NCERT Solutions | Class 12 Chemistry Chapter 10 | Haloalkanes and Haloarenes

CBSE Solutions | Chemistry Class 12
Check the below NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes Pdf free download. NCERT Solutions Class 12 Chemistry were prepared based on the latest exam pattern. We have Provided Haloalkanes and Haloarenes Class 12 Chemistry NCERT Solutions to help students understand the concept very well.
NCERT | Class 12 Chemistry
Book: | National Council of Educational Research and Training (NCERT) |
---|---|
Board: | Central Board of Secondary Education (CBSE) |
Class: | 12th |
Subject: | Chemistry |
Chapter: | 10 |
Chapters Name: | Haloalkanes and Haloarenes |
Medium: | English |
Haloalkanes and Haloarenes | Class 12 Chemistry | NCERT Books Solutions
NCERT Exercises
NCERT Solutions for Class 12 Chemistry Chapter 10, Question 1.
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcydohexane
(iii) 4-terf-Butyl-3-iodoheptane
(iv) 1, 4-Dibromobut-2-ene
(v) 1 -Bromo-4-sec-butyl-2-methylbenzene
Solution.

NCERT Solutions for Class 12 Chemistry Chapter 10, Question 2.
Solution.
H2SO4 is a strong oxidising agent. Therefore, when it is used in presence of KI, it tends to convert KI to HI and finally oxidises it to I2.NCERT Solutions for Class 12 Chemistry Chapter 10, Question 3.
Solution.
The structures of all possible dihalogen derivatives of propane are
NCERT Solutions for Class 12 Chemistry Chapter 10, Question 4.
(i) A single monochloride
(ii) Three isomeric monochlorides
(iii) Four isomeric monochlorides.
Solution.
An alkane with molecular formula C5H12 can exist in the following isomeric forms :


NCERT Solutions for Class 12 Chemistry Chapter 10, Question 5.


Solution.
The major haloderivatives formed in the given reactions are
NCERT Solutions for Class 12 Chemistry Chapter 10, Question 6.
(i) Bromomethane, bromoform, chloromethane, dibromomethane.
(ii) 1 – Chloropropane, isopropyl chloride, 1 – chlorobutane.
Solution.
The boiling points of organic compounds depend on the strength of the intermolecular forces in them. These forces are :(a) van der Waals forces and
(b) dipole-dipole interactions These forces are dependent on the
(i) molecular mass and
(ii) surface area of the molecules
(i) As the molecular mass of the compound increases, the boiling point also increases. Therefore the correct order is
chloromethane < bromomethane < dibromomethane < bromoform
(ii) Amongst molecules with same mass, it is the size of the molecule that determines the boiling point. Branched compounds are more compact and therefore have less surface area as compared to their straight chain counterparts and therefore lower boiling point. The order of boiling point is
iso-propyl chloride < 1-chloropropane < 1-chlorobutane
NCERT Solutions for Class 12 Chemistry Chapter 10, Question 7.

Solution.
We know that SN2 mechanism involves a transition state wherein both, the incoming nucleophile as well as the leaving group are present around the carbon atom. There are 5 atoms simultaneously bonded to it.Thus, for such a transition state to be possible, there should be minimum steric hindrance. Hence, 1° alkyl halides are most reactive towards SN2 followed by 2° and finally 3°.
1° RX > 2° RX > 3° RX
NCERT Solutions for Class 12 Chemistry Chapter 10, Question 8.

Solution.
SN1 reaction proceeds via the formation of a carbocation intermediate. This intermediate is formed by the cleavage of the C — X bond. More stable is the resultant carbocation faster is the SN1 reaction.Order of stability of carbocation is
3° carbocation > 2° carbocation >1° carbocation

NCERT Solutions for Class 12 Chemistry Chapter 10, Question 9.

Solution.



