CBSE Class 12 Chemistry Chapter 10: Haloalkanes and Haloarenes NCERT Solutions

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This section provides comprehensive NCERT Solutions for Chapter 10 of CBSE Class 12 Chemistry, focusing on Haloalkanes and Haloarenes. It covers the structures of various organic compounds, including substituted alkanes, cycloalkanes, and aromatic compounds, as well as understanding the reactivity and limitations of reagents in specific reactions. The solutions explain why certain acids are not suitable for particular reactions, detailing the oxidizing properties that interfere with the desired product formation. Additionally, it explores the different types of dihalogen derivatives of propane, illustrating their structural variations. These solutions are designed to help students grasp the fundamental concepts of alkyl and aryl halides, their nomenclature, and their chemical behavior, aiding in effective exam preparation and revision.

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BoardCBSE
ClassClass 12
SubjectChemistry
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 19

Chapter summary

Chapter 10 of the NCERT Class 12 Chemistry syllabus deals with Haloalkanes and Haloarenes. This solution set provides answers to in-text questions, focusing on drawing the structures of complex halogenated organic compounds and understanding reaction mechanisms. It clarifies the role of reagents, explaining why sulfuric acid is unsuitable for reactions with KI and suggesting alternatives. The chapter also delves into isomerism by asking for different dihalogen derivatives of propane. These solutions offer step-by-step guidance to reinforce learning.

Learning outcomes

  • Understand the nomenclature and structure of haloalkanes and haloarenes.
  • Draw the structures of complex halogenated organic compounds.
  • Explain the role and limitations of reagents in organic reactions.
  • Identify and draw different dihalogen derivatives of propane.
  • Analyze the chemical properties and reactivity of alkyl and aryl halides.

Topics covered

Paper topics

  • Nomenclature of Haloalkanes
  • Nomenclature of Haloarenes
  • Structure of Haloalkanes
  • Structure of Haloarenes
  • Reactions of Alcohols with KI
  • Role of Sulfuric Acid in Reactions
  • Role of Phosphoric Acid in Reactions
  • Dihalogen Derivatives of Propane
  • Structural Isomerism in Dihalogenated Alkanes
  • Tert-butyl group
  • sec-butyl group
  • Cyclohexane derivatives

Important topics

  • Drawing structures of complex haloalkanes and haloarenes
  • Understanding reagent limitations (e.g., H2SO4 vs. H3PO4 with KI)
  • Identifying and drawing isomers of dihalogenated alkanes
  • Nomenclature of substituted haloalkanes and haloarenes

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Questions and Solutions

Question 10.1

Write structures of the following compounds:
  1. 2-Chloro-3-methylpentane
  2. 1-Chloro-4-ethylcyclohexane
  3. 4-tert. Butyl-3-iodoheptane
  4. 1,4-Dibromobut-2-ene
  5. 1-Bromo-4-sec. butyl-2-methylbenzene
Solution:

The structures of the given compounds are as follows:

  1. 2-Chloro-3-methylpentane: This is a five-carbon chain (pentane). A chlorine atom is attached to the second carbon, and a methyl group is attached to the third carbon. The structure is:

    Structure of 2-Chloro-3-methylpentane

  2. 1-Chloro-4-ethylcyclohexane: This is a cyclohexane ring. A chlorine atom is attached to the first carbon, and an ethyl group (C2H5) is attached to the fourth carbon of the ring. The structure is:

    Structure of 1-Chloro-4-ethylcyclohexane

  3. 4-tert. Butyl-3-iodoheptane: This is a seven-carbon chain (heptane). An iodine atom is attached to the third carbon, and a tert-butyl group is attached to the fourth carbon. A tert-butyl group has a central carbon atom bonded to three methyl groups. The structure is:

    Structure of 4-tert. Butyl-3-iodoheptane

  4. 1,4-Dibromobut-2-ene: This is a four-carbon chain (but) with a double bond between the second and third carbons (-2-ene). Bromine atoms are attached to the first and fourth carbons. The structure is:

    Structure of 1,4-Dibromobut-2-ene

  5. 1-Bromo-4-sec. butyl-2-methylbenzene: This is a benzene ring. A bromine atom is at position 1, a methyl group at position 2, and a sec-butyl group at position 4. A sec-butyl group is a four-carbon chain attached to the ring via its second carbon. The structure is:

    Structure of 1-Bromo-4-sec. butyl-2-methylbenzene

Question 10.2

Why is sulphuric acid not used during the reaction of alcohols with KI?
Solution:

