Chemistry Class 12 NCERT Solutions Chapter 10 Haloalkanes and Haloarenes – Important Questions

Introduction

Chemistry, often dubbed the central science, plays a pivotal role in understanding the world around us. For students pursuing a Class 12 education under the NCERT curriculum, Chapter 10 of the Chemistry textbook is crucial as it dives into significant concepts that are essential for mastering both theoretical and practical aspects of chemistry. This chapter typically covers the topic of “Haloalkanes and Haloarenes,” a fundamental subject in organic chemistry that deals with the structure, properties, and reactions of compounds containing halogen atoms.

In this article, we will explore important questions from Chapter 10 of the Class 12 NCERT Chemistry textbook, providing detailed explanations and solutions to help students excel in their studies. These questions are selected based on their relevance, difficulty level, and their tendency to appear in board exams and competitive tests.

Key Concepts of Chapter 10: Haloalkanes and Haloarenes

Before diving into specific questions, let’s review some core concepts from this chapter:

1. Haloalkanes (Alkyl Halides): These are organic compounds where a halogen atom (fluorine, chlorine, bromine, or iodine) replaces a hydrogen atom in an alkane. The general formula for haloalkanes is CnH2n+1X, where X represents the halogen.

2. Haloarenes (Aryl Halides): These are compounds where a halogen atom is directly bonded to an aromatic ring. The general formula for haloarenes is C6H5X, where X is a halogen.

3. Nomenclature and Classification: Understanding the systematic naming of haloalkanes and haloarenes is crucial. For example, chlorobenzene refers to a benzene ring with a chlorine substituent.

4. Preparation Methods: Various methods for synthesizing haloalkanes and haloarenes include halogenation of alkanes and arenes, and the use of reagents like phosphorus trihalides.

5. Reactions: This includes nucleophilic substitution reactions, elimination reactions, and reactions specific to the halogen group.

6. Physical Properties: Understanding the solubility, boiling points, and density of haloalkanes and haloarenes.

7. Chemical Properties: Reactions with nucleophiles, bases, and electrophiles.

Important Questions and Detailed Solutions

Question 1: Explain the preparation of haloalkanes from alkanes.

Solution:

Haloalkanes can be prepared from alkanes through a process known as halogenation. The general reaction involves the replacement of one or more hydrogen atoms in the alkane with halogen atoms. This process typically occurs via a free radical mechanism and is initiated by the presence of UV light or heat.

Steps Involved:

1. Initiation: The halogen molecule (Cl2 or Br2) undergoes homolytic cleavage under UV light, generating two halogen radicals: $\text{Cl}_2 \xrightarrow{\text{UV light}} 2 \text{Cl}^\cdot$

2. Propagation: The halogen radical reacts with the alkane to form an alkyl radical and a hydrogen halide: $\text{CH}_4 + \text{Cl}^\cdot \rightarrow \text{CH}_3\text{Cl} + \text{H}^\cdot$

   The alkyl radical then reacts with another halogen molecule to produce the haloalkane and another halogen radical: $\text{CH}_3^\cdot + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\cdot$

3. Termination: The chain reaction ends when two radicals combine to form a stable product or when a radical reacts with a molecule that does not generate further radicals.

Example: The chlorination of methane is a common example: $\text{CH}_4 + \text{Cl}_2 \xrightarrow{\text{UV light}} \text{CH}_3\text{Cl} + \text{HCl}$

Question 2: Discuss the differences in reactivity between haloalkanes and haloarenes towards nucleophilic substitution reactions.

Solution:

Haloalkanes and haloarenes differ significantly in their reactivity towards nucleophilic substitution reactions due to the nature of the carbon-halogen bond and the electronic environment of the carbon atom.

1. Haloalkanes:

   • Reactivity: Haloalkanes are generally more reactive towards nucleophilic substitution reactions compared to haloarenes. This is due to the fact that the carbon-halogen bond in haloalkanes is more polarized, making the carbon atom more electrophilic.
   • Mechanism: Haloalkanes undergo nucleophilic substitution via both the $S_N1$ and $S_N2$ mechanisms. For example, the reaction of methyl chloride with a nucleophile like hydroxide ion: $\text{CH}_3\text{Cl} + \text{OH}^- \rightarrow \text{CH}_3\text{OH} + \text{Cl}^-$

2. Haloarenes:

   • Reactivity: Haloarenes are less reactive towards nucleophilic substitution because the carbon-halogen bond is less polarized due to the involvement of the halogen atom in the aromatic π-system. This results in a more stable carbon-halogen bond.
   • Mechanism: Nucleophilic substitution in haloarenes generally requires specific conditions such as the presence of a strong nucleophile or activation of the aromatic ring. The reaction often proceeds via an $S_NAr$ (nucleophilic aromatic substitution) mechanism, which involves the formation of an intermediate Meisenheimer complex.

