case study questions haloalkanes and haloarenes

Case Study Questions Haloalkanes and Haloarenes Class 12 Chemistry

        Case Study Questions Haloalkanes and Haloarenes Class 12 Chemistry

1. Read the passage given below and answer the following questions: Nucleophilic substitution reactions are of two types; substitution nucleophilic bimolecular (S N 2) and substitution nucleophilic unimolecular (S N 1) depending on molecules taking part in determining the rate of reaction. The reactivity of alkyl halide towards S N 1 and S N 2 reactions depends on various factors such as steric hindrance, stability of intermediate or transition state, and polarity of the solvent. S N 2 reaction mechanism is favoured mostly by primary alkyl halide or transition state and polarity of the solvent, S N 2 reaction mechanism is favoured mostly by primary alkyl halide then secondary and then tertiary. This order is reversed in the case of S N 1 reactions. (i) Which of the following is most reactive towards nucleophilic substitution reaction? (a) C 6 H 5 Cl (b) CH 2 = CHCl (c) ClCH 2 CH = CH 2 (d) CH 3 CH = CHCl

(ii) Isopropyl chloride undergoes hydrolysis by (a) S N 1 mechanism (b) S N 2 mechanism (c) S N 1 and S N 2 mechanism (d) Neither S N 1 nor S N 2 mechanism

(iii) Tertiary alkyl halides are practically inert to substitution by S N 2 mechanism because of (a) Insolubility (b) Instability (c) Inductive effect (d) Steric Hindrance

(iv) Which of the following is the correct order of decreasing S N 2 reactivity? (a) RCH 2 X > R 2 CHX > R 3 CX (b) R 3 CX > R 2 CHX >RCH 2 X (c) R 2 CHX > R 3 CX > RCH 2 X (d) RCH 2 X > R 3 CX > R 2 CHX

(v) An organic molecule necessarily shows optical activity if it- a) Contains asymmetric carbon atoms b) Is non-polar c) Is non-superimposable on its mirror image d) Is superimposable on its mirror image.

2. Read the passage given below and answer the following questions: The replacement of hydrogen atom in a hydrocarbon, aliphatic or aromatic results in the formation of haloalkanes and haloarenes respectively. Haloalkanes contain a halogen atom attached to sp 3 hybridized carbon atom of an alkyl group whereas haloarenes contain a halogen atom attached to sp 2 hybridized carbon atom of an aryl group. Haloalkanes and haloarenes may be classified on the basis of the number of halogen atoms in their structures as mono, di, or poly halogen compounds and also on the basis of the state of hybridization of the carbon atom to which the halogen atom is bonded. (i) Which of the following halide is 2°? (a) Isopropyl chloride (b) Isobutyl chloride (c) n-propyl chloride (d) n-butyl chloride

(ii) Which of the following is a Gem-dibromide is: (a) CH 3 CH(Br)CH 2 (Br) (b) CH 3 CBr 2 CH 3 (c) CH 2 (Br)CH 2 CH 2 (d) CH 2 BrCH 2 Br

(iii) IUPAC name of (CH 3 ) 3 CCl is: (a) 3-Chlorobutane (b) 2-Chloro-2-methylpropane (c) t-butyl chloride (d) n-butyl chloride

(iv) Which of the following is a primary halide? (a) Isopropyl iodide (b) Secondary butyl iodide (c) Tertiarybutyl bromide (d) Neohexyl chloride

(v) Which one of the following is not an allylic halide? (a) 4-Bromopent-2-ene (b) 3-Bromo-2-methylbut-1-ene (c) 1-Bromobut-2-ene (d) 4-Bromobut-1-ene

3. Read the passage given below and answer the following questions: Alkyl halides are prepared by the free radical halogenation of alkanes, addition of halogen acids to alkenes, replacement of -OH group of alcohols with halogens using phosphorus halides, thionyl chloride, or halogen acids. Aryl halides are prepared by electrophilic substitution to arene. Fluorine and iodides are best prepared by the halogen exchange method. These compounds find wide applications in industry as well as in day-to-day life. These compounds are generally used as solvents and as starting materials for the synthesis of a large number of organic compounds. (i) The best method for the conversion of an alcohol into an alkyl chloride is by treating the alcohol with (a) PCl 5 (b) Dry HCl in the presence of anhydrous ZnCl 2 (c) SOCl 2 in presence of pyridine (d) None of these

(ii) The catalyst used in the preparation of an alkyl chloride by the action of dry HCl on alcohol is (a) anhydrous AlCl 3 (b) FeCl 3 (c) anhydrous ZnCl 2 (d) Cu

(iii) An alkyl halide reacts with metallic sodium in dry ether. The reaction is known as: (a) Frankland’s reaction (b) Sandmeyer’s reaction (c) Wurtz reaction (d) Kolbe’s reaction

(iv) Fluorobenzene (C 6 H 5 F) can be synthesized in the laboratory (a) By direct fluorination of benzene with F 2 gas (b) By reacting bromobenzene with NaF solution (c) By heating phenol with HF and KF (d) From aniline by diazotization followed by heating the diazonium salt with HBF 4

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Case Study Questions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes

• Last modified on: 3 years ago
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There is Case Study Questions in class 12 Chemistry in session 2020-21. For the first time, the board has introduced the case study questions in the board exam. The first two questions in the board exam question paper will be based on Case Study and Assertion & Reason. The first question will have 5 MCQs out of which students will have to attempt any 4 questions. The second question will carry 5 Assertion & Reason type questions with the choice to attempt any four. Here are the questions based on case study.

Case Study Question 1:

Read the passage given below and answer the following questions:

Nucleophilic substitution reactions are of two types; substitution nucleophilic bimolecular (S N 2) and substitution nucleophilic unimolecular (S N 1) depending on molecules taking part in determining the rate of reaction. Reactivity of alkyl halide towards S N 1 and S N 2 reactions depends on various factors such as steric hindrance, stability of intermediate or transition state and polarity of solvent. S N 2 reaction mechanism is favoured mostly by primary alkyl halide or transition state and polarity of solvent, S N 2 reaction mechanism is favoured mostly by primary alkyl halide then secondary and then tertiary. This order is reversed in case of SN1 reactions.

The following questions are multiple choice question. Choose the most appropriate answer:

(i) Which of the following is most reactive towards nucleophilic substitution reaction? (a) C 6 H 5 Cl (b) CH 2 =CHCl (c) ClCH 2 CH=CH 2 (d) CH 3 CH=CHCl

(ii) Isopropyl chloride undergoes hydrolysis by (a) S N 1 mechanism (b) S N 2 mechanism (c) S N 1 and S N 2 mechanism (d) neither S N 1 nor S N 2 mechanism

(iii) The most reactive nucleophile among the following is (a) CH 3 O- (b) C 6 H 5 O- (c) (CH 3 ) 2 CHO- (d) (CH 3 ) 3 CO-

(iv) Tertiary alkyl halides are practically inert to substitution by S N 2 mechanism because of (a) insolubility (b) instability (c) inductive effect (d) steric hindrance

Which of the following is the correct order of decreasing S N 2 reactivity? (a) RCH 2 X > R 2 CHX > R 3 CX (b) R 3 CX > R 2 CHX >RCH 2 X (c) R 2 CHX >R 3 CX > RCH 2 X (d) RCH 2 X >R 3 CX >R 2 CHX

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Class 12 Chemistry Case Study Questions Chapter 10 Haloalkanes and Haloarenes

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In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Download of CBSE Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes Case Study and Passage Based Questions with Answers were Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Chemistry Case Study Questions Haloalkanes and Haloarenes  to know their preparation level.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

In CBSE Class 12 Chemistry Paper, There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Haloalkanes and Haloarenes Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 12 Chemistry  Chapter 10 Haloalkanes and Haloarenes

Case Study/Passage-Based Questions

Case Study 1: A chlorocompound (A) on reduction with Zn-Cu and ethanol gives the hydrocarbon (B) with five carbon atoms. When (A) is dissolved in dry ether and treated with sodium metal it gave 2,2,5,5 tetramethylhexane. The treatment of (A) with alcoholic KCN gives compound (C).

(i) The compound (A) is

Answer: (a) 1-chloro- 2, 2-dimethylpropane.

(ii) The reaction of (C) with Na, C 2 H 5 OH gives

Answer: (c) (CH)3C CH2CH2NH2

(iii) The reaction of (C) with Na, C 2 H 5 OH is called

Answer: (b) Mendius reaction

(iv) Compound (B) is

Answer: (b) 2, 2-dimethylpropane

Case Study 2: Nucleophilic substitution reactions are of two types; substitution nucleophilic bimolecular (SN2) and substitution nucleophilic unimolecular (SN1) depending on molecules taking part in determining the rate of reaction. The reactivity of alkyl halide towards SN1 and SN2 reactions depends on various factors such as steric hindrance, stability of intermediate or transition state and polarity of solvent. SN2 reaction mechanism is favored mostly by primary alkyl halide or transition state and polarity of the solvent, SN2 reaction mechanism is favored mostly by primary alkyl halide then secondary and then tertiary. This order is reversed in the case of SN1 reactions.

(i) Which of the following is most reactive towards nucleophilic substitution reaction? (a) C 6 H 5 Cl (b) CH 2 =CHCl (c) ClCH 2 CH=CH 2 (d) CH 3 CH=CHCl

Answer: (c) ClCH2CH=CH2

(ii) Isopropyl chloride undergoes hydrolysis by (a) S N 1 mechanism (b) S N 2 mechanism (c) S N 1 and S N 2 mechanism (d) neither S N 1 nor S N 2 mechanism

Answer: (c) SN1 and SN2 mechanism

(iii) The most reactive nucleophile among the following is (a) CH 3 O- (b) C 6 H 5 O- (c) (CH 3 ) 2 CHO- (d) (CH 3 ) 3 CO-

Answer: (a) CH3O-

Which of the following is the correct order of decreasing S N 2 reactivity? (a) RCH 2 X > R 2 CHX > R 3 CX (b) R 3 CX > R 2 CHX >RCH 2 X (c) R 2 CHX >R 3 CX > RCH 2 X (d) RCH 2 X >R 3 CX >R 2 CHX

Answer: (a) RCH2X > R2CHX > R3CX

Case Study 3: A primary alkyl halide (A) C 4 H 9 Br reacted with alcoholic KOH to give compound (B). Compound (B) is reacted with HBr to give compound (C) which is an isomer of (A). When (A) reacted with sodium metal, it gave a compound (D) C 8 H 18  that is different than the compound obtained when n-butyl bromide reacted with sodium metal

(i) Which type of isomerism is present in compounds (A) and (C)?

Answer: (c) Chain

(ii) IUPAC name of compound (D) is

Answer: (c) 2-methyl heptane

Hope the information shed above regarding Case Study and Passage Based Questions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 12 Chemistry Haloalkanes and Haloarenes Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate

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Case Study on Haloalkanes And Haloarenes Class 12 Chemistry PDF

Better preparation of Case Study on Haloalkanes And Haloarenes Class 12 Chemistry can help students score good marks in the CBSE Class 12 Board examination. Additionally, it helps build confidence and enables students to deepen their knowledge of Haloalkanes And Haloarenes. Because case-based questions are equally important for learning and board exam preparation, our team has prepared Case Study on Haloalkanes And Haloarenes Class 12 Chemistry in a PDF file for free distribution among students.

