NCERT Solutions Class 12 Chemistry Chapter 10
NCERT Solutions Class 12 Chemistry Chapter 10 – Haloalkanes and Haloarenes
NCERT Solutions Class 12 Chemistry Chapter 10: Haloalkanes and Haloarenes are based on the most recent NCERT recommendations. Students can quickly obtain NCERT Solutions Class 12 Chemistry Chapter 10 from the relevant official website as a study module. The NCERT Solutions Class 12 Chemistry Chapter 10 list is available on Extramarks, one of the significant e-learning websites. Each questions’ extensive analysis and explanation is exclusively aimed to aid students in understanding the concepts and theories in a better way.
Haloalkanes and haloarenes are chemical compounds made up of one or more halogens that are formed from alkanes and arenes. NCERT Solutions Class 12 Chemistry Chapter 10 cover the definitions, Classifications, examples, applications, and other subtopics of Haloalkanes and Haloarenes. These solutions also teach students about the properties and preparations of organic halogen compounds. The list of solutions consists of a defined number of questions that students must answer to understand the concepts. They will also learn to write answers precisely as per the requirements of CBSE.
Answers to sample problems, HOTS (Higher Order Thinking Skills), MCQs exercises, assignments, and worksheets are also included in the NCERT Solutions Class 12 Chemistry Chapter 10 to help you study the concept thoroughly and make notes of the chapter haloalkanes and haloarenes for first-term exam preparation.
Key Topics Covered in NCERT Solutions Class 12 Chemistry Chapter 10
- Explain the reactions involved in the creation of haloalkanes and haloarenes and the reactions they undergo
- Correlate the structural composition of haloalkanes and haloarenes with different types of reactions
- Use Stereochemistry as an object for learning reaction mechanisms
- Make good use of the applications of organometallic compounds and highlight the environmental effects of Polyhalogen compounds.
These are some of the key topics covered in Chapter 10 of the CBSE Syllabus for Class 12 Chemistry Term I.
Subtopics in Class 12 Chemistry Chapter 10
- Nature of C-X Bond
- Methods of Preparation
- Physical Properties
- Chemical Reactions
- Polyhalogen Compounds
NCERT Solutions Class 12 Chemistry Chapter 10: Marks Weightage
|Unit Name||Marks Weightage|
|Haloalkanes and Haloarenes||4|
Despite the fact that the chapter has fewer marks than the others, students must study every detail in order to ace the exam.
Important Topics for NCERT Class 12 Chemistry Chapter 10
Why chlorobenzene has a smaller dipole moment than cyclohexyl chloride
- The dipole moment is considered to be a Vector Quantity. In chlorobenzene, the Cl-atom links with an sp2 hybridised carbon atom and The Cl-atom links to an sp3 hybridised carbon atom in cyclohexyl chloride. Since sp2 hybridised carbon has more s-character than sp3 hybridised carbon atom, the former is more electronegative than the latter. Hence, the electron density of the C − Cl bond near the Cl-atom is less in chlorobenzene and less in cyclohexyl chloride.
- Moreover, the -R effect created by the benzene ring of chlorobenzene decreases the density of electrons of the C − Cl bond near the Cl-atom. Therefore, the polarity of the C − Cl bond in chlorobenzene goes down. This is the reason why the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
What are Ambident Nucleophiles?
Nucleophiles with two nucleophilic sites are called Ambident Nucleophiles. They have two sites through which they can attack. A nitrite ion is an example of an ambident nucleophile.
Nitrite ion can either attack through oxygen resulting in the formation of alkyl nitrites, or it can attack through nitrogen resulting in the formation of nitroalkanes.
NCERT Solutions Class 12 Chemistry Chapter 10: Exercises & Solutions
The NCERT Solutions Class 12 Chemistry Chapter 10: Haloalkanes and Haloarenes presented here are based on the CBSE’s most recent term – I Syllabus. The solutions to the exercises and critical problems in the NCERT textbook make up most of the NCERT Solutions Class 12 Chemistry Chapter 10. Refer to these solutions to understand how to respond to and solve such queries correctly.
Extramarks subject specialists have compiled and solved the NCERT Solutions for Class 12 Chemistry Chapter 10. These solutions are written in plain English to simplify comprehension. NCERT Solutions Class 12 Chemistry Chapter 10 are available as a list of links and can be accessed for free. It can be used as resource material by students available anywhere, anytime.
