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NCERT Exemplar · Class 12 · Chemistry

Chapter 10: Haloalkanes and Haloarenes

Practice each question, then open the answer or worked solution.

Questions

55 questions
Q51Written answer

Which of the products will be a major product in the reaction given below?

Explain.

CH3CH=CH2 + HI → CH3CH2CH2I + CH3CHICH30

(A) (B)

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Worked solution

The molecule (B) will be the major product in the reaction. This addition reaction is carried out by following Markovnikoff’s rule, wherein a double bond, the hydrogen from the hydrogen halide is added to the carbon atom with the most hydrogen atoms attached to it,

Q52Written answer

Why is the solubility of haloalkanes in water very low?

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Worked solution

Haloalkanes are very slightly soluble in water because, to dissolve a haloalkane in water, energy is required to overcome the attractions between the haloalkane molecules and also to break the hydrogen bonds between water molecules.

Q53Written answer

Draw other resonance structures related to the following structure and find out whether the functional group present in the molecule is ortho, para directing or meta directing.NCERT Exemplar Solutions of Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes-22

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Worked solution

NCERT Exemplar Solutions of Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes-23

The functional group present in these molecules are ortho-para directing as electron density is more at ortho and para positions.

Q54Written answer

Classify the following compounds as primary, secondary and tertiary halides.

(i) 1-Bromobut-2-ene

(ii) 4-Bromopent-2-ene

(iii) 2-Bromo-2-methylpropane

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Worked solution

(i) 1-Bromobut-2-ene is a primary halide

(ii) In 4-Bromopent-2-ene, the bromine is attached to the secondary carbon atom, hence this a secondary halide.

(iii) In 2-Bromo-2-methylpropane, the bromine is attached to the tertiary carbon atom, making it a tertiary halide.

Q55Written answer

Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution.

The rate of this reaction depends upon the concentration of the compound ‘A’

only. When another optically active isomer ‘B’ of this compound was treated

with aq. KOH solution, the rate of reaction was found to be dependent on

the concentration of compound and KOH both.

(i) Write down the structural formula of both compounds ‘A’ and ‘B’.

(ii) Out of these two compounds, which one will be converted to the product

with an inverted configuration.

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Worked solution

NCERT Exemplar Solutions of Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes-24

(ii) Compound ‘B’ will undergo inversion of configuration and give an inverted product because it undergoes SN2 reaction.

The Class 12 NCERT Exemplar Problems on Haloalkanes and Haloarenes

Studying the NCERT Class 12 Chemistry Exemplar for Chapter 10 PDF aids you in preparing notes on the chapter. This page has answers to exemplar questions in the textbook, question papers and sample papers. MCQs, short answer questions, match the following and other exercises help you in preparing for the graduate entrance exams, such as NEET, AIPMT, JEE, AIEE, and state entrance exams.
Key concepts to remember

  1. Classification – Based on the Number of Halogen Atoms, Compounds Containing sp3 C—X Bond (X= F, Cl, Br, I), Compounds Containing sp2 C—X Bond
  2. Nomenclature – Structure, Common Name and IUPAC Name
  3. Nature of C-X Bond- Carbon-Halogen (C—X) Bond Lengths, Bond Enthalpies and Dipole Moments
  4. Methods of Preparation – From Alcohols and Hydrocarbons
  5. Physical Properties- Melting Points, Boiling Points, Melting Points and Solubility
  6. Chemical Reactions – Nucleophilic Substitution, Elimination Reactions and Reaction with Metals
  7. Polyhalogen Compounds

Download The NCERT Class 12 Chemistry Exemplar for Chapter 10 Haloalkanes and Haloarenes from the links below.

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