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Haloalkanes And Haloarenes Case Study for Class 12 Chemistry with Solutions in PDF

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The solutions are in the hint manner as well as contain full examples too, refer to the link to access the Case Study on Haloalkanes And Haloarenes Class 12 Chemistry with Solutions in PDF - it’s free.

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4 Tips to Answer Class 12 Chemistry Haloalkanes And Haloarenes Case Study Questions?

There are 4 solid tips to answer Class 12 Chemistry Haloalkanes And Haloarenes Case Study questions that we are discussing in this section.

• Read the Case Carefully: To start gathering insights from the given case-based questions, it is vital to read the Haloalkanes And Haloarenes case carefully and identify the key facts, figures, and units of measurement. Pay attention to any diagrams or graphs related to Haloalkanes And Haloarenes provided, as they may contain important information.
• Identify the Problem: While reading the Haloalkanes And Haloarenes case it is essential to consider the possible causes and effects. This will help you determine the appropriate approach to solving the problem in Class 12 Chemistry Haloalkanes And Haloarenes.
• Use Elimination Methods Too: Since case study questions of Class 12 Chemistry Haloalkanes And Haloarenes, are often framed in Multiple choice questions, students should have the knowledge of elimination methods in MCQs to better answer the questions.
• Before All, Master the Concept of Haloalkanes And Haloarenes: If the above two methods are not working for you to answer Case Study on Haloalkanes And Haloarenes Class 12 Chemistry then you need to revisit the lesson and master the concepts explained in the Haloalkanes And Haloarenes Class 12 Chemistry.

What is the Benefit of Practising Class 12 Chemistry Haloalkanes And Haloarenes Case Study Questions?

When a student decides to practise Class 12 Chemistry Haloalkanes And Haloarenes case study questions before the board examination they generally get these 3 benefits:

• Quick Conceptual Revision: Nothing is better for revision than solving relevant questions and therefore, those who involve in solving the Haloalkanes And Haloarenes Case study questions before the Class 12 Chemistry exam are able to better revise their conceptual learnings from the lesson.
• Better Board Exam Preparation: No doubt, the more you practise Haloalkanes And Haloarenes case study questions the better your exam preparation will be so, solving Case-based questions prior to the board examination helps a lot in the preparation. It also enables the students to know what are the Haloalkanes And Haloarenes questions which have the possibility to be asked in the board examination.
• Confident in Using Analytical or Critical Thinking Skills: The Case-based questions on Haloalkanes And Haloarenes are all about using analytical or critical thinking skills where students are required to solve problems based on the situations or data given. Thus, practising Case Study on Haloalkanes And Haloarenes Class 12 Chemistry benefits students to feel confident and comfortable in using analytical skills.

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CBSE Class 12 Chemistry Syllabus

• Solid State
• Electrochemistry
• Chemical Kinetics
• Surface Chemistry
• General Principles and Processes of Isolation of Elements
• p-Block Elements
• d- and f-Block Elements
• Coordination Compounds
• Haloalkanes and Haloarenes.
• Alcohols, Phenols and Ethers
• Aldehydes, Ketones and Carboxylic Acids
• Organic compounds containing Nitrogen
• Biomolecules
• Chemistry in Everyday life

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Actinoids - Electronic configuration, oxidation states and comparison with lanthanoids.

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Unit X: Haloalkanes and Haloarenes. 15 Periods Haloalkanes: Nomenclature, nature of C–X bond, physical and chemical properties, optical rotation mechanism of substitution reactions.

Haloarenes: Nature of C–X bond, substitution reactions (Directive influence of halogen in monosubstituted compounds only). Uses and environmental effects of - dichloromethane, trichloromethane, tetrachloromethane, iodoform, freons, DDT.

Unit XI: Alcohols, Phenols and Ethers 14 Periods

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Phenols: Nomenclature, methods of preparation, physical and chemical properties, acidic nature of phenol, electrophilic substitution reactions, uses of phenols.

Ethers: Nomenclature, methods of preparation, physical and chemical properties, uses.

Unit XII: Aldehydes, Ketones and Carboxylic Acids 15 Periods

Aldehydes and Ketones: Nomenclature, nature of carbonyl group, methods of preparation, physical and chemical properties, mechanism of nucleophilic addition, reactivity of alpha hydrogen in aldehydes, uses.

Carboxylic Acids: Nomenclature, acidic nature, methods of preparation, physical and chemical properties; uses.

Unit XIII: Amines 14 Periods

Amines: Nomenclature, classification, structure, methods of preparation, physical and chemical properties, uses, identification of primary, secondary and tertiary amines.

Diazonium salts: Preparation, chemical reactions and importance in synthetic organic chemistry.

Unit XIV: Biomolecules 18 Periods

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Vitamins - Classification and functions. Nucleic Acids: DNA and RNA.

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There are many ways to ascertain whether a student has understood the important points and topics of a particular chapter and is he or she well prepared for exams and tests of that particular chapter. Apart from reference books and notes, Question Banks are very effective study materials for exam preparation. When a student tries to attempt and solve all the important questions of any particular subject , it becomes very easy to gauge how much well the topics have been understood and what kind of questions are asked in exams related to that chapter.. Some of the other advantaging factors of Question Banks are as follows

• Since Important questions included in question bank are collections of questions that were asked in previous exams and tests thus when a student tries to attempt them they get a complete idea about what type of questions are usually asked and whether they have learned the topics well enough. This gives them an edge to prepare well for the exam.Students get the clear idea whether the questions framed from any particular chapter are mostly either short or long answer type questions or multiple choice based and also marks weightage of any particular chapter in final exams.
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• Although students cover up all the chapters included in the course syllabus by the end of the session, sometimes revision becomes a time consuming and difficult process. Thus, practicing important questions from Question Bank allows students to check the preparation status of each and every small topic in a chapter. Doing that ensures quick and easy insight into all the important questions and topics in each and every individual. Solving the important questions also acts as the revision process.

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CBSE Chemistry Chapter 10 Haloalkanes and Haloarenes Class 12 Important Questions

• Class 12 Important Question
• Chapter 10: Haloalkanes And Haloarenes

Haloalkanes and Haloarenes Class 12 Important Questions - Free PDF Download

In the realm of CBSE Class 12 Chemistry, Chapter 10 - Haloalkanes and Haloarenes, plays a pivotal role in understanding the world of organic compounds. This chapter delves into the intriguing world of haloalkanes and haloarenes, shedding light on their structures, nomenclature, properties, and reactions. To excel in this subject, it is imperative to have a strong grasp of the fundamental concepts and the ability to tackle a variety of questions effectively. Vedantu's Important Questions PDF for this chapter serves as an invaluable tool, meticulously curated to aid students in their preparation. These questions not only reinforce core concepts but also provide insight into the types of questions that can be expected in examinations, making learning engaging and effective.

The important questions from haloalkanes and haloarenes are now available for FREE in PDF format on Vedantu. These questions and answers are prepared by the highly experienced teachers at Vedantu according to the latest NCERT curriculum .

Important Topics Covered in CBSE Class 12 Haloalkanes and Haloarenes  

The important questions of Haloalkanes and Haloarenes of Class 12 cover the topics of:

Classification

Nomenclature

Nature of C-X Bond

Methods of Preparation

Types of Reactions (Elimination, substitution, etc)

Physical and Chemical properties, the order of reactivity.

Polyhalogen Compounds

Extra questions of Haloalkanes and Haloarenes are very helpful to streamline the preparation and revision of these topics.

Download CBSE Class 12 Chemistry Important Questions 2024-25 PDF

Also, check CBSE Class 12 Chemistry Important Questions for other chapters:

Important Questions For Class 12 Chemistry Chapter 10

1. Write the IUPAC names of the following compounds.

Ans: The IUPAC name of the compound will be 3-Bromo-3-Methylhexane.

Ans: The IUPAC name of the compound will be 1-Bromo-2,3-dimethylbut-2-ene.

Ans: The IUPAC name of the compound is Benzyl chloride.

Ans: The IUPAC name of the compound is 1-Bromo-2-ethyl-3,3,4-trimethylpentane.

x$\text{C}{{\text{H}}_{\text{2}}}\text{Br-CH=CH-C}{{\text{H}}_{\text{2}}}\text{-C}\equiv \text{CH}$ 

Ans: The IUPAC name of the compound is 6-Bromohex-4-ene-1-yne,

Ans: The IUPAC name of the compound is 1-Bromo-1-methylcyclohexane.

Ans: The IUPAC name of the compound is 3-sec-propyl pent-2,4-diol.

${{\text{(CC}{{\text{l}}_{\text{3}}}\text{)}}_{\text{3}}}\text{CCl}$ 

Ans: The IUPAC name of the compound is tris-(trichloromethyl) chloromethane.

2. Write the structure of the following halogen compounds

(i) 2-chloro-3-methylpentane

Ans: The structure of the compound is given below:

(ii) 2-(2-chlorophenyl)-1-iodooctane

Ans:  The structure of the compound is given below:

(iii) 1-bromo-4-sec-butyl-2–methylbenzene.

(iv) p-bromotoluene.

(v) Chlorophenylmethane

3. Arrange the following in the increasing order of properly indicated :

i. bromomethane, chloromethane, dichloromethane. (Increasing order of boiling points).

Ans: As we can see that all the compounds given above are haloalkanes. The order will be:

Chloromethane < Bromomethane < Dichloromethane

This is due to the fact that as the halogen size increases the boiling point will increase and as the number of halogen atoms increases in the same chain, the boiling point will increase.

ii. 1-chloropropane, isopropyl chloride, 1-chlorobutane  (Increasing order of boiling point)

Ans: In all the compounds there is a chlorine atom present and the size of the alkyl chain is different. The order will be:

Isopropyl chloride < 1- Chloropropane < 1 – Chlorobutane 

This is due to the fact that as the branching of the chain increases the boiling point will decrease and as the size of the chain increase the boiling point will increase.

iii. dichloromethane, chloroform, carbon tetrachloride. (Increasing order of dipole moment.

Ans: Below is the three-dimensional structures of the three compounds, as well as the direction of each bond's dipole moment:

$\text{CC}{{\text{l}}_{\text{4}}}$ has no dipole moment since it is symmetrical. When two C-Cl dipole moments are added to $\text{CHC}{{\text{l}}_{\text{3}}}$, the C-H and C-Cl bonds oppose each other. $\text{CHC}{{\text{l}}_{\text{3}}}$ has a limited dipole moment (1.03 D) because the dipole moment of the second resultant is anticipated to be less than that of the first. This means that in $\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$, the resulting dipole moment of C-Cl pairs is greater than in $\text{CHC}{{\text{l}}_{\text{3}}}$. Due to its dipole moment, $\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$ is the strongest. The order will be:

Carbon tetrachloride < Chloroform < Dichloromethane

iv. $\text{C}{{\text{H}}_{\text{3}}}\text{F}$, $\text{C}{{\text{H}}_{\text{3}}}\text{Cl}$, $\text{C}{{\text{H}}_{\text{3}}}\text{Br}$, $\text{C}{{\text{H}}_{\text{3}}}\text{I}$  (Increasing reactivity towards Nucleophilic substitution and increasing order of dipole moment)

Ans: The reaction is known as a nucleophilic substitution reaction when a nucleophile, which is an electron pair giver, interacts with an electron pair acceptor.