10. A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.
Ans. The hydrocarbon with molecular formula C5H10 can either a cycloalkane or an alkene.Since the compound does not react with Cl2 in the dark, therefore it cannot be an alkene but must be a cycloalkane. Since the cycloalkane reacts with Cl2 in the presence of bright sunlight to give a single monochloro compound, C5H9Cl, therefore, all the ten hydrogen atoms of the cycloalkanes must be equivalent. Thus, the cycloalkane is cyclopentane.
NCERT Exercises
NCERT Solutions for Class 12 Chemistry Chapter 10, Question 1.
(i) (CH3)2CHCH(Cl)CH3
(ii) CH3CH2CH (CH3)CH(C2H5)Cl
(iii) CH3CH2C(CH3)2CH2l
(iv) (CH3)3CCH2CH(Br)C6H5
(v) CH3CH(CH3)CH(Br)CH3
(vi) CH3C(C2H5)2CH2Br
(vii) CH3C(CI)(C2H5)CH2CH3
(viii) CH3CH = C(Cl)CH2CH(CH3)2
(ix) CH3CH = CHC(Br)(CH3)2
(x) p-ClC6H4CH2CH(CH3)2
(xi) m-CICH2C6H4CH2C(CH3)3
(xii) o-BrC6H4CH(CH3)CH2CH3
Solution.



NCERT Solutions for Class 12 Chemistry Chapter 10, Question 2.
(i) CH3CH(Cl)CH(Br) (CH3)
(ii) CHF2CBrClF
(iii) ClCH2C = CCH2Br
(iv) (CCl3)3CCl
(v) CH3C(p-ClC6H4)2CH(Br)CH3
(vi) (CH3)3CCH = C(CI) C6H4 l-p
Solution.


NCERT Solutions for Class 12 Chemistry Chapter 10, Question 3.
(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
Solution.
Structures of the given compounds are :

NCERT Solutions for Class 12 Chemistry Chapter 10, Question 4.
(i) CH2Cl2
(ii) CHCl3
(iii) CCl4
Solution:


NCERT Solutions for Class 12 Chemistry Chapter 10, Question 5.
Solution.
A number of structural isomers are possible for molecular formula C5H10. But, the given compound gives a single monochloro derivative when reacted with Cl2 in sunlight suggests that, all the H-atoms in the compound are equivalent. This is possible only if the compound is a cyclic alkane.
NCERT Solutions for Class 12 Chemistry Chapter 10, Question 6.
Solution.
The possible isomers of C4H9Br are
NCERT Solutions for Class 12 Chemistry Chapter 10, Question 7.
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-1-ene
Solution.

NCERT Solutions for Class 12 Chemistry Chapter 10, Question 8.
Solution.
Ambident nucleophiles are nucleophiles that are capable of attacking the substrate (alkyl halide) through two different atoms.It so ‘happens due to the presence of two nucleophilic centres which arise from the contributing (resonance) structures that are possible for the ion.
e.g., In NO–2 ion, there is a lone pair of electrons on N and therefore makes it nucleophilic while oxygen by virtue of the negative charge acts as a nucleophile. Thus, NO–2 can attack via O or N atom thereby making it ambidentate.
NCERT Solutions for Class 12 Chemistry Chapter 10, Question 9.
(i) CH3Br or CH3l
(ii) (CH3)3CCI or CH3Cl
Solution.
(i) Between CH3Br and CH3I, CH3I will react faster via the SN2 mechanism. In SN2 mechanism, C – X bond breaks and the faster it breaks faster is the reaction.I- is a better leaving group. Owing to its large size, the C – I bond breaks faster than the C – Br bond and reaction proceeds further at a greater rate.
(ii) The order of reactivity in an SN2 reaction depends on minimal steric hindrance around the carbon involved in the C – X bond. Lesser the steric hindrance felt by the incoming nucleophile, more reactive will be the alkyl halide towards SN2 reaction.
Based on this, CH3Cl will react faster than (CH3)3CCl.
NCERT Solutions for Class 12 Chemistry Chapter 10, Question 10.
(i) 1-Bromo-1-methylcyclohexane
(ii) 2-Chloro-2-methylbutane
(iii) 2, 2, 3-Trimethyl-3-bromopentane.
Solution.


