(Unit: 10.

Haloalkanes and Haloarenes)

1. Nomenclature of Haloalkanes and Haloarenes:- The general formula of alkyl halide is CnH2n+1–X
where X is a halogen atom. Alkyl halides are usually represented by R – X where R is an alkyl
group.
2. Classification: Alkyl halides can be classified as methyl halide, primary alkyl halide, secondary alkyl
halide (2°) and tertiary alkyl halide (3°), according to the number of other carbon atoms attached
to the carbon bearing the halogen atom.
H H R' R'
H X R X R X R X
H H H R"
Methyl halide Primary alkyl Secondary alkyl Tertiary alkyl
halide (1°) halide (2°) halide (3°)
R, R and R may be same or different.
3. IUPAC names of some halides:-
CH3CH2CH(Cl)CH3 2-Chlorobutane
(CH3)3CCH2Br 1-Bromo-2,2-dimethylpropane
(CH3)3CBr 2-Bromo-2-methylpropane
CH2 = CHCl 1-Chloroethene
CH2 = CHCH2Br 3-Bromopropene
CH2Cl2 1,1-Dichloromethane
CHCl3 1,1,1-Trichloromethane
CHBr3 1,1,1-Tribromomethane
CCl4 1,1,1,1-Tetrachloromethane
CH3CH2CH2F 1-Fluoropropane
o-Chlorotoluene(2-Chlorotoluene)
1-Chloro-2-methylbenzene
Benzyl chloride
1-Chloro1-phenylmethane
Answer the following
1. Write structures of the following compounds: (i) 2-Chloro-3-methylpentane (ii) 1-Chloro-4-
ethylcyclohexane (iii) 4-tert. Butyl-3-iodoheptane (iv) 1,4-Dibromobut-2-ene (v) 1-Bromo-4-sec.
butyl-2-methylbenzene.
2. Write the structures of the following organic halogen compounds. (i) 2-Chloro-3-methylpentane
(ii) p-Bromochlorobenzene (iii) 1-Chloro-4-ethylcyclohexane (iv) 2-(2-Chlorophenyl)-1-iodooctane
(v) Perfluorobenzene (vi) 4-tert-Butyl-3-iodoheptane (vii) 1-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2-ene
3. Give the IUPAC names of the following compounds:(i) CH3CH(Cl)CH(Br)CH3 (ii) CHF2CBrClF
(iii) ClCH2C≡CCH2Br (iv) (CCl3)3CCl (v) CH3C(p-ClC6H4)2CH(Br)CH3 (vi) (CH3)3CCH=ClC6H4I-p
4. Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl
(primary, secondary, tertiary), vinyl or aryl halides:
(i) (CH3)2CHCH(Cl)CH3 (ii) CH3CH2CH(CH3)CH(C2H5)Cl (iii) CH3CH2C(CH3)2CH2I
(iv) (CH3)3CCH2CH(Br)C6H5(v) CH3CH(CH3)CH(Br)CH3 (vi) CH3C(C2H5)2CH2Br
(vii) CH3C(Cl)(C2H5)CH2CH3 (viii) CH3CH=C(Cl)CH2CH(CH3)2(ix) CH3CH=CHC(Br)(CH3)2
(x) p-ClC6H4CH2CH(CH3)2 (xi) m-ClCH2C6H4CH2C(CH3)3 (xii) o-Br-C6H4CH(CH3)CH2CH3

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 5. Write IUPAC names of the following:

4. Methods of Preparation:- (Haloalkanes)

a) From Alcohol (i) Alkyl chlorides can be prepared from alcohols by treating it with PCl5 or SOCl2.
CH3CH2OH + PCl5 ⎯→ CH3CH2Cl + POCl3 + HCl
CH3CH2OH + SOCl2 ⎯→ CH3CH2Cl + SO2 + HCl
Note :Preparation of alkyl chloride from alcohols by treating it with SOCl2 is the best method as it
gives almost pure alkyl chloride since the by products of the reaction i.e. SO 2 and HCl are in
gaseous phase.
(ii) For the preparation of iodoalkane, we take a mixture of red phosphorus and I2.
CH 3CH 2OH ⎯R⎯ ⎯⎯→CH 3CH 2I
ed P +I 2