It is well known that as one moves down the group, the size of the components grows larger. As a result, the larger the element, the easier it will serve as a leaving group, allowing the nucleophile to connect more quickly. The order is given below:

$\text{C}{{\text{H}}_{\text{3}}}\text{F}$ < $\text{C}{{\text{H}}_{\text{3}}}\text{Cl}$ < $\text{C}{{\text{H}}_{\text{3}}}\text{Br}$ < $\text{C}{{\text{H}}_{\text{3}}}\text{I}$ 

v. o,m,p-dichlorobenzenes (Increasing order of melting points)

Ans: Because of its symmetry and structure, p-dichlorobenzene has the greatest melting point, followed by ortho, and finally meta.

The melting point of a compound is related to its symmetry. As a result, the symmetry of the compound follows the same pattern as the melting point. The order is given below:

m-Dichlorobenzene < o-Dichlorobenzen < p-Dichlorobenzene

4. Complete the following reactions: 

Ans: The complete reaction is given below:

(ii) $\text{C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-Cl + AgN}{{\text{O}}_{\text{2}}}\text{ }\to $ 

\[\text{C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-Cl + AgN}{{\text{O}}_{\text{2}}}\text{ }\to \text{ C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-N}{{\text{O}}_{\text{2}}}\text{ + AgCl}\] 

Ans: In this reaction, the amine group will be replaced with methyl bromine. The reaction is given below:

Ans: In this reaction, the iodine from the hydrogen iodide will attach to the carbon atom having the double bond as well as a methyl group. The reaction is given below:

Ans: In this reaction, the hydroxyl group with the methyl group will be replaced with the chlorine atom. The reaction is given below:

(viii) $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{Br + NaI }\xrightarrow{\text{acetone}}$ 

Ans: The Chlorine atom from the alkyl halide by iodine. The major product of the reaction is 1-Iodoethane. The reaction is given below:

\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{Br + NaI }\xrightarrow{\text{acetone}}\text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{I + NaBr}\]

Ans: In this reaction, the bromine atom will attack the alpha-carbon atom of the double bond. The reaction is given below:

(x) ${{\text{(C}{{\text{H}}_{\text{3}}}\text{)}}_{\text{3}}}\text{CBr + KOH }\xrightarrow{\text{Ethanol}}$ 

Ans: There will be Dehydrohalogenation and the major product of the reaction is 2-methylpropene. The reaction is given below:

(xi) $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{Br + KCN}\xrightarrow{\text{aq}\text{.ethanol}}$  

Ans: The bromide ion will be replaced with the cyanide ion. The major product will be propanenitrile. The reaction is given below:

\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{Br + KCN }\xrightarrow{\text{aq}\text{.ethanol}}\text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CN + KBr}\]  

Ans: There will be the formation of a Grignard reagent. The reaction is given below:

(xiii) ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{ONa + }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{Cl }\to $ 

Ans: The major product in the above reaction will be Phenetole. The reaction is given below:

\[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{ONa + }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{Cl }\to {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{-O-}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{ + NaCl}\]

Ans: The product formed in this reaction will be 3,4-Dimethylhexane. The reaction is given below:

(xv) $\text{C}{{\text{H}}_{\text{3}}}\text{CH(Br)C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}\text{ + NaOH }\xrightarrow{\text{Water}}$ 

Ans: The bromine atom will be replaced with the hydroxyl ion. The major product will be Butan-2-ol. The reaction is given below:

(xvi) ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{N}_{\text{2}}^{\text{+}}\text{C}{{\text{l}}^{\text{-}}}\text{ + KI }\to $ 

Ans: There will be the formation of iodobenzene in this case. The reaction is given below:

\[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{N}_{\text{2}}^{\text{+}}\text{C}{{\text{l}}^{\text{-}}}\text{ + KI }\to \text{ }{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{I + }{{\text{N}}_{\text{2}}}\text{ + KCl}\] 

5. How will you bring about the following conversions?

i. benzene to 3-bromonitrobenzene

Ans: Benzene will first react with concentrated sulfuric acid and nitric acid to form nitrobenzene. Now, this nitrobenzene will react with bromine in the presence of $\text{FeB}{{\text{r}}_{\text{3}}}$ to form 3-Bromonitrobenzene. The reaction is given below:

ii. ethanol to but-1-yne

Ans: Ethanol will react with $\text{SOC}{{\text{l}}_{\text{2}}}$ and pyridine to Chloroethane. Acetylene will react with $\text{NaN}{{\text{H}}_{\text{2}}}$ to form sodium acetylide. Now Chloroethane and Sodium acetylide will react to form But-1-yne. The reactions are given below:

\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{OH }\xrightarrow{\text{SOC}{{\text{l}}_{\text{2}}}\text{, Pyridine}}\text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{-Cl}\] 

\[\text{CH}\equiv \text{CH + NaN}{{\text{H}}_{\text{2}}}\xrightarrow{\text{Liq}\text{. N}{{\text{H}}_{\text{3}}}\text{, 196K}}\text{ HC}\equiv {{\text{C}}^{\text{-}}}\text{N}{{\text{a}}^{\text{+}}}\] 

\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{Cl + HC}\equiv {{\text{C}}^{\text{-}}}\text{N}{{\text{a}}^{\text{+}}}\text{ }\to \text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{-C}\equiv \text{CH + NaCl}\]

iii. 1-bromopropane to 2-bromopropane

Ans: 1-Bromopropane will react with alcoholic KOH to form propene. Propene will react with HBr to form 2-Bromopropane. The reaction is given below:

\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{Br}\xrightarrow{\text{Alc}\text{.KOH}}\text{C}{{\text{H}}_{\text{3}}}\text{CH=C}{{\text{H}}_{\text{2}}}\xrightarrow{\text{HBr}}\text{C}{{\text{H}}_{\text{3}}}\text{-CH(Br)-C}{{\text{H}}_{\text{3}}}\]

iv. benzene to 4-bromo-1-nitrobenzene

Ans: Benzene will react with Bromine in the presence of $\text{FeB}{{\text{r}}_{\text{3}}}$ to form Bromobenzene. Bromobenzen will react with concentrated nitric acid and concentrated sulfuric acid to form 4-Bromonitrobenzene. The reaction is given below:

v. aniline to chlorobenzene

Ans: Aniline will undergo diazotization to form Benzene diazonium chloride. The Benzene diazonium chloride will react with copper chloride in the presence of hydrochloric acid to given Chlorobenzene. The reaction is given below:

vi. 2-methyl-1-propene to 2-chloro-2-methylpropane

Ans: 2-Methyl-1-propene will react with Hydrogen chloride to give 2-Chloro-2-methylpropane. The reaction is given below:

vii. ethyl chloride to propanoic acid

Ans: Ethyl chloride will react with KCN to give propanenitrile. Propanenitrile on hydrolysis will give propanoic acid. The reaction is given below:

\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{Cl}\xrightarrow{\text{KCN}}\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CN}\xrightarrow{{{\text{H}}^{\text{+}}}\text{/}{{\text{H}}_{\text{2}}}\text{O}}\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{COOH}\]  

viii. but-1-ene to n-butyl iodide

Ans: But-1-ene will react with HBr in the presence of peroxide to form 1-Bromobutane. 1-Bromobutane will react with NaI in the presence of Acetone to give n-butyl iodide. The reaction is given below:

\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CH=C}{{\text{H}}_{\text{2}}}\xrightarrow[\text{peroxide}]{\text{HBr}}\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{Br}\]

\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{Br}\xrightarrow[\text{Acetone}]{\text{NaI}}\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{I}\] 

ix. benzene to phenyl chloromethane.

Ans: Benzene will first react with chloromethane in the presence of ferric chloride to form toluene. Now, the toluene will react with chlorine in the presence of sunlight. The reaction is given below:

x. tert-butyl bromide to isobutyl bromide.

Ans: Tert-butyl bromide will react with alcoholic KOH to form 2-Methyl-1-propene. 2-Methyl-1-propene will react with HBr in the presence of peroxide to form isobutyl bromide. The reaction is given below:

6. Identify the products formed in the following sequence :

ii. $\text{Br-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-Br }\xrightarrow{\text{alc}\text{.KOH}}\text{ A }\xrightarrow{\text{NaN}{{\text{H}}_{\text{2}}}}\text{ B}$ 

\[\text{Br-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-Br }\xrightarrow{\text{alc}\text{.KOH}}\text{ C}{{\text{H}}_{\text{2}}}\text{=CHBr }\xrightarrow{\text{NaN}{{\text{H}}_{\text{2}}}}\text{CH}\equiv \text{CH}\] 

iii.${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{C}{{\text{H}}_{\text{2}}}\text{CHBrC}{{\text{H}}_{\text{3}}}\xrightarrow{\text{alc}\text{.KOH}}\text{A }\xrightarrow{\text{HBr}}\text{B}$ 

v. $\text{C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{CH=C}{{\text{H}}_{\text{2}}}\text{+B}{{\text{r}}_{\text{2}}}\xrightarrow{\text{CC}{{\text{l}}_{\text{4}}}}\text{A}$ 

vi.$\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CH=C}{{\text{H}}_{\text{2}}}\text{ + B}{{\text{r}}_{\text{2}}}\xrightarrow[\text{UV light}]{\text{heat}}\text{ B}$ 

Ans: In this reaction, the reactant is an alkyl halide while the product is an alkene and there is the elimination of HBr from the reactant so, the catalyst will be alcoholic KOH. Therefore, A will be an alcoholic KOH.

(viii) $\text{C}{{\text{H}}_{\text{3}}}\text{Br}\xrightarrow{\text{KCN}}\text{A}\xrightarrow{{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}}\text{B}\xrightarrow{\text{LiAl}{{\text{H}}_{\text{4}}}}\text{C}$ 

\[\text{C}{{\text{H}}_{\text{3}}}\text{Br}\xrightarrow{\text{KCN}}\text{C}{{\text{H}}_{\text{3}}}\text{CN}\xrightarrow{{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}}\text{C}{{\text{H}}_{\text{3}}}\text{COOH}\xrightarrow{\text{LiAl}{{\text{H}}_{\text{4}}}}\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{OH}\] 

7. Explain the following reactions with suitable examples:

(i) Finkelstein reaction.

Ans: In the traditional Finkelstein reaction, an alkyl bromide or alkyl chloride is converted to an alkyl iodide, which is then treated with a sodium iodide solution in acetone. The reaction is given below:

\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{Br + NaI }\to \text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{I + NaBr}\] 

(ii) Swarts reaction.

Ans: Swarts' reaction is commonly used to make alkyl fluorides from alkyl chlorides or alkyl bromides. This is accomplished by heating the alkyl chloride / bromide in the presence of fluoride in certain heavy metals. The reaction is given below:

\[\text{C}{{\text{H}}_{\text{3}}}\text{-Br + AgF }\to \text{ C}{{\text{H}}_{\text{3}}}\text{F + AgBr}\] 

(iii) Wurtz reaction.

