Chapter 16

Haloalkanes and Haloarenes

Chapter Contents
Introduction
 Introduction
Alkyl halides or haloalkanes and aryl halides or haloarenes are organic
 Classification compounds obtained by replacement of one or more H-atoms of aliphatic
and aromatic hydrocarbons respectively by halogen atom(s).
 IUPAC Nomenclature
Haloalkanes have each of the halogen atom attached to sp3 hybridised
 Methods of Preparation of C-atom of the alkyl group. Haloarenes have each of halogen atom
Haloalkanes attached to sp2 hybridized C-atom of aryl group.

 Physical Properties
CLASSIFICATION
 Chemical Properties These compounds may be classified as :
 Stereochemical Aspects of
Based on the Number of Halogen Atoms
Nucleophilic Substitution
Reactions (a) Monohaloalkanes and monohaloarenes are compounds containing one
halogen atom only, e.g., chloroethane, chlorobenzene etc.
 Reactions of Haloarenes
CH3 CH2 Cl C6H5Cl
 Polyhalogen Compounds (Chloroethane) (Chlorobenzene)

Monohaloalkanes may be further classified as primary (1°), secondary

(2°) or tertiary (3°) depending on whether halogen atom is bonded to
1°, 2° or 3° C-atom, e.g.,

CH3 CH2 CH2 Cl CH3 CHCl CH3

2-Chloropropane (2°)
1-Chloropropane (1°)
or Isopropylchloride

CH3
CH3 C Cl
CH3
2-Chloro-2-methylpropane (3°)
or Tert. butyl chloride

(b) Dihaloalkanes and dihaloarenes are compounds containing two halogen

atoms. Dihaloalkanes may be further classified as
(i) Gem-dihalides containing two halogen atoms attached to the
same C-atom and

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(ii) Vicinal dihalides containing two halogen atoms attached to the adjacent C-atoms. Gem-dihalides
are also known as alkylidene halides and vicinal dihalides are also known as alkylene halides. e.g.,

Cl
Cl
CH2Cl
CH2Cl CH3 CHCl2
1, 2-Dichloroethane 1, 1-Dichloroethane 1, 2-Dichlorobenzene
or Ethylene chloride or Ethylidene chloride or o-Dichlorobenzene
(vic-dihalide) (gem-dihalide)

(c) Polyhaloalkanes and polyhaloarenes are compounds containing three or more halogen atoms. They may
be attached to the same or different C-atoms, e.g.,
Cl
Cl

CHCl3 CHBr3 CHI3 CCl4 Cl

Chloroform Bromoform Iodoform Carbon tetrachloride 1, 2, 3-Trichlorobenzene

Based on the type of sp3 Hybridised C-atom containing Halogen Atom(s)

(a) Allylic halides are compounds containing halogen atom bonded to sp3 hybridised C-atom next to carbon,
carbon double bond, e.g.,

CH2 CH CH2 Cl CH3 CH CH CH 2 Cl

3-Chloropropene (1°) 1-Chlorobut-2-ene (1°)

(b) Benzylic halides are compounds containing halogen atom bonded to sp3 hybridised C-atom next to an
aromatic ring. For example
CH2Cl CH3 CH Cl

Benzyl chloride (1°) 1-Chloro-1-phenylethane (2°)

Based on sp2 hybridised C-atom containing Halogen Atom(s)

(a) Vinylic halides are compounds containing halogen atom bonded to sp2 hybridised C-atom of an aliphatic
compound, e.g.,

Cl
CH2 CH Cl
Chloroethene 1-Chlorocyclopentene

(b) Aryl halides are compounds containing halogen atom bonded to sp2 hybridised C-atom of an aromatic
ring. For example

Cl Cl

CH3
1-Chloronaphthalene (m-Chlorotoluene)
1-Chloro-3-methylbenzene

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IUPAC NOMENCLATURE
Haloalkanes and haloarenes are named as halogen-substituted hydrocarbons in the IUPAC system of
nomenclature. The rules of naming such compounds are same as used for hydrocarbons considering halogens
as substituents. The common names of these compounds are derived by writing the appropriate name of alkyl
or aryl group followed by halide, e.g.,
CH3
CH3 CH CH2Cl CH3 C CH2Br
CH3 CH3
Common name : Isobutyl chloride Neopentyl bromide
IUPAC : 1-Chloro-2-methylpropane 1-Bromo-2-2-dimethylpropane

CH3
CH3 CH CH CH3
Cl CH3 CF2 CH2 CH3
IUPAC : 2-Chloro-3-methylbutane 2, 2-Difluorobutane (gem)

CH3 CHF CHF CH3 (CH3)2C CH CH2Cl

IUPAC : 2, 3-Difluorobutane (vic) 1-Chloro-3-methylbut-2-ene

H CH3
Cl
H3C CH2Cl
IUPAC : 1-Chloromethyl-1, 3-dimethyl 3-Chlorocyclohex-1-ene
cyclopentane

Example 1 : Draw all possible geometrical isomers of molecular formula C2FClBrI.

Solution : There are six possible geometrical isomers of C2FClBrI as shown below.
F Br F I F Cl
C C ; C C ; C C
Cl I Cl Br Br I

F I F Cl F Br
C C ; C C ; C C
Br Cl I Br I Cl

Nature of C-X Bond

Alkyl halides are polar molecules having permanent dipole moment due to higher electronegativity of halogen
than carbon. The carbon-halogen bond length increases and hence bond energy decreases with the size of
halogen atom. In vapour phase dipole moment of Me–X varies as Me–Cl > Me–F > Me–Br > Me–I because
dipole moment depends on electronegativity of halogen atom as well as on C–X bond length as shown in the
following table.

–1
C–X bond Bond length (pm) Bond energy (kJ mole ) Dipole moment (D)
Me–F 139 452 1.847
Me–Cl 178 351 1.860
Me–Br 193 293 1.830
Me–I 214 234 1.636

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METHODS OF PREPARATION OF HALOALKANES

Haloalkanes may be prepared by any one of the following methods.

From Alcohols
(a) By reaction with halogen acids : Alcohols react with halogen acids under anhydrous conditions to form
haloalkanes. The reactivity of halogen acids follows the order HI > HBr > HCl because the strength of
HI bond is less than that of HBr which in turn is less than that of HCl. The reactivity order of alcohols
towards a given hydrogen halide is 3°ROH > 2°ROH > 1°ROH. This is based on the polarity of C—O
bond. The polarity of C—O bond increases with the increase in number of electron-releasing groups
attached to the carbon-containing OH group.
Chloroalkanes are synthesised by treating alcohol with HCl in presence of anhydrous ZnCl2. This is known
as Groove’s process. Primary alcohols react with HCl in presence of anhydrous ZnCl2 only on heating,
whereas secondary and tertiary alcohols react at room temperature.

Anhydrous ZnCl
CH3CH2OH  HCl 

2
 CH3 CH2Cl  H2O

Bromoalkanes and iodoalkanes can also be obtained in a similar manner by treating alcohol with HBr
and HI respectively. HBr or HI is generated in the reaction mixture as follows :

NaBr  H2SO4 
 NaHSO4  HBr
(Conc.)

NaI  H3PO4 
 NaH2PO4  HI

To synthesize HI, phosphoric acid is used instead of H2SO4 because HI gets oxidised by H2SO4 to I2.

CH3CH2OH  HBr 
 CH3 CH2Br  H2O

CH3CH2OH  HI 
 CH3CH2I  H2O

Primary alcohols undergo these reactions by SN2 mechanism. But secondary and tertiary alcohols follow
SN1 mechanism in which carbocation is formed as intermediate. Whenever there is a possibility of
rearrangement, rearranged product is the major product, e.g.,

CH3 CH3
CH3 CH CH CH3 + HBr CH3 C CH2 CH3 + H2O
OH Br
(Major)
Mechanism :

CH3 CH3 CH3

H –H2O
CH3 CH CH CH3 CH3 CH CH CH3 CH3 C CH CH3
1 2
OH OH2 H
2°-Carbocation

CH3 CH3
1, 2-Hydride shift Br
CH3 C CH2 CH3 CH3 C CH2 CH3
3°-Carbocation
Br
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(b) By reaction with phosphorus halides :
CH3CH2OH  PCl5 
 CH3CH2Cl  POCl3  HCl

3CH3CH2OH  PX3 
 3CH3CH2 X  H3PO3

PX3 may be PCl3, PBr3 or PI3 and is prepared by treating red phosphorus with Cl2, Br2 or I2. These
reactions follow SN2 mechanism which is always accompanied by inversion of configuration. The
advantage of using phosphorus halides is that, no rearranged product is formed because the reaction does
not involve the formation of carbocation.

(c) By reaction with thionyl chloride (Darzen’s Method) :

CH3 CH2 OH  SOCl 2 

 CH3 CH2 Cl  SO2   HCl 
(Better method to prepare alkyl chloride because the side products are gases)

H H
CH3 C OH + SOCl2 CH3 C Cl + SO2 + HCl 
D D
(Configuration retained)

Alcohols react with SOCl2 to give alkyl chloride with the retention of configuration as the reaction follows
intramolecular nucleophilic substitution mechanism (SNi). Both the by-products SO2 and HCl are gases and
escape.

Mechanism :
H O H O H O
–HCl 
CH3 C OH + S Cl CH3 C O S Cl CH3 C O S
D Cl D H Cl D Cl
H
SNi
CH3 C Cl + SO2
D
However, when the same reaction is carried out in presence of pyridine which traps HCl forming pyridinium
chloride, the reaction follows SN2 mechanism. The product formed in this case has inverted configuration.

+ HCl Cl
N N
H
H O H
Cl + CH3 C O S Cl C CH3 + SO2 + Cl
D Cl D
(Configuration inverted)

Example 2 : Identify all the products formed in the following reaction and indicate the major product.
CH2OH


+ HBr

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Solution : Three products will be formed in the given reaction as shown below :
CH2OH CH2 OH2 Br CH2

H Br

–H2O

Br CH2 CH3 H3C Br

Ring
Br expansion 1,2-Hydride Br
shift
(Major)

From Hydrocarbons

(a) From Alkanes : Alkanes react with halogens in presence of sunlight and/or high temperature to give a
mixture of mono and polyhaloalkanes. Monohaloalkane is the major product if alkane is used in excess.
If halogen is used in excess, one by one all hydrogens are substituted by halogens. For a given alkane,
the reactivity order of halogens is F2 > Cl2 > Br2 > I2. And for a given halogen the reactivity order of
H-atom in alkane is tertiary > secondary > primary > CH4.

e.g., free radical halogenation of alkanes

h
CH3 – CH2 – CH3  Cl2 
Room temp.
 CH3 – CHCl – CH3  CH3 – CH2 – CH2Cl
(Excess) (Major)

h
CH3 CH CH3  Br2 CH3 CBr CH3  CH3 CH CH2Br
125°C
CH3 CH3 CH3
(Main) (Traces)

The reaction with I2 is reversible and is carried out in presence of an oxidising agent e.g., HIO3 or HNO3
to oxidise HI to I2.


CH4  I2 
 
 CH3I  HI

HIO3  5HI 
 3I2  3H2O

Example 3 : A hydrocarbon of molecular formula C6H14 on monochlorination gives two products. Identify the
structure of hydrocarbon.
Solution : Hydrocarbon of molecular formula C6 H14 is an alkane. Since it gives only two products on
monochlorination, it has two different types of H-atom. Such an alkane is 2,3-dimethylbutane.

CH3 CH3 CH2Cl CH3 CH3

h
CH3 CH CH CH3 + Cl2 CH3 CH CH CH3 + CH3 C CH CH3
CH3 Cl

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(b) From Alkenes :

(i) Hydrogen halides (HCl, HBr or HI) add to alkenes to form alkyl halides. The reaction involves
electrophilic addition of H+ to alkene resulting in the formation of carbocation as intermediate. If there
is a possibility of rearrangement, the carbocation undergoes rearrangement. This is followed by the
attack of halide ion to the carbocation giving the final addition product. Thus addition of hydrogen
halide to unsymmetrical alkene follows Markovnikov’s rule with the possibility of rearrangement.