(iii) When alcohols are treated with PX3 or HX they give alkyl halides.
→ RCH 2 X (X= Cl,Br) RCH OH ⎯⎯→ RCH X (X= Cl,Br)
HX
RCH 2 OH ⎯P⎯ X3
2 2
ZnCl2
b) From Alkenes:-
i. Addition of X2 on alkenes: Addition of X2 on alkenes gives dihaloalkane.

ii. Addition of HX on alkenes give the alkyl halides.

a. Markovnikov’s rule : During addition of a polar molecule to unsymmetrical alkene, the negative
part of addendum attaches to that double bonded carbon which has least no. of hydrogen atom.
CH3 CH = CH2 + HBr ⎯ ⎯→ CH3CHBrCH3 (2-Bromo Propane)
b. Peroxide effect (kharasch effect) when a polar molecule is added to unsymmetrical alkene in the
presence of organic peroxide then addition occurs against the Markovnikov’s rule (only for
addition of H Br)
CH3 CH = CH2 + H Br ⎯ P⎯ ⎯
eroxide
CH3-CH2-CH2-Br (1-Bromo Propane)
c) Halide Exchange: →
i) Finkelstein reaction: Alkyl chlorides/ bromides on reaction with NaI in dry acetone to give Alkyl
iodides. R – Cl +Na I ⎯ a⎯ ⎯→ R – I + NaCl
cetone

R – Br +Na I ⎯⎯⎯→ R – I + NaBr

acetone

ii) Swarts reaction Heating of alkyl chloride/bromide in the presence of a metallic fluoride such as
AgF, Hg2F2, CoF2 or SbF3 to give alkyl fluorides
R – X ⎯A⎯ 2 2
⎯ ⎯ ⎯ ⎯ ⎯→ R – F
gF, Hg F , C oF or S bF
2 3

CH3 –Br + AgF ⎯ ⎯→ CH3 –F + AgBr

d) From halogenation of Alkanes:
• By free radical halogenations Free radical chlorination or bromination of alkanes gives a
complex mixture of isomeric mono- and polyhaloalkanes, which is difficult to separate as pure
compounds. Consequently, the yield of any one compound is low

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 Answer the following
1.

2. Write the equations for the preparation of 1-iodobutane from (i) 1-butanol (ii) 1-chlorobutane (iii)
but-1-ene.
3. Complete the reaction :
(a) CH3CH2CH2Cl – Cl +Na I ⎯ a⎯ ⎯→
cetone

(b) CH3CH2CH2Cl – Br +Na I ⎯ a⎯ ⎯→

cetone

(c)CH3CH2Br + AgF→
4.

5. Alkyl halides are generally not prepared in laboratory by free radical halogenations of
alkanes.Why
6. Identify all the possible monochloro structural isomers expected to be formed on free radical
monochlorination of (CH3)2CHCH2CH3.
7. Why is sulphuric acid not used during the reaction of alcohols with KI?
8. Write structures of different dihalogen derivatives of propane.
9. Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical
chlorination yields (i) A single monochloride. (ii) Three isomeric monochlorides. (iii) Four isomeric
monochlorides.
Physical Properties of Haloalkanes:
a) Alkyl halides are colourless when pure. However, bromides and iodides develop colour when
exposed to light. Many volatile halogen compounds have sweet smell
b) Physical State : Lower members of alkyl halides (CH3F, CH3Cl, CH3Br and C2H5Cl) are colourless
gases at room temperature. The higher members up C18 are coloulress liquids where as other
members are colourless solids.
c) Solubility: Despite of polar nature of alkyl halides, they are insoluble in water due to inability to
form hydrogen bond but they are soluble in non-polar solvents..
d) Melting and Boiling Point:
➢ Due to greater polarity as well as higher molecular mass as compared to the parent hydrocarbon,
the intermolecular forces of attraction (dipole-dipole and van der Waals) are stronger in the
halogen derivatives. That is why the boiling points of chlorides, bromides and iodides are
considerably higher than those of the hydrocarbons of comparable molecular mass.
➢ The attractions get stronger as the molecules get bigger in size and have more electrons.. For the
same alkyl group, the boiling points of alkyl halides decrease in the order: RI> RBr> RCl> RF. This is

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 because with the increase in size and mass of halogen atom, the magnitude of van der Waal
forces increases.
➢ The boiling points of isomeric haloalkanes decrease with increase in branching. For example, 2-
bromo-2-methylpropane has the lowest boiling point among the three isomers.