Ans: To generate higher alkanes, alkyl halides are treated with sodium metal in a dry ethereal (moisture-free) solution. It can also be utilized to create higher alkanes with an even number of carbon atoms. The reaction is given below:

\[\text{2 R-X + 2Na }\to \text{ R-R + 2 NaX}\] 

(iv) Wurtz-Fitting reaction

Ans: The Wurtz–Fitting reaction is a chemical reaction that produces substituted aromatic compounds by combining aryl halides with alkyl halides and sodium metal in the presence of dry ether. The reaction is given below:

(v) Friedel-Crafts alkylation reaction.

Ans: This reaction permitted alkylbenzenes to be formed from alkyl halides, but it was hampered with undesired supplementary activity, which decreased its efficiency. The reaction is given below:

(vi) Friedel-Crafts acylation reaction

Ans: The Friedel–Crafts acylation uses a strong Lewis acid catalyst to react an arene with acyl chlorides or anhydrides. This process produces monoacetylated compounds by electrophilic aromatic substitution. The reaction is given below:

(vii) Sandmeyer reaction.

Ans: The Sandmeyer reaction is a chemical process that uses copper salts as reagents or catalysts to synthesize aryl halides from aryl diazonium salts. The reaction is given below:

8. Write the major products and name the rule responsible for the formation of the product.

Ans: In this reaction, the alkyl halide is treated with KOH and ethanol so, first, there will be the formation of alkene and then there will be the addition of water molecules. This is called a substitution reaction. The reaction is given below:

Ans: In this reaction, the alkyl halide is treated with HBr in the presence of peroxide so, the Br will attack the carbon atom having larger number of hydrogen atoms and the hydrogen will attack the carbon atom having less number of hydrogen atoms. This is called Anti-Markovnikov’s rule. Therefore, the product will be Bromobutane. The reaction is given below:

9. Write the Difference Between

(i) Enantiomers and Diastereomers

Ans: This is tabulated below:

(ii) Retention and Inversion of configuration.

(iii) Electrophilic and Nucleophilic substitution reactions.

10. Give a chemical test to distinguish between the following pairs of compounds:

(i) Chlorobenzene and Cyclohexylchloride.

Ans: Cyclohexylchloride can react with KOH to give Cyclohexanol and KCl. The KCl from the product can be reacted with $\text{AgN}{{\text{O}}_{\text{3}}}$ to give white ppt. AgCl. Whereas, chlorobenzene will not give this reaction.

(ii) Vinyl chloride and ethyl chloride.

Ans: Bromine water can be used to tell the difference. When vinyl chloride reacts with bromine water, it becomes decolorized, but ethyl chloride does not.

(iii) N-propyl bromide and isopropyl bromide.

Ans: When isopropyl bromide reacts with KOH, 2-propanol is produced. 2-propanol reacts with HCl and $\text{ZnC}{{\text{l}}_{\text{2}}}$  to generate 2-propyl chloride, which causes turbidity after 5 minutes, whereas n-propyl bromide reacts with KOH to produce propanol, and propanol reacts with HCl and $\text{ZnC}{{\text{l}}_{\text{2}}}$ to make 2-propyl chloride, which does not cause turbidity.

11. Give mechanism of the following reactions:

(i) ${{\text{(C}{{\text{H}}_{\text{3}}}\text{)}}_{\text{3}}}\text{C-Cl +}{{\text{ }}^{\text{-}}}\text{OH }\to \text{ (C}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{3}}}\text{-OH}$ 

Ans: This reaction will follow the unimolecular Nucleophilic substitution reaction mechanism (${{\text{S}}_{\text{N}}}\text{1}$). The mechanism is given below:

(ii) $\text{C}{{\text{H}}_{\text{3}}}\text{-Cl + O}{{\text{H}}^{\text{-}}}\text{ }\to \text{ C}{{\text{H}}_{\text{3}}}\text{-OH}$ 

Ans: This reaction will follow the bimolecular Nucleophilic substitution reaction mechanism (${{\text{S}}_{\text{N}}}2$). The mechanism is given below:

Ans: In this reaction, there will be an elimination mechanism, as there is the formation of an alkene by the elimination of HCl molecules. The mechanism is given below:

Ans: In this reaction, there is a substitution mechanism as the hydrogen atom from the benzene is substituted with the chlorine atom thus forming chlorobenzene.

Ans: In this reaction, there is a substitution mechanism as the chlorine atom from the benzene is substituted with the hydroxyl ion thus forming p-nitrophenol.

12. Which compound in each of the following pairs will react faster in ${{\text{S}}_{\text{N}}}\text{2}$  reaction with $\text{O}{{\text{H}}^{\text{-}}}$?

(i) $\text{C}{{\text{H}}_{\text{3}}}\text{Br}$ or $\text{C}{{\text{H}}_{\text{3}}}\text{I}$ 

Ans: Both the compounds are alkyl halide but the iodide ion is a larger atom than bromide ion. So, ${{\text{I}}^{\text{-}}}$ ion is better leaving group than $\text{B}{{\text{r}}^{\text{-}}}$ ion. Therefore, $\text{C}{{\text{H}}_{\text{3}}}\text{I}$ will react faster than $\text{C}{{\text{H}}_{\text{3}}}\text{Br}$ towards ${{\text{S}}_{\text{N}}}\text{2}$ reaction with hydroxyl ion.

(ii) ${{\text{(C}{{\text{H}}_{\text{3}}}\text{)}}_{\text{3}}}\text{CCl}$ or $\text{C}{{\text{H}}_{\text{3}}}\text{Cl}$ 

Ans: In ${{\text{S}}_{\text{N}}}\text{2}$reaction the steric hindrance should be very less. ${{\text{(C}{{\text{H}}_{\text{3}}}\text{)}}_{\text{3}}}\text{CCl}$ has very high steric hindrance and $\text{C}{{\text{H}}_{\text{3}}}\text{Cl}$ has less steric hindrance. So, $\text{C}{{\text{H}}_{\text{3}}}\text{Cl}$ will react faster to the ${{\text{S}}_{\text{N}}}\text{2}$ reaction with hydroxyl ion.

13. In the following pairs of halogen compounds, which compound undergoes faster ${{\text{S}}_{\text{N}}}\text{1}$  reaction?

Ans: For the ${{\text{S}}_{\text{N}}}1$ reaction the order of reactivity is ${{3}^{\circ }} > {{2}^{\circ }} > {{1}^{\circ }}$, so we can solve the question according to the order of the reactivity.

Ans: The first compound is a tertiary compound and the second compound is a secondary compound. So, the first compound will undergo faster ${{\text{S}}_{\text{N}}}1$reaction.

Ans: The first compound is a secondary compound and the second compound is the primary compound. So, the first compound will undergo a faster ${{\text{S}}_{\text{N}}}1$reaction.

(iii) ${{\text{(C}{{\text{H}}_{\text{3}}}\text{)}}_{\text{3}}}\text{C-Cl}$ and ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{C}{{\text{H}}_{\text{2}}}\text{Cl}$ 

Ans: The first compound is a tertiary compound and the second compound is primary compound. So, the first compound will undergo a faster ${{\text{S}}_{\text{N}}}1$reaction.

(iv)${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{C}{{\text{H}}_{\text{2}}}\text{Cl}$ and ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{C(Cl)}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}$ 

Ans: The first compound is the primary compound and the second compound is the secondary compound. So, the second compound will undergo faster ${{\text{S}}_{\text{N}}}1$reaction.

(v) $\text{C}{{\text{H}}_{\text{2}}}\text{=CH-Cl}$ and $\text{C}{{\text{H}}_{\text{2}}}\text{=CH-C}{{\text{H}}_{\text{2}}}\text{Cl}$ 

Ans: The first compound is the vinylic primary compound and the second compound is the primary compound. So, the second compound will undergo faster ${{\text{S}}_{\text{N}}}1$reaction because in the first there will be resonance.

14. Give reasons for the following :

(i) The bond length of the C–Cl bond is larger in haloalkanes than that in haloarenes.

Ans: The C linked to the halogen atom in haloalkanes is $\text{s}{{\text{p}}^{\text{3}}}$  hybridized, whereas it is $\text{s}{{\text{p}}^{2}}$  hybridized in haloarenes.

In haloarene, the halogen is in resonance with the benzene ring. Due to this, there is a partial double bond character between halogen and carbon which results in a shorter bond length.

(ii) Although alkyl halides are polar in nature but are not soluble in water.

Ans: Dipole-dipole attractions exist in alkyl halides, whereas hydrogen bonding and van der Waals force exist in water. Alkyl halides have a hard time breaking those hydrogen bonds, thus they're nearly insoluble in water.

(iii) tert-butyl bromide has a lower boiling point than n-Butyl bromide.

Ans: Because the boiling point of n-butyl bromide increases as branching decreases, it has a higher boiling point than tert butyl bromide. This is due to the fact that when the surface area of the molecule decreases, the molecule becomes more spherical. Intermolecular forces weaken, as a result, resulting in a lower boiling point.

(iv) haloalkanes react with KCN to form alkyl cyanide as the main product while with AgCN alkyl isocyanide is the main product.

Ans: KCN is primarily an ionic compound that produces cyanide ions in solution. Although both the carbon and nitrogen atoms of CN have the ability to contribute electron pairs, the assault is focused on the carbon atom rather than the nitrogen atom because the C-C bond is more stable than the C-N link. However, in nature, AgCN is mostly covalent, and nitrogen is free to contribute electron pairs, resulting in isocyanide as the primary product.

(v) Sulphuric acid is not used in the reaction of alcohol with Kl.

Ans: Because sulfuric acid is an oxidizing agent, it will convert KI to equivalent HI and oxidize the HI if it is used in the process. As a result, no sulfuric acid is needed in the reaction of alcohols with KI.

(vi) Thionyl chloride is the preferred reagent for converting ethanol to chloroethane.

Ans: Thionyl chloride is favored for preparing alkyl chlorides from alcohols because the reaction's by-products include SO 2 and HCl, both of which are gaseous and escape into the atmosphere, leaving only pure alkyl chlorides behind.

(vii) Haloalkanes undergo nucleophilic substitution reaction easily but haloarenes do not undergo nucleophilic substitution under ordinary conditions.

Ans: Because haloalkanes are more polar than haloarenes, they are more prone to nucleophilic substitution reactions. The reason for this is that the carbon atom linked to the halogen atom in haloalkanes is $\text{s}{{\text{p}}^{\text{3}}}$ hybridized, whereas it is $\text{s}{{\text{p}}^{2}}$ hybridised in haloarenes. The electronegativity difference in the CX bond of haloalkanes is higher because an $\text{s}{{\text{p}}^{\text{3}}}$ hybridized carbon is less electronegative than a $\text{s}{{\text{p}}^{2}}$ hybridised carbon, making them more polar.

(viii) Chlorobenzene on reaction with fuming sulphuric acid gives ortho and para chlorosulphonic acids.

Ans: Although chlorine is an electronegative molecule, it possesses a lone pair of electrons that have a tendency to be added to the benzene ring's resonance structure, resulting in partial negative charges on ortho and para positions, with the electron density being lowest at meta.