NCERT Solutions for Class 12 Chemistry Chapter 10, Question 11.
- Ethanol to but-1 -yne
- Ethane to bromoethene
- Propeneto 1-nitropropane
- Toluene to benzyl alcohol
- Propene to propyne
- Ethanol to ethyl fluoride
- Bromomethane to propanone
- But-1 -ene to but-2-ene
- 1-Chlorobutane ton-octane
- Benzene to biphenyl
Solution.




NCERT Solutions for Class 12 Chemistry Chapter 10, Question 12.
- the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
- alkyl halides, though polar, are immiscible with water?
- Grignard reagents should be prepared under anhydrous conditions?
Solution.
(i) (a) In order to understand the lower dipole moment of chlorobenzene we need to look into the contributing structures of the molecules.
(b) From the above structures we find that the C – Cl bond in chlorobenzene has a partial double bond character (structure II, III and IV). As a result, the C – Cl bond length here is shorter than the C – Cl single bond but longer than the C – Cl double bond.
(c) Also evident is the positive charge on Cl atom which reduces the partial negative (δ-) charge which it is expected to carry by the virtue of its electronegativity.
(d) Consequently, the dipole moment, which is a product of bond length and partial negative charge on Cl atom, reduces. However, in cyclohexyl chloride this does not happen. It is an alkyl halide and carbon is purely sp3 hybridised and C – Cl bond has the bond length of a single bond and 8“ appearing on Cl is also higher, thus, the greater dipole moment.
(ii) Only those compounds which can form hydrogen bonds with water are miscible with it. Alkyl halides, though polar due to the presence of electronegative halogen atom, are immiscible since they cannot form hydrogen bonds.
(iii) Grignard reagents R – Mg – X is a class of highly reactive compounds which can extract a proton even from water molecule. They thus, turn into the corresponding alkanes and render any other desired reaction ineffective.
This is why Grignard reagents are prepared in the absolute absence of water (anhydrous conditions), (e.g.,)
NCERT Solutions for Class 12 Chemistry Chapter 10, Question 13.
Solution.
(i) Freon 12 (CCl2F2) is- used in aerosol propellants
- refrigeration
- air-conditioning.
(ii) DDT (p, p’- dichlorodiphenyltrichloroethane) is
- used as an insecticide,
- mainly used against mosquitoes.
(iii) Carbontetrachloride (CCl4) is used
- in manufacture of refrigerants and propellants for aerosol cans
- in synthesis of chlorofluorocarbons
- as degreasing agent
- as cleansing agent
- as a solvent in laboratories
(iv) Iodoform (CHI3) is used as an antiseptic.
NCERT Solutions for Class 12 Chemistry Chapter 10, Question 14.

Solution.



NCERT Solutions for Class 12 Chemistry Chapter 10, Question 15.

Solution.

NCERT Solutions for Class 12 Chemistry Chapter 10, Question 16.
- 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
- 1-Bromo-3-methylbutane, 2-Bromo-2- methylbutane, 3-Bromo-2-methylbutane
- 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methyl butane, 1-Bromo-3- methylbutane.
Solution.
SN2 reaction proceeds via the formation of transition state where the carbon atom is surrounded by 5 other atoms (groups). Thus, for such a transition state to form, the steric interactions have to be minimum. Therefore, the most favourable substrates for SN2 reactions are 1° alkyl halides followed by 2° and 3° alkyl halide. Order of reactivity towards SN2 : 1° > 2° > 3° > aryl halide.