➢ Boiling points of isomeric dihalobenzenes are very nearly the same. However, the para-isomers
are high melting as compared to their orthoand meta-isomers. It is due to symmetry of para-
isomers that fits in crystal lattice better as compared to ortho- and meta-isomers.

e) Density :Bromo, iodo and polychloro derivatives of hydrocarbons are heavier than water. The
density increases with increase in number of carbon atoms, halogen atoms and atomic mass of
the halogen atoms.
Answer the following
1. The boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF.
2. Alkyl halides, though polar, are immiscible with water.
3. Haloalkanes have higher boiling points as compared to those of corresponding alkanes.
4. P-Dichlorobenzene has higher m.p and solubility than those of o- and m- isomers.
5. Arrange each set of compounds in order of increasing boiling points.
(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Chemical Properties of Haloalkanes:
1. Nucleophilic Substitution Reactions In this type of reaction, a nucleophile reacts with haloalkane
(the substrate) having a partial positive charge on the carbon atom bonded to halogen. A
substitution reaction takes place and halogen atom, called leaving group departs as halide ion.
Since the substitution reaction is initiated by a nucleophile, it is called nucleophilic substitution

reaction. or
a. CH3CH2Cl + KOH(aq) ⎯→ CH3CH2OH (Alcohol) + KCl
b. CH3CH2Br + NaOH(aq) ⎯→ CH3CH2OH (Alcohol) +NaBr
c. CH3CH2I + NaOR ⎯→ CH3CH2OR (Alkoxy alkane)+NaI
d. CH3CH2I + AgCN ⎯→ CH3CH2NC (Alkyl Isocyanide) +Ag I
e. CH3CH2I + KCN ⎯→ CH3CH2CN(Alkyl cyanide) +KI
f. CH3CH2Cl + AgNO2 ⎯→ CH3CH2 NO2 (Nitroalkane) +AgCl
g. CH3CH2Cl + KNO2 ⎯→ CH3CH2ONO(Alkylnitrite) +KI
h. CH3CH2Br +NH3 ⎯→ CH3CH2NH2 (Primary Amines) +HBr
i. CH3CH2Br +RNH2 ⎯→ CH3CH2NHR +HBr
j. CH3CH2Br ⎯ L⎯ ⎯→ CH3CH2_H
IAlH4

➢ Ambident nucleophiles :Groups like cyanides and nitrites possess two nucleophilic centres and
are called ambident nucleophiles. Actually cyanide group is a hybrid of two contributing
structures and therefore can act as a nucleophile in two different ways [-C≡N ↔ :C=N-], i.e.,
linking through carbon atom resulting in alkyl cyanides and through nitrogen atom leading to
isocyanides. Similarly nitrite ion also represents an ambident nucleophile with two different
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 points of linkage The linkage through oxygen results in alkyl nitrites while through nitrogen atom,
it leads to nitroalkanes.
➢ Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms
isocyanides as main product.since KCN is predominantly ionic and provides cyanide ions in
solution. Although both carbon and Nitrogen can donate electron pair but carbon donates
electron pair instead of Nitrogen to form more stable C-C bond. How ever, AgCN is mainly
covalent in nature and Nitrogen is free to donate electron pair forming isocyanide as the main
product.
Answer the following
1. What are ambident nucleophiles? Explain with an example.
2. What happens when (i) Ethyl chloride is treated with aqueous KOH, (ii) Methyl chloride is treated
with KCN?
3. Write the structure of the major organic product in each of the following reactions:
a. (CH3)3CBr + KOH
b. CH3CH(Br)CH2CH3 + NaOH
c. CH3CH2Br + KCN
d. C6H5ONa + C2H5Cl
Mechanism Nucleophilic Substitution Reactions
1. SN1 mechanism
• Substitution Nucleophilic Unimoecular (SN1)
• In this reaction, the rate of reaction depends only on the concentration of alkyl halide ie
rate=K [ R X ]
• It is mainly shown by tertiary alkyl halids eg. Tertiary butyl halide.
• Example: The reaction between tert-butyl bromide and hydroxide ion yields tert-butyl
alcohol and follows the first order kinetics

• It is Two step reactions.