The NCERT Solutions Class 12 Chemistry Chapter 10: Haloalkanes and Haloarenes are available as a list of links and include a variety of questions from the in-text and NCERT exercises. Students will understand every topic and subtopic of Haloalkanes and Haloarenes by following the different short answer and long answer type questions in these links.
- NCERT Solutions Class 12 Chemistry Chapter 10 Ex 10.1
- NCERT Solutions Class 12 Chemistry Chapter 10 Ex 10.2
- NCERT Solutions Class 12 Chemistry Chapter 10 Ex 10.3
- NCERT Solutions Class 12 Chemistry Chapter 10 Ex 10.4
- NCERT Solutions Class 12 Chemistry Chapter 10 Ex 10.5
- NCERT Solutions Class 12 Chemistry Chapter 10 Ex 10.6
- NCERT Solutions Class 12 Chemistry Chapter 10 Ex 10.7
Our subject specialists create NCERT Solutions to help students comprehend ideas more quickly and correctly. NCERT Solutions offer detailed, step-by-step explanations of textbook problems. NCERT Solutions are available for all Classes.
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Advantages of NCERT Solutions Class 12 Chemistry Chapter 10
The following are the main advantages of NCERT Solutions.
- These NCERT Solutions for Class 12 Chemistry Chapter 10 comply with the latest NCERT Syllabus and will help students understand and write accurate answers to the tests.
- Logical explanations accompany each question as these NCERT Solutions were prepared under the supervision of experienced teachers at Extramarks.
- Furthermore, these NCERT Solutions follow precise CBSE Board guidelines, allowing students to ace the correct answer patterns for their board exams.
As a result, Chemistry Class 12 students looking for dependable study resources can access and study the Haloalkanes and Haloarenes Class 12 NCERT Solutions available at the Extramarks website.
NCERT Exemplar Class 12 Chemistry
The NCERT Class 12 Chemistry Exemplars are offered here to help students succeed in their CBSE Class 12 exams and graduate entrance exams. NCERT Exemplar for Class 12 Chemistry contains solved chapter-by-chapter questions to assist students in completing a rapid review of the entire Syllabus and performing well in their exams.
Students can obtain a more extensive understanding of the subject by practising problems from this guide. Extramarks subject expert team uses all of the questions to answer the questions by following the NCERT syllabus guidelines (2022-2023).
Q.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
(9) CH3CH = CHC(Br)(CH3)2
Q.2 Give the IUPAC names of the following compounds:
Q.3 Write the structures of the following organic halogen compounds.
Q.4 Which one of the following has the highest dipole moment?
In case of CH2Cl2, the resultant of the dipole moments of two C-Cl bonds is strengthened by the resultant of the dipole moments of two C-H bonds.Hence, CH2Cl2 has a high dipole moment of 1.60 D.
In CHCl3, the resultant of dipole moments of two C-Cl bonds is opposed by the resultant of dipole moments of one C-H bond and one C-Cl bond. Since the resultant of one C-H bond and one C-Cl bond dipole moments is smaller than two C-Cl bonds, the opposition is to a small extent. Hence, CHCl3 has a small dipole moment of 1.08 D.
CCl4 is a symmetrical molecule. Therefore, the dipole moments of all four C-Cl bonds cancel each other. As a result, its resultant dipole moment is zero.
Hence, the given compounds can be arranged in the increasing order of their dipole moments as:
CCl4 < CHCl3 < CH2Cl2
Q.5 A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.
A hydrocarbon with the molecular formula, C5H10 belongs to the group with a general molecular formula CnH2n. Therefore, it may either be an alkene or a cycloalkane. Since, the hydrocarbon does not react with chlorine in the dark, it cannot be an alkene. Thus, it should be a cycloalkane.
Moreover, the hydrocarbon gives a single monochloro compound, C5H9Cl by 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 compound is cyclopentane.
The reactions involved in the question can be represented as shown below:
Q.6 Write the isomers of the compound having formula C4H9 Br.
There are four isomers of the compound having the formula C4H9Br. These isomers are given below.
Q.7 Write the equations for the preparation of 1-iodobutane from
Q.8 What are ambident nucleophiles? Explain with an example.
