. sp2
hybridized orbital is smaller in size as compared to sp 3 orbital of carbon, therefore, C-
Cl bond in haloarenes is shorter and stronger. Hence, bond is difficult to break.
(ii) sp2 hybrid carbon has greater s-character and hence is more
electronegative, therefore sp2 hybrid carbon of C-X bond in aryl or vinyl halides has
less tendency to release electrons to the halogen than a sp 3 carbon atom in alkyl
halides. As a result aryl halides are less polar than alkyl halides. Hence halogen atom
cannot be displaced easily.
(iii) Phenyl or vinyl cation formed as a result of self ionization is not
stabilized by resonance because the sp 2 hybridized orbital of carbon having the
positive charge is perpendicular to the p-orbitals of the phenyl ring. Hence phenyl
cation formed is unstable.

2. Replacement by hydroxyl group:

The presence of an electron withdrawing group (-NO2) at ortho and para positions
increases the reactivity of haloarenes. The presence of NO2 group in arenes activates
only ortho and para positions.

Explanation:
a) The presence of NO2 group at ortho and para position withdraws the electron
density from the benzene ring and then facilitates the attack of nucleophile on
haloarene.
b) The carboanion formed is stabilized through resonance
c) The negative charge appeared at ortho and para position with respect to halogen
substituent is stabilized by NO2 group.
d) In case of meta nitrobenzene, none of the resonating structures has negative
charge on the carbon atom bearing NO2 group. Therefore NO2 group at meta
position does not stabilize the negative charge at meta position.
e) NO2 group takes part in resonance.

NO2 group at para position

NO2 Group at ortho position

Note: NO2 group at meta position does not stabilize the negative charge and no effect
on reactivity is observed.

3. Electrophilic substitution reaction

 Haloarenes undergo electrophilic substitution reaction benzene.

 Halogen atom besides deactivating is ortho and para directing. Hence
electrophilic substitution is favoured at ortho and para position.
 Due to resonance, the electron density increases at ortho and para position.
 Halogen atom exhibits –I inductive effect i.e. withdraws electron from
benzene ring. Hence benzene ring gets deactivated.
 Substitution reactions occurs slowly electrophilic substitution reaction.

i) Halogenation
ii) Nitration

iii) Sulphonation

iv) Friedel crafts reaction

Mechanism of Friedel craft’s Reaction:
Step 1: formation of electrophile

Step 2: The electrophile attacks the benzene ring to form a carbonium ion.

Step 3: Proton abstraction

Inductive effect

(i) It is an electron displacement effect.

(ii) It is a weak but permanent effect.
(iii) The effect involves the displacement of sigma bond electrons towards more
electronegative atoms.
(iv) The effect is fast 4th carbon atom.

Types of inductive effect

i) –I effect ii) +I effect

i) –I effect
It is shown by electron attracting or with drawing species. Sigma electrons are
displaced away from carbon atom. Halogen is electron attracting group and
shows –I effect. NO2> CN- > COOH> X- > NH2 > OH > OR > Ph.
Influence of –I effect
a) it increases acidic nature and decreases basic nature.
b) It increases the stability of carbonium ion and decreases the stability of
carbocation.
ii) +I effect
It is shown by electron releasing species. Sigma electrons are displaced
towards carbon atom. +I effect showing groups in decreasing order
O- > COO- > 3o Alkyl > 2o Alkyl > 1o Alkyl > Hydrogen
Influence of +I effect
a) it increases basic nature and decreases acidic character.
 b) It increases the stability of carbocation C+ and decreases the syability of
carboanion C-.
Inductive effect is responsible for high melting point, boiling point and dipole moments.

4. Wurtz Fitting reaction

5. Fittig Reaction

Practice questions:

6. Which alkyl halide from the following pairs would you expect to react more
rapidly by SN2 mechanism? Explain your answer.

Solution: (i) CH3CH2CH2CH2Br is a 1o alkyl halide while CH3-CH2-CHBr-CH3 is 2o

alkyl halide. Since there will be some steric hindrance in 2 o alkyl halides
than in 1o alkyl halides, therefore, CH3CH2CH2CH2Br will react faster than
CH3-CH2-CHBr-CH3 in SN2 reaction.
(ii) CH3-CH2-CHBr-CH3 is a 2o alkyl halide while (CH3)3C-Br is 3o alkyl
halide. Since, due to lesser steric hindrance, 2 o alkyl halides than in 3o
alkyl halides, therefore, CH3-CH2-CHBr-CH3 will react faster than
(CH3)3C-Br in SN2 reaction.
(iii)

Both the alkyl halides are 2o alkyl halides. But in alkyl halide (II), the CH 3
group at C2 is closer to the Br atom while in alkyl halide (I), the CH3 group
 at C3 is little away from the Br atom. As a result alkyl halide (I) will suffer
greater steric hindrance than alkyl halide (I). Therefore, (CH3)2CHCH2Br
will react faster than CH3CH2CH(CH3)CH2Br in SN2 reaction.

7. Identify A, B, C, D, E, R and R1 in the following.

Solution:

Since D gets attached to the same carbon atomon which MgX was present, therefore,

The third part of the question is wrong because tert-alkyl halides do not undergo
wurtz reaction but undergo dehydrohalogenation to give alkenes. Thus R 1 = (CH3)3C-
and hence D and E are

Important alkyl and aryl halides and their applications

1. Chloroamphenicol: chlorine containing antibiotic used to cure typhoid fever.

2. Thyroxin: hormone containing iodine.

3. Chloroquine: used for treatment of malaria.

4. Dichloromethane (Methyl chloride):

Uses: as solvents, as metal cleaning
Adverse effects: Causes dizziness, burning of skin.

5. Trichloro methane (Chloroform):

Uses: as a solvent for fats and oils, as a refrigerant, in anesthesia
Effects: causes dizziness, headache.

Chloroform is kept in dark coloured bottles.

Reason: Chloroform is slowly oxidized by air in presence of light to form an
extremely poisonous gas, carbonyl chloride and phosgene. Phosgene is poisonous

Ethyl alcohol is added to chloroform while it is packed.

Reason: this is done that if any COCl2 is formed alcohol reacts with it to form
diethyl carbonate.

6. Triiodomethane: used as antiseptic

7. Tetrachloromethane:
Uses: as refrigerant, for preparing flourochlorocarbons, as a general solvent,
degreasing agent.
Effects: causes liver cancer, dizziness, nausea, headache, vomiting, makes heart
beta irregular when inhaled in excess, depletion of ozone layer.

8. Freon: Used in refrigerator and effects ozone layer.

9. p,p’-Dichoro-diphenyl trichloroethane:
Uses: as chlorinated agent, as insecticides
Effect: insects developed immunity and caused adverse effects on human beings
and animals