• Step I: In the first step slow dissociation of alkyl halide takes place by reversible reaction
forming a carbocation.

• Step II: The carbocation at once combines with the nucleophile to form final product.

• Order of reactivity: 30 > 20 > 10 alkyl halide

• Greater the stability of carbocation, greater will be its ease of formation from alkyl halide
and faster will be the rate of reaction. In case of alkyl halides, 3 0 alkyl halides undergo SN1
reaction very fast because of the high stability of 30 carbocations.
• For a given alkyl group, the reactivity of the halide, R-X, follows the order R–I> R–Br>R–
Cl>>R–F.( As iodine is a better leaving group because of its large size, it will be released at a
faster rate in the presence of incoming nucleophile.)
2. SN2 mechanism
• Substitution Nucleophilic bimolecular (SN2)

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 • In this reaction, the rate of reaction depend on the concentration of both the alkyl halide and
Nucleophile ie rate = [ R X ] [ Nu- ]
• It is mainly given by primary alkyl halides e.g. n-alkyl halide
• Example: The reaction between CH3Cl and hydroxide ion to yield methanol and chloride ion
• CH3Cl + OH- → CH3 OH + Cl-
• It is One step reaction

• It results in complete inversion of configuration

• Order of reactivity of alkyl halide : Primary halide10 > Secondary halide20 > Tertiary halide 30
• Of the simple alkyl halides, methyl halides react most rapidly in SN2 reactions because there
are only three small hydrogen atoms. Tertiary halides are the least reactive because bulky
groups hinder the approaching nucleophiles.
Answer the following
1. Explain the SN1 & SN2 mechanism with suitable examples.
2. In the following pairs of halogen compounds, which would undergo SN2 reaction faster?

3. Predict the order of reactivity of the following compounds in SN1 and SN2 reactions:(i) The four
isomeric bromobutanes (ii) C6H5CH2Br, C6H5CH (C6H5) Br, C6H5CH (CH3) Br, C6H5C(CH3)(C6H5)Br
4. Which alkyl halide from the following pairs would you expect to react more rapidly by an S N2
mechanism? Explain your answer.

5. In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?

6. Which compound in each of the following pairs will react faster in S N2 reaction with –OH? (i)
CH3Br or CH3I (ii) (CH3)3CCl or CH3Cl
7. Write the mechanism of the following reaction:

8. Arrange the compounds of each set in order of reactivity towards SN2 displacement (i)2-Bromo-
2-methylbutane, 1-Bromopentane , 2-Bromopentane
(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo-2-methylbutane
(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-
methylbutane
2. Elimination reaction :- When a halo alkane with ß-hydrogen atom is treated with alcoholic
solution of potassium hydroxide there is an elimination of hydrogen atom from  -carbon and a
halogen atom from the ά-carbon atom forming an alkene.
CH3CH2Br + KOH(alcoholic) ⎯→ CH2 = CH2 + H2O +KBr
➢ Saytzeff rule: If in an elimination reaction there is availability of more than one  -
hydrogen atom then that alkene is the major product which is highly alkylated ie
containing greater number of alkyl groups attached to the doubly bonded carbon atoms.

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Question: Predict all the alkenes that would be formed by dehydrohalogenation of the following halides
with sodium ethoxide in ethanol and identify the major alkene:(i) 1-Bromo-1-methylcyclohexane (ii) 2-
Chloro-2-methylbutane (iii) 2,2,3-Trimethyl-3-bromopentane.
3. Reaction with Metal.
i) Reaction with Sodium
➢ Wurtz reaction : Alkyl halides react with sodium in dry ether to give hydrocarbons
(Alkanes)containing double the number of carbon atoms present in the halide

.
ii) Reaction with Magnesium
➢ Grignard Reagent: alkyl or aryl magnesium halide is called grignard reagent. It is obtained
by treating alkyl or aryl bromide with magnesium in the presence of ether.