Ambident nucleophiles are the 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.
Q.9 Which compound in each of the following pairs will react faster in SN2 reaction with OH–?
- CH3Br or CH3I
- (CH3)3CCl or CH3Cl
1. In the SN2 mechanism, the reactivity of halides for the same alkyl group increases in the order:
R-F << R-Cl < R-Br < R-I
This happens because as the size increases, the halide ion becomes a better leaving group. Therefore, CH3I will react faster than CH3Br in SN2 reactions with OH–.
2. The Sn2 mechanism involves the attack of the nucleophile at the atom bearing the leaving group. But, in case of (CH3)3CCl, the attack of the nucleophile at the carbon atom is hindered because 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 CH3Cl. Hence, CH3Cl reacts faster than (CH3)3CCl in SN2 reaction with OH–.
Q.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:
In the given compound, all β-hydrogen atoms are equivalent. Thus, dehydrohalogenation of this compound gives only one alkene.
In the given compound, there are two different sets of equivalent β-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.
In the given compound, there are two different sets of equivalent β-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.
Q.11 How will you bring about the following conversions?
- Ethanol to but-1-yne
- Ethane to bromoethene
- Propene to 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 to n-octane
- Benzene to biphenyl.
Q.12 Explain why
- 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?
In chlorobenzene, the Cl-atom is linked to a sp2 hybridized carbon atom. In cyclohexyl chloride, the Cl-atom is linked to a sp3 hybridized carbon atom. Now, sp2 hybridized carbon has more s-character than sp3 hybridized carbon atom. Therefore, the former is more electronegative than the latter. Therefore, the density of electrons of C-Cl bond near the Cl-atom is less in chlorobenzene than in cydohexyl chloride. Moreover, the -R effect of the benzene ring of chlorobenzene decreases the electron density of the C-Cl bond near the Cl-atom. As a result, the polarity of the C-Cl bond in chlorobenzene decreases. Hence, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
2. 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.
Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes. Therefore, Grignard reagents should be prepared under anhydrous conditions.
Q.13 Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
Uses of Freon – 12
Freon-12 (dichlorodifluoromethane, CF2Cl2) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners. It is also used in aerosol spray propellants such as body sprays, hair sprays, etc.
Uses of DDT
DDT (p, p’-Dichloro-diphenyl-trichloro-ethane) is one of the best known insecticides. It is very effective against mosquitoes and lice.
Uses of carbontetrachloride (CCl4)
1. It is used for manufacturing refrigerants and propellants for aerosol cans.
2. It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals.
3. It is used as a solvent in the manufacture of pharmaceutical products.
Until the mid 1960’s, carbon tetrachloride was widely used as a cleaning fluid, a degreasing agent in industries, a spot reamer in homes, and a fire extinguisher.
Uses of iodoform (CHI3)
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.
Q.14 Write the structure of the major organic product in each of the following reactions:
Q.15 Write the mechanism of the following reaction:
The given reaction is:
The given reaction is an SN2 reaction. In this reaction, CN– acts as the nucleophile and attacks the carbon atom to which Br 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.
Q.16 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.
An SN2 reaction involves the approaching of the nucleophile to the carbon atom to which the leaving group is attached. When the nucleophile is sterically hindered, then the reactivity towards SN2 displacement decreases. Due to the presence of substituents, hindrance to the approaching nucleophile increases in the following order.
1-Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane
Hence, the increasing order of reactivity towards SN2 displacement is:
2- Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane
Since steric hindrance in alkyl halides increases in the order of 1° < 2° < 3°, the increasing order of reactivity towards S N2 displacement is
3° < 2° < 1°.
Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards SN2 displacement as:
2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane
[2-Bromo-3-methylbutane is incorrectly given in NCERT]
The steric hindrance to the nucleophile in the SN2 mechanism increases with a decrease in the 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:
1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane
< 1-Bromo-2, 2-dimethylpropane
Hence, the increasing order of reactivity of the given compounds towards SN2 displacement is:
1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3-methylbutane < 1-Bromobutane
Q.17 Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH?
Hydrolysis by aqueous KOH proceeds through the formation of carbocation. If carbocation is stable, then the compound is easily hydrolyzed by aqueous KOH. Now, C6H5CH2Cl forms 1º carbocation, while C6H5CHCLC6H forms 2º carbocation, which is more stable than 1º carbocation. Hence, C6H5CHCIC6H5 is hydrolyzed more easily than C6H5CH2Cl by aqueous K0H.