CH3 CH CH2 + HX CH3 CH CH3 + CH3CH2CH2X

(Minor)
X
(Major)

H X
CH3 CH CH2 CH3 CH CH3 CH3 CHX CH3
2° carbocation (Major)
(Markovnikov’s product)
CH3 CH3 CH3
H Hydride
CH3 CH CH CH2 CH3 C CH CH3 CH3 C CH2 CH3
shift
3° carbocation
H
X
2° carbocation

CH3

CH3 C CH2 CH3

X
(Major)
(Rearranged product)

The addition of HBr (only) to unsymmetrical alkenes in presence of peroxide follows

anti-Markovnikov’s rule. This is known as Kharasch effect, e.g.,

Peroxide
CH3 CH CH + HBr CH3CH2CH2Br

Note :
In certain cases the intermediate carbonium ion may undergo rearrangement through 1, 2 shift
(hydride or methyl shift) to form stable carbocation.

(ii) Alkenes decolourise the brown colour of Br2 dissolved in CCl4 due to addition of Br2 to give vicinal
dibromide which are colourless. This is a test of unsaturation. The addition of Br2 to alkene is
anti-addition. It helps us to predict the stereochemistry of the product, e.g.,
CH3 CH3 CH3 CH3
CCl4
C C + Br2 Br C C Br
H H H H
cis-But-2-ene ( )
Racemic mixture

CH3 H CH3 H
CCl4
C C + Br2 Br C C Br
H CH3 H CH3
trans-But-2-ene (Meso)

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(iii) When alkenes are heated with Cl2 or Br2 at high temperature of about 800 K, the H-atom of allylic
carbon is substituted with halogen atom resulting in the formation of allyl halide.
800 K
CH 3 CH CH 2 + Cl2 CH2Cl CH CH2 + HCl

Note :
The allylic substitution with bromine can also be carried out by treating the alkene with
N-bromosuccinimide (NBS).
O Br O

+ N Br + N H
H
O O

Example 4 : What are the major products formed in the following reactions?
(i) C 6H 5 CH2 CH CH2 + HBr

(ii) CH3 + HBr

H 1,2-Hydride
Solution : (i) C6H5 CH2 CH CH2 C6H5 CH2 CH CH2
shift
Br
C6H5 CH CH2 CH3 C6H5 CHBr CH2 CH3
(Major)

H CH3 Hydride Br Br
(ii) CH3
shift
H CH3 CH3
(Major)

(c) From Alkynes:

Alkynes react with Cl2 or Br2 dissolved in CCl4 to give vicinal tetrahalides, e.g.,

Br Br
CCl4
CH3 C CH + 2Br2 CH3 C C H
Br Br

Note :
Alkynes react with hydrogen halides in presence of mercuric chloride (HgCl2) as a catalyst to give
vinyl halides, e.g.,
HgCl2
HC CH + HCl H2C CH Cl
Mechanism :
2+
Hg +
Hg
2+ – +
Hg Cl H3O
HC CH HC CH HC CH H2C CH Cl
Cl

Halogen exchange :
Alkyl iodides are generally prepared by halide exchange in Finkelstein Reaction. In this reaction, alkyl
chloride or alkyl bromide is treated with NaI in acetone or methanol to give alkyl iodide in fairly good yield.
NaCl or NaBr formed in this reaction as by-product being insoluble in solvent and gets precipitated.
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Acetone
CH3 CH2Cl + NaI ⎯⎯⎯⎯→ CH3 CH2I + NaCl ↓

H H
NaI
CH3 CH2 C Br I C CH2 CH3
Acetone
CH3 CH3
(2°-alkylhalide) (Walden inversion product)
(S-configuration) (R-configuration)

The reaction proceeds by SN2 mechanism resulting in inversion of configuration.

Halide exhange is also used for the preparation of alkyl fluorides by Swarts Reaction. Alkyl chloride/bromide
is heated in presence of AgF, Hg2F2, CoF2 or SbF3 to given alkyl fluoride.

CH3 Br + AgF CH3 F + AgBr

From silver salts of fatty acids (Borodine Hunsdiecker Reaction) :

Silver salts of carboxylic acid in carbon tetrachloride solution can be decomposed by chlorine or bromine to
form alkyl halides with one carbon atom less than in acid.
CCl
4  C H Br  AgBr  CO
C6H5 COOAg  Br2  6 5 2
Silver benzoate Reflux Phenyl
bromide

Bromine is generally used because chlorine gives a poor yield of alkyl chloride.

From Arenes
(a) Arenes undergo electrophilic aromatic substitution with Cl2, Br2 or I2 in presence of Lewis acid to give
haloarenes.

Br
FeBr3
+ Br2 + HBr

CH3 CH3 CH3

Fe
+ Cl2 +
dark
Cl Cl
o-chlorotoluene p-chlorotoluene

The ortho and para isomers can be easily separated due to large difference in their melting points.
(b) When alkyl-substituted arenes are treated with NBS in presence of CCl4 and peroxides, benzyl bromides
are mainly formed, e.g.,

CH2 CH3 O Br CH CH3 O

+ N Br + NH

O (Main product) O

The substitution of benzylic hydrogen with chlorine is effected by treating alkyl substituted arenes with
Cl2 in presence of sunlight and/or heat. This reaction is also accompanied by substitution of other
hydrogen atoms of the alkyl part only (and not the arene ring) along with polysubstitution.
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CH2 CH3 CHCl CH3 CH2 CH2Cl

h
+ Cl2 + + Polysubstituted product

From Diazonium Salt

Aromatic primary amine is first converted into diazonium salt by treating it with a mixture of NaNO2 and HCl
at 0–5°C. The diazonium salt is used to prepare aryl fluoride by first converting it into diazonium fluoroborate
and then heating it. Aryl chloride/bromide is obtained by treating diazonium salt with Cu2Cl2/Cu2Br 2
(Sandmeyer reaction) or Cu powder and HCl/HBr (Gatterman reaction).
Reaction of diazonium salt with KI results in the formation of aryl iodide.

 –
N2 BF4 F

HBF4 
+ BF3 + N2

Cl

 Cu2Cl2
– + N2
NH2 N NCl or
Cu/HCl
NaNO2 + HCl
(0-5°C) Br
Diazonium salt Cu2Br2
+ N2
or
Cu/HBr
I

KI
+ N2

Example 5 : Convert benzene to 1-bromo-3-ethylbenzene.

COCH3 COCH3 CH2CH3

AlCl3 Br2 /Fe Zn(Hg)
Solution : + CH3COCl
anhy. HCl
Br Br

Example 6 : Synthesize m-dibromobenzene from benzene.

NO2 NO2
Br2/Fe
Solution : + HNO3 + Conc. H2SO4
Dark
Br
 –
NH2 N2 Cl Br
Sn/HCl NaNO2/HCl Cu2Br2
(0-5°C)
Br Br Br

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EXERCISE

1. Which of the following is an example of aryl alkyl halide?

(1) p-chlorotoluene (2) Chlorobenzene

(3) Allyl chloride (4) Benzyl chloride

2. Which of the following alkyl halides is iso-butyl bromide?

(1) CH3CH2CHCH3 (2) CH3CH2CH2CH2Br

Br

CH3
(3) CH3 – C – Br (4) CH 3 – CH – CH 2Br
CH3 CH3
3. How many isomeric halogen derivatives including stereoisomers are possible for C2H2Br2?

(1) 2 (2) 3

(3) 4 (4) 5

4. Hunsdiecker reaction is used to prepare alkyl chloride and alkyl bromide starting from

(1) Diazonium salt

(2) Silver salt of carboxylic acids

(3) Sodium salt of carboxylic acids

(4) Alcohols

5. The best reagent for converting an alcohol into the corresponding chloride is

(1) PCl5 (2) PCl3

(3) HCl/ZnCl2 (4) SOCl2

6. Iso-butylene can be converted into tert-butyl bromide by its reaction with

(1) HBr (2) Br2

(3) HBr in the presence of peroxides (4) HOBr

7. The number of isomers possible for the molecular formula C2FClBrI is

(1) 2 (2) 3

(3) 4 (4) 6

8. Which of the following will react most readily with HI?

(1) CH3OH (2) (CH3)3COH

(3) CH3CH2OH (4) (CH3)2CHOH

9. The isomer of C6H14, which will give maximum number of monochloroderivative is

(1) 2, 3-dimethylbutane (2) 2-methylpentane

(3) 3-methylpentane (4) Hexane

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10. Which of the following is most reactive towards electrophilic aromatic substitution for halogen?

CH3 NO2
(1) (2)

C2H5 Cl
(3) (4)

PHYSICAL PROPERTIES
(i) Colour. Lower members like methyl chloride, methyl bromide and ethyl chloride are colourless
gases while methyl iodide and some higher members are sweet smelling liquids; still higher members
are colourless solids.
 –
(ii) Polarity. Alkyl halides are moderately polar due to polar C  X bond. (About dipole moment)
(iii) Solubility. In spite of their polarity, all alkyl halides are insoluble in water.
(iv) Density. Alkyl chlorides are generally lighter than water, however, alkyl bromides and iodides are generally
heavier.
Their densities follow the order :
RI > RBr > RCl
All aryl halides are, however, heavier than water
(v) Melting and boiling points.
For monohalogen compounds boiling point increases with molecular mass. For fixed hydrocarbon it follows
order as
Fluoride < Chloride < Bronide < Iodide
e.g.,
(i) CH4 CH3F CH3Cl CH3Br CH3I
BP(K) 353 358 405 429 462
(ii) With given halogen, the boiling point increases with increasing size and decreases with increasing
branching.
CH3 – Cl CH3CH2Cl CH3CH2CH2Cl CH3CH2CH2CH2Cl
BP(K) 249 285.5 320 351.5
(iii) BP. of chloro, bromo and iodo compounds increases with increasing number of halogen atoms
CH3Cl CH2Cl2 CHCl3 CCl4
BP(K) 249 313 334 350
(iv) Symmetry or closed packing is responsible for rise in melting point

Cl Cl Cl
Cl
e.g.,
Cl
Cl

BP(K) 448 453 446

MP(K) 323 256 249
They are generally toxic and cause anaesthesia when inhaled in larger amounts
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CHEMICAL PROPERTIES

Reactions of Haloalkanes
The C—X bond of haloalkanes is polar with partial positive charge on carbon and partial negative charge on
halogen. Any nucleophile stronger than halide ion can attack at the C-atom due to positive charge causing
nucleophilic substitution. Halide ions being weak bases and are good leaving groups, thus haloalkanes undergo
elimination reaction with a strong base. Reactive metals cause cleavage of C—X bond forming organometallic
compounds.

Nucleophilic Substitution Reactions

In nucleophilic substitution reactions, the incoming nucleophile having at least one atom with a lone pair of
electrons attacks at the carbon atom bonded to halogen. The halide ion leaves with the bond pair of electrons.

–
  –
Nu C X Nu C X

The incoming nucleophile may be neutral or negatively charged and the substrate on which nucleophile attacks
may be neutral or positively charged.
Incoming nucleophile  Substrate Product  Outgoing nucleophile

HO  CH3 — X CH3 — OH  X


(CH3 )3 N  CH3 — X CH3 — N(CH3 )3  X

HO  CH3 — N(CH3 )3 CH3 — OH  N(CH3 )3
 
H2S  CH3 — N(CH3 )3 CH3 — SH2  N(CH3 )3
Some of the nucleophilic substitution reactions are given below.
Aq. KOH or Moist Ag2O
R OH + KX or AgX

RO Na
– +

R O R + NaX

KSH
R SH + KX

RS Na
– +

R S R + NaX
NH3 (excess)
R NH2 + NH4X

RC C Na
– +

R X R C C R + NaX
–
I /Acetone –
R I + X
KCN
R C N + KX
NaNO2/AgNO2
R NO2 + R O N O + NaX/AgX
R2CuLi
R R + R Cu + LiX
RCOOAg
RCOOR + AgX
AgCN
R N C + AgX
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Ambident Nucleophiles
Some nucleophiles have lone pair of electrons on more than one atom and can attack through more than one
site. Such nucleophiles are called ambident nucleophiles. For example, cyanide ion has two attacking sites
viz., C and N. Attack through C leads to the formation of alkyl cyanide while attack through N leads to the
formation of isocyanide. Alkyl halides react with NaCN/KCN to form alkyl cyanides as the major product while
AgCN forms isocyanides as major product. KCN is largely ionic and provides CN– ions in solution. The attack
through C–atom of CN– is preferred since C—C bond is more stable than C—N bond. With AgCN being mainly
covalent in nature, N-atom is free to attack alkyl halide to form isocyanide. Similarly, alkyl halides react with
NaNO2/KNO2 to from R O N O (nitritoalkane) as major product due to attack by O-atom of NO2 ion. With
AgNO2, the reaction proceeds through attack by the N-atom forming R—NO2 (nitroalkane) as major product.