➢ Grignard reagents are highly reactive and react with any source of proton to give
hydrocarbons. Even water, alcohols, amines are sufficiently acidic to convert them to
corresponding hydrocarbons. It is therefore necessary to avoid even traces of moisture
from a Grignard reagent.
➢ Grignard reagents should be prepared under anhydrous conditions.because Grignard
reagent reacts with water and get decomposed and form alkanes. RMgX +
H2O ⎯ ⎯→ RH + Mg(OH)X
Answer the following
1. What happens when methyl bromide is treated with sodium in the presence of dry ether?
2. How the following conversions can be carried out? (a) Chloroethane to butane(b) Benzene to
diphenyl (c) 2-Chlorobutane to 3, 4-dimethylhexane (d) 1-Chlorobutane to octane.

Haloarenes
Nomenclature of Haloarenes: According to IUPAC system, aryl halides are named as Haloarenes. If more
than one halogen is present their positions in the ring are indicated by numbers or appropriate prefixes,
ortho, meta, para.

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 CH3
Br CH3 Cl
Cl

Br CH3
1,3-dibromobenzene or 1- chloro-2-methylbenzene 2- chloro-1,4-dimethylbenzene
m-dibromobenzene or o-chlorotoluene

General Methods of Preparation

i) By direct halogenation: This method is used for the preparation of chloro and bromo derivatives.
Halgoens react with aromatic hydrocarbons in presence of catalysts or halogen carriers such as
iron, iodine or aluminium chloride at room temperature in absence of direct sunlight.

C6 H 6 + Cl 2 ⎯⎯
Fe
→C 6H 5Cl + HCl

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

⎯ C⎯ ⎯⎯→
uC l/ HCl
heat

chlorobenzene
Br

+ -
N2 Cl ⎯ C⎯ ⎯⎯→
uB r/ HBr
heat

bromobenzene
I
Benzene
diazonium
⎯ ⎯KI
→
chloride heat

iodobenzene
F

⎯⎯ ⎯→
NaB F4
heat

fluorobenzene

Physical Properties:
• Aryl halides are colourless stable liquids with pleasant odour.
• These are insoluble in water but readily miscible with organic solvents. Most of them are steam
volatile, heavier than water.
• Their boiling points are higher than corresponding alkyl halides.
• The boiling points rise gradually from fluoro to iodo compounds.
Chemical Properties of Haloarenes:
(a) Nucleophilic Substitution reactions: Aryl halides are extremely less reactive towards Nucleophilic
Substitution reactions.because C—X bond acquires a partial double bond character due to
resonance. As a result, the bond cleavage in haloarene is difficult than haloalkane and therefore,
they are less reactive towards nucleophilic substitution reaction.

Chlorobenzene to Phenol

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(b) Electrophilic substitution reactions

(c) Reaction with Metal

Fittig reaction: Aryl halides(Haloarenes) also give analogous compounds when treated with
sodium in dry ether, in which two aryl groups are joined together..

Wurtz-Fittig reaction : A mixture of an alkyl halide(Haloalkanes) and aryl halide(Haloarenes.)

gives an alkylarene when treated with sodium in dry ether.

