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

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This comprehensive set of NCERT Solutions for CBSE Class 12 Chemistry, Chapter 10, focuses on Haloalkanes and Haloarenes. It provides detailed explanations and step-by-step solutions for the exercises, covering the IUPAC nomenclature, classification of halides (alkyl, allyl, benzyl, vinyl, aryl), and their structural identification. The solutions guide students through naming complex organic structures and classifying them based on the carbon atom attached to the halogen. This resource is designed to help students build a strong foundation in organic chemistry, understand the fundamental concepts of halogenated hydrocarbons, and prepare effectively for their board examinations by clarifying common doubts and reinforcing learning through practice.

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BoardCBSE
ClassClass 12
SubjectChemiry
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 10: Haloalkanes and Haloarenes - NCERT Exercises Solutions

Chapter summary

Chapter 10 of the NCERT Class 12 Chemistry textbook delves into Haloalkanes and Haloarenes. The NCERT Solutions for this chapter offer clear explanations for naming these compounds using the IUPAC system and classifying them based on their structure and the type of carbon atom bonded to the halogen. The exercises focus on identifying and naming various haloalkanes and haloarenes, reinforcing the understanding of structural features and nomenclature rules essential for organic chemistry.

Learning outcomes

  • Understand the IUPAC nomenclature for haloalkanes and haloarenes.
  • Classify haloalkanes and haloarenes as primary, secondary, or tertiary.
  • Identify and classify alkyl, allyl, benzyl, vinyl, and aryl halides.
  • Apply nomenclature rules to complex halogenated organic structures.
  • Differentiate between various types of halogenated hydrocarbons based on their structure.

Topics covered

Paper topics

  • IUPAC Nomenclature of Haloalkanes
  • IUPAC Nomenclature of Haloarenes
  • Classification of Halides
  • Alkyl Halides
  • Allyl Halides
  • Benzyl Halides
  • Vinyl Halides
  • Aryl Halides
  • Primary Halides
  • Secondary Halides
  • Tertiary Halides
  • Structure-based Classification

Important topics

  • IUPAC Nomenclature
  • Classification of Haloalkanes
  • Classification of Haloarenes
  • Identifying Primary, Secondary, and Tertiary Halides
  • Distinguishing Benzyl and Allyl Halides

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Questions and Solutions

Question 10.1:

Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
  1. (CH<sub>3</sub>)<sub>2</sub>CHCH(Cl)CH<sub>3</sub>
  2. CH<sub>3</sub>CH<sub>2</sub>CH(CH<sub>3</sub>)CH(C<sub>2</sub>H<sub>5</sub>)Cl
  3. CH<sub>3</sub>CH<sub>2</sub>C(CH<sub>3</sub>)<sub>2</sub>CH<sub>2</sub>I
  4. (CH<sub>3</sub>)<sub>3</sub>CCH<sub>2</sub>CH(Br)C<sub>6</sub>H<sub>5</sub>
  5. CH<sub>3</sub>CH(CH<sub>3</sub>)CH(Br)CH<sub>3</sub>
  6. CH<sub>3</sub>C(C<sub>2</sub>H<sub>5</sub>)<sub>2</sub>CH<sub>2</sub>Br
  7. CH<sub>3</sub>C(Cl)(C<sub>2</sub>H<sub>5</sub>)CH<sub>2</sub>CH<sub>3</sub>
  8. CH<sub>3</sub>CH=C(Cl)CH<sub>2</sub>CH(CH<sub>3</sub>)<sub>2</sub>
  9. CH<sub>3</sub>CH=CHC(Br)(CH<sub>3</sub>)<sub>2</sub>
  10. p-ClC<sub>6</sub>H<sub>4</sub>CH<sub>2</sub>CH(CH<sub>3</sub>)<sub>2</sub>
  11. m-ClCH<sub>2</sub>C<sub>6</sub>H<sub>4</sub>CH<sub>2</sub>C(CH<sub>3</sub>)<sub>3</sub>
  12. o-Br-C<sub>6</sub>H<sub>4</sub>CH(CH<sub>3</sub>)CH<sub>2</sub>CH<sub>3</sub>
Solution:

We need to determine the IUPAC name and classify each of the given halogenated organic compounds.

  1. Structure: (CH<sub>3</sub>)<sub>2</sub>CHCH(Cl)CH<sub>3</sub>

    The longest carbon chain has 4 carbons. The chlorine atom is on the second carbon, and a methyl group is on the third carbon. Numbering from the end closer to the chlorine gives the lowest locant for the substituent.

    IUPAC Name: 2-Chloro-3-methylbutane

    Classification: The chlorine atom is attached to a secondary carbon atom (a carbon bonded to two other carbon atoms). Therefore, it is a secondary alkyl halide.

  2. Structure: CH<sub>3</sub>CH<sub>2</sub>CH(CH<sub>3</sub>)CH(C<sub>2</sub>H<sub>5</sub>)Cl

    To find the longest carbon chain, we can trace it through the ethyl group. The longest chain has 6 carbons. The chlorine atom is on the 3rd carbon, and a methyl group is on the 4th carbon. Numbering from the end closer to the chlorine gives the lowest locant.