Basicity and Nucleophilicity

A negatively charged species can function as a base as well as nucleophile. When it attacks at the C–atom
of substrate, it behaves as a nucleophile. But when it removes proton from the substrate, it behaves as a base.

– –
Z C X Z C X
(Nucleophile)

– –
Z  H A Z H A
(Base)

The basic strength of anion is directly related to its charge density. Higher the charge density of anion, higher
is it its basic strength. For example, the order of basic strength of halide ions is F– > Cl– > Br– > I–.

The nucleophilic strength of anion depends on the nature of solvent used. Broadly the solvents can be divided
into four groups, namely

1. Polar protic solvents e.g., H2O, CH3OH, CH3COOH etc.

2. Polar aprotic solvents e.g., DMF, DMSO etc.

3. Partially polar aprotic solvents e.g., CH3COCH3 etc.

4. Non-polar solvents e.g., CCl4, CS2, C6H6 etc.

In polar solvent the salts dissociate into ions. The extent of ionisation for a given salt depends on concentration
and nature of solvent. In partially polar or non-polar solvents, the salts have greater tendency to remain as
ion-pairs. The ion pairing is strong if the ions have high charge and/or small size. As the charge of anion decreases,
weaker will be ion pair effect and stronger will be its nucleophilicity. Thus nucleophilic strength of halide ions in
partially polar and non-polar solvents is I– > Br– > Cl– > F–. In polar protic solvents both the cation and anion
get solvated. Higher the charge density of anion, higher will be the solvation and weaker its nucleophilicity.
Thus nucleophilic strength of halide ions in polar protic solvents varies as I– > Br– > Cl– > F–. In polar aprotic
solvents only the cations get solvated. The positive end of each solvent molecule is sterically hindered and
unable to solvate the anion. Thus in such solvents nucleophilicity order runs parallel to basic strength order
i.e., for halide ions F– > Cl– > Br– > I–.

STEREOCHEMICAL ASPECTS OF NUCLEOPHILIC SUBSTITUTION REACTIONS

Optical Isomerism
When a compound is non super imposable on its mirror image. The compound would be chiral and it will exhibit
optical isomerism. All chiral objects rotate plane polarized light.

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Br Br

I C Cl Cl C I
X F F Y

Non-superimposable mirror images

 Chiral and hence X and Y are non-identical. They are enantiomers

Br Br

H C H H C H
A Cl Cl B

Superimposable mirror images

 A and B are identical .
X is chiral while A is achiral. X will exhibit optical isomerism.

Note : Ultimate criterion for chirality is NON-SUPERIMPOSABILITY of object on its mirror image.

If a substance rotates the plane of plane polarised light in clock wise direction then it is dextrorotatory, (+)
form or (d) form. If it rotates the plane in anticlockwise direction then it is laevorotatory, (–) form or (l ) form.

Chiral Carbon
If all the four valencies of carbon are satisfied by 4 different group the carbon is chiral carbon or asymmetric
carbon. Presence of chiral carbon never ensure the molecule is optically active.
Compound with one chiral carbon is always chiral. While compounds containing more than one chiral carbon
may or may not be chiral.
Any molecule is optically active if
(a) it is non-superimposable on its mirror image.
(b) does not contain any element of symmetry

Elements of Symmetry
Elements of symmetry offer a simple device to decide whether a molecule is chiral or achiral, i.e., whether
the molecule is superimposable or non-superimposable on its mirror image. When a molecule lacks all element
of symmetry it is chiral.

(i) Plane of symmetry () : Any imaginary plane which divides the full molecule into two halves which are
mirror images of each other.
(ii) Centre of inversion (i) : Any imaginary point, through which when inversion will be carried out it will repeat
equivalent configuration.
Ph HO

H
Centre of
H Symmetry

HO Ph
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(iii) Alternating axis of symmetry (Sn) : A molecule posses an alternating axis of symmetry, if when rotated
through an angle of 360°/n about the axis and then followed by reflection in a plane perpendicular to the
axis, molecule comes to its equivalent form.

Enantiomers
Molecules which are non-superimposable mirror image of each other are called enantiomers.

(i) Enantiomers have the same physical properties except their optical rotation which is equal in magnitude
but in opposite direction.

(ii) Enantiomers have the same chemical behaviour towards the molecules which are optically inactive, but
they have different chemical behaviour towards the molecules which are optically active.

Diastereomers
Stereoisomers which are not mirror images of each other are called diastereomers. Diastereomers have different
physical as well as chemical properties.

Racemic Mixture
1 : 1 molar mixture of dextro and laevo forms of the same enantiomeric pair is optically inactive because the
optical rotation due to dextro form is compensated by the optical rotation due to laevo form. It is the case
of external compensation.

Resolution
The process of separation of racemic mixture into enantiomers (d, l form) is called resolution.

Racemization
It is reverse of resolution, i.e., the conversion of (+) or (–) isomer into its racemic mixture (±) is termed
Racemization. This involves the change of half of the active compound to the isomer of opposite rotation,
resulting the formation of racemic mixture i.e.,

2 (+) A  (+) A and (–) A or (±) mixture

or 2(–) A  (+) A and (–) A or (±) mixture

e.g., when a solution of (+) or (–) tartaric acid in water is heated under pressure, it is transformed into a
completely inactive mixture of racemic (or also meso) tartaric acid.

Retention
Retention of configuration is the preservation of the spatial arrangement of bonds to an asymmetric centre during
a chemical reaction or transformation.

In general, if during a reaction, no bond to the stereocentre is broken, the product will have the same general
configuration of groups around the stereocentre as that of reactant. Such a reaction is said to proceed with
retention of the configuration. Consider as an example, the reaction that takes place when (–)-2-methylbutan-
1-ol is heated with concentrated hydrochloric acid.
CH3 heat
OH H Cl H CH 2 Cl
H CH2 H OH
CH2 CH2

CH3 CH 3
(–)–2-Methylbutan-1-ol (+)-1-Chloro-2-methylbutane

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It is important to note that configuration at a symmetric centre in the reactant and product is same but the
sign of optical rotation has changed in the product. This is so because two different compounds with same
configuration at asymmetric centre may have different optical rotation. One may be dextrorotatory (plus sign
of optical rotation) while other may be laevorotatory (negative sign of optical rotation).

Mechanism of SN reaction :
The nucleophilic substitution proceeds mainly by two different mechanisms as described below:

(a) Substitution nucleophilic bimolecular SN2: Nucleophilic substitution bimolecular or SN2 is a single
step bimolecular reaction in which the incoming nucleophile attacks the C–atom of substrate in a
direction opposite to the outgoing nucleophile.

The reaction passes thorugh a transition state in which both the incoming and outgoing nucleophiles are
bonded to the same C–atom.

H
H H
– RDS   –
HO C
H Cl HO C Cl HO C H + Cl
H H H H
Incoming (Substrate) (Transition state) (Product) Outgoing
nucelophile nucelophile

In the transition state, the C-atom is sp2 hybridised with a p-orbital whose one lobe overlaps with an
orbital of incoming nucleophile and the other lobe overlaps with an orbital of outgoing nucleophile. The
three non-reacting atoms or groups attached to the C–atom are nearly co-planar at an angle of 120°.
The reaction is completed when the outgoing nucleophile leaves with the bond pair of electrons and
simultaneously the incoming nucleophile binds to the C–atom. As the reaction progresses, the
configuration of C–atom under attack is inverted. An SN2 reaction is always accompained by
inversion of configuration. The inversion in configuration implies change in configuration from R to S
or S to R (provided the incoming nucleophile and outgoing nucleophile have same priority) and not from
(+) to (–) or (–) to (+).
Steric hindrance plays a very vital role in an SN2 reaction. As steric hindrance increases the rate of
SN2 reaction decreases. Thus for the same halogen the reactivity order of alkyl halides towards SN2
reaction is as under.
Methyl halide > 1°-Alkyl halide > 2°-Alkyl halide > 3°-Alkyl halide

H H H
H H H
H C C H C
H H
Nu X X Nu X Nu X
H C
H H H H
H H C H C
H
H H
Methyl Ethyl 1° Isopropyl 2° Tert-butyl 3°
(30) (1) (0.02) (0)

For the same alkyl group, the reactivity of alkyl halides increases with the decrease in C—X bond
dissociation energy. Therefore, R—I > R—Br > R—Cl > R—F.

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(b) Substitution nucleophilic unimolecular (SN1) : Nucleophilic substitution unimolecular or SN1 is a two
step unimolecular reaction. The first step is the slow ionisation of substrate and is the rate-determining
step. The second step is the rapid reaction between the carbocation (formed in the first step) and the
nucleophile. SN1 reactions generally proceed in polar protic solvents such as H2O, CH3OH, CH3COOH
etc.
RDS
R—X  R   X (Step 1)

Fast
R  Nu  R — Nu (Step 2)

The energy required to break C—X bond in step 1, is compensated to a large extent by the solvation
of ionic intermediates formed in this step.
Since first step is the rate-determining step, the rate of reaction depends upon the concentration
of alkyl halide only and is independent of the concentration of nucleophile. Further, greater the stability
of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of reaction.
Thus reactivity order of alkyl halides towards SN1 reaction is
3°-Alkyl halide > 2°-Alkyl halide > 1°-Alkyl halide > Methyl halide
3°-Alkyl halides undergo nucleophilic substitution mainly by SN1 mechanism whereas 2°-Alkyl halides
undergo partly by SN2 and partly by SN1. If the solvent is non-polar, the SN1 mechanism stops working.
1°-Alkyl and methyl halides undergo nucleophilic substitution mainly by SN2.
For the same alkyl group, the reactivity order of alkyl halides is same towards SN1 as well as SN2 i.e.,
R—I > R—Br > R—Cl > R—F.
For the same reasons, allyl and benzyl halides follow SN1 mechanism because the carbocation formed
in each case is stabilised by resonance.

CH2 CH CH2 CH2 CH CH2

CH2 CH2 CH2 CH2 CH2

However, if the solvent is non-polar they also undergo nucleophilic substitution by SN2 mechanism.
Furthermore, the carbocation formed in the first step is sp2 hybridised and planar. It has a vacant
unhybridised p-orbital perpendicular to this plane. So, the attack of nucleophile has almost equal chance of
attack from either side of the plane. Thus, any optically active 3° alkyl halide gives nearly racemic mixture since
attack of the nucleophile from the side of the outgoing nucleophile is slightly sterically hindered.

R R R R
– –
–X Nu
R C X C R C Nu + Nu C R
(RDS) (Fast)
R R R R R
The general energy diagram of an SN1 reaction is as given below
TS1
TS2

(Intermediate)
Energy

(Reactant)

(Product)
Reaction co-ordinate

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Example 7 : What happens when optically active 3-bromo-3-methyl hexane is hydrolysed at room temperature?
Solution : 3-Bromo-3-methyl hexane undergoes hydrolysis by SN1 mechanism which leads to the formation of
a partially racemic mixture of 3-methyl-3-hexanol.

CH3 CH3
(RDS)
CH3CH2 C CH2CH2CH3 C + Br
Br CH3CH2 CH2CH2CH3
(Optically active)

CH3 CH3 CH3

H2O
C CH3CH2 C OH2 + H2O C CH2 CH3
CH3CH2 CH2CH2CH3 CH2CH2CH3 CH2CH2CH3

CH3
+
–H
CH3CH2 C OH
CH2CH2CH3
( )

EXERCISE
11. Amongst the C – X bond, the correct bond energy order is
(1) C – Cl > C – Br > C – I (2) C – I > C – Cl > C – Br
(3) C – Br > C – Cl > C – I (4) C – I > C – Br > C – Cl
12. Which of the following has highest boiling point?
(1) 1-chloropentane (2) 2-chloropentane
(3) 3-chloropentane (4) All have equal boiling point
13. Out of the following compounds which one will have zero dipole moment?
(1) Chloromethane (2) Dichloromethane
(3) Trichloromethane (4) Tetrachloromethane
14. Treatment of ammonia with excess ethyl chloride will yield
(1) Triethyl amine (2) Diethyl amine
(3) Ethyl amine (4) Tetraethyl ammonium chlorides
15. Identify Z in the following sequence
KCN dil. HCl
CH3CH2Br X Z

(1) CH3COCl (2) CH3CONH2
(3) CH3COOH (4) CH3CH2COOH
16. Most reactive halide towards SN1 reaction is
(1) n-butyl chloride (2) sec-butyl chloride
(3) tert-butyl chloride (4) Allyl chloride

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17. Which of the following is most reactive towards nucleophilic substitution reaction?
(1) CH2 = CH – Cl (2) C6H5Cl
(3) CH3 – CH = CH – Cl (4) Cl – CH2 – CH = CH2
18. NBS is a specific reagent for
(1) Nucleophilic substitution reaction (2) Electrophilic substitution reaction
(3) Allylic substitution (4) Electrophilic addition
19. For the reaction, C2H5OH + HX  C2H5X + H2O
The order of reactivity is
(1) HCl > HBr > HI (2) HI > HBr > HCl
(3) HBr > HCl > HI (4) HI > HCl > HBr
20. Which of the following is not true for SN1 reactions?
(1) They occur through a single step concerted reaction
(2) They are favoured by polar solvents
(3) 3° alkyl halides generally react through this mechanism
(4) Concentration of nucleophile does not affect the rate of such reaction

Elimination Reactions of Haloalkanes

Haloalkanes having -hydrogen atom when heated in presence of alc. KOH undergo dehydrohalogenation
forming alkene. The reaction may be brought about by any one of the following mechanisms.