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Stereoisomerism : Isomerism exihibited by two or more compounds with the same molecular and
structural formulae but different spatial arrangements of atoms or groups in space .
1. Certain compounds rotate the plane polarised light (produced by passing ordinary light through
Nicol prism) when it is passed through their solutions. Such compounds are called optically active
compounds.
2. The angle by which the plane polarised light is rotated is measured by an instrument called
polarimeter.
3. If the compound rotates the plane polarised light to the right, i.e., clockwise direction, it is called
dextrorotatory (Greek for right rotating) or the d-form and is indicated by placing a positive (+)
sign before the degree of rotation.
4. If the light is rotated towards left (anticlockwise direction), the compound is said to be
laevorotatory or the l-form and a negative (–) sign is placed before the degree of rotation. Such
(+) and (–) isomers of a compound are called optical isomers and the phenomenon is termed as
optical isomerism.
5. if all the substituents attached to that carbon are different, such a carbon is called asymmetric
carbon or stereocentre. The resulting molecule would lack symmetry and is referred to as
asymmetric molecule. The asymmetry of the molecule is responsible for the optical activity in
such organic compounds.
6. The objects which are non superimposable on their mirror image (like a pair of hands) are said to
be chiral and this property is known as chirality. While the objects, which are, superimposable on
their mirror images are called achiral.
7. Butan-2-ol has four different groups attached to the tetrahedral carbon and as expected is chiral.
Some common examples of chiral molecules such as 2-chlorobutane, 2, 3-dihyroxypropanal,
(OHC–CHOH–CH2OH), bromochloro-iodomethane (BrClCHI), 2-bromopropanoic acid (H3C–CHBr–
COOH), etc.
8. The stereoisomers related to each other as nonsuperimposable mirror images are called
enantiomers
9. Enantiomers possess identical physical properties namely, melting point, boiling point, solubility,
refractive index, etc. They only differ with respect to the rotation of plane polarised light. If one of
the enantiomer is dextro rotatory, the other will be laevo rotatory.
10. A mixture containing two enantiomers in equal proportions will have zero optical rotation, as the
rotation due to one isomer will be cancelled by the rotation due to the other isomer. Such a
mixture is known as racemic mixture or racemic modification.
11. A racemic mixture is represented by prefixing dl or before the name, for example
butan-2-ol. The process of conversion of enantiomer into a racemic mixture is known as
racemisation.
12. Inversion, retention and racemisation: There are three outcomes for a reaction at an asymmetric
carbon atom. Consider the replacement of a group X by Y in the following reaction.

• If (A) is the only compound obtained, the process is called retention of configuration.
• If (B) is the only compound obtained, the process is called inversion of configuration.

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 • If a 50:50 mixture of the above two is obtained then the process is called racemisation and the
product is optically inactive, as one isomer will rotate light in the direction opposite to another.
• Question :Identify chiral and achiral molecules in each of the following pair of compounds:

REASONING BASED QUESTIONS

1. n- butyl bromide has higher point than tert.- butyl bromide.
Answer: n- butyl bromide being a straight chain alkyl halide has larger surface area than tert. butyl
bromide. Larger the surface area, larger the magnitude of the vanderwaal’s forces and higher is the
boiling point.
2. Racemic mixture is optically inactive.
Answer: A Racemic mixture contains two enantiomers in d & l equal proportions. As the rotation due to
one enatiomer is cancelled by equal and opposite rotation of another isomer. Therefore it is optically
inactive.
3. The presence of nitro group (–NO2 ) at ortho and para postions in haloarenes increases the
reactivity of haloarenes towards Nucleophilic substitution reaction.
Answer: The presence of nitro group (–NO2 ) at ortho and para postions in haloarenes helps in
stabilization of resulting carbanion by –R and –I effects and hence increases the reactivity of haloarenes
towards Nucleophilic substitution reaction.
4. Alkyl halides, though polar, are immiscible with water.
Answer: Alkyl halides, though polar, are immiscible with water.because alkyl haides are unable to form
H-bond with water as well as unable to break the existing H-bonds among water molecules.
5. Butan-1-ol optically inactive but Butan-2-ol is optically active.
Answer: Butan-2-ol has four different groups attached to the tetrahedral carbon and has chiral carbon,
therefore it is optically active.
6. 1-Bromobutane optically inactive but 2-Bromobutane is optically active.
Answer: 2-Bromobutane has four different groups attached to the tetrahedral carbon and has chiral
carbon, therefore it is optically active
7. Electrophilic aromatic substitution reactions in haloarenes occur slowly.
Answer:Chlorine withdraws electrons through inductive effect and releases electrons through
resonance. The inductive effect is stronger than resonance and causes net electron withdrawal as a
result Electrophilic aromatic substitution reactions in haloarenes occur slowly
8. Although chlorine is an electron withdrawing group, yet it is ortho-, para- directing in
electrophilic aromatic substitution reactions.
Answer: As the weaker resonance (+R) effect of Chlorine which stabilize the carbocation formed tends
to oppose the inductive effect of Chlorine which destabilize the carbocation at ortho and para postions
and makes deactivation less for ortho and para postion.
9. Haloalkanes easily dissolve in organic solvents.
Answer: because the new intermolecular attractions between haloalkanes and organic solvents have
much the same strength as ones being broken in the separate haloalkanes and solvent molecules.