Q.18 p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.
p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, it fits 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 than o- and m-isomers.
Q.19 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
Q.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.
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.
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 β-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.
OH- ion is a much weaker base than RO– ion. Also, OH– ion is highly solvated in an aqueous solution and as a result, the basic character of OH– ion decreases. Therefore, it cannot abstract a hydrogen from the β-carbon.
Q.21 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.
There are two primary alkyl halides having the formula, C4H9Br. They are n – butyl bromide and isobutyl bromide.
Therefore, compound (a) is either n-butyl bromide or isobutyl bromide. Now, compound (a) reacts with Na metal to give compound (b) of molecular formula, C8H18, which is different from the compound formed when n-butyl bromide reacts with Na metal. Hence, compound (a) must be isobutyl bromide.
Thus, compound (d) is 2, 5-dimethylhexane.
It is given that compound (a) reacts with alcoholic KOH to give compound (b). Hence, compound (b) is 2-methylpropene.
Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a). Hence, compound (c) is 2-bromo-2-methylpropane.
Q.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.
(i) When n-butyl chloride is treated with alcoholic KOH, the formation of but-1-ene takes place. This reaction is a dehydrohalogenation reaction.
(ii) When bromobenzene is treated with Mg in the presence of dry ether, phenylmagnesium bromide is formed.
(iii) Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm to form phenol.
(iv) When ethyl chloride is treated with aqueous KOH, it undergoes hydrolysis to form ethanol.
(v) When methyl bromide is treated with sodium in the presence of dry ether, ethane is formed. This reaction is known as the Wurtz reaction.
(vi) When methyl chloride is treated with KCN, it undergoes a substitution reaction to give methyl cyanide.
FAQs (Frequently Asked Questions)
1. Why should I use the NCERT Solutions Class 12 Chemistry Chapter 10?
Classifications, definitions, properties of halogen compounds, along with the examples based on them are covered in Chapter 10 of the NCERT Solutions for Class 12 Chemistry. Students will have a good understanding of multiple techniques of preparation, physical and chemical properties, and the uses of organo-halogen compounds from these solutions. Students will also be able to name Haloarenes and Haloalkanes based on the IUPAC system of nomenclature using the structures presented by referring to these solutions.
2. Is it plausible to score well with the NCERT Solutions for Class 12 Chemistry Chapter 10?
The NCERT Solutions for Class 12 Chemistry Chapter 10 can help students achieve excellent grades by guiding them through the test preparation. All of the solutions are presented step-by-step, allowing students to grasp the concepts quickly. Extramarks faculty has supplied diagrams in each significant idea from the exam standpoint, where necessary, to promote visual learning. The solutions are provided straightforwardly and concisely to meet the exam standards.
3. According to Chapter 10 of Class 12 Chemistry, explain why alkyl halides, despite being polar, are immiscible (insoluble) in water.
- Alkyl halides are polar molecules. This is why their molecules are held together close by dipole-dipole forces. At the same time, the molecules of H20 are held together by hydrogen bonds. Hence, when alkyl halides are mixed into water, the new forces of attraction between water and haloalkane molecules become less robust than the forces of attraction existing between alkyl halide-alkyl halide molecules and water-water molecules. Hence, alkyl halides, although polar, are not soluble in water.
4. How should one respond to challenging problems in Class 12 Chemistry Chapter 10?
Students must be well prepared with all of the basic concepts to solve the challenging questions in Chapter 10 of Class 12 Chemistry. A strong understanding of the core concepts will help students answer any difficult questions. Students should place a greater emphasis on chapters with higher weightage and importance in exam preparation. They must clear all their doubts regarding any Chapter and thoroughly understand it.
5. How do the NCERT Solutions for Class 12 Chemistry Chapter 12 Haloalkanes and Haloarenes help students?
- Class 12 Chemistry NCERT Solutions, important synthesis methods, applications, and physical and chemical properties of organo-halogen compounds are discussed in Haloalkanes and Haloarenes. Furthermore, each NCERT solution follows the most recent CBSE Board guidelines. Students can use the solutions to comprehend the concepts better and perform well in their 12th-grade boards and other competitive exams.