E2 Elimination or -Elimination
It is a single-step bimolecular elimination reaction in which both the leaving groups i.e., halogen atom and
-hydrogen atom are lost simultaneously. The two leaving groups align themselves so that they are in the
same plane at 180° to each other before they are lost. The reaction passes through a transition state, as
shown below
–
H OC2H5
C2H5O– –
CH3 CH CH3 CH3 CH CH2 CH3CH CH2 + C2H5OH + Br

–
Br Br
Haloalkanes having two or more type of -hydrogen atoms on dehydrohalogenation gives more than one
alkenes. Major product is decided on the basis of Saytzeff’s rule which states that haloalkanes on heating
with alc. KOH gives more substituted alkene as the major product.
However, if bulky base like (CH3)3CO– is used then less substituted alkene is the major product (Hofmann’s
product) because bulky base finds it easier to remove proton from less sterically hindered -carbon atom. For
example, 2-bromobutane gives 2-butene as major product with C2H5OK  and 1-butene as major product with

(CH3 )3 COK  .
C2H5OK/C2H5OH
CH3 CH CH CH3 + CH2 CH CH2 CH3
(Major)
CH3 CH CH2 CH3
(CH3)3COK
Br CH3 CH CH CH3 + CH2 CH CH2 CH3

(Major)

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E1cB Elimination
If the leaving group is a strong base like F–, the reaction proceeds in two steps. In the first step, base removes
proton in the first-step forming carbanion as intermediate. The carbanion attacks the carbon atom containing
F atom with the elimination of F– ion in the second step. This is known as E1cb elimination unimolecular
through conjugate base, e.g.,

–
C2H5O
CH3 CH CH2 CH3 CH2 CH CH2 CH3 + CH3 CH CH CH3

F F F
(More stable) (less stable)

–
CH2 CH CH2 CH3 CH2 CH CH2 CH3 + F
(Major)
F

E1 Elimination
It is a two-step unimolecular elimination reaction. In first step haloalkane undergoes ionisation in the rate-
determining step with the formation of carbocation as intermediate. Carbocation looses proton from the adjacent
C-atom in the second-step forming alkene. For example, tert. butyl bromide on heating with H2O gives
isobutene as major product.

CH3 CH3
(RDS)
CH3 C Br C + Br
CH3 H3C CH3

CH2 H CH2

OH2 +
C C + H3O

H3C CH3 H3C CH3

Example 8 : Identify the final product (C) formed in the following sequence of reactions
alc.KOH HBr aq. KOH
CH3 CH2CH2Br  (A) 
 (B)  (C)

alc. KOH HBr aq. KOH

Solution : CH3CH2CH2Br CH3 CH CH2 CH3 CH CH3 CH3 CH CH3
(A)
Br OH
(B) (C)

Reaction with Metals

Haloalkanes react with metals in presence of dry ether forming organometallic compounds. The reactivity of
organometallic compounds depends on electropositivity of the metal.
Dry
 C2H5Br  2Li ether
C2H5Li  LiBr

2C2H5Li + Cul  (C2H5)2 CuLi + Lil

(Gilman reagent)

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Dry
 CH3I  Mg ether
CH3MgI
(Grignard's reagent)

Dry ether
 2CH3Cl + Na  CH3CH3 +2NaCl

This reaction is known as wurtz reaction which is not given by 3° alkyl halide.
Reactivity of haloalkanes to form organometallic compounds increases as R—F < R—Cl < R—Br < R—I.
In fact fluoroalkanes do not respond to this reaction.

Reduction of Haloalkanes
Haloalkanes can be reduced to alkanes by number of reducing agents like (i) Zn/CH3COOH, (ii) Zn/HCl,
(iii) Zn/NaOH, (iv) Zn-Cu couple/ethanol, (v) LiAlH4, (vi) NaBH4 etc.
Reduction
R X R H
The number of C-atoms as present in haloalkane remains the same in alkane.

EXERCISE

21. 2-chloro-2-methyl butane, on reaction with aq. KOH gives X as the major product. X is
(1) 2-butene (2) 2-methyl-1-butene
(3) 2-methyl-2-butene (4) 2-methyl-2-butanol
22. Which of the following does not form Grignard reagent on reaction with Mg in the presence of ether?
(1) Chloroethane (2) 1-chloropropane
(3) Bromobenzene (4) Vinyl chloride
23. 1-phenyl-2-chloropropane when treated with alcoholic KOH gives ______ as major product
(1) 1-phenyl propene (2) 3-phenyl propene
(3) 1-phenyl-2-propanol (4) 3-phenyl-1-propanol
24. An alkyl halide on reaction with sodium in the presence of ether gives 2, 2, 5, 5-tetramethyl hexane. The alkyl
halide possibly is
(1) 1-chloropentane
(2) 1-chloro-2, 2-dimethyl propane
(3) 3-chloro-2, 2-dimethyl butane
(4) 2-chloro-2-methyl butane
25. Ethyl bromide reacts with sodium lead alloy to form
(1) Tetraethyl lead (2) Ethyl sodium
(3) Ethane (4) Ethene
26. Allybromide on dehydrobromination gives
(1) Propadiene (2) Propylene
(3) Allyl alcohol (4) Acetone

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27. Identify the major product

CH3 alc KOH

Br

CH3
CH2 CH3
(1) (2) (3) (4)

28. Identify the most suitable reagent for the following conversion?

Reagent
CH3 – CH2 – CH – CH3 CH3 – CH = CH – CH3

Br
(1) Aqueous KOH (2) (CH3)3CO–K+ / 
(3) alc. KOH /  (4) All of these
29. Which of the following is an example of 1,2-elimation?

Reagent
(1) CH3 – CH – CH3 CH3 – CH = CH2
Br

Reagent
(2) CH2 – CH2 – CH2
X Y

Reagent
(3) H2C – CH2 – CH2 – CH2 H3C – CH = CH – CH3
X Y
(4) All of these
30. The carbocation is the intermediate in
(1) E2 (2) E1 (3) SNAr (4) Both (2) & (3)

REACTIONS OF HALOARENES
1. Nucleophilic Substitution Reactions: Haloarenes are very much less reactive than haloalkanes towards
nucleophilic substitution reactions for the following reasons :
(i) In haloalkanes the halogen is bonded to sp3 hybridised C-atom whereas in haloarenes, the halogen is
bonded to sp2 hybridised C-atom. The sp2 hybridised C-atom being more electronegative than sp3
hybridised C-atom, the C—X bond in haloarenes is less polar and shorter than in haloalkanes.
X H
sp2-hybrid carbon
sp3-hybrid carbon
C
R H X

(ii) In haloarenes the lone pair of electrons on halogen atom is delocalised with the -electrons of benzene
ring as shown below :

Cl Cl Cl Cl Cl

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It acquires partial double bond character in carbon halogen bond due to resonance. Therefore, it is more
difficult to break carbon-halogen bond in haloarenes than in haloalkanes.
(iii) The incoming nucleophile is repelled by the -cloud of benzene ring.
(iv) The self ionisation of haloarenes results in the formation of phenyl cations which is not stable. Therefore,
haloarenes cannot undergo nucleophilic substitution by SN1 mechanism.
But aryl halides undergo nucelophilic substitution reactions only in extreme conditions. For example,
phenol, alkyl aryl ether, aromatic amine and aryl cyanide can be synthesised from haloarenes under
drastic conditions. Aryl halides cannot be used in Friedel-Crafts reaction to generate an electrophilie.
+
aq. NaOH, 350°C – + H
300 atm
C6H5O Na C6H5 OH

aq. NH3, 200°C, Cu2O

C6H5 Cl C6H5 NH2
60 atm

CuCN in DMF, heat

C6H5 CN
However, by introducing at least one strongly electron-withdrawing group like –NO2, –CF3 either at the
ortho or at para position to the halogen atom makes aryl halides more susceptible to nucelophilic
substitution.

Cl OH
(i) aq. NaOH, 170°C
+
(ii) H

NO2 NO2

Cl OH
NO2 NO2
(i) aq. NaOH, 95°C
+
(ii) H

NO2 NO2

Cl OH
O2N NO2 O2N NO2
warm H2O

NO2 NO2

2. Electrophilic Aromatic Substitution Reactions: Haloarenes are somewhat less reactive than benzene
towards electrophilic substitution reactions of the benzene ring due to –I effect of halogen atom which
decreases the electron density of benzene. The halogen atom attached to benzene is o-, p- directing as it
stabilises the arenium ion (an intermediate) by resonance when an electrophile attacks the benzene ring either
at ortho or para position with respect to halogen atom.

Cl Cl Cl Cl Cl
H H H H
+
+ E E E E E

Cl
E
+
+ H

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Haloarenes undergo halogenation, nitration, sulphonation, and Friedel-Crafts reactions.

Cl Cl Cl
Br
Anh. FeBr3
(i) Halogenation: + Br2 +

(Minor)
Br
(Major)

Cl Cl Cl
NO2
Conc. HNO3 + Conc. H2SO4
(ii) Nitration: +

(Minor)
NO2
(Major)

Cl Cl Cl
SO3H
Conc. H2SO4, 
+
(iii) Sulphonation :
(Minor)
SO3H
(Major)

(iv) Friedel-Crafts reaction:

Cl Cl Cl
CH3
Anh. AlCl3
+ CH3Cl +

(Minor)
CH3
(Major)

Cl Cl Cl
O COCH3
Anh. AlCl3
+ CH3 C Cl +

(Minor)
COCH3
(Major)

3. Reaction with Metals:

Fittig reaction : Haloarenes when treated with sodium in presence of dry ether react in a manner simillar
to haloalkanes to give diarenes.

Cl
Dry ether
2 + 2Na + 2NaCl

(Diphenyl)

This is known as Fittig reaction.

Wurtz-Fittig reaction : When a mixture of haloalkane and haloarene is treated with sodium under identical
conditions we get alkyl arene.

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Cl CH3
Dry ether
+ CH3Cl + 2Na + 2NaCl

(Methylbenzene)

Ullmann reaction : It is a coupling reaction between aryl halides and copper.

473 K
I + 2Cu + l + Cu 2 l2

Diphenyl

Example 9 : Which one of the following compounds readily reacts with NaOH?
F F

CH3 NO2
NO2 CH3
(A) (B)

Solution : Compound (A) will readily react with NaOH because it has a strongly electron-withdrawing group (NO2)
at the para position with respect to fluorine.
F OH

+ NaOH
CH3 CH3
NO2 NO2
Compound (B), however, does not react with NaOH under normal conditions as it does not have NO2
group either at ortho or para with respect to fluorine.

Example 10 : Convert chlorobenzene to 4-nitrophenol.

Solution : Nitration of chlorobenzene results in the formation of 4-nitrochlorobenzene as the major product. This
is subjected to Dow’s process to give 4-nitrophenol.
Cl Cl OH
Conc. HNO3 + Conc. H2SO4 (i) 15% NaOH, 433 K
+
(ii) H

NO2 NO2
(Major)

POLYHALOGEN COMPOUNDS
Carbon compounds containing two or more than two halogen atoms are generally referred to as polyhalogen
compounds. A compound containing two halogen atoms is called dihalide and if it contains three halogen
atoms, it is called trihalide and so on.

