NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

NCERT
SOLUTIONS
CLASS - 12th

aglase .co

Class : 12th
Subject : Chemistry
Chapter : 10
Chapter Name : Haloalkanes and Haloarenes

Q10.1 Write structures of the following compounds:
(i) 2-Chloro-3-methylpentane
(ii) I-Chloro-4-ethylcyclohexane
(iii) 4-tert. Butyl-3-iodoheptane
(iv) 1,4-Dibromobut-2-ene
(v) I-3romo-4-sec. butyl-2-methylbenzene

Answer.

Page : 293 , Block Name : Intext Questions

Q10.2 Why is sulphuric acid not used during the reaction of alcohols with KI?

Answer. In the presence of sulphuric acid H SO , KI produces HI
2 4

2KI + H2 SO4 ⟶ 2KHSO4 + 2HI

Since H SO is an oxidizing agent, it oxidizes HI (produced in the reaction to 12).
2 4

2HI + H2 SO4 ⟶ I2 + SO2 + H2 O

As a result, the reaction between alcohol and HI to produce alkyl iodide cannot occur. Therefore,
sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidizing acid
such as H PO is used.
3 4

Page : 297 , Block Name : Intext Questions

Q10.3 Write structures of different dihalogen derivatives of propane.

Answer. There are four different dihalogen derivatives of propane. The structures of these
derivatives are shown below.

Page : 297 , Block Name : Intext Questions

Q10.4 Among the isomeric alkanes of molecular formula C H , identify the one that on
5 12

photochemical chlorination yields
(i) A single monochloride.
(ii) Three isomeric monochlorides.
(iii) four isomeric monochlorides.

Answer. (i) To have a single monochloride, there should be only one type of H-atom in the isomer
of the alkane of the molecular formula C H . This is because, replacement of any H-atom leads
5 12

to the formation of the same product. The isomer is neopentane.

(ii) To have three isomeric monochlorides, the isomer of the alkane of the molecular formula
C H
5 12 should contain three different types of H-atoms. Therefore, the isomer is n-pentane. It can
be observed that there are three types of H atoms labelled as a, b and cin n-pentane.
CH3 − CH2 − CH2 − CH2 − CH3

n -Pentane

(iii) To have four isomeric monochlorides, the isomer of the alkane of the molecular formula
C H
5 12 should contain four different types of H-atoms. Therefore, the isomer is 2- methylbutane.
It can be observed that there are four types of H-atoms labelled as a, b, c, and d in 2-
methylbutane.
CH3

CH3 − CH − CH2 − CH3

Page : 297 , Block Name : Intext Questions

Q10.5 Draw the structures of major monohalo products in each of the following reactions:

Answer.

Page : 297 , Block Name : Intext Questions

Q10.6 Arrange each set of compounds in order of increasing boiling points.
(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(ii) I-Chloropropane, Isopropyl chloride, I-Chlorobutane.

Answer. (i)

For alkyl halides containing the same alkyl group, the boiling point increases with an increase in

the atomic mass of the halogen atom.
Since the atomic mass of Br is greater than that of Cl, the boiling point of bromomethane is higher
than that of chloromethane. Further, for alkyl halides containing the same alkyl group, the boiling
point increases with an increase in the number of halides. Therefore, the boiling point of
Dibromomethane is higher than that of chloromethane and bromomethane, but lower than that of
bromoform. Hence, the given set of compounds can be arranged in the order of their increasing
boiling points as:
Chloromethane < Bromomethane < Dibromomethane < Bromoform.
(ii)

For alkyl halides containing the same halide, the boiling point increases with an increase in the
size of the alkyl group. Thus, the boiling point of I-chlorobutane is higher than that of isopropyl
chloride and I-chloropropane. Further, the boiling point decreases with an increase in branching
in the chain. Thus, the boiling point of isopropyl alcohol is lower than that of I-chloropropane.
Hence, the given set of compounds can be arranged in the increasing order of their boiling points
as:
Isopropyl chloride < I-Chloropropane < I-Chlorobutane

Page : 299 , Block Name : Intext Questions

Q10.7 Which alkyl halide from the following pairs would you expect to react more rapidly by an
S 2 mechanism? Explain your answer.
N

Answer. (i)

2-bromobutane is a 2 alkylhalide whereas I-bromobutane is a 1 alkyl halide. The approaching of
∘ ∘

nucleophile is more hindered in 2-bromobutane than in I-bromobutane. Therefore, I-
bromobutane reacts more rapidly than 2-bromobutane by an S 2 mechanism.
N

(ii)