Now, we know that when sulphuric acid is introduced to this molecule, the mechanism is electrophile ($\text{S}{{\text{O}}_{\text{3}}}$)attacks, and since it will approach the position with greater electron density, it will attack the negative, ortho/para positions, yielding ortho and para chloro sulphonic acid.

(ix) 2, 4-dinitro chlorobenzene is much more reactive than chlorobenzene towards hydrolysis reaction with NaOH.

Ans: On the benzene ring, the lone pair of electrons on chlorobenzene are delocalized. As a result, the C-Cl bond takes on a partial double bond appearance. As a result, the C-Cl bond in chlorobenzene is extremely strong and difficult to break. However, in the case of 2,4-dinitro chlorobenzene, the presence of $\text{N}{{\text{O}}_{\text{2}}}$ groups at ortho and para positions pulls electrons away from the benzene ring, making the nucleophile attack on para chlorobenzene easier. The resonance stabilizes the carbanion that results. In comparison to chlorobenzene, 2,4-dinitrochlorobenzene is more reactive towards nucleophilic substitution processes.

(x) The Grignard reagent should be prepared under anhydrous conditions.

Ans: Grignard reagents have a high degree of reactivity. When they come into contact with moisture, they react to form alkanes. Grignard reagents should thus be produced in anhydrous circumstances.

(xi) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

Ans: Chlorobenzene has a lower dipole potential than cyclohexyl chloride because the C-Cl bond in chlorobenzene is $\text{s}{{\text{p}}^{2}}$ hybridised, whereas the C-Cl link in cyclohexyl chloride is $\text{s}{{\text{p}}^{\text{3}}}$ hybridized. Because $\text{s}{{\text{p}}^{2}}$ has a higher s character and is more electronegative than $\text{s}{{\text{p}}^{\text{3}}}$, it is more electronegative. Compared to cyclohexyl chloride, chlorobenzene is less polar.

(xii) Neopentyl bromide undergoes nucleophilic substitution reactions very slowly.

Ans: Bromine is sterically hindered because it is linked to carbon where the surrounding carbon group contains a significant number of alkyl substituents ( like a shielding effect to prevent nucleophilic attack). This is why this nucleophilic substitution takes so long.

(xiii) Vinyl chloride is unreactive in nucleophilic substitution reactions.

Ans: Due to resonance, vinyl chloride is unreactive in nucleophilic substitution reactions. The lone pair of electrons on chlorine in vinyl chloride is in resonance with the C-C double bond, giving the C-Cl bond a partial double bond nature. The C-Cl connection grows stronger and more difficult to break as a result of its double bond nature.

(xiv) An optically inactive product is obtained after the hydrolysis of optically active 2- bromobutane.

Ans: The ${{\text{S}}_{\text{N}}}\text{1}$  reaction is used to hydrolyze 2-bromobutane. The ${{\text{S}}_{\text{N}}}\text{1}$  reaction is triggered by the creation of a carbocation, in which the OH-attack the carbocation from both sides, culminating in the production of 2-butanol, a racemic product. As a result of the production of the racemic product, 2-butanol is optically inactive.

15.  Write the different products and their number formed by the monochlorination of following compounds:

(i) $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}$

Ans: There will be the formation of 2 products. 

When this compound is monochlorinated then one product will be 1-Chlorobutane. The reaction is given below:

\[\text{C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{3}}}\text{ + C}{{\text{l}}_{\text{2}}}\text{ }\to \text{ C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{Cl}\] 

Another product when the compound is chlorinated will be 2-Chlorobutane. The reaction is given below:

\[\text{C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{3}}}\text{ + C}{{\text{l}}_{\text{2}}}\text{ }\to \text{ C}{{\text{H}}_{\text{3}}}\text{-CHCl-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{3}}}\] 

(ii) ${{\text{(C}{{\text{H}}_{\text{3}}}\text{)}}_{\text{2}}}\text{CHC}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}$ 

Ans: There will be the formation of 4 products.

First, the product will be 2-Chloro-2-methylbutane. The reaction is given below:

Second, the product will be 2-Chloro-3-methylbutane. The reaction is given below:

Third, the product formed will be 1-Chloro-3-methylbutane. The reaction is given below:

Fourth, the product formed will be 1-Chloro-2-methylbutane. The reaction is given below:

(iii) ${{\text{(C}{{\text{H}}_{\text{3}}}\text{)}}_{\text{2}}}\text{CHCH(C}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{2}}}$ 

Ans: There will be the formation of 2 products.

First, there will be the formation of 1-Chloro-2,3-dimethyl butane. The reaction is given below:

Second, there will be the formation of 2-Chloro-2,3-dimethyl. The reaction is given below:

(a)  When 3-methylbutan-2-ol is treated with HBr, the following reaction takes places :

Give the mechanism of this reaction.

Ans: Below is the process for the reaction between 3-methyl butane-2-ol and HBr.

The -OH group is protonated in the first step. A water molecule is lost in the second stage, resulting in secondary carbocation.

The third stage is the 1,2-hydride shift, which converts less stable secondary carbocation to more stable tertiary carbocation.

The nucleophilic assault of bromide ion on the tertiary carbocation to produce 2-Bromo-2-methyl butane is the last step.

(b) In the following reaction:

Major and minor products are:

(iii) 

Ans: The terminal carbon atom will take the electrons of the double which will form a secondary carbocation. Now there will be a 1,2-methyl shift to form tertiary carbocation because of more stability. The hydroxyl will attack the carbocation and the product will be (iii). In another case, the electrons of the double bond will be taken by the second carbon and there will be the formation of primary carbocation and the hydroxyl ion will attack the carbocation which will form the product as (iv). The major product will be (iii) and the minor product will be (iv). The reaction is given below:

17. Give one use of each of the following:

(i) Freon-12

Ans: Industry uses Freon-12 ($\text{CC}{{\text{l}}_{\text{2}}}{{\text{F}}_{\text{2}}}$), which is the most prevalent form of the refrigerant.

Refrigerant or air-conditioning components, aerosol propellants.

Ans: Its efficacy against mosquitoes that carry malaria and other insects that harm crops led to a dramatic increase in its usage worldwide following World War II.

DDT, on the other hand, has been widely used since the 1940s. Toxic for fishes, DDT acquired tolerance in many insect species. When it comes to animals, DDT is not readily metabolized, but instead accumulates and is retained in fatty tissues. As long as the animals continue to eat DDT at the same rate, it builds up in their bodies.

(iii) Carbon tetrachloride

Ans: For oil, fats, and resins in the industrial sector, as well as in dry cleaning.

In addition, $\text{CC}{{\text{l}}_{\text{4}}}$ vapours are extremely inflammable, according to the manufacturer. As a result, $\text{CC}{{\text{l}}_{\text{4}}}$ is sold as pyrene, a fire extinguishing agent.

Used in the production of aerosol can refrigerants and propellants.

(iv) Iodoform

Ans: Early on it was considered to be an antiseptic, however, the characteristics are attributable to the free iodine that is released, not the substance itself. It has been superseded by other iodine-containing formulations due to its offensive odor.

18. An optically active compound having molecular formula ${{\text{C}}_{\text{7}}}{{\text{H}}_{\text{15}}}\text{Br}$ reacts with aqueous KOH to give ${{\text{C}}_{\text{7}}}{{\text{H}}_{\text{15}}}\text{OH}$, which is optically inactive. Give a mechanism for the reaction.

Ans: Because the intermediate carbocation produced is $\text{s}{{\text{p}}^{2}}$ hybridized and planar, racemization occurs when an optically alkyl halide undergoes the ${{\text{S}}_{\text{N}}}\text{1}$  process. The nucleophile ($\text{O}{{\text{H}}^{\text{-}}}$ ) has an equal chance of attacking it from both sides, resulting in the production of dextro and laevo-rotatory alcohols in equal quantities.

Although the compound ${{\text{C}}_{\text{7}}}{{\text{H}}_{\text{15}}}\text{Br}$ can have a variety of structures, the ${{\text{S}}_{\text{N}}}\text{1}$  mechanism is based on a tertiary alkyl halide. The reaction is given below:

19. An organic compound ${{\text{C}}_{\text{8}}}{{\text{H}}_{\text{9}}}\text{Br}$ has three isomers A, B and C. A is optically active. Both A and B gave the white precipitate when warmed with alcoholic $\text{AgN}{{\text{O}}_{\text{3}}}$ solution in alkaline medium. Benzoic acid, terephthalic and p- bromobenzoic acid were obtained on oxidation of A, B and C respectively. Identify A, B and C.

Ans: Because all three chemicals produce molecules with a benzene nucleus when oxidized, they must all be benzene compounds. Because A and B produce white ppt. when reacting with $\text{AgN}{{\text{O}}_{\text{3}}}$, Br must be in the free form, i.e. Br is not directly linked to benzene in A and B but is in C. All the structures of A, B and C are given below:

20. An alkyl halide X having molecular formula ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{13}}}\text{Cl}$ on treatment with potassium tert-butoxide gives two isomeric alkenes Y and Z but alkene y is symmetrical. Both alkenes on hydrogenation give 2, 3-dimethylbutane.

Identify X, Y, and Z.

Ans: The given formula ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{13}}}\text{Cl}$  shows that it is an alkyl halide. Since the treatment of tert-butoxide gives alkene and both these on hydrogenation give 2, 3-Dimethylbutane, so the X compound will be 2-Chloro-2,3-dimethyl butane. Y is a symmetrical alkene so, its name will be 2,3-Dimethylbut-2-ene and Z is unsymmetrical alkene so, its name will be 2,3-Dimethylbut-1-ene. The structures of X, Y, and Z are given below:

21. An organic compound (A) having molecular formula ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{Cl}$ on reaction with alcoholic solution of KCN gives compound B. The compound B on hydrolysis with dilute HCl gives compound C. C on reduction with ${{\text{H}}_{\text{2}}}$ / Ni gives 1-aminobutane. Identify A, B and C.

Ans: The given formula ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{Cl}$  shows that is it an alkyl halide. This reaction with KCN gives B and the compound B on hydrolysis with KCN gives C. When compound C is reduced with hydrogen and nickel it gives 1-aminobutane which means that all the compounds in the question are straight-chain compounds. So, compound A will be 1-Chloropropane, compound B will be Propionitrile, compound C will be Butanamide. The reactions are given below:

\[\text{C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-Cl + KCN }\to \text{ C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-CN + KCl}\] 

\[\text{C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-CN }\xrightarrow{{{\text{H}}_{\text{2}}}\text{O/HCl}}\text{ C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-CON}{{\text{H}}_{\text{2}}}\] 

\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{-CON}{{\text{H}}_{\text{2}}}\xrightarrow{{{\text{H}}_{\text{2}}}\text{/Ni}}\text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{N}{{\text{H}}_{\text{2}}}\] 

22. Identify A, B, C, D, E, R and ${{\text{R}}^{\text{1}}}$ in the following:

Ans: The compound A in the reaction is cyclohexylmagnesium bromide, compound B is Cyclohexane. 

Compound R –Br will be 2-Bromopropane and the compound C is given below:

The third part of the question is incorrect because the tertiary-alkyl halides do not undergo Wurtz reaction but they undergo dehydrohalogenation to give alkenes. 