Dihalides
 1, 1-Dihalides are commonly called gem dihalides or alkylidene dihalides.
They can be synthesised as :
(i) By the addition of halogen acids to alkynes e.g., ethylidene dibromide from acetylene and HBr.
HBr
HC CH + HBr H2C CHBr H 3C CHBr2

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NEET Haloalkanes and Haloarenes 165
(ii) By the action of phosphorus pentahalides on aldehydes or ketones. e.g., acetone gives
isopropylidene dichloride when treated with PCl5.
CH3COCH3 + PCl5 CH3CCl2CH3 + POCl3
 1, 2-Dihalides are commoly called vicinal dihalides or alkylene dihalides.
They can be synthesised by the addition of Cl2 or Br2 to alkenes in CCl4.
CCl4
CH3 CH CH2 + Cl2 CH3 CHCl CH2Cl
Addition of I2 to alkenes generally does not take place because 1,2-diiodides are not stable and they
tend to eliminate I2 particularly at raised temperature.
Reactions :

(i) 1,1-Dihalides are hydrolysed by aqueous alkali to carbonyl compounds. Terminal gem dihalides on
hydrolysis give aldehydes whereas internal gem halides on hydrolysis give ketones.

CH3CH2CHCl2 + aq. KOH [CH3CH2CH(OH) 2] CH3CH2CH O

CH3CCl2CH3 + aq. KOH [CH3CH(OH)2CH3] CH3COCH3

(ii) 1,1-Dihalides undergo dehydrohalogenation when treated with NaNH2 to give alkynes.

CH3CH2CHCl2 + NaNH2 CH3C CH

or
CH3CCl2CH3

(iii) 1,2-Dihalides are hydrolysed by aqueous alkali to vicinal diols.

CH3 CHCl CH2Cl + aq. 2KOH CH3 CH(OH) CH2(OH) + 2KCl
(iv) 1,2-Dihalides also undergo dehydrohalogenation to give alkyne.
CH3CHCl CHCl CH3 + NaNH2 CH3C CCH3

Trihalogen Derivatives
The most important trihalogen derivatives are those of methane i.e., CHCl3 (Chloroform), CHBr3 (Bromoform)
and CHI3 (Iodoform). They may be prepared by heating ethanol or acetone with X2 (Cl2, Br2 or I2) and alkali.

NaOH
CH3CH2OH + X2 CHX3 + HCOONa
NaOH
CH 3COCH 3 + X 2 CH3COONa + CHX3

Reactions :
(i) Chloroform when heated in presence of silver yields acetylene.

2CHCl3 + 6Ag HC CH + 6AgCl

(ii) It is oxidised by oxygen present in air in presence of sunlight to a deadly poisonous gas phosgene.
h
2CHCl 3 + O2 2COCl2 + 2HCl

(iii) With conc. HNO3, chloroform gives chloropicrin.

CHCl3 + HNO3 CCl3NO2 + H2O

Chloropicrin is used as an insecticide.

(iv) Aliphatic as well as aromatic primary amines react with chloroform and alc. KOH to give a foul smelling
isocyanide. This is known as carbylamine test.

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166 Haloalkanes and Haloarenes NEET

R NH2 R N C
or + CHCl3 + 3KOH or + 3KCl + 3H2O
Ar NH2 Ar N C

(v) On heating with chloral in the presence of concentrated sulphuric acid, chlorobenzene forms
(p, p' - dichlorodiphenyl trichloroethane (commonly known as DDT-a powerful insecticide.) But it is not
biodegradable because it contains halogen atoms attached to the benzene rings.

Cl H H Cl
Cl H Cl
Cl – C – C = O + Conc. H2SO4
  Cl – C – C
Cl H Cl H2O
Chloral Cl Cl
Chlorobenzen e
DDT
(2 molecules)
2, 2–bis–(p–chlorophenyl )
1, 1, 1– trichloroethane

Tetrahalogen Derivatives
Carbon tetrachloride is synthesised by the action of Cl2 on CS2 in presence of AlCl3 as a catalyst.
AlCl3
CS2 + 3Cl2 CCl4 + S2Cl2
Carbon tetrachloride is a liquid insoluble in water but readily soluble in ethanol or ether. It reacts with water
vapours at 500°C to form phosgene.
500°C
CCl4 + H2O COCl2 + 2HCl

EXERCISE
31. IUPAC name of Gammexene is

(1) Hexachlorobenzene (2) Benzene hexachloride

(3) 1, 2, 3, 4, 5, 6-hexachlorocyclohexane (4) All of these

32. ‘Pyrene’ is the trade name of _____ when used as fire extinguisher

(1) CO2 (2) CHCl3

(3) CCl4 (4) CH2Cl2

33. Which of the following with aqueous KOH will give acetaldehyde?
(1) 1, 2-dichloroethane (2) 1, 1-dichloroethane
(3) Chloroacetic acid (4) Ethyl chloride
34. Carbylamine test is performed in alcoholic KOH by heating a mixture of
(1) Chloroform and silver powder (2) Chloroform and a primary amine
(3) An alkyl halide and a primary amine (4) An alkyl cyanide and a primary amine
35. What happens when CCl4 is treated with AgNO3 solution?
(1) NO2 will be evolved (2) A white ppt. of AgCl will form

(3) CCl4 will dissolve in AgNO3 solution (4) Nothing will happen

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NEET Haloalkanes and Haloarenes 167
36. DDT is formed from

(1) Benzene and chlorobenzene (2) Chloral and chlorobenzene

(3) Chloral and benzene (4) Chlorobenzene and chlorine

37. Iodoform in medicines is used as

(1) Antibiotic (2) Antiseptic

(3) Analgesic (4) Antipyretic

38. If chloroform is left open in the air in the presence of sunlight

(1) Explosion takes place (2) Polymerisation takes place

(3) Poisonous gas phosgene is formed (4) No reaction takes place

39. Non-sticking frying pans are coated with teflon which is polymer of

(1) Ethylene (2) Styrene

(3) Tetrafluoro ethylene (4) Chloro-fluoromethane

40. The correct decreasing order of dipole moment?

(1) CH3Cl, CH3Br, CH3F (2) CH3Cl, CH3F, CH3Br

(3) CH3Br, CH3Cl, CH3F (4) CH3Br, CH3F, CH3Cl

  

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 t
en
nm nment
sig ssig
As A Assignment

Assignment
6. Which among the given compounds has highest
SECTION - A dipole moment? [NCERT Pg. 294]
NCERT Based MCQs
(1) CH3F
1. The substrate which will racemise in presence of
aqueous NaOH is [NCERT Pg. 308] (2) CH3Br
CH3 Ph (3) CH3l
(1) Br (2) CH 3
(4) CH3Cl
H 3C H H 3C Br
Ph Ph 7. Major product (P) of the given reaction is
[NCERT Pg. 309]
(3) C2H5 (4) CH3
Br CH3 Ph Br C 2H 5OH/KOH
Ph CH 2 CH CH 3 P

2. The compound which will undergo SN1 reaction Cl
most readily is [NCERT Pg. 303]
(1) Ph CH2 CH = CH2
Br
(1) (2)
Br (2) Ph CH2 CH CH 3

Br OH
(3) (4) Br
(3) Ph – CH = CH – CH3
3. Consider the following reaction [NCERT Pg. 295]

CH3 – CH = CH2
HBr
P(Major)
(4) Ph – CH – CH2 CH3
OH
Major product P is
(1) CH3CH2CH2Br (2) BrCH2 – CH = CH2 8. When 3-phenylprop-1-ene is treated with HBr in
presence of peroxide then the major product of the
(3) CH2 – CH – Br (4) CH3 – CH – CH 3
reaction among the following is [NCERT Pg. 297]
CH2 Br

4. Which among the following is an ambidented (1) (2)

Br
nucleophile? [NCERT Pg. 300]
Br
(1) NO2
Br
(2) NH3 (3) (4)
Br
(3) RCOO
9. For the given alkyl halides CH3 – F, CH3 – Cl, CH3 –
(4) CH3NH2
Br and CH 3 – I the correct order of reactivity
5. The incorrect statement among the following is towards SN2 reaction is [NCERT Pg. 304]
[NCERT Pg. 308]
(1) CH3 – F > CH3 – Cl > CH3 – Br > CH3 – I
(1) Chiral molecules are optically active
(2) CH3 – Cl > CH3 – F > CH3 – Br > CH3 – I
(2) Achiral molecules are optically inactive
(3) Enantiomers have same refrective index (3) CH3 – I > CH3 – Cl > CH3 – Br > CH3 – F

(4) Enantiomers have different melting point (4) CH3 – I > CH3 – Br > CH3 – Cl > CH3 – F

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NEET Haloalkanes and Haloarenes 169
10. Which among the following is Darzen’s method of
preparation of alkyl chloride? [NCERT Pg. 294]
SECTION - B

PCl5 Objective Type Questions

(1) CH3CH2OH   CH3CH2Cl
1. Which one is not example of allylic chloride?
HCl
(2) CH3CH2OH 
ZnCl2
 CH3CH2Cl CH3
(1) CH3 – CH = CH – CH – Cl
PCl
(3) CH3CH2OH 
3  CH CH Cl
3 2
(2) CH2 = CH – CH2 – Cl
SOCl2
(4) CH3CH2OH 
 CH3CH2Cl Cl
(3)
11. The correct reagent that gives vic-dibromide on
reaction with but-2-ene is [NCERT Pg. 295]
Cl
(1) NBS
(4)
(2) Cu2Br2
(3) HBr
2. Which will have highest boiling point?
(4) Br2/CCl4 (1) CH3–CH3 (2) CH3–CH2–I
12. Enantiomers are chiral molecules (3) CH3–CH2–Cl (4) CH3–CH2–Br
[NCERT Pg. 307]
3. One having highest melting point is
(1) Superimposable mirror image
Cl Cl
(2) Non-superimposable mirror images Cl
(1) (2)
Cl
(3) Symmetric molecules
Cl
(4) Achiral molecules
CH3
13. Best reagent for converting an alcohol into (3) (4)
corresponding chloride is [NCERT Pg. 294] CH3 –CH–CH2 –CH3
Cl
(1) HCl/ZnCl2
CH 3
(2) PCl3 Fe
4. + Cl 2 dark A ; Compound "A" is
(3) PCl5 CH3
(4) SOCl2
CH 3 CHCl2
14. IUPAC name of neopentyl bromide is
[NCERT Pg. 291] (1) (2)
CH 2Cl CH 2Cl
(1) 1-Bromo-4, 4-dimethylpentane
CH3
(2) 1-Bromo-3-methylbutane CH3

(3) 1-Bromo-2, 2-dimethylpropane (3) Cl (4) CH3

CH3 Cl
(4) 2-Bromo-2-methylbutane
5. One which is not an ambident nucleophile is
15. Swart’s reaction is used for the preparation of
[NCERT Pg. 296] (1) – C  N
(1) CH3 – Cl
(2) O – N = O
(2) CH3 – Br
(3) CH3 – I (3) SCN

(4) CH3 – F (4) Cl

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170 Haloalkanes and Haloarenes NEET

6. Which shows Swarts Reaction? 10. t-butyl chloride combine with hydroxy anion by
which mechanism?
(1) CH3CH2Br + AgF  CH3CH2F + AgBr
(1) SN1 (2) SN2
(2) 3CH3CH–CHCH3+2SbF3  (3) SE1 (4) SE2
| |
Cl Cl 11. The intermediate involved during the chlorination of
propylene to allyl chloride is
3CH3CH–CHC H3+2SbCl3
| |  
(1) C H2 – CH = CH2 (2) CH2 – CH = CH2
F F

(3) CH2 – CH = CH2 (4) None of these
UV light
(3) CH3 CHCl2  Cl2  CH3 CCl3  HCl
Br 2 (I) Mg/ether
12. A B
(4) Both (1) & (2) Fe (II) D2O

7. The decreasing order of rate of SN1 reaction is in Compound "B" is

D
Br Br
Ph – C – Ph Ph – CH – Ph
(1) (2)
CH3
(I) (II)
D
Ph – CH – CH3 PhCH2 Br (3) (4)
Br
(III) (IV) 13. Which reagent cannot be used to prepare an alkyl
chloride from an alcohol?
(1) (I) > (II) > (III) > (IV)
(1) HCl + anhydrous ZnCl2
(2) (II) > (I) > (III) > (IV) (2) NaCl
(3) (IV) > (III) > (II) > (I) (3) PCl5
(4) (III) > (IV) > (II) > (I) (4) SOCl2

8. Predict the most reactive towards SN1 reaction 14. Chlorobenzene reacts with Mg in dry ether to give
a compound (A) which further reacts with ethanol
Cl to yield
CH3
(1) (2) Br (1) Phenol (2) Benzene
(3) Ethyl benzene (4) Phenyl