2-3romobutane is 2 alkylhalide whereas 2-bromo-2-methylpropane is 30 alkyl halide. Therefore,
∘

greater numbers of substituents are present in 3 alkyl halide than in 2 alkyl halide to hinder the
∘ ∘

approaching nucleophile. Hence, 2-bromobutane reacts more rapidly than 2-bromo-2-
methylpropane by an S 2 mechanism.
N

(iii)

Both the alkyl halides are primary. However, the substituent —CH, is at a greater distance to the
carbon atom linked to Br in I-bromo-3-methylbutane than in I-bromo-2-
methylbutane. Therefore, the approaching nucleophile is less hindered in case of the former than
in case of the latter. Hence, the former reacts faster than the latter by S 2mechanism.
N

Page : 316 , Block Name : Intext Questions

Q10.8 In the following pairs of halogen compounds, which compound undergoes faster S 1 N

reaction?

Answer. (i)

SN 1reaction proceeds via the formation of carbocation. The alkyl halide (I) is 3 while (11) is 2 .
∘ ∘

Therefore, (I) forms 3 carbocation while (II) forms 2 carbocation. Greater the stability of the
∘ ∘

carbocation, faster is the rate of S 1 reaction. Since 30 carbocation is more stable than 20
N

carbocation. (I), i.e. 2—chloro-2-methylpropane, undergoes faster S 1 reaction than (II) i.e., 3-
N

chloropentane.
(ii)

The alkyl halide (I) is 2 while (II) is 1 . 2 carbocation is more stable than 1 carbocation.
∘ ∘ ∘ ∘

Therefore, (I), 2—chloroheptane, undergoes faster S reaction than (II), 1 chlorohexane.
N1

Page : 316 , Block Name : Intext Questions

Q10.9

Answer.

When an alkyl halide is treated with Na in the presence of ether, a hydrocarbon containing double
the number of carbon atoms as present in the original halide is
obtained as product. This is known as Wurtz reaction. Therefore, the halide, R —X, is
1

Page : 316 , Block Name : Intext Questions

Q10.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl,
benzyl (primary, secondary, tertian'), vinyl or aryl halides:
(i)
(CH3 ) CHCH(Cl)CH3
2

(ii)
CH3 CH2 CH (CH3 ) CH (C2 H5 ) Cl

(iii)
CH3 CH2 C(CH3 ) CH2 I
2

(iv)
(CH3 ) CCH2 CH(Br)C6 H5
3

(v)
CH3 CH (CH3 ) CH(Br)CH3

(vi)
CH3 C(C2 H5 ) CH2 Br
2

(vii)
CH3 C(Cl) (C2 Hs ) CH2 CH3

(viii)
CH3 CH = C(Cl)CH2 CH(CH3 )
2

(ix)
CH3 CH = CHC(Br)(CH3 )
2

(x)
p − ClC6 H4 CH2 CH(CH3 )
2

(xi)
m − ClCH2 C6 H4 CH2 C(CH3 )
3

(xii)
o − Br − C6 H4 CH (CH3 ) CH2 CH3

Answer.

Page : 319 , Block Name : Exercise

Q10.2 Give the IUPAC names of the following compounds:
(i)
CH3 CH(Cl)CH(Br)CH3

(ii)
CHF2 CBrClF

(iii)
ClCH2 C ≡ CCH2 Br

(iv)
(CCl3 ) CCl
3

(v)
CH3 C(p − ClC6 H4 ) CH(Br)CH3
2

(vi)
(CH3 ) CCH = CClC6 H4 I − p
3

Answer.

Page : 319 , Block Name : Exercise

Q10.3 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)
per uorobenzene
(vi)
4-tert-Butyl-3-iodoheptane
(vii)
1-Bromo-4-sec-butyl-2-methylbenzene
(viii)
1,4-Dibromobut-2-ene

Answer.

Page : 319 , Block Name : Exercise

Q10.4 Which one of the following has the highest dipole moment?
(i) CH2 Cl2

(ii) CHCl3

(iii) CCl4

Answer.