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 10. Halogen compounds used in industry as solvents are alkyl chlorides rather than bromides and
iodides.
Answer: because alkyl chlorides are more stable and more volatile than bromides and iodides.
11. Ethyl iodide undergoes SN2 undergoes reaction faster than ethyl bromide.
Answer:Since I- ion is better leaving group than Br- ion, Ethyl iodide reacts faster than ethyl bromide in
SN2.
12. C-X bond length in halobenzene is smaller than C-X bond length in CH3-X
Answer: because C—X bond in halobenzene acquires a partial double bond character due to resonance
13. .P-Dichlorobenzene has higher m.p and solubility than those of o- and m- isomers.
Answer: P-Dichlorobenzene has higher m.p and solubility than those of o- and m- isomers. Since p-
dichlorobenzene has more symmetrical structure than other two isomers, hence its molecules fit more
closely in the crystal lattice and consequently stronger intermolecular attractive forces.
14. Benzyl chloride undergoes SN1 reaction faster than cyclohexyl methyl chloride.
Answer: because in case of benzyl chloride the carbocation formed is stablised by resonance.
15. The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
Answer: The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.because in
chlorobenzene electron withdrawing inductive effect is opposed by electron releasing resonance effect
there fore it is relatively less polar. On the other hand in cyclohexyl chloride there is only electron with
drawing is inductive effect of -Cl atom due to which more polar.
16. Grignard reagents should be prepared under anhydrous conditions.
Answer: Grignard reagents should be prepared under anhydrous conditions.because Grignard reagent
reacts with water and get decomposed and form alkanes RMg X + H2O→ R-H+Mg (OH) X.
17. p - nitro chlorobenzene undergoes nucleophilic substitution faster than chlorobenzene
Answer: p - nitro chlorobenzene undergoes nucleophilic substitution faster than chlorobenzene,due to
presence of electron withdrawing groups such as nitro group in p - nitro chlorobenzene.,it forms more
stable carbanion
16. The treatment of alkyl chlorides with aq.KOH leads to the formation of alcohols but in the
presence of alc.KOH alkenes are major products.
Answer: The treatment of alkyl chlorides with aq.KOH leads to the formation of alcohols but in the
presence of alc.KOH alkenes are major products.because in presence of aq.KOH(more polar),
nucleophilic substitution reaction takes place ,thus alcohols are formed while in presence alc.KOH(less
polar),elimination reaction takes place.thus alkenes are major products.
17. Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms
isocyanides as main product.
Answer: Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms
isocyanides as main product.since KCN is predominantly ionic and provides cyanide ions in solution.
Although both carbon and Nitrogen can donate electron pair but carbon donates electron pair instead of
Nitrogen to form more stable C-C bond. How ever, AgCN is mainly covalent in nature and Nitrogen is free
to donate electron pair forming isocyanide as the main product.
18. Allyl chloride is more reactive than n - propyl chloride towards nucleophilic substitution
reaction.
Answer: Allyl chloride is more reactive than n - propyl chloride towards nucleophilic substitution
reaction.because allyl carbocation is stabilized by resonance whereas n propyl carbocation is stabilized
by +I effect only.
19. Preparation of alkyl chloride from alcohols by treating it with Thionyl chloride SOCl 2 is the best
method.
Answer: Because it gives almost pure alkyl chloride since the by products of the reaction i.e. SO2 and
HCl are in gaseous phase.
20. Aryl halides are extremely less reactive towards Nucleophilic Substitution reactions.