I CH3 Cl 15. The iodoform test is not given by

O
(3) (4)
(1) I 3C – C – H
9. The correct order of nucleophilic substitution in OH
(i) Chlorobenzene (2) CH3 – CH 2 – CH – Ph
(ii) o-nitrochlorobenzene OH
(iii) 1, 3-dinitrochlorobenzene (3) CH3 – C – CH2 – CH 3
(iv) 1, 3, 5-trinitrochlorobenzene H

(1) (iv) > (ii) > (iii) > (i) (2) (iv) > (iii) > (ii) > (i) O
(3) (iii) > (iv) > (ii) > (i) (4) (iii) > (iv) > (i) > (ii) (4) Ph – CH2 – C – CH3

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NEET Haloalkanes and Haloarenes 171
16. Chloropicrin is not
CH2 – CH = CH – CH3
(1) Insecticide (2) War gas
22. NBS
 "A" . Compound "A" is
(3) Nitrochloroform (4) Hypnotic drug Major

17. The decreasing order of rate of SN1 reaction _____ in

(1) Ph – CH2 – CH = CH – CH2Br
CH3
(CH3 )3 C – Cl Ph – C – Cl Br
CH3 (2) Ph – CH – CH = CH – CH 3
(I) (II)

(CH3 )2 CH – Cl CH3 – CH 2 – CH2 – Cl (3) Ph – CH2 – CH 2 – CH – CH3

(III) (IV) Br

(1) (II) > (I) > (IV) > (III) (2) (II) > (I) > (III) > (IV)
(4) Ph – CH2 – CH – CH2 – CH3
(3) (IV) > (III) > (II) > (I) (4) (III) > (II) > (IV) > (I) Br

Br 23. Arrange CH 3F, CH 3Cl, CH 3Br and CH3I in the

18. + KSH A ; Compound "A" is
CH2Br decreasing order of their dipole moment.

(1) CH3Cl > CH3F > CH3Br > CH3I

SH SH
(1) (2) (2) CH3F > CH3Cl > CH3Br > CH3I
CH2SH CH2Br
(3) CH3F > CH3Br > CH3Cl > CH3I
Br
(4) CH3I > CH3Br > CH3Cl > CH3F
(3) (4)
CH2SH
24. Arrange CCl4, CHCl3, CH2Cl2 and CH3Cl in the
19. C–Cl bond of chlorobenzene in comparision to decreasing order of density.
C–Cl bond in methyl chloride is
(1) CH3Cl > CH2Cl2 > CHCl3 > CCl4
(1) Longer and weaker (2) Shorter and weaker
(2) CCl4 > CHCl3 > CH2Cl2 > CH3Cl
(3) Shorter and stronger (4) Longer and stronger
(3) CCl4 > CHCl3 > CH3Cl > CH2Cl2
CH3
Alc. KCN (4) CHCl3 > CCl4 > CH2Cl2 > CH3Cl
20. CH3 – C – Br A ; Compound "A" is
CH3 Major 25. Arrange the following in decreasing order of
reactivity CH3F, CH3Br, CH3Cl, CH3I.
CH3
(1) (CH3)3C – CN (2) C = CH 2 (1) CH3I > CH3Br > CH3Cl > CH3F
CH3
CH3
(2) CH3F > CH3Cl > CH3Br > CH3I
(3) CH–CH2–CN (4) CH3–CH2–CH = CH2
CH3
(3) CH3Cl > CH3F > CH3Br > CH3I

21. The compound which react faster in S N 2 (4) CH3I > CH3Br > CH3F > CH3Cl
mechanism is 26. Arrange the following in order of reactivity for SN2
(1) (CH3)3 C – Cl reaction in aprotic DMSO.
(A) CH3CH(CH3)CH2CH2I
(2) CH2 = CH – Cl
(B) CH3CH(CH3)CHICH3
(3) CH2 = CH – CH2 – Cl
(C) CH3CI(CH3)CH2CH3
Cl
(1) A > B > C (2) C > B > A
(4)
(3) B > A > C (4) B > C > A
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172 Haloalkanes and Haloarenes NEET

27. What will be the ratio of 1, 2 and 3 chloropentane 34. Which of the following possess highest melting
if n-pentane is made to undergo chlorination? point?
Assume that reactivity of all H-atoms is same.
(1) Chlorobenzene (2) o-dichlorobenzene
(1) 33.33% each
(3) m-dichlorobenzene (4) p-dichlorobenzene
(2) 16.67% : 33.33% : 50%
35. Aryl halides are less reactive towards nucleophilic
(3) 25% : 25% : 50% substitution reaction as compared to alkyl halides
(4) 50% : 33.33% : 16.67% due to
28. When 2-chlorobutane is treated with alcoholic (1) The formation of less stable carbonium ion
KOH, the major product is
(2) Resonance stablization
(1) 1-Butene (2) 2-Butene
(3) Longer carbon-halogen bond
(3) 1-Butyne (4) 2-Butyne
(4) The inductive effect
29. When 1-chloropropane is treated with alcoholic
36. Silver benzoate reacts with bromine to form
KOH it forms an alkene. This reaction is
COOBr
(1) Substituion reaction (2) Elimination reaction
(1) (2)
(3) Addition reaction (4) Dehydration reaction
30. Which of the following methods is used to prepare COOAg
chlorobenzene?
(3) (4) C6H5Br
(1) Action of PCl5 on phenol
Br
(2) Action of HCl on phenol in presence of ZnCl2
37. For chloro-alkanes containing the same halogen,
(3) Action of diazonium chloride with CuCl/HCl reactivity is maximum for
(4) Direct chlorination of toluene (1) Primary halide (2) Secondary halide
31. The halogen atom in aryl halides is (3) Tertiary halide (4) All of these
(1) o-, p- directing 38. An alkyl halide by formation of Grignard’s reagent
(2) m-directing followed by hydrolysis yield butane. What is the
original alkyl halide?
(3) Both o-, p- and m-directing
(1) Ethyl halide (2) Propyl halide
(4) None of these
(3) Butyl halide (4) Methyl halide
32. Which of the following compounds is formed in the
given reaction? 39. How many dibromo derivatives of an aromatic
hydrocarbon with molecular mass 78 can be?
Sunlight
C6H6 + Cl2  
(1) 4 (2) 3
(excess)
(3) 2 (4) 5
(1) Chlorobenzene (2) Dichlorobenzene
40. For the reaction
(3) Hexachlorobenzene (4) Benzene hexachloride
ZnCl
33. Iodoform is formed on warming iodine and sodium C 2H 5 OH + HX 
2
 C 2 H 5X the order of
hydroxide with reactivity is
(1) C2H5OH (1) HBr > HI > HCl
(2) CH3OH (2) Hl > HCl > HBr

(3) CH3CH2CHO (3) Hl > HBr > HCl

(4) CH3CH2CHOHCH2CH3 (4) HCl > HBr > Hl

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NEET Haloalkanes and Haloarenes 173
41. Which one of the following is known as westrosol? 47. The intermediate during the addition of HCl to
propene in presence of peroxide is
(1) Acetylene tetrachloride
(1) CH 3 CH CH2 Cl
(2) Acetylene dichloride
(2) CH3 CH CH3
(3) Trichloroethylene
(4) Vinylchloride (3) CH 3 CH 2 CH 2

42. Identify Z in the following series. (4) CH 3 CH 2 CH 2

alc KOH
2 Y KCN Br Br2/heat
C2H5I 
 X   Z 48. P ; P is

(1) CH – CH2 – CN
Br
(2) CN – CH2 – CH2 – CN (1)
(3) Br – CH2 – CH2 – CN
(4) Br – CH = CHCN Br

43. Clevage of an ether with excess of HI gave only (2)

one alkyl iodide which contains 74.7% iodine. The
hydrolysis of alkyl halides gives 2° alcohol. The
ether is Br
(1) Diethyl ether (2) Diisopropyl ether (3)
Br
(3) Diisobutyl ether (4) None of these
44. The best reagent to distinguish between ethylidene Br Br
chloride and ethylene chloride is to treat them (4)
separately with
(1) Aq. KOH 49. The addition of propene with HOCl proceeds via the
addition of
(2) Alc. KOH
(1) H+ in the first step
(3) Zn/CH3OH
(2) Cl+ in the first step
(4) Zinc dust
(3) OH– in the first step
45. Allyl bromide and n-propyl bromide can be
(4) Either H+ or OH– in first step
distinguished by
(1) AgNO3 CH3

(2) NaOH Cl2/h

50. P; Product (P) is
(3) Tollen's reagent
(4) Baeyer's reagent CH3 CH3
46. CH2 CH NO2 + HBr P ; The major Cl
product ‘P’ is (1) (2)

(1) CH2 CH2 NO2 (2) CH3 CH NO2 Cl

Br Br
CH2 Cl CH3
(3) CH2 C NO2 (4) CH 2 CH Br
(3) (4)
Br Cl
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174 Haloalkanes and Haloarenes NEET

CH3 Cl Cl
Br2/h
51. P (Major) , Product (P) is
(1) (2)
NH2
CH2Br CH3 NH2
(1) (2) Cl NH2
Br NH2
(3) (4)
CH3 CH3
(3) Br (4) 55. The reaction,
Br
CH3Br + OH CH3OH + Br
CH3
KOH (alcoholic) HBr/R2O2 proceeds mainly by the mechanism
52. CH3 C Br (A) (B) ,

(Major) (1) SN1 (2) SN2
CH3
(B) is
(3) SE1 (4) SE2

CH3 CH3 H aq. KOH H

56. Cl HO
H H
(1) CH3 C CH2 Br (2) CH3 C CH3 H H
H Br The reaction goes through

(1) SN1 (2) SN2

CH3 CH2 H (3) E2 (4) E1
(3) CH3 C OH (4) CH3 C Br
CH2 Br CH3
SECTION - C
Previous Years Questions
CH2 OH 1. The alkane that gives only one mono-chloro product
SOCl2 KCN H
+
on chlorination with Cl2 in presence of diffused
53. (A) (B) (C), sunlight is [NEET-2019 (Odisha)]
(1) Isopentane (2) 2, 2-dimethylbutane
the compound (C) is
(3) neopentane (4) n-pentane
2. The hydrolysis reaction that takes place at the
COOH CH3
slowest rate, among the following is

(1) (2) [NEET-2019 (Odisha)]

COOH aq. NaOH

(1) CH2Cl CH2OH
CH2 COOH CH3

aq. NaOH
– +
(3) (4) (2) Cl ONa
COOH
CH3 CH3

Cl
(3) H3 C – CH2 – Cl 
aq.NaOH
 H3C – CH2 –OH
Liq. NH3
54. + KNH2 (P). Product (P) is aq. NaOH
(4) H2C  CH–CH2Cl H2C = CH – CH2OH

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NEET Haloalkanes and Haloarenes 175
3. Among the following, the reaction that proceeds OCH3
through an electrophilic substitution, is: Br
[NEET-2019] (3) and cine substitution reaction

+ –
Cu 2Cl2
(1) N2CI CI + N2 OCH3

(4) and cine substitution reaction

AlCl 3
(2) + Cl2 CI + HCl

6. Consider the reaction

CI CI CH3CH2CH2Br + NaCN  CH3CH2CH2CN + NaBr

This reaction will be the fastest in
UV light
(3) + Cl2 CI CI [NEET-(Phase-2)-2016]
(1) Ethanol
CI CI (2) Methanol
(3) N, N-dimethylformamide (DMF)
heat
(4) CH 2OH + HCl (4) Water
7. Which of the following biphenyl is optically active?
CH2Cl + H2O
[NEET-2016]

4. The compound C 7H 8 undergoes the following CH3 O2N

reactions: [NEET-2018] (1) (2)

C7H8 
2 3Cl / 
 A 
2 Zn/HCl
 B  C
Br /Fe
CH3 I

Br Br I
The product 'C' is
(1) m-bromotoluene (3) (4)

(2) o-bromotoluene I I I
8. Two possible stereo-structures of CH3CHOH.COOH,
(3) p-bromotoluene
which are optically active, are called
(4) 3-bromo-2,4,6-trichlorotoluene
[Re-AIPMT-2015]
5. Identify A and predict the type of reaction
(1) Enantiomers
OCH3
(2) Mesomers
NaNH2 (3) Diastereomers
A [NEET-2017]
Br (4) Atropisomers
OCH3 9. In an SN1 reaction on chiral centres, there is
[Re-AIPMT-2015]
(1) and substitution reaction
(1) 100% retention
NH2
(2) 100% inversion
OCH3
NH2 (3) 100% racemization
(2) and elimination addition reaction (4) Inversion more than retention leading to partial
racemization
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176 Haloalkanes and Haloarenes NEET