In example (iii), although all the given alkyl halides are 1° but the steric hindrance around the carbon bearing the -Br atom decides the order of reactivity. More the number of bulky groups around this carbon lower will be its reactivity towards SN2.
NCERT Solutions for Class 12 Chemistry Chapter 10, Question 17.
Solution.
C6 H5 CHClC6 H5 is hydrolysed faster.(a) Hydrolysis of an alkyl halide is an example of nucleophilic substitution reaction. In case of aryl halides this follows the SN1 pathway i.e., via the formation of carbocation.
(b) C6 H5 CH2 Cl or benzyl chloride gives

(c) Out of I & II, carbocation II is more stable. The reason is the presence of two phenyl rings attached to the carbon carrying the positive charge.
(d) As a result, the delocalisation of the +ve charge is greater and the carbocation is more stable. Due to this, (II) is formed faster and the corresponding halide is hydrolysed with greater ease as compared to benzyl chloride.
NCERT Solutions for Class 12 Chemistry Chapter 10, Question 18.
Solution.
The para-isomers have high melting points and solubility as compared to their ortho and meta isomers due to symmetry of para-isomers that fits into crystal lattice better than ortho and para isomers.NCERT Solutions for Class 12 Chemistry Chapter 10, Question 19.
- Propene to propan-1 -ol
- Ethanol to but-1 -yne
- 1 -Bromopropane to 2-bromopropane
- Toluene to benzyl alcohol
- Benzene to 4-bromonitrobenzene
- Benzyl alcohol to 2-phenylethanoic acid
- Ethanol to propanenitrile
- Aniline to chlorobenzene
- 2-Chlorobutane to 3,4-dimethylhexane
- 2-Methyl-1 -propene to 2-chloro-2- methylpropane
- Ethyl chloride to propanoic acid
- But-1-ene to n-butyliodide
- 2-Chloropropaneto 1-propanol
- Isopropyl alcohol to iodoform
- Chlorobenzene to p-nitrophenol
- 2-Bromopropane to 1-bromopropane
- Chloroethane to butane
- Benzene to diphenyl
- ferf-Butyl bromide to isobutyl bromide
- Aniline to phenylisocyanide
Solution.







NCERT Solutions for Class 12 Chemistry Chapter 10, Question 20.
Solution.
Formation of alcohols from the reaction between alkyl chlorides and aqueous KOH is an example of simple nucleophilic substitution.
But when aqueous KOH is replaced by alcoholic KOH, alkenes are formed instead of alcohols due to elimination of HCl from an alkyl halide.



NCERT Solutions for Class 12 Chemistry Chapter 10, Question 21.
Solution.


NCERT Solutions for Class 12 Chemistry Chapter 10, Question 22.
- n-butyl chloride is treated with alcoholic KOH,
- bromobenzene is treated with Mg in the presence of dry ether,
- chlorobenzene is subjected to hydrolysis,
- ethyl chloride is treated with aqueous KOH,
- methyl bromide is treated with sodium in the presence of dry ether,
- methyl chloride is treated with KCN?
Solution.


Chlorobenzene is highly unreactive towards nucleophilic substitution. However, it can be hydrolysed to phenol by heating in aqueous sodium hydroxide solution at a temperature of 623 K and 300 atm pressure. The presence of an electron withdrawing group increases the reactivity of haloarenes.

NCERT Class 12 Chemistry
Class 12 Chemistry Chapters | Chemistry Class 12 Chapter 10
Chapterwise NCERT Solutions for Class 12 Chemistry
-
NCERT Solutions For Class 12 Chemistry Chapter 1 The Solid State
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions
NCERT Solutions For Class 12 Chemistry Chapter 3 Electro chemistry
NCERT Solutions For Class 12 Chemistry Chapter 4 Chemical Kinetics
NCERT Solutions For Class 12 Chemistry Chapter 5 Surface Chemistry
NCERT Solutions For Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements
NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements
NCERT Solutions For Class 12 Chemistry Chapter 8 The d and f Block Elements
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds
NCERT Solutions For Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes
NCERT Solutions For Class 12 Chemistry Chapter 11 Alcohols Phenols and Ethers
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers
NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life
NCERT Solutions for Class 12 All Subjects | NCERT Solutions for Class 10 All Subjects |
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