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Answer: Aryl halides are extremely less reactive towards Nucleophilic Substitution reactions.because C—
X bond acquires a partial double bond character due to resonance. As a result, the bond cleavage in
haloarene is difficult than haloalkane and therefore, they are less reactive towards nucleophilic
substitution reaction.(draw the resonating structures of chlorobenzene)
21. Sulphuric acid is not used during the reaction of alcohols with KI.
Answer:.H2SO4 cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it
converts KI to corresponding HI and then oxidizes it to I2.
22. Alkyl halides are generally not prepared in laboratory by free radical halogenations of alkanes.
Answer: Alkyl halides are generally not prepared in laboratory by free radical halogenations of
alkanes.since Free radical chlorination or bromination of alkanes gives a complex mixture of isomeric
mono- and polyhaloalkanes, which is difficult to separate as pure compounds. Consequently, the
yield of any one compound is low.
23. The boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF.
Answer: The boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF.This is because
with the increase in size and mass of halogen atom, the magnitude of van der Waal forces increases.
24. Haloalkanes have higher boiling points as compared to those of corresponding alkanes.
Answer:Haloalkanes have higher boiling points as compared to those of corresponding alkanes.Due to
greater polarity as well as higher molecular mass as compared to the parent hydrocarbon,
25. Chloroform is stored in closed dark coloured bottles.
Answer: Chloroform is stored in closed dark coloured bottles.because it is slowly oxidized by air in the
presence of light to an extremely poisonous gas, phosgene.
CHEMICAL TEST TO DISTINGUISH BETWEEN PAIR OF ORGANIC COMPOUND
Haloalkanes are more reactive Haloalkenes & Haloarenes towards nucleophilic substitution reactions
,therefore Haloalkanes on heating with aq.KOH undergoes hydrolysis to produce alcohol & KCl.This
mixture on acidification with dil.HNO3 followed by treatment with AgNO3 solution produces a white
ppt [AgCl]or Yellow ppt AgBr]

Note: Haloalkenes & Haloarenes does not undergoes hydrolysis with KOH

For example:
1. CH3CH2Cl + KOH (aq) → CH3CH2OH + KCl (This mixture on acidification with dil.HNO3
followed by treatment with AgNO3 solution produces a white ppt AgCl.
2. CH3CH2Br + KOH (aq) → CH3CH2OH + KBr (This mixture on acidification with dil.HNO3
followed by treatment with AgNO3 solution produces a AgBr .

Give one chemical test to distinguish between the following pairs of compounds:
1. Ethyl bromide & Bromobenzene
2. Ethyl bromide & Vinyl chloride
3. Chlorobenzene & n-hexyl chloride
4. Chlorobenzene & chlorocyclohexane
5. Chlorobenzene and Benzylchloride
6. Bromobenzene and Benzylbromide

Name-reactions Based Questions (with suitable examples)

(i)Wurtz reaction (ii) Wurtz-Fittig reaction (iii) Fittig reaction (iv)Finkelstein reaction (v) Swarts
reaction (vi) Diazotization (vii) Sandmeyers reaction

Some reaction based questions:

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 How the following conversions can be carried out? (i) Propene to propan-1-ol (ii) Ethanol to but-1-
yne (iii) 1-Bromopropane to 2-bromopropane (iv) Toluene to benzyl alcohol (v) Benzene to 4-
bromonitrobenzene (vi) Benzyl alcohol to 2-phenylethanoic acid (vii) Ethanol to propanenitrile (viii)
Aniline to chlorobenzene (ix) 2-Chlorobutane to 3, 4-dimethylhexane (x) 2-Methyl-1-propene to 2-
chloro-2-methylpropane (xi) Ethyl chloride to propanoic acid (xii) But-1-ene to n-butyliodide (xiii) 2-
Chloropropane to 1-propanol (xiv) Isopropyl alcohol to iodoform (xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane (xvii) Chloroethane to butane (xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide (xx) Aniline to phenylisocyanide
3.What happens when (i) n-butyl chloride is treated with alcoholic KOH, (ii) bromobenzene is treated
with Mg in the presence of dry ether, (iii) chlorobenzene is subjected to hydrolysis, (iv) ethyl chloride
is treated with aqueous KOH, (v) methyl bromide is treated with sodium in the presence of dry ether,
(vi) methyl chloride is treated with KCN?

.Word Problems
(a) Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound
(b).Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with
sodium metal it gives compound (d), C8H18 which is different from the compound formed when n-butyl
bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the
reactions.
(b) A hydrocarbon C5H10 does not react with chlorine in dark but gives a single
monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.

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