Chapter 6 Haloalkanes and Haloarenes

NCERT Solutions Class 12 Chemistry