Sulfuric acid (H2SO4) is not used during the reaction of alcohols with potassium iodide (KI) because concentrated sulfuric acid is a strong oxidizing agent. When sulfuric acid reacts with KI, it produces hydroiodic acid (HI) as an intermediate:

2KI + H_2SO_4 \longrightarrow 2KHSO_4 + 2HI

However, the concentrated sulfuric acid then oxidizes this HI to iodine (I2):

2HI + H_2SO_4 \longrightarrow I_2 + SO_2 + 2H_2O

This oxidation consumes the HI, which is necessary for the reaction with the alcohol to form the alkyl iodide. Therefore, the desired reaction between the alcohol and HI to produce alkyl iodide cannot proceed effectively. To overcome this, a non-oxidizing acid like phosphoric acid (H3PO4) is used instead, which facilitates the formation of HI without oxidizing it.

Question 10.3

Write structures of different dihalogen derivatives of propane.
Solution:

Propane is a three-carbon alkane (CH3-CH2-CH3). Dihalogen derivatives are formed when two hydrogen atoms are replaced by halogen atoms (let's use bromine, Br, for illustration). There are four possible structural isomers for dihalogen derivatives of propane, depending on the positions of the two halogen atoms:

  1. 1,1-Dihalopropane: Both halogen atoms are attached to the first carbon atom.

    Structure of 1,1-Dibromopropane

  2. 1,2-Dihalopropane: One halogen atom is attached to the first carbon, and the other is attached to the second carbon.

    Structure of 1,2-Dibromopropane

  3. 1,3-Dihalopropane: One halogen atom is attached to the first carbon, and the other is attached to the third carbon.

    Structure of 1,3-Dibromopropane

  4. 2,2-Dihalopropane: Both halogen atoms are attached to the second carbon atom.

    Structure of 2,2-Dibromopropane

These represent all the unique ways to attach two halogen atoms to a propane molecule.

Common mistakes

  • Incorrectly drawing the structures of complex substituted haloalkanes and haloarenes.
  • Misunderstanding the oxidizing properties of reagents like sulfuric acid and their impact on reaction outcomes.
  • Failing to identify all possible structural isomers for dihalogenated alkanes.
  • Errors in IUPAC nomenclature leading to incorrect structural representations.

Revision tips

  • Practice drawing structures for all given compounds to solidify understanding of IUPAC naming and substituent placement.
  • Focus on the 'why' behind reagent choices, especially understanding the oxidizing nature of H2SO4.
  • Draw out all possible dihalogen derivatives of propane systematically to avoid missing any isomers.
  • Review the functional groups and their positions carefully when interpreting chemical names.

Practice MCQs

Q1. Which of the following is a correct structure for 2-Chloro-3-methylpentane?

Q2. Why is concentrated sulfuric acid not used when reacting alcohols with potassium iodide (KI)?

Q3. How many different dihalogen derivatives of propane exist?

Q4. What is the role of H3PO4 in the reaction of alcohols with KI?

Q5. The structure 1,4-Dibromobut-2-ene contains:

Frequently asked questions

What are haloalkanes and haloarenes?

Haloalkanes are organic compounds where one or more hydrogen atoms in an alkane are replaced by halogen atoms (F, Cl, Br, I). Haloarenes are compounds where a halogen atom is directly attached to an aromatic ring.

Why is sulfuric acid avoided when reacting alcohols with KI?

Concentrated sulfuric acid is an oxidizing agent that oxidizes the HI formed during the reaction to I2, thus preventing the formation of the desired alkyl iodide. A non-oxidizing acid like H3PO4 is preferred.

How can I draw the structure of 4-tert. Butyl-3-iodoheptane?

First, draw a seven-carbon chain (heptane). Number it from one end. Place an iodine atom at the 3rd carbon. At the 4th carbon, attach a tert-butyl group, which is a central carbon bonded to three methyl groups.

What are the different dihalogen derivatives of propane?

The different dihalogen derivatives of propane include 1,1-dibromopropane, 1,2-dibromopropane, 1,3-dibromopropane, and 2,2-dibromopropane, differing in the positions of the two bromine atoms on the propane chain.

How do these NCERT solutions help in exam preparation?

These solutions provide clear, step-by-step explanations for complex questions, helping students understand the underlying concepts, practice drawing structures, and learn about reaction mechanisms and reagent choices, which are crucial for exams.

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