So, the compound ${{\text{R}}^{\text{1}}}\text{-Br}$ is given below:

Compound D is Tertiary butyl magnesium bromide and compound E is 2-Methylpropane.

The complete reaction is given below

23. Which nomenclature is not according to the IUPAC system.

(i) $\text{Br-C}{{\text{H}}_{\text{2}}}\text{CH=C}{{\text{H}}_{\text{2}}}$ : 1-Bromoprop-2-ene

Ans: This is the wrong name because the numbering preference will be given to the double bond, therefore, the numbering will start from the double bond. Hence, the correct name will be 3-Bromoprop-1-ene.

4-Bromo-2,4-dimethylhexane 

Ans: The name of the compound given is correct as it follows the IUPAC rules.

2-Methyl-3-phenylpentane

5-Oxohexanoic acid

Ans: The name of the compound given is correct as it follows the IUPAC rules. 

Important Questions for Class 12 Chemistry Chapter 10 - Extra Questions for Practice

Distinguish between C 2 H 5 Br and C 6 H 5 Br with a chemical test.

Arrange the following according to the increasing order of their boiling points:

I. CH 3 CH 2 CH 2 CH 2 Br

II. (CH 3 ) 3 .Br

III. (CH 3 ) 2 C.Br 

Why Chlorobenzene is less reactive with a nucleophilic substitution reaction?

Benzyl chloride and Chlorobenzene - Distinguish between the two with the help of a chemical test.

Why is the dipole moment of chlorobenzene lower than cyclohexyl chloride?

Explain ambident nucleophiles.

Salient Features of Class 12 Chemistry Chapter 10 Important Questions

Take a look at the important features of this study material which will help the students to prepare for their Class 12 Chemistry examination :

Important questions of Haloalkanes and Haloarenes of Class 12 PDF can be downloaded easily from our website and students can learn from it anywhere and at any time. 

These PDFs include various questions and answers along with the problems given in the exercise, for the better practice of students. 

These questions and solutions are carefully prepared by our subject experts in such a way that by referring to them students can clear their doubts and boost their exam preparation.

All the important concepts of this chapter are included in this content in the form of important questions.

The important questions are presented in the form of a free PDF, which helps in convenient learning of this chapter. 

Important Questions for Class 12 Chemistry Chapter 10 - Summary of the Chapter

Haloalkanes and Haloarenes are formed by the replacement of a hydrogen atom with a halogen atom in an aromatic compound or aliphatic hydrocarbons. Haloalkanes are formed by substitution of H atom in aliphatic hydrocarbon and the same replacement in aromatic compounds gives rise to Haloarenes. The chemical is used as a solvent in the case of non-polar compounds. Its application can be seen in refrigerants, pharmaceuticals, etc. 

Important Questions for Class 12 Chemistry Chapter 10 - Study With Vedantu 

The solutions to important questions from Haloalkanes and Haloarenes along with NCERT solutions , MCQs, assignments, and worksheets on Vedantu will help students to learn the concepts comprehensively. Also, it will help them in making notes for the competitive examinations. How else Vedantu will guide you?

In addition to the solutions, Vedantu also provides some of the best subject experts who can guide students to learn the subject in a more simplified and conceptual manner. 

Along with that, we provide free access to all our study materials.  

The study materials have been set in a manner to help students learn quickly and remember the concepts for longer. 

So, go through the set of important questions of Haloalkanes and Haloarenes of Class 12 and prepare well for your exams.

Well, this content is specially targeted to the CBSE Class 12 Chemistry students who are willing to know and study the important questions and answers from the chapter - Haloalkanes and Haloarenes. 

We have provided the pdf of the important questions from this chapter in order to have a convenient study. Additionally, we also have provided some extra questions for practice and the students are suggested to refer to the added links at the end of this content. 

Important Related Links for CBSE Class 12 Chemistry 

Conclusion .

The compilation of Important Questions for CBSE Class 12 Chemistry Chapter 10 - Haloalkanes and Haloarenes serves as a vital tool for students' exam preparation. These questions are carefully curated to cover key concepts, reactions, and topics within the chapter, ensuring comprehensive revision. They provide students with a strategic approach to studying, helping them identify crucial areas to focus on and assess their knowledge. These questions also mirror the exam pattern and difficulty level, thus aiding in building confidence and readiness for the final assessment. Overall, these important questions are an indispensable resource for Class 12 Chemistry students, offering a structured pathway towards academic success in this challenging subject.

FAQs on CBSE Chemistry Chapter 10 Haloalkanes and Haloarenes Class 12 Important Questions

1. What are the important subtopics in Haloalkanes and Haloarenes of Class 12 Chemistry for NEET?

According to past observations, the question papers for NEET over the years consisted of 3% of the total question based on Haloalkanes and Haloarenes. So, it is necessary to prepare the important subtopics from the chapter. These include:

Preparation of haloalkanes

Stereochemistry

Physical and chemical properties of haloalkanes

Preparation of haloarenes

Nucleophilic substitution mechanism

Uses and environmental effects of some haloalkanes and haloarenes

Physical and chemical properties of haloarenes

2. What are Haloalkanes and Haloarenes according to Chapter 10 of Class 12 Chemistry?

As discussed in Chapter 10 of Class 12 Chemistry, Haloalkanes are hydrocarbons containing aliphatic alkane with one or more hydrogen atoms replaced by halogens while Haloarenes are hydrocarbons containing aromatic alkane with halogens replacing one or more hydrogen atoms. Haloalkanes are aliphatic hydrocarbons, however, haloarenes are aromatic hydrocarbons. Students can find further explanations and differences between the two available on Vedantu’s e-platform.

3. Where can I find important questions for Chapter 10 Class 12 Chemistry?

Vedantu is an e-platform providing the students with one of the best study materials meant to help them enhance their understanding, practice in a more efficient way, revise easily, and make their preparation stronger for their Class 12 exams. You can access such study material like Important Questions for CBSE Class 12 Chemistry Chapter 10  available on Vedantu’s online website as well as the mobile app.

4. What are the rules of nomenclature of Haloalkanes?

The following are the rules of nomenclature of Haloalkanes 

Look for the longest carbon chain.

Put the longest carbon chain in numerical order in a way that the carbon atoms to which the halogens are attached get the lowest numbers. 

In case there is more than one halogen atom attached to the same carbon atom, the numeral is repeated many times. But if there are different types of halogens attached, name them alphabetically.

Denote the position of the halogen atom by writing the name and position of the halogen right before the name of the parent hydrocarbon.

5. Do I need to practice all questions covered in Important Questions for Class 12 Chemistry Chapter 10?

Questions that have been covered by Vedantu in the Important Questions for CBSE Class 12 Chemistry Chapter 10 - Haloalkanes and Haloarenes all need to be practised well before your Class 12 Chemistry Exam. This is necessary because it cannot be predicted which questions are asked in the exam and ignoring any questions during your preparation may lead to a loss of marks. Hence, students must ensure that they thoroughly prepare all the important questions provided.

CBSE Class 12 Chemsitry Important Questions

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Unit 7: Haloalkanes & haloarenes

About this unit.

Let's explore the beauty of haloalkanes & haloarenes. We will also dive into the concepts of nucleophilicity, SN1, SN2, E1, E2, reaction mechanisms, chirality, enantionomers and many more!

Free radical reaction

• Free radical reactions (Opens a modal)

Nucleophilicity & basicity

• Nucleophilicity (nucleophile strength) (Opens a modal)
• Nucleophilicity vs. basicity (Opens a modal)

Sn2/Sn2/E1/E2

• Identifying nucleophilic and electrophilic centers (Opens a modal)
• SN1 mechanism: Kinetics and substrates (Opens a modal)
• Sn1 mechanism: stereochemistry (Opens a modal)
• Carbocation stability: Recap (Opens a modal)
• Effect of substrate on the rate of an SN1 reaction- Part 1 (Opens a modal)
• Effect of substrate on the rate of SN1 reaction-Part 2 (Opens a modal)
• Identifying the major product- Carbocation rearrangement (Opens a modal)
• Factors affecting SN1 reaction: leaving group and solvent effects. (Opens a modal)
• Sn2 mechanism: kinetics and substrate (Opens a modal)
• Factors affecting SN2 reactions: substrate effect. (Opens a modal)
• Factors affecting SN2 reactions: leaving group- Part 1 (Opens a modal)
• Factors affecting SN2 reactions: leaving group-Part 2 (Opens a modal)
• Factors affecting SN2 reactions: Leaving group- Part 3 (Opens a modal)
• Factors affecting SN2 reactions: Strength of a nucleophile (Opens a modal)
• Sn1 vs Sn2: Summary (Opens a modal)
• Comparing E2, E1, Sn2, Sn1 reactions (Opens a modal)
• E2 E1 Sn2 Sn1 reactions example 2 (Opens a modal)
• E2 E1 Sn2 Sn1 reactions example 3 (Opens a modal)
• Effect of substrate on SN1 reactions (BASIC) 4 questions Practice
• Effect of substrate on SN1 reactions (INTERMEDIATE) 4 questions Practice
• Effect of substrate of SN2 reactions 4 questions Practice
• Factors affecting leaving group ability - I 4 questions Practice
• Effect of leaving group on SN2 reactions - II 4 questions Practice
• Strength of a nucleophile in SN2 reactions 4 questions Practice
• Practice questions on SN2 reactions 4 questions Practice
• Kinetics of SN1 vs SN2 reactions 4 questions Practice
• Introduction to chirality (Opens a modal)
• Chiral examples 1 (Opens a modal)
• Chiral examples 2 (Opens a modal)
• Chiral vs achiral (Opens a modal)
• Stereoisomers, enantiomers, and chirality centers (Opens a modal)
• Identifying chirality centers (Opens a modal)

Enantiomers

• Drawing enantiomers (Opens a modal)
• Cahn-Ingold-Prelog system for naming enantiomers (Opens a modal)
• R,S system (Opens a modal)
• R,S (Cahn-Ingold-Prelog) naming system example 2 (Opens a modal)
• R,S system practice (Opens a modal)
• More R,S practice (Opens a modal)
• Fischer projection introduction (Opens a modal)
• Fischer projection practice (Opens a modal)
• Optical activity (Opens a modal)
• Optical activity calculations (Opens a modal)

Stereoisomeric relationships

• Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds (Opens a modal)
• Enantiomers and diastereomers (Opens a modal)
• Meso compounds (Opens a modal)

Case Based Questions Test: Haloalkanes & Haloarenes - 1 - NEET MCQ

10 questions mcq test - case based questions test: haloalkanes & haloarenes - 1, read the passage given below and answer the following questions: nucleophilic substitution reactions are of two types; substitution nucleophilic bimolecular (s n 2) and substitution nucleophilic unimolecular (s n 1) depending on molecules taking part in determining the rate of reaction. reactivity of alkyl halide towards s n 1 and s n 2 reactions depends on various factors such as steric hindrance, stability of intermediate or transition state and polarity of solvent. s n 2 reaction mechanism is favoured mostly by primary alkyl halide or transition state and polarity of solvent, s n 2 reaction mechanism is favoured mostly by primary alkyl halide then secondary and then tertiary. this order is reversed in case of s n 1 reactions. q. isopropyl chloride undergoes hydrolysis by.