CCl4 is a symmetrical molecule. Therefore, the dipole moments cancel each other. Hence, its
resultant dipole moment is zero.
As shown in the above gure, in CHCl , the resultant of dipole moments of two C Cl bonds is
3

opposed by the resultant of dipole moments of one C—H bond and one C—CI bond. Since the
resultant of one bond and one C—CI bond dipole moments is smaller than two C—CI bonds, the
opposition is to a small extent. As a result, CHCl has a small dipole moment of 1.08 D.
3

On the other hand, in case of CH Cl , the resultant of the dipole moments of two C—CI bonds is
2 2

strengthened by the resultant of the dipole moments of two C—H bonds. As a result, CH Cl has2 2

a higher dipole moment of 1.60 D than CHCl i.e., CH Cl has the highest dipole moment.
3 2 2

Hence, the given compounds can be arranged in the increasing order of their dipole moments as:
CCl4 < CHCl3 < CH2 Cl2

Page : 319 , Block Name : Exercise

Q10.5 A hydrocarbon C H does not react with chlorine in dark but gives a single monochloro
5 10

compound C H Cl in bright sunlight. Identify the hydrocarbon.
5 9

Answer. A hydrocarbon with the molecular formula, C H belongs to the group with a general
5 10

molecular formula C H . Therefore, it may either be an alkene or a cycloalkane.
n 2n

Since hydrocarbon does not react with chlorine in the dark, it cannot be an alkene. Thus, it should
be a cycloalkane. Further, the hydrocarbon gives a single monochloro compound, C H Cl by
5 9

reacting with chlorine in bright sunlight. Since a single monochloro compound is formed, the
hydrocarbon must contain H—atoms that are all equivalent. Also, as all H—atoms of a cycloalkane

are equivalent, the hydrocarbon must be a cycloalkane. Hence, the said compound is
cyclopentane.

Page : 319 , Block Name : Exercise

Q10.6 Write the isomers of the compound having formula C H Br.
4 9

Answer. There are four isomers of the compound having the formula C H Br These isomers are
4 9

given below.

Page : 319 , Block Name : Exercise

Q10.7 Write the equations for the preparation of I—iodobutane from
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-1-ene.

Answer.

Page : 319 , Block Name : Exercise

Q10.8 What are ambident nucleophiles? Explain with an example.

Answer. Ambident nucleophiles are nucleophiles having two nucleophilic sites. Thus, ambident
nucleophiles have two sites through which they can attack. For example, nitrite ion is an ambident
nucleophile.

Nitrite ion can attack through oxygen resulting in the formation of alkyl nitrites. Also, it can
attack through nitrogen resulting in the formation of nitroalkanes.

Page : 319 , Block Name : Exercise

Q10.9 Which compound in each of the following pairs will react faster in S 2 reaction with OH ?
N

(i) CH3 Br or CH3 I

(ii) (CH3 ) CCl or CH3 Cl
3

Answer. (i) In the S 2 mechanism, the reactivity of halides for the same alkyl group increases in
N

the order. This happens because as the size increases, the halide ion becomes a better leaving
group.
R − F << R − Cl < R − Br < R − I

Therefore, CH I will react faster than CH Br in S 2 reactions with OH¯.
3 3 N

(ii)

The S 2 mechanism involves the attack of the nucleophile at the atom bearing the leaving group.
N

But, in case of (CH ) CCl, the attack of the nucleophile at the carbon atom is hindered because
3 3

of the presence of bulky substituents on that carbon atom bearing the leaving group. On the other
hand, there are no bulky substituents on the carbon atom bearing the leaving group in
(CH ) CCl. Hence, (CH ) CCl reacts faster than (CH ) CCl in S 2 reaction with OH
3 3 3 N
3 3 3

Page : 320 , Block Name : Exercise

Q10.10 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) I-Bromo-l-methylcyclohexane
(ii) 2-Chloro-2-methylbutane
(iii) 2,2,3-Trimethyl-3-bromopentane

Answer. (i)

In the given compound, all ß-hydrogen atoms are equivalent. Thus, dehydrohalogenation of this
compound gives only one alkene.

(ii)

In the given compound, there are two different sets of equivalent ß hydrogen atoms labelled as a
and b. Thus, dehydrohalogenation of the compound yields two alkenes.

Saytzeff's rule implies that in dehydrohalogenation reactions, the alkene having a greater number
of alkyl groups attached to a doubly bonded carbon atoms is preferably
produced. Therefore, alkene (I) i.e., 2-methylbut-2-ene is the major product in this reaction.

(iii)

In the given compound, there are two different sets of equivalent ß-hydrogen atoms labelled as a
and b. Thus, dehydrohalogenation of the compound yields two alkenes.

According to Saytzeff's rule, in dehydrohalogenation reactions, the alkene having a greater
number of alkyl groups attached to the doubly bonded carbon atom is preferably formed.
Hence, alkene (I) i.e., 3,4,4-trimethylpent-2-ene is the major product in this reaction.