10. In which of the following compounds, the C – Cl (1) (i) and (ii) (2) (ii) and (iv)
bond ionisation shall give most stable carbonium
(3) (iii) and (iv) (4) (iv) only
ion? [AIPMT-2015]
13. In the replacement reaction
H H3C H
(1) C — Cl (2) C — Cl C –  + MF C – F + M
O2NH2C H H3C
The reaction will be most favourable if M happens
H to be [AIPMT (Mains)-2012]
H3C CH — Cl
C — Cl (1) Na (2) K
(3) H C (4)
3
CH3 (3) Rb (4) Li

11. What products are formed when the following 14. Which of the following compounds undergoes
compound is treated with Br2 in the presence of nucleophilic substitution reaction most easily ?
FeBr3? [AIPMT (Mains)-2011]
CH3
Cl
Cl
(1) (2)
CH3
OCH3
[AIPMT-2014]
Cl
CH3 CH 3
Cl
Br
(3) (4)
(1) and NO2
CH3 CH3 CH3
Br 15. Consider the reactions
CH3 CH3
Br Br (i) (CH3)2CH – CH2Br
(2) and
 (CH3)2CH–CH2OC2H5 + HBr
C H OH
CH3 CH3 
2 5

CH3 CH3 (ii) (CH3)2CH – CH2Br

Br

(3) and C2H5 O
  (CH3)2CH–CH2OC2H5 + Br
–

CH3 CH3
Br The mechanisms of reactions (i) and (ii) are
CH3 respectively [AIPMT (Mains)-2011]
CH3
(1) SN2 and SN2 (2) SN2 and SN1
(4) and
CH3 (3) SN1 and SN2 (4) SN1 and SN1
Br CH3
Br
16. The correct order of increasing reactivity of C–X
12. Which of the following compounds will undergo
bond towards nucleophile in the following compound
racemisation when solution of KOH hydrolyses ?
is
CH2Cl X
(i) X
NO2
(ii) CH3CH2CH2Cl I. II.

CH3 NO2
(iii) H3C–CH–CH2Cl III. (CH3)3C – X IV. (CH3)2CH – X
CH3 [AIPMT (Prelims)-2010]
(iv) H C Cl (1) I < II < IV < III (2) II < III < I < IV
C2H5
[AIPMT-2014] (3) IV < III < I < II (4) III < II < I < IV

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NEET Haloalkanes and Haloarenes 177
17. Which one is most reactive towards SN1 reaction ? 21. In a SN2 substitution reaction of the type
[AIPMT (Prelims)-2010] DMF 
R–Br+Cl–  R – Cl + Br–
(1) C6H5CH(C6H5)Br
Which one of the following has the highest relative
(2) C6H5CH(CH3)Br rate? [AIPMT (Prelims)-2008]

(3) C6H5C(CH3) (C6H5)Br (1) CH3CH2Br

(4) C6H5CH2Br (2) CH3–CH2–CH2Br

1. Mg,Ether (3) CH3 – CH – CH2Br

18. In the following reaction, C6H5CH2Br 
2. H O
 X,
3
CH3
the product 'X' is [AIPMT (Mains)-2010]
CH3
(1) C6H5CH2OCH2C6H5
(4) CH3 – C – CH2Br
(2) C6H5CH2OH
CH3
(3) C6H5CH3

(4) C6H5CH2CH2C6H5 22. How many stereoisomers does the molecules

have? CH3CH = CHCH2CHBrCH3
19. Which of the following reactions is an example of
[AIPMT (Prelims)-2008]
nucleophilic substitution reaction?
(1) 2
[AIPMT (Prelims)-2009]
(2) 4
(1) 2 RX + 2 Na  R–R + 2 NaX
(3) 6
(2) RX + H2  RH + HX
(4) 8
(3) RX + Mg  RMgX
23. If there is no rotation of plane polarized light by a
(4) RX + KOH ROH + KX compound in a specific solvent, thought to be
chiral, it may mean that [AIPMT (Prelims)-2007]
20. H3C–CH–CH=CH2 + HBr  A
(1) The compound may be a racemic mixture
CH3
(2) The compound is certainly a chiral
A (predominantly) is [AIPMT (Prelims)-2008] (3) The compound is certainly meso

(1) CH3–CH–CH–CH3 (4) There is no compound in the solvent

CH3 Br 24. For the following

a. I–
(2) CH3–CH–CH2–CH2Br
b. Cl–
CH3
c. Br–

Br The increasing order of nucleophilicity would be

[AIPMT (Prelims)-2007]
(3) CH3–C–CH2CH3
(1) Br– < Cl– < I–
CH3
(2) I– < Br– < Cl–
(4) CH3–CH–CH–CH3 (3) Cl– < Br– < l–
Br CH3 (4) I– < Cl– < Br–

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178 Haloalkanes and Haloarenes NEET

25. Which of the following undergoes nucleophilic 29. In Finkelstein reaction when acetone is replaced by
substitution exclusively by SN1 mechanism ? water then
[AIPMT (Prelims)-2005] (1) Reaction occurs in forward direction via SN1
(1) Benzyl chloride pathway

(2) Ethyl chloride (2) Reaction occurs in forward direction via SN2
pathway
(3) Chlorobenzene
(4) Isopropyl chloride (3) Reaction occurs in backward because NaCl
formed in right hand side is soluble in water
Br and cannot ppt. out
C
26. The chirality of the compound H3C H is (4) Reaction is not possible
Cl
[AIPMT (Prelims)-2005] 30. Which graph is incorrect?
(1) R (2) S
(3) Z (4) E

Rate
Questions asked Prior to Medical Ent. Exams. 2005 (1)
+
H3O CaO
27. C5H11Br + NaCN A B + NaOH C Base
‘C’ has the formula C5H12 which can give four
structural isomeric monochloro derivative. What is For Ph3CBr + OH–  Product
the structure of C5H11Br?

(1) CH 3—CH — CH2—CH2—Br

Rate

CH3 (2)

Br Base

(2) CH3 CH 2 CH CH 2 CH3

For (CH3)2CHBr + OH– (in acetone) Product

(3) CH3—CH2 — CH2—CH—CH3

Br
Ea
CH3 (3)

(4) CH2—C — CH2—Br Reaction

Progress
CH3

28. CD2 = CH – CH2 – Br is subjected to SN1 and SN2 For PhCHDBr + KOH  Product
reactions separately, which of the following statement
is correct?
(1) Both SN1 and SN2 give two products Ea
(2) Both SN1 and SN2 give only one product (4)

(3) SN1 gives two products but SN2 gives only Reaction
one product Progress
(4) SN1 gives one product but S N2 gives two
products For Me2(Et) COTs + NH3  Product

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NEET Haloalkanes and Haloarenes 179
31. Which will undergo fastest SN2 substitution reaction 35. When CH3CH2CHCl2 is treated with NaNH2, the
when treated with NaOH? product formed is
(1) CH3 – CH = CH2
CH3
(2) CH3 – C  CH
(1) H5C2 C Br NH2
(3) CH 3CH 2CH
H NH2
CH3 Cl
(4) CH3CH2CH
(2) CH3 C Br NH2
36. Grignard reagent is prepared by the reaction between
CH3 (1) Magnesium and alkane
CH3 (2) Magnesium and aromatic hydrocarbon
(3) Zinc and alkyl halide
(3) D C Br
(4) Magnesium and alkyl halide
C2H5 37. The following reaction is described as

H H3C – (CH2)5 (CH2)5CH3

H H
–
(4) H C OH
CH2 — CH2 — CH3 C Br HO C

Br
H3C CH3
32. Given reaction
(1) SN2 (2) SN0
ether (i) H O (3) SE2 (4) SN1
Br Mg X (ii) HCl
2
Y(main product)
38. 2-Bromopentane is heated with potassium ethoxide
Y in the reaction is in ethanol. The major product obtained is
(1) trans-pentene-2 (2) Pentene-1
(1) Hexane
(3) 2-ethoxypentane (4) cis-pentene-2
(2) Cyclohexane
39. Which of the following compounds is not chiral?
(3) Cyclohexylcyclohexane
(1) CH3CHDCH2Cl (2) CH3CH2CHDCl
(4) Cyclohexylether
(3) DCH2CH2CH2Cl (4) CH3CHClCH2D
33. Which one of the following alkyl bromides undergoes
most rapid solvolysis in methanol solution to give 40. An organic compound A(C4H9Cl) on reaction with Na/
corresponding methyl ether? diethyl ether gives a hydrocarbon which on
monochlorination gives only one chloro derivative
CH2 Br
(1) =C (2) then, A is
Br
(1) t-butyl chloride
(3) (4) (2) Secondary butyl chloride
Br
(3) Isobutyl chloride
Br
(4) n-butyl chloride
34. Monobromination of 2-methylbutane gives how many
distinct isomers? 41. Reactivity order of halides for dehydrohalogenation is
(1) One (1) R – F > R – Cl > R – Br > R – I
(2) Two (2) R – I > R – Br > R – Cl > R – F

(3) Three (3) R – I > R – Cl > R – Br > R – F

(4) Four (4) R – F > R – l > R – Br > R – Cl

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180 Haloalkanes and Haloarenes NEET

NaCN Ni / H2 4. Consider the following groups

42. CH3CH2Cl X Y
OAC; OMe; O SO2 Me and
Y in the above reacting sequence is
I II III
(1) CH3CH2CH2NHCOCH3 O SO2 CF3
(2) CH3CH2CH2NH2 IV
(3) CH3CH2CH2CONHCH3 The order of leaving power is

(4) CH3CH2CH2CONHCOCH3 (1) I > II > III > IV (2) IV > III > I > II

43. Which of the following is least reactive in a (3) III > II > I > IV (4) II > III > IV > I
nucleophilic substitution reaction? 5. Identify the correct order of reactivity in nucleophilic
(1) (CH3)3C – Cl (2) CH2 = CHCl substitution reactions of the following compounds
(3) CH3CH2Cl (4) CH2 = CHCH2Cl Cl
Cl Cl Cl
44. Which of the following is not chiral? NO2
NO2
(1) 2-hydroxypropanoic acid
(2) 2-butanol
O2N NO2
(3) 2,3-dibromopentane A B C D
(4) 3-bromopentane
45. Which of the following is used as an anti-knocking (1) A > B > C > D (2) D > C > A > B
material? (3) B > A > C > D (4) B > C > A > D
(1) Glyoxal (2) Freon 6. Which of the following will hydrolyse most readily?
(3) TEL (4) Ethyl alcohol Cl
(1) H3C (2) CH3Cl
SECTION - D
CH3
NEET Booster Questions Cl
1. Freon used as refrigerant is H 3C
(3) (4)
(1) CF2 = CF2 (2) CH2F2 Cl
(3) CCl2F2 (4) CF4 7. In which reaction major product formation takes
2. Which of the following nucleophiles will show place by Saytzeff rule?
minimum reactivity towards SN2 reaction? – +
CH3O Na
(1) Me3CO (2) MeO (1) CH3 CH2 CH CH3
(3) CH3CH2O – (4) Me2CHO Br
3. Which of the following is the correct order for the
formation of carbocation by given compounds? CH3

O K ,


(2) CH3 C CH2 CH CH3

CH 3Br Br
CH3 Cl
I II

Br Br KOH(Alc)
(3) CH 3 CH2 CH CH3
III IV
F
(1) I > II > III > IV (2) III > IV > II > I
C2H5ONa/
(3) II > I > III > IV (4) II > IV > III > I (4) CF3 CHCl2
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NEET Haloalkanes and Haloarenes 181
8. SN1 reaction is fastest in 13. Which alkene will give meso form with Br2?
CH3 CH3 CH3
CH Cl (1) C C
(1) CH3CH2Cl (2)
CH3 H H
CH3 CH3 H
(2) C C
(3) CH3 C Cl (4) CH3 CH Cl H CH3
C6H5 COOH
CH3 CH2 (3) C C
H H
CH3 C6H5 H
9. Which is best leaving group among the following? (4) C C
H COOH
(1) CH3O– (2) OH–
14. Which of the following has the highest
(3) C2H5O– (4) NH2 nucleophilicity?