S N 1 mechanism

S N 2 mechanism

S N 1 and S N 2 mechanism

neither S N 1 nor S N 2 mechanism  

Isopropyl chloride undergoes hydrolysis via both SN-1 and SN-2 mechanisms. Both of them yield the same result/ products. CH 3 - CH(CI) - CH 3 + HO - → CH 3 + CH(OH)  CH 3 + CI - Hence the correct option is C. 

Read the passage given below and answer the following questions: Nucleophilic substitution reactions are of two types; substitution nucleophilic bimolecular (S N 2) and substitution nucleophilic unimolecular (S N 1) depending on molecules taking part in determining the rate of reaction. Reactivity of alkyl halide towards S N 1 and S N 2 reactions depends on various factors such as steric hindrance, stability of intermediate or transition state and polarity of solvent. S N 2 reaction mechanism is favoured mostly by primary alkyl halide or transition state and polarity of solvent, S N 2 reaction mechanism is favoured mostly by primary alkyl halide then secondary and then tertiary. This order is reversed in case of S N 1 reactions. Q. Tertiary alkyl halides are practically inert to substitution by S N 2 mechanism because of  

insolubility

instability

inductive effect

steric hindrance

The repulsion between the alkyl or any other groups present on an carbon atom, if the distance between the two is less than vander waals radius, then it is said to be the steric hindrance. That is the reason why tertiary alkyl halides are practically inert to substitution by S N 2 mechanism as there is steric hindrance. 

Read the passage given below and answer the following questions: Nucleophilic substitution reactions are of two types; substitution nucleophilic bimolecular (S N 2) and substitution nucleophilic unimolecular (S N 1) depending on molecules taking part in determining the rate of reaction. Reactivity of alkyl halide towards S N 1 and S N 2 reactions depends on various factors such as steric hindrance, stability of intermediate or transition state and polarity of solvent. S N 2 reaction mechanism is favoured mostly by primary alkyl halide or transition state and polarity of solvent, S N 2 reaction mechanism is favoured mostly by primary alkyl halide then secondary and then tertiary. This order is reversed in case of S N 1 reactions. Q. Which of the following is most reactive towards nucleophilic substitution reaction?

ClCH 2 - CH = CH 2

CH 3 CH=CHCl

Order of reactivity of different halo compounds towards nucleophilic substitution reaction are: Allyl chloride > Vinyl chloride > Chlorobenzene

Read the passage given below and answer the following questions: Nucleophilic substitution reactions are of two types; substitution nucleophilic bimolecular (S N 2) and substitution nucleophilic unimolecular (S N 1) depending on molecules taking part in determining the rate of reaction. Reactivity of alkyl halide towards S N 1 and S N 2 reactions depends on various factors such as steric hindrance, stability of intermediate or transition state and polarity of solvent. S N 2 reaction mechanism is favoured mostly by primary alkyl halide or transition state and polarity of solvent, S N 2 reaction mechanism is favoured mostly by primary alkyl halide then secondary and then tertiary. This order is reversed in case of S N 1 reactions.

Q. The most reactive nucleophile among the following is  

(CH 3 ) 2 CHO-

(CH 3 ) 3 CO-

Smaller size the of the nucleophile (i.e., CH 3 O - more reactive it is.

Read the passage given below and answer the following questions: Nucleophilic substitution reactions are of two types; substitution nucleophilic bimolecular (S N 2) and substitution nucleophilic unimolecular (S N 1) depending on molecules taking part in determining the rate of reaction. Reactivity of alkyl halide towards S N 1 and S N 2 reactions depends on various factors such as steric hindrance, stability of intermediate or transition state and polarity of solvent. S N 2 reaction mechanism is favoured mostly by primary alkyl halide or transition state and polarity of solvent, S N 2 reaction mechanism is favoured mostly by primary alkyl halide then secondary and then tertiary. This order is reversed in case of S N 1 reactions. Q. Which of the following is the correct order of decreasing S N 2 reactivity?

RCH 2 X > R 2 CHX > R 3 CX

R 3 CX > R 2 CHX >RCH 2 X

R 2 CHX >R 3 CX > RCH 2 X

RCH 2 X >R 3 CX >R 2 CHX

Larger the number of alkyl groups at alpha-carbon atom, more is the steric hindrance and hence lesser the reactivity towards S N 2 mechanism.

Read the  passage given below and answer the following questions: Nucleophilic substitution reaction of haloalkane can be conducted according to both S N 1 and S N 2 mechanisms. However, which mechanism it is based on is related to such factors as the structure of haloalkane, and properties of leaving group, nucleophilic reagent and solvent. Influences of halogen: No matter which mechanism the nucleophilic substitution reaction is based on, the leaving group always leave the central carbon atom with electron pair. This is just the opposite of the situation that nucleophilic reagent attacks the central carbon atom with electron pair. Therefore, the weaker the alkalinity of leaving group is , the more stable the anion formed is and it will be more easier for the leaving group to leave the central carbon atom; that is to say, the reactant is more easier to be substituted. The alkalinity order of halogen ion is I − < Br − < Cl − < F − and the order of their leaving tendency should be I − > Br − > Cl − > F − . Therefore, in four halides with the same alkyl and different halogens, the order of substitution reaction rate is RI > RBr > RCl > RF. In addition, if the leaving group is very easy to leave, many carbocation intermediates are generated in the reaction and the reaction is based on S N 1 mechanism. If the leaving group is not easy to leave, the reaction is based on S N 2 mechanism. Influences of solvent polarity: In S N 1 reaction, the polarity of the system increases from the reactant to the transition state, because polar solvent has a greater stabilizing effect on the transition state than the reactant, thereby reduce activation energy and accelerate the reaction. In S N 2 reaction, the polarity of the system generally does not change from the reactant to the transition state and only charge dispersion occurs. At this time, polar solvent has a great stabilizing effect on Nu than the transition state, thereby increasing activation energy and slow down the reaction rate. For example, the decomposition rate (S N 1) of tertiary chlorobutane in 25° water (dielectric constant 79) is 300000 times faster than in ethanol (dielectric constant 24). The reaction rate (S N 2) of 2-bromopropane and NaOH in ethanol containing 40% water is twice slower than in absolute ethanol. In a word, the level of solvent polarity has influence on both S N 1 and S N 2 reactions, but with different results. Generally speaking, weak polar solvent is favorable for S N 2 reaction, while strong polar solvent is favorable for S N 1 reaction, because only under the action of polar solvent can halogenated hydrocarbon dissociate into carbocation and halogen ion and solvents with a strong polarity is favorable for solvation of carbocation, increasing its stability. Generally speaking, the substitution reaction of tertiary haloalkane is based on S N 1 mechanism in solvents with a strong polarity (for example, ethanol containing water). Q. Nucleophilic substitution will be fastest in case of:

1-Chloro-2,2-dimethyl propane

1-Iodo-2,2-dimethyl propane

1-Bromo-2,2-dimethyl propane

1-Fluoro-2,2-dimethyl propane

Read the  passage given below and answer the following questions: Nucleophilic substitution reaction of haloalkane can be conducted according to both S N 1 and S N 2 mechanisms. However, which mechanism it is based on is related to such factors as the structure of haloalkane, and properties of leaving group, nucleophilic reagent and solvent. Influences of halogen: No matter which mechanism the nucleophilic substitution reaction is based on, the leaving group always leave the central carbon atom with electron pair. This is just the opposite of the situation that nucleophilic reagent attacks the central carbon atom with electron pair. Therefore, the weaker the alkalinity of leaving group is , the more stable the anion formed is and it will be more easier for the leaving group to leave the central carbon atom; that is to say, the reactant is more easier to be substituted. The alkalinity order of halogen ion is I − < Br − < Cl − < F − and the order of their leaving tendency should be I − > Br − > Cl − > F − . Therefore, in four halides with the same alkyl and different halogens, the order of substitution reaction rate is RI > RBr > RCl > RF. In addition, if the leaving group is very easy to leave, many carbocation intermediates are generated in the reaction and the reaction is based on S N 1 mechanism. If the leaving group is not easy to leave, the reaction is based on S N 2 mechanism. Influences of solvent polarity: In S N 1 reaction, the polarity of the system increases from the reactant to the transition state, because polar solvent has a greater stabilizing effect on the transition state than the reactant, thereby reduce activation energy and accelerate the reaction. In S N 2 reaction, the polarity of the system generally does not change from the reactant to the transition state and only charge dispersion occurs. At this time, polar solvent has a great stabilizing effect on Nu than the transition state, thereby increasing activation energy and slow down the reaction rate. For example, the decomposition rate (S N 1) of tertiary chlorobutane in 25° water (dielectric constant 79) is 300000 times faster than in ethanol (dielectric constant 24). The reaction rate (S N 2) of 2-bromopropane and NaOH in ethanol containing 40% water is twice slower than in absolute ethanol. In a word, the level of solvent polarity has influence on both S N 1 and S N 2 reactions, but with different results. Generally speaking, weak polar solvent is favorable for S N 2 reaction, while strong polar solvent is favorable for S N 1 reaction, because only under the action of polar solvent can halogenated hydrocarbon dissociate into carbocation and halogen ion and solvents with a strong polarity is favorable for solvation of carbocation, increasing its stability. Generally speaking, the substitution reaction of tertiary haloalkane is based on S N 1 mechanism in solvents with a strong polarity (for example, ethanol containing water). Q. Polar solvents make the reaction faster as they:

destabilize transition state and decrease the activation energy

destabilize transition state and increase the activation energy

stabilize transition state and increase the activation energy

stabilize transition state and decrease the activation energy

Polar protic solvents have large dipole moments. They lower the energy of both the transition state and the starting material. But they stabilize the transition state more because it is more polar. This lowers the activation energy, so the reaction goes faster.