    IUPAC Name: 3-Chloro-4-methylhexane

    Classification: The chlorine atom is attached to a secondary carbon atom (a carbon bonded to two other carbon atoms). Therefore, it is a secondary alkyl halide.

  3. Structure: CH<sub>3</sub>CH<sub>2</sub>C(CH<sub>3</sub>)<sub>2</sub>CH<sub>2</sub>I

    The longest carbon chain has 4 carbons (butane). The iodine atom is attached to the first carbon. There are two methyl groups on the second carbon.

    IUPAC Name: 1-Iodo-2,2-dimethylbutane

    Classification: The iodine atom is attached to a primary carbon atom (a carbon bonded to only one other carbon atom). Therefore, it is a primary alkyl halide.

  4. Structure: (CH<sub>3</sub>)<sub>3</sub>CCH<sub>2</sub>CH(Br)C<sub>6</sub>H<sub>5</sub>

    The longest chain containing the functional group and the phenyl ring is considered. The bromine is attached to a carbon that is part of a 4-carbon chain (butane), and this chain is attached to a phenyl group. The bromine is on the 1st carbon of this chain, and the phenyl group is on the 4th carbon. However, the question implies the phenyl group is part of the main structure. Let's re-evaluate based on the common representation of benzyl halides.

    The bromine is attached to a carbon atom which is directly attached to the phenyl ring (C<sub>6</sub>H<sub>5</sub>). This carbon is also attached to two other carbon atoms (one from the tert-butyl group's CH<sub>2</sub> and one from the phenyl ring). The structure is better interpreted as a substituted benzyl system.

    Let's consider the chain attached to the phenyl group. The carbon attached to the phenyl group is CH(Br). This carbon is also attached to a CH<sub>2</sub> group, which is part of a tert-butyl group. The longest chain including the carbon attached to the phenyl group is 4 carbons long (if we consider the chain starting from the CH(Br) carbon). The bromine is on the first carbon of this chain. The phenyl group is attached to the first carbon. The tert-butyl group is attached to the first carbon via a CH2. This interpretation is complex. Let's assume the question intends to classify based on the carbon attached to Br.

    The bromine is attached to a carbon atom that is bonded to the phenyl ring and two other carbon atoms (one in the CH2 and one in the tert-butyl group). This makes it a secondary carbon. Since this carbon is attached to the phenyl ring, it's a benzyl system.

    IUPAC Name: 1-Bromo-3,3-dimethyl-1-phenylbutane

    Classification: The bromine is attached to a secondary carbon atom which is also attached to a phenyl group. Thus, it is a secondary benzyl halide.

  5. Structure: CH<sub>3</sub>CH(CH<sub>3</sub>)CH(Br)CH<sub>3</sub>

    The longest carbon chain has 4 carbons. The bromine atom is on the 2nd carbon, and a methyl group is on the 3rd carbon. Numbering from the end closer to the bromine gives the lowest locant.

    IUPAC Name: 2-Bromo-3-methylbutane

    Classification: The bromine atom is attached to a secondary carbon atom (a carbon bonded to two other carbon atoms). Therefore, it is a secondary alkyl halide.

  6. Structure: CH<sub>3</sub>C(C<sub>2</sub>H<sub>5</sub>)<sub>2</sub>CH<sub>2</sub>Br

    The longest carbon chain has 4 carbons (butane). The bromine atom is attached to the first carbon. There are two ethyl groups attached to the second carbon.

    IUPAC Name: 1-Bromo-2-ethyl-2-methylbutane

    Classification: The bromine atom is attached to a primary carbon atom (a carbon bonded to only one other carbon atom). Therefore, it is a primary alkyl halide.

  7. Structure: CH<sub>3</sub>C(Cl)(C<sub>2</sub>H<sub>5</sub>)CH<sub>2</sub>CH<sub>3</sub>

    The longest carbon chain has 5 carbons (pentane). The chlorine atom and an ethyl group are attached to the 3rd carbon. Numbering from either end gives the same locant for the principal chain substituents.

    IUPAC Name: 3-Chloro-3-ethylpentane

    Classification: The chlorine atom is attached to a tertiary carbon atom (a carbon bonded to three other carbon atoms). Therefore, it is a tertiary alkyl halide.

  8. Structure: CH<sub>3</sub>CH=C(Cl)CH<sub>2</sub>CH(CH<sub>3</sub>)<sub>2</sub>

    This is an alkene with a chlorine substituent. The longest chain containing the double bond has 6 carbons. The double bond is between C2 and C3. The chlorine is on C3, and there are two methyl groups on C5. Numbering starts from the end closer to the double bond.

    IUPAC Name: 3-Chloro-5-methylhex-2-ene

    Classification: The chlorine atom is attached to a carbon atom that is part of a double bond (vinylic carbon). Therefore, it is a vinyl halide.