Example: The reaction of chlorobenzene with sodium hydroxide under high pressure and temperature to form phenol is an example of nucleophilic aromatic substitution: $\text{C}_6\text{H}_5\text{Cl} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{NaCl}$

Question 3: Derive the mechanism of the reaction between haloalkanes and aqueous sodium hydroxide.

Solution:

The reaction between haloalkanes and aqueous sodium hydroxide typically proceeds via a nucleophilic substitution mechanism. The general reaction can be represented as:

$\text{R-X} + \text{NaOH} \rightarrow \text{R-OH} + \text{NaX}$

Here’s the mechanism for an $S_N2$ (bimolecular nucleophilic substitution) reaction:

1. Nucleophilic Attack: The hydroxide ion (OH⁻) acts as the nucleophile. It attacks the carbon atom bonded to the halogen (X) in a concerted fashion, which means both the nucleophile and the leaving group (halogen) interact simultaneously. $\text{R-X} + \text{OH}^- \rightarrow \text{R-OH} + \text{X}^-$

2. Transition State: A transition state is formed where the carbon is partially bonded to both the nucleophile and the halogen. The bond between carbon and the halogen is breaking, and a new bond between carbon and the hydroxide ion is forming.

3. Formation of Products: The transition state collapses, leading to the formation of the alcohol (R-OH) and the halide ion (X⁻) as products.

Example: The reaction of methyl bromide with sodium hydroxide to form methanol and sodium bromide: $\text{CH}_3\text{Br} + \text{NaOH} \rightarrow \text{CH}_3\text{OH} + \text{NaBr}$

Question 4: What are the methods for preparing haloarenes? Explain with examples.

Solution:

Haloarenes can be prepared using several methods, primarily through the halogenation of aromatic compounds. Here are some common methods:

1. Halogenation of Benzene:

   • Reagents: Chlorine (Cl2) or bromine (Br2) in the presence of a Lewis acid catalyst like FeCl3 or FeBr3.
   • Mechanism: The halogen reacts with the aromatic ring to form the haloarene.
   • Example: Chlorination of benzene: $\text{C}_6\text{H}_6 + \text{Cl}_2 \xrightarrow{\text{FeCl}_3} \text{C}_6\text{H}_5\text{Cl} + \text{HCl}$ This reaction produces chlorobenzene.

2. Finkelstein Reaction:

   • Reagents: This method involves the exchange of halogens. For example, converting chlorobenzene to bromobenzene using sodium bromide in the presence of a catalyst like copper(I) bromide.
   • Example: $\text{C}_6\text{H}_5\text{Cl} + \text{NaBr} \rightarrow \text{C}_6\text{H}_5\text{Br} + \text{NaCl}$

3. Nucleophilic Substitution in Aromatic Compounds:

   • Reagents: Strong nucleophiles such as hydroxide ions, especially in the presence of activating groups.
   • Example: The preparation of phenol from chlorobenzene: $\text{C}_6\text{H}_5\text{Cl} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{NaCl}$

Question 5: Compare the reactivity of different haloalkanes towards nucleophilic substitution reactions.

Solution:

The reactivity of haloalkanes towards nucleophilic substitution reactions depends on the type of haloalkane (primary, secondary, or tertiary) and the strength of the carbon-halogen bond. Here’s a comparison:

1. Primary Haloalkanes:

   • Reactivity: Primary haloalkanes are generally more reactive towards nucleophilic substitution due to the less steric hindrance around the carbon atom.
   • Mechanism: They typically undergo $S_N2$ reactions where the nucleophile attacks the carbon atom from the opposite side of the halogen.