Page : 320 , Block Name : Exercise

Q10.11 How will you bring about the following conversions?
(i) Ethanol to but-I-yne
(ii) Ethane to bromoethene
(iii) Propene to I-nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl uoride
(vii) Bromomethane to propanone
(viii) But-I-ene to but-2-ene
(ix) I-Chlorobutane to n-octane
(x) Benzene to biphenyl

Answer.

Page : 320 , Block Name : Exercise

Q10.12 Explain why
(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
(ii) alkyl halides, though polar, are immiscible with water?
(iii) Grignard reagents should be prepared under anhydrous conditions?

Answer. (i)

In chlorobenzene, the Cl-atom is linked to a sp hybridized carbon atom. In cyclohexyl chloride,
2

the Cl-atom is linked to a sp hybridized carbon atom. Now, sp hybridized carbon has more s-
3 2

character than sp hybridized carbon atom. Therefore, the former is more electronegative than the
2

latter. Therefore, the density of electrons of C—CI bond near the Cl-atom is less in chlorobenzene
than in cyclohexyl chloride.
Moreover, the —R effect of the benzene ring of chlorobenzene decreases the electron density of
the C—CI bond near the Cl-atom. As a result, the polarity of the C—CI bond in chlorobenzene
decreases. Hence, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(ii) To be miscible with water, the solute-water force of attraction must be stronger than the
solute-solute and water-water forces of attraction. Alkyl halides are polar molecules and so held
together by dipole-dipole interactions. Similarly, strong H-bonds exist between the water

molecules. The new force of attraction between the alkyl halides and water molecules is weaker
than the alkyl halide alkyl halide and water-water forces of attraction. Hence, alkyl halides
(though polar) are immiscible with water.
(iii) Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes.
R MgX + H2 O ⟶ R − H + Mg(OH)X

Therefore, Grignard reagents should be prepared under anhydrous conditions.

Page : 320 , Block Name : Exercise

Q10.13 Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

Answer. Uses of Freon — 12
Freon-12 (dichlorodi uoromethane, CF Cl ) is commonly known as CFC. It is used as a
2 2

refrigerant in refrigerators and air conditioners. It is also used in aerosol spray propellants such as
body sprays, hair sprays, etc. However, it damages the ozone layer. Hence, its manufacture was
banned in the United States and many other countries in 1994.
uses of DDT :
DDT (p, p dichlorodiphenyltrichloroethane) is one of the best known insecticides. It is very
effective against mosquitoes and lice. But due its harmful effects, it was banned in the United
States in 1973.

Uses of carbontetrachloride (CC14) :
(i) It is used for manufacturing refrigerants and propellants for aerosol cans.
(ii) It is used as feedstock in the synthesis of chloro uorocarbons and other chemicals.
(iii) It is used as a solvent in the manufacture of pharmaceutical products.
(iv) Until the mid 1960's, carbon tetrachloride was widely used as a cleaning uid, a degreasing
agent in industries, a spot reamer in homes, and a re extinguisher.

Uses Of iodoform (CHI,)
Iodoform was used earlier as an antiseptic, but now it has been replaced by other formulations-
containing iodine-due to its objectionable smell. The antiseptic property of iodoform is only due
to the liberation of free iodine when it comes in contact with the skin.

Page : 320 , Block Name : Exercise

Q10.14 Write the structure of the major organic product in each of the following reactions:

Answer.

Page : 320 , Block Name : Exercise

Q10.15 Write the mechanism of the following reaction:

Answer. The given reaction is

The given reaction is an S 2 reaction. In this reaction, CN-acts as the nucleophile and attacks the
N

carbon atom to which ar is attached. CN ion is an ambident nucleophile and can attack through
−

both C and N. In this case, it attacks through the C-atom.

Page : 320 , Block Name : Exercise

Q10.16 Arrange the compounds of each set in order of reactivity towards SN2 displacement:
(i) 2-3romo-2-methylbutane, I-Bromopentane, 2-Bromopentane
(ii) I-Bromo-3-methylbutane, 2-3romo-2-methylbutane, 3-3romo-2- methylbutane
(iii) I-aromobutane, I-Bromo-2,2-dimethylpropane, I-Bromo-2-methylbutane, bromo-3-
methylbutane.