  9. Structure: CH<sub>3</sub>CH=CHC(Br)(CH<sub>3</sub>)<sub>2</sub>

    This is an alkene with a bromine substituent. The longest chain containing the double bond has 5 carbons. The double bond is between C2 and C3. The bromine is on C4, and there are two methyl groups on C4. Numbering starts from the end closer to the double bond.

    IUPAC Name: 4-Bromo-4-methylpent-2-ene

    Classification: The bromine atom is attached to a tertiary carbon atom (C4) which is also attached to two methyl groups and is adjacent to the double bond. However, the classification is based on the carbon atom directly attached to the halogen. Since this carbon is part of the alkene structure (allylic position relative to the double bond if it were saturated, but here it's directly on the double bond system), it's a vinyl halide. Let's re-examine. The bromine is attached to C4, which is saturated. The double bond is at C2-C3. The carbon C4 is allylic to the double bond. Therefore, it is an allylic halide (specifically, a tertiary allylic halide).

  10. Structure: p-ClC<sub>6</sub>H<sub>4</sub>CH<sub>2</sub>CH(CH<sub>3</sub>)<sub>2</sub>

    This compound has a chlorine atom attached to a benzene ring (aryl halide) and also an alkyl chain attached to the ring. The question asks to name and classify. The chlorine is directly attached to the benzene ring at the para position. The side chain is an isopropyl group attached via a methylene bridge.

    IUPAC Name: 1-Chloro-4-(2-methylpropyl)benzene (or p-chloroisopropylbenzene is a common name, but IUPAC prefers naming the benzene as parent).

    Classification: The chlorine atom is directly attached to the benzene ring. Therefore, it is an aryl halide.

  11. Structure: m-ClCH<sub>2</sub>C<sub>6</sub>H<sub>4</sub>CH<sub>2</sub>C(CH<sub>3</sub>)<sub>3</sub>

    This compound has a chlorine atom attached to a methylene group (-CH<sub>2</sub>) which is attached to a benzene ring at the meta position. The other side chain is a tert-butyl group attached via a methylene bridge.

    IUPAC Name: 1-(Chloromethyl)-3-(2,2-dimethylpropyl)benzene

    Classification: The chlorine atom is attached to a methylene group (-CH<sub>2</sub>), which is then attached to the benzene ring. This makes it a primary alkyl halide (specifically, a primary benzyl halide type structure, but the chlorine is on a CH2 group attached to the ring).

  12. Structure: o-Br-C<sub>6</sub>H<sub>4</sub>CH(CH<sub>3</sub>)CH<sub>2</sub>CH<sub>3</sub>

    This compound has a bromine atom attached to a benzene ring at the ortho position. The side chain is a sec-butyl group attached to the ring.

    IUPAC Name: 1-Bromo-2-(1-methylpropyl)benzene

    Classification: The bromine atom is directly attached to the benzene ring. Therefore, it is an aryl halide.

Common mistakes

  • Incorrectly identifying the parent carbon chain for IUPAC naming.
  • Misclassifying halides (e.g., confusing secondary alkyl with secondary benzyl).
  • Errors in applying numbering rules to find the lowest locant for substituents.
  • Difficulty in distinguishing between vinyl and aryl halides.

Revision tips

  • Practice drawing structures from IUPAC names and vice versa.
  • Create flashcards for different types of halides and their classifications.
  • Focus on identifying the carbon atom directly attached to the halogen for classification.
  • Review the IUPAC naming rules specifically for branched and substituted haloalkanes.

Practice MCQs

Q1. What is the IUPAC name for the compound CH3CH(CH3)CH(Cl)CH3?

Q2. Classify the halide CH3CH2CH(CH3)CH(C2H5)Cl.

Q3. Which type of halide is (CH3)3CCH2CH(Br)C6H5?

Q4. The compound CH3CH2C(CH3)2CH2I is classified as:

Q5. What is the IUPAC name for CH3CH2C(Cl)(C2H5)CH2CH3?

Frequently asked questions

What is the main focus of Chapter 10 NCERT Solutions for Class 12 Chemistry?

Chapter 10 focuses on Haloalkanes and Haloarenes, and the NCERT Solutions provide detailed explanations for their IUPAC nomenclature and classification based on structure.

How do these solutions help in classifying halides?

The solutions guide students to identify the type of carbon atom (primary, secondary, tertiary) attached to the halogen, and to recognize specific types like alkyl, allyl, benzyl, vinyl, and aryl halides based on the structure.

Are the IUPAC names provided in the solutions accurate?

Yes, the solutions provide accurate IUPAC names for the given haloalkane and haloarene structures, following the standard nomenclature rules.

What is the difference between an alkyl halide and a benzyl halide?

An alkyl halide has the halogen attached to an aliphatic carbon atom. A benzyl halide has the halogen attached to a carbon atom that is directly bonded to a benzene ring.

How can I use these solutions for exam preparation?

You can use these solutions to practice naming and classifying various haloalkanes and haloarenes, understand the reasoning behind each classification, and reinforce your knowledge of organic chemistry concepts for exams.

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