10. In which alkyl halide S N 2 reaction is most (1) F– (2) OH–

favourable?
(3) CH3 (4) NH2
CH3
15. Which of the following compounds would react
(1) CH3 C CH2 Br
most rapidly with iodide ion by SN2 reaction?
CH3
CH3 Br (1) CH3Cl (2) CH3 Cl

(2) CH3 C CH CH3 CH3 Cl

CH3 (3) (CH3)2CHCH2Cl (4)

Br CH3 H
16. Total number of asymmetric centre in the following
(3) CH 3 CH H compound will be
CH3
Cl
(4) CH3 CH2 C Br H3C

CH3 H3C
11. Which one of the following is most reactive towards
(1) Three (2) Zero
nucleophilic substitution reaction?
(3) Four (4) One
(1) CH 2 CH Cl
17. Cis-but-2-ene on bromination in CCl4 gives which
(2) C6H5 Cl one of the following as major product?
(3) CH 3 CH CH Cl (1) Meso form (2) Racemic mixture
(4) CH 2 CH CH2 Cl (3) d-form (4) l-form
12. The given reaction 18. Alkyl iodides can be best prepared by
C2H5 C2H5 CCl
NaI (1) RCH2 COOAg  I2  4

H C Cl acetone I C H 

H3C CH3 (2) CH4  I2 


is an example of which reaction?
acetone
(1) SN1 (2) SN2 (3) RCH2Br  NaI 

 RCH2I  NaBr

(3) E2 (4) E1 (4) All of these

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182 Haloalkanes and Haloarenes NEET

19. Which of the following statement is/are correct? 24. The chloro compound which is used as a fire
(1) NBS is a specific reagent for allylic extinguisher is
bromination reaction (1) CCl4 (2) COCl2
(2) Allylic bromination occurs through free radical (3) CH2Cl2 (4) CHCl3
intermediates.
25. Which of the following statements is incorrect?
(3) NBS is known as N-Bromo Succinimide
(1) A strong nucleophile in an aprotic solvent
(4) All of these increases the rate or favours SN2 reaction
20. Which of the following is the example of S N2
(2) When an alkyl halide is heated with dry Ag2O,
reaction?
it produces ether
H
(1) CH3—CH2OH CH2 CH2 + H2O (3) A weak nucleophile and a protic solvent

increases the rate or favours SN1 reaction
DMSO
(2) CH3—CH—CH3 + OH CH3—CH—CH3 + Br (4) Benzyl bromide on treatment with aq. KOH
Br OH followed by addition of AgNO3 gives white ppt.
26. Which of the following pairs of compounds is/are
H
enantiomers?
N
(3) CH3—C—OH + SOCl2
CH3 CH3
D (1) Cl H Cl H
H &
H Cl Cl H
Cl—C—CH3 + SO2 + HCl CH3 CH3

D CH3 CH3
(4) Both (2) & (3) (2) H Cl Cl H
&
Br Cl H H Cl
* CH3 CH3
21. Reaction of bromobenzene with sodamide in
CH3 CH3
presence of liquid ammonia gives (3) H Cl Cl H
&
* NH2 Cl H Cl H
(1) (2) CH3 CH3
NH2
NH2 (4) All of these
(3) (4) Both (1) and (2) 27. Chlorination of CH4 involves following steps
*
h
22. Which of the following is correct about (i) Cl — Cl  2Cl
‘Dichlorocarbene’? 
(ii) CH4 + Cl CH3 + HCl
(1) It can be prepared by CHCl3 and alcoholic 
KOH (iii) CH 3 + Cl2 CH 3Cl + Cl
(2) It may be diamagnetic or paramagnetic Which of the following is rate determining step?
(3) It may be act as electrophile or nucleophile (1) Step (ii) (2) Step (i)
(4) Both (1) & (2) (3) Step (iii) (4) Both step (i) & (ii)
23. The number of monochloro derivatives obtained with 28. A hydrocarbon ‘A’ on mono-chlorination gives ‘B’ in
Cl 2 /h from the following compounds are presence of ‘h’ which on heating with alcoholic
respectively potassium hydroxide changes into another
hydrocarbon ‘C’. The latter decolourises Baeyer’s
reagent and on ozonolysis forms carbonyl
compound which on reaction with red P/HI gives
(I) (II) (III) propane. The hydrocarbon A is

(1) 1, 3, 2 (2) 1, 2, 1 (1) Cyclopentane (2) Cyclopropane

(3) 2, 3, 1 (4) 2, 3, 2 (3) Propene (4) Methane

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NEET Haloalkanes and Haloarenes 183
29. Which of the following statements is correct?
H H
(1) Addition of HBr in buta-1,3-diene gives 3-bromo- NaI
34. C2H5 C Cl acetone
I C C2H5
but-1-ene as major product at high temperature
CH3 CH3
(2) Addition of HBr in buta-1,3-diene gives 1-bromo-
pent-2-ene as major product at low Which of the following statements is/are correct for
temperature above reaction?
(3) Addition of Br 2 in trans-but-2-ene in the (1) It is known as Finkelstein reaction
presence of CCl4 gives meso form through (2) The above reactant and product have no chiral
‘syn’ addition carbon
(4) The reaction between benzene and (3) If reactant has S-configuration then product
(CH3)3CCOCl in the presence of anhydrous also having S-configuration
AlCl3 gives tert-butyl benzene as major product
(4) Both (1) & (3)
30. CHCl3 CH3
KOH
P
(Major) Cl2/h CH3Cl aq.KOH
35. A B C
excess FeCl3
The major product ‘P’ is (anhy.)

Cl Cl The product ‘C’ in the above reaction sequence is

(1) (2)
Cl CH3 CH3
Cl
(1) (2)
Cl COO K
(3) (4) Cl
COO K
31. Which of the following is correct order of dipole COOH
moment? CH3
(1) CH3F > CH3Cl > CH3Br > CH3I (3) (4)
COO K CH3
(2) CH3Cl > CH3F > CH3Br > CH3I
(3) CH3I > CH3Br > CH3Cl > CH3F 36. Which of the following statements is not correct?

(4) CH3Cl > CH3Br > CH3I > CH3F (1) The reaction of tert-butylbromide with aqueous
KOH follows first order kinetics
32. Which of the following compounds has the highest
(2) The SN2 reaction follows second order kinetics
melting point and boiling point respectively?
(3) The SN2 reaction proceeds with stereochemical
(i) o-dichlorobenzene (ii) m-dichlorobenzene
inversion
(iii) p-dichlorobenzene (4) Bond breaking and forming simultaneously
(1) (iii) & (i) (2) (iii) & (ii) occur in SN1 reaction mechanism

(3) (i) & (ii) (4) (i) & (iii) NaCN

37. CH3CH2 Cl (A)
33. The most reactive compound towards formation of KNO2
(B)
alkene when treated with alcoholic KOH is
Cl The products ‘A’ and ‘B’ respectively are
Cl
(1) CH3CH2CN & CH3CH2ONO
(1) (2)
(2) CH 3CH2NC & CH3—CH2—CH2NO 2

Cl (3) CH3CH2NC & CH3CH2NH2

(3) (4) (4) CH 3CH2CN & CH3—N— H

Cl CH3

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184 Haloalkanes and Haloarenes NEET

38. Chloropicrin is obtained by the reaction of 43. Which of the following is least reactive towards SN2
mechanism?
(1) Nitric acid on chloroform
Cl
(2) Nitrous acid on chlorobenzene
(3) Ammonia on chloromethane (1) (2) CH2==CH—CH2—Cl

(4) Steam on tetrachloromethane

Cl CH2 Cl
39. In an SN2 reaction of the type
(3) (4)
DMSO
R—Br + Cl R—Cl + Br
Which one of the following alkyl bromide has the CH3
Cl2/h NaOH
highest relative rate? 44. excess A B
CH3
(1) CH3CH2CH2Br (2) CH3—CH—CH2Br
How many moles of CH3OH required to react
CH3
completely with B ?
CH3
(1) One (2) Two
(3) CH3—C—CH2—Br (4) CH2Br
(3) Three (4) Four
CH3
40. Chloral reacts with chlorobenzene in presence of Cl
sulphuric acid produces Na Br2/
45. dry ether
A B
1 eq.
(1) BHC alc. KCN
Cl H3O
+

(2) DDT D 1 mol

C
(3) TEL
(4) BTX Product D in the reaction is

41. Which one of the following is a free-radical (1) Optically active cyanide
substitution reaction?
(2) Optically inactive acid

CH3 Boiling CH2Cl (3) Optically active acid

(1) + Cl2
(4) Optically active aldehyde
46. Which of the following is the least volatile?
Anh. AlCl3 CH3
(2) + CH3Cl (1) CH3—CH2—CH2—F
(2) CH3—CH2—CH2—Cl
CH2Cl CH2NO2
(3) + AgNO2 (3) CH3—CH—CH3

Br
(4) CH3CHO + HCN  CH3CH(OH)CN
(4) CH3—CH2—CH2—Br
42. The reaction of toluene with Cl2 in presence of
FeCl3 gives X and reaction in presence of light 47. Which of the following will be readily soluble in
gives Y. Thus, X and Y are water?
(1) X = Benzyl chloride, Y = m-chlorotoluene
(2) X = Benzal chloride, Y = o-chlorotoluene (1) Cl (2) Cl

(3) X = m-chlorotoluene, Y = p-chlorotoluene

Cl Cl
(4) X = o-and p-chlorotoluene, Y = Trichloromethyl (3) (4)
benzene
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NEET Haloalkanes and Haloarenes 185
48. Which of the following compounds undergoes 52. The correct order of reactivity of the following
nucleophilic substitution reaction most easily? compounds towards aqueous NaCN will be
Cl
Cl

Cl I.
(1) (2)
II. CH2 = CH—CH2—Cl
OCH3
Cl Cl

Cl III. CH3—CH — CH3

(3) (4) IV. CH2 = CH—CH2—Br

NO2
CH3 (1) I > II > III > IV

49. Replacement of Cl of chloro-benzene to give phenol (2) I > IV > II > III
requires drastic conditions. But chlorine of (3) IV > II > I > III
2, 4-dinitro-chloro-benzene is readily replaced
because (4) IV > II > III > I

(1) NO2 donates electron at meta position Cl OH

(2) NO2 withdraws electron from ortho/para positions
53. + aq. KOH
(3) NO2 make ring electron rich at ortho & para
(4) NO2 withdraws electron from meta position G G
For which of the following G, above reaction will be
CH2OH the fastest?
dry ether (1) –OH (2) –CH3
50. + HCl A + Mg B
(excess) (3) –NO2 (4) –CHO
OH 54. Which of the following is least reactive towards
nucleophilic substitution?
What is B?
F Cl
NO2 NO2
CH2MgCl CH2OH (1) (2)

(1) (2) Br I
MgCl NO2 NO2
MgCl
(3) (4)

C H3 CH 2MgCl
55. Which of the following does not give yellow
(3) (4) precipitate with AgNO3?
(1) (CH3)3C—CH2 —I (2) CH2 ==CH—CH2—I
OMgCl
Cl
I
51. Ethylidene chloride reacts with aq. KOH to form
which of the following compound? CH CH3
(3) CH2 C CH3 (4)
(1) Acetaldehyde (2) Ethylene glycol
(3) Ethyl alcohol (4) Acetic acid I

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186 Haloalkanes and Haloarenes NEET

56. The correct orders of reactivity towards SN1 reaction 59. In the given reaction
is Br
Cl S
CH3 H Zn/ether
Cl CH2 Cl R Product (Major)
H Br
CH3
CH3
I II III The product will be
(1) I > II > III (2) II > III > I (1) Cis-2-butene (2) Trans-2-butene
(3) III > II > I (4) I > III > II (3) 2-butyne (4) Buta-1, 3-diene
Acetone
57. CH3—Cl + NaI CH3—I + NaCl 60. Under identical conditions, solvolysis of which of
the following substrate would lead to maximum
Above equilibrium is more towards right because
racemisation?
(1) NaI is more reactive than NaCl
CH3
(2) CH3I is more reactive than CH3Cl
(1) H D
(3) NaCl is less soluble than NaI in acetone
Br
(4) It is Finkelstein’s reaction
CH3
CH3 H 1% cold
Br2 Zn/ether KMnO4/OH H Br
58. C C CCl4
A B C
H CH3 (2)

The product (C) is

CH3 CH3 OCH3

H OH HO H CH3
(1) (2)
HO H H OH
(3) Br NO2
CH3 CH3
H
CH3
Br
H OH
(3) (4) Mixture of (1) & (2) (4) C6H5 D
H OH
CH3 CH3

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