Answer. (i)

An S 2 reaction involves the approaching of the nucleophile to the carbon atom to which the
N

leaving group is attached. When the nucleophile is sterically hindered, then the reactivity towards
S 2 displacement decreases. Due to the presence of substituents, hindrance to the approaching
N

nucleophile increases in the following order.
I-Bromopentane < 2-bromopentane < 2-3romo-2-methylbutane
Hence, the increasing order of reactivity towards S 2 displacement is:
N

2-3romo-2-methylbutane < 2-Bromopentane < I-Bromopentane
(ii)

Since steric hindrance in alkyl halides increases in the order of 1 , the increasing order
0 0 0
< 2 < 3

of reactivity towards SN2 displacement is
0 0 0
3 < 2 < 1

Hence, the given set of compounds can be arranged in the increasing order of their reactivity
towards S displacement as:
2

N

2-aromo-2-methylbutane < 2-Bromo-3-methylbutane < I-aromo-3-methylbutane
[2-3romo-3-methylbutane is incorrectly given in NCERT]
(iii)

The steric hindrance to the nucleophile in the S 2 mechanism increases with a decrease in the
N

distance of the substituents from the atom containing the leaving group. Further, the steric
hindrance increases with an increase in the number of substituents. Therefore, the increasing
order of steric hindrances in the given compounds is as below:
I-Bromobutane < I-Bromo-3-methylbutane < I-Bromo-2-methylbutane
< I-Bromo-2, 2-dimethylpropane

Hence, the increasing order of reactivity of the given compounds towards SN2 displacement is:
I-Bromo-2, 2-dimethylpropane < I-Bromo-2-methylbutane < I-aromo-3 methylbutane < I-
Bromobutane

Page : 320 , Block Name : Exercise

Q10.17 Out of C H CH Cl and C H CHClC H , which is more easily hydrolysed by aqueous
6 5 2 6 5 6 5

KOH?

Answer.

Hydrolysis by aqueous KOH proceeds through the formation of carbocation. If carbocation is
stable, then the compound is easily hydrolyzed by aqueous KOH. Now,
C H CH Cl
6 5 2 forms 1
∘ While C H CHCLC H forms 2 -carbocation, which is more
-carbocation 6 5 6 5
∘

stable than 2 -carbocation. Hence, C H CHCIC H 6 is hydrolyzed more easily than
∘
6 5 6 5

C H CH Cl by aqueous KOH.
e 3 2

Page : 320 , Block Name : Exercise

Q10.18 p-Dichlorobenzene has higher m.p. and lower solubility than those of o- and m-isomers.
Discuss.

Answer.

p-Dichlorobenzene is more symmetrical than 0-and m-isomers. For this reason, it ts more
closely than O_and m_isomers in the crystal lattice. Therefore, more energy is required to break
the crystal lattice of p-dichlorobenzene. As a result, p-dichlorobenzene has a higher melting point
and lower solubility than 0-and m-isomers.

Page : 320 , Block Name : Exercise

Q10.19 How the following conversions can be carried out?
(i) Propene to propan-l-ol

(ii) Ethanol to but-I-yne
(iii) I-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-I-ene to n-butyliodide
(xiii) 2-Chloropropane to I-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-3romopropane to I-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide

Answer.

Page : 320 , Block Name : Exercise

Q10.20 The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but
in the presence of alcoholic KOH, alkenes are major products. Explain.

Answer. In an aqueous solution, KOH almost completely ionizes to give OH ions. OH ion is a
− −

strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form
alcohol.
R − Cl + KOH(aq) ⟶ R − OH + KCl

On the other hand, an alcoholic solution of KOH contains alkoxide RO ion, which is a strong
−

base. Thus, it can abstract a hydrogen from the ß-carbon of the alkyl chloride and form an alkene
by eliminating a molecule of HCI.
R − CH2 − CH2 − Cl + KOH(alc) ⟶ R − CH = CH2 + KCl + H2 O

ion is a much weaker base than RO¯ ion. Also, OH¯ ion is highly solvated in an aqueous
−
OH

solution and as a result, the basic character of ion decreases. Therefore, it cannot abstract a
hydrogen from the ß-carbon.

Page : 321 , Block Name : Exercise

Q10.21 Primary alkyl halide C H Br (a) reacted with alcoholic KOH to give compound
4 9

(b).Compound (b) is reacted with Har to give (c) which is an isomer of (a). When (a) is reacted with
sodium metal it gives compound (d), C H which is different from the compound formed when n-
8 18

butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations
for all the reactions.

Answer. There are two primary alkyl halides having the formula, C4H9Br. They are n —bulyl
bromide and isobutyl bromide.

Therefore, compound (a) is either II—butyl bromide or isobutyl bromide.
Now, compound (a) reacts with Na metal to give compound (b) of molecular formula,
C H
8 18which is different from the compound formed when n—butyl bromide reacts with Na metal.
Hence. compound (a) must be isobutvl bromide.

Page : 321 , Block Name : Exercise

Q10.22 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.

Answer.

Page : 321 , Block Name : Exercise