Read the  passage given below and answer the following questions: Nucleophilic substitution reaction of haloalkane can be conducted according to both S N 1 and S N 2 mechanisms. However, which mechanism it is based on is related to such factors as the structure of haloalkane, and properties of leaving group, nucleophilic reagent and solvent. Influences of halogen: No matter which mechanism the nucleophilic substitution reaction is based on, the leaving group always leave the central carbon atom with electron pair. This is just the opposite of the situation that nucleophilic reagent attacks the central carbon atom with electron pair. Therefore, the weaker the alkalinity of leaving group is , the more stable the anion formed is and it will be more easier for the leaving group to leave the central carbon atom; that is to say, the reactant is more easier to be substituted. The alkalinity order of halogen ion is I − < Br − < Cl − < F − and the order of their leaving tendency should be I − > Br − > Cl − > F − . Therefore, in four halides with the same alkyl and different halogens, the order of substitution reaction rate is RI > RBr > RCl > RF. In addition, if the leaving group is very easy to leave, many carbocation intermediates are generated in the reaction and the reaction is based on S N 1 mechanism. If the leaving group is not easy to leave, the reaction is based on S N 2 mechanism. Influences of solvent polarity: In S N 1 reaction, the polarity of the system increases from the reactant to the transition state, because polar solvent has a greater stabilizing effect on the transition state than the reactant, thereby reduce activation energy and accelerate the reaction. In S N 2 reaction, the polarity of the system generally does not change from the reactant to the transition state and only charge dispersion occurs. At this time, polar solvent has a great stabilizing effect on Nu than the transition state, thereby increasing activation energy and slow down the reaction rate. For example, the decomposition rate (S N 1) of tertiary chlorobutane in 25° water (dielectric constant 79) is 300000 times faster than in ethanol (dielectric constant 24). The reaction rate (S N 2) of 2-bromopropane and NaOH in ethanol containing 40% water is twice slower than in absolute ethanol. In a word, the level of solvent polarity has influence on both S N 1 and S N 2 reactions, but with different results. Generally speaking, weak polar solvent is favorable for S N 2 reaction, while strong polar solvent is favorable for S N 1 reaction, because only under the action of polar solvent can halogenated hydrocarbon dissociate into carbocation and halogen ion and solvents with a strong polarity is favorable for solvation of carbocation, increasing its stability. Generally speaking, the substitution reaction of tertiary haloalkane is based on S N 1 mechanism in solvents with a strong polarity (for example, ethanol containing water). Q. S N 1 mechanism is favoured in which of the following solvents:

carbon tetrachloride

acetic acid

carbon disulphide

Read the  passage given below and answer the following questions: Nucleophilic substitution reaction of haloalkane can be conducted according to both S N 1 and S N 2 mechanisms. However, which mechanism it is based on is related to such factors as the structure of haloalkane, and properties of leaving group, nucleophilic reagent and solvent. Influences of halogen: No matter which mechanism the nucleophilic substitution reaction is based on, the leaving group always leave the central carbon atom with electron pair. This is just the opposite of the situation that nucleophilic reagent attacks the central carbon atom with electron pair. Therefore, the weaker the alkalinity of leaving group is , the more stable the anion formed is and it will be more easier for the leaving group to leave the central carbon atom; that is to say, the reactant is more easier to be substituted. The alkalinity order of halogen ion is I − < Br − < Cl − < F − and the order of their leaving tendency should be I − > Br − > Cl − > F − . Therefore, in four halides with the same alkyl and different halogens, the order of substitution reaction rate is RI > RBr > RCl > RF. In addition, if the leaving group is very easy to leave, many carbocation intermediates are generated in the reaction and the reaction is based on S N 1 mechanism. If the leaving group is not easy to leave, the reaction is based on S N 2 mechanism. Influences of solvent polarity: In S N 1 reaction, the polarity of the system increases from the reactant to the transition state, because polar solvent has a greater stabilizing effect on the transition state than the reactant, thereby reduce activation energy and accelerate the reaction. In S N 2 reaction, the polarity of the system generally does not change from the reactant to the transition state and only charge dispersion occurs. At this time, polar solvent has a great stabilizing effect on Nu than the transition state, thereby increasing activation energy and slow down the reaction rate. For example, the decomposition rate (S N 1) of tertiary chlorobutane in 25° water (dielectric constant 79) is 300000 times faster than in ethanol (dielectric constant 24). The reaction rate (S N 2) of 2-bromopropane and NaOH in ethanol containing 40% water is twice slower than in absolute ethanol. In a word, the level of solvent polarity has influence on both S N 1 and S N 2 reactions, but with different results. Generally speaking, weak polar solvent is favorable for S N 2 reaction, while strong polar solvent is favorable for S N 1 reaction, because only under the action of polar solvent can halogenated hydrocarbon dissociate into carbocation and halogen ion and solvents with a strong polarity is favorable for solvation of carbocation, increasing its stability. Generally speaking, the substitution reaction of tertiary haloalkane is based on S N 1 mechanism in solvents with a strong polarity (for example, ethanol containing water). Q. SN1 reaction will be fastest in which of the following solvents?

Acetone (dielectric constant 21)

Ethanol (dielectric constant 24)

Methanol (dielectric constant 32)

Chloroform (dielectric constant 5)

Read the  passage given below and answer the following questions: Nucleophilic substitution reaction of haloalkane can be conducted according to both S N 1 and S N 2 mechanisms. However, which mechanism it is based on is related to such factors as the structure of haloalkane, and properties of leaving group, nucleophilic reagent and solvent. Influences of halogen: No matter which mechanism the nucleophilic substitution reaction is based on, the leaving group always leave the central carbon atom with electron pair. This is just the opposite of the situation that nucleophilic reagent attacks the central carbon atom with electron pair. Therefore, the weaker the alkalinity of leaving group is , the more stable the anion formed is and it will be more easier for the leaving group to leave the central carbon atom; that is to say, the reactant is more easier to be substituted. The alkalinity order of halogen ion is I − < Br − < Cl − < F − and the order of their leaving tendency should be I − > Br − > Cl − > F − . Therefore, in four halides with the same alkyl and different halogens, the order of substitution reaction rate is RI > RBr > RCl > RF. In addition, if the leaving group is very easy to leave, many carbocation intermediates are generated in the reaction and the reaction is based on S N 1 mechanism. If the leaving group is not easy to leave, the reaction is based on S N 2 mechanism. Influences of solvent polarity: In S N 1 reaction, the polarity of the system increases from the reactant to the transition state, because polar solvent has a greater stabilizing effect on the transition state than the reactant, thereby reduce activation energy and accelerate the reaction. In S N 2 reaction, the polarity of the system generally does not change from the reactant to the transition state and only charge dispersion occurs. At this time, polar solvent has a great stabilizing effect on Nu than the transition state, thereby increasing activation energy and slow down the reaction rate. For example, the decomposition rate (S N 1) of tertiary chlorobutane in 25° water (dielectric constant 79) is 300000 times faster than in ethanol (dielectric constant 24). The reaction rate (S N 2) of 2-bromopropane and NaOH in ethanol containing 40% water is twice slower than in absolute ethanol. In a word, the level of solvent polarity has influence on both S N 1 and S N 2 reactions, but with different results. Generally speaking, weak polar solvent is favorable for S N 2 reaction, while strong polar solvent is favorable for S N 1 reaction, because only under the action of polar solvent can halogenated hydrocarbon dissociate into carbocation and halogen ion and solvents with a strong polarity is favorable for solvation of carbocation, increasing its stability. Generally speaking, the substitution reaction of tertiary haloalkane is based on S N 1 mechanism in solvents with a strong polarity (for example, ethanol containing water). Q. S N 1 reaction will be fastest in case of:

1-Chloro-2-methyl propane

1-Iodo-2-methyl propane

1-Chlorobutane

1-Iodobutane

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12th Chemistry Haloalkanes and Haloarenes Chapter Case Study Question with Answers CBSE

By QB365 on 13 Jan, 2024

Haloalkanes and Haloarenes

Haloalkanes and Haloarenes Chapter Case Study Question with Answers

12th Chemistry CBSE case study questions for Haloalkanes and Haloarenes - 2024

CBSE 12th Chemistry Haloalkanes and Haloarenes Case Study Question & Answers 

CBSE 12th Chemistry Haloalkanes and Haloarenes Chapter Case Study Question with Answers

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Haloalkanes and Haloarenes

What are Haloalkanes and Haloarenes?

Haloalkanes and haloarenes are the hydrocarbons in which one or more hydrogen atoms have been replaced with halogen atoms. The primary difference between haloalkanes and haloarenes is that haloalkanes are derived from open-chain hydrocarbons (alkanes) whereas haloarenes are derived from aromatic hydrocarbons .

Table of Content

• Overview of Haloalkanes and Haloarenes
• Related Topic of Haloalkanes and Haloarenes

Classification of Haloalkanes and Haloarenes

Uses of haloalkanes and haloarenes, frequently asked questions-faqs, overview of haloalkanes and haloarenes.

Haloalkanes are commonly referred to as alkyl halides whereas haloarenes are commonly referred to as aryl halides. These compounds can contain multiple halogen atoms, as illustrated below.

Generally, In haloalkanes, the halogen atoms are attached to sp 3 hybridized carbon atoms whereas in haloarenes, the point of attachment is sp2 hybridized carbon atom. The difference in the hybridization state of the carbon atom in C-X bond is responsible for the different characteristics which the two families have.  The presence of halogens makes haloalkanes and haloarenes more chemically reactive than the parent alkanes and aromatic compounds. These compounds have many medicinal uses as well.

Related Topics on Haloalkanes and Haloarenes

• Nomenclature Of Haloalkanes and Haloarenes
• Nature of C-X bond
• Haloarenes: Nature of C-X bond
• Physical and Chemical properties of Haloalkanes
• S N 1 and S N 2 Substitution Reactions of Haloalkanes
• Optical Rotation
• Polyhalogen Compounds

Alkyl halides and aryl halides can be classified based on the following parameters:

• Number of halogen atoms in the molecule
• sp 3 hybridized carbon-halogen bond.
• sp 2 hybridized carbon-halogen bond.

The classification of haloalkanes and haloarenes is described in the tabular column provided below.

Thus, the different ways in which haloalkanes and haloarenes are classified are discussed with the help of a tabular column. Click here to learn about the nomenclature of haloalkanes as per IUPAC guidelines.

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Some important applications of these compounds are listed below.

• These organic compounds can dissolve non-polar compounds and are therefore used as solvents.
• Many derivatives of alkyl and aryl halides are used in medicine. One such example is the compound chloramphenicol, which is used to treat cases of typhoid.
• Another example is chloroquine, which is very useful in the treatment of malaria.
• Dichlorodiphenyltrichloroethane (commonly referred to as DDT) is used as an insecticide.

Some haloalkanes and haloarenes have adverse effects on the environment and are labelled as pollutants. One such example is chlorofluorocarbons (or CFCs), which lead to the depletion of the ozone layer which protects the Earth from the harmful radiation coming from the sun. To learn more about these compounds, register with BYJU’S and download the mobile application on your smartphone.

1. What is the difference between haloalkanes and Haloarenes?

When hydrogen atoms in aliphatic hydrocarbons(alkanes) are replaced by halogen atoms ,the compounds formed are known as haloalkanes. Similarly, when hydrogen atoms attached to benzene rings are replaced by halogen atoms the compounds that are formed are known as haloarenes.

2. What are haloalkanes reactions?

Haloalkanes react with magnesium metal in the presence of completely anhydrous ether to form organomagnesium halide which is popularly known as Grignard reagents. Haloalkane undergoes nucleophilic substitution reaction by the presence of aqueous alkali solution or moist silver oxide solution form alcohols.

3. Which is the example of Haloarenes?

When hydrogen atoms attached to benzene rings are replaced by halogen atoms the compounds that are formed are known as haloarenes. Examples of haloarenes are chlorobenzene, bromobenzene, iodobenzene, 2-Chlorotoluene etc.

4. What are the uses of Haloarenes?

Haloarenes used to prepare several compounds like DDT, Picric acid, Phenol etc. DDT was used as insecticide to destroy anopheles mosquitoes which spread malaria but due to its toxic nature it was banned in the year 1973. Picric acid is used in the production of explosives, matches, electric batteries, manufacturing colored glass and synthesis of dyes. Phenol is used in the manufacture of nylon and other synthetic fibers.

5. What is Mono Haloalkanes?

The haloalkane which contains only one halogen atom is known as mono haloalkane

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