2. Secondary Haloalkanes:

   • Reactivity: Secondary haloalkanes have moderate reactivity. They can undergo both $S_N1$ and $S_N2$ mechanisms, but steric hindrance can affect the rate of $S_N2$ reactions.
   • Mechanism: Secondary haloalkanes often show a preference for $S_N1$ reactions in the presence of strong nucleophiles and under suitable conditions.

3. Tertiary Haloalkanes:

   • Reactivity: Tertiary haloalkanes are the most reactive towards nucleophilic substitution reactions due to the high steric hindrance around the carbon atom.
   • Mechanism: They primarily undergo $S_N1$ reactions where the formation of a stable carbocation intermediate facilitates the nucleophilic attack.

Example:

• Primary Haloalkane: Methyl chloride reacts rapidly with hydroxide ion: $\text{CH}_3\text{Cl} + \text{OH}^- \rightarrow \text{CH}_3\text{OH} + \text{Cl}^-$

• Secondary Haloalkane: 2-Bromo-2-methylpropane can react via $S_N1$ mechanism: $\text{(CH}_3\text{)}_3\text{CBr} + \text{OH}^- \rightarrow \text{(CH}_3\text{)}_3\text{COH} + \text{Br}^-$

• Tertiary Haloalkane: Tert-butyl bromide reacts via $S_N1$ mechanism with nucleophiles like water: $(\text{CH}_3\text{)}_3\text{CBr} + \text{H}_2\text{O} \rightarrow (\text{CH}_3\text{)}_3\text{COH} + \text{HBr}$

Question 6: Describe the physical properties of haloalkanes and haloarenes.

Solution:

The physical properties of haloalkanes and haloarenes are influenced by the presence of the halogen atom, which affects their boiling points, melting points, and solubility.

1. Haloalkanes:

   • Boiling and Melting Points: Haloalkanes generally have higher boiling and melting points compared to their parent alkanes due to the presence of the halogen atom which increases intermolecular forces. The boiling points increase with the size of the halogen.
   • Solubility: Haloalkanes are less soluble in water due to their inability to form hydrogen bonds with water molecules. However, they are soluble in organic solvents.

   Example:

   • Methyl chloride: Boiling point = -24.2°C
   • Chloroform (CHCl3): Boiling point = 61.2°C

2. Haloarenes:

   • Boiling and Melting Points: Haloarenes generally have lower boiling points compared to haloalkanes with similar molecular weights due to the delocalization of electrons in the aromatic ring which reduces the strength of intermolecular interactions.
   • Solubility: Haloarenes are also sparingly soluble in water but are soluble in organic solvents.

   Example:

   • Chlorobenzene: Boiling point = 132°C
   • Bromobenzene: Boiling point = 156°C

Question 7: Explain the mechanism of nucleophilic aromatic substitution.

Solution:

Nucleophilic aromatic substitution (S_NAr) is a reaction where a nucleophile replaces a leaving group on an aromatic ring. This process is different from the standard nucleophilic substitution reactions seen in aliphatic compounds and typically involves the following mechanism:

1. Formation of Meisenheimer Complex:

   • The nucleophile attacks the carbon atom of the aromatic ring that is bonded to the leaving group, forming a negatively charged intermediate known as the Meisenheimer complex. $\text{Ar-X} + \text{Nu}^- \rightarrow \text{Ar-Nu}^- \text{(Meisenheimer complex)}$

2. Elimination of Leaving Group:

   • The Meisenheimer complex then collapses, resulting in the elimination of the leaving group and formation of the substituted aromatic compound. $\text{Ar-Nu}^- \rightarrow \text{Ar-Nu} + \text{X}^-$

Example:

• The reaction of chlorobenzene with sodium hydroxide: $\text{C}_6\text{H}_5\text{Cl} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{NaCl}$

In this reaction, the hydroxide ion attacks the carbon atom in the chlorobenzene ring, forming the Meisenheimer complex, which then loses the chlorine atom to produce phenol.

Conclusion

Mastering Chapter 10 of the Class 12 NCERT Chemistry textbook requires a deep understanding of the preparation, reactivity, and properties of haloalkanes and haloarenes. By focusing on these important questions and their detailed solutions, students can build a solid foundation in organic chemistry, preparing them for both board exams and future studies in chemistry.

Feel free to refer back to this guide for a thorough review and practice with these key questions to enhance your understanding and performance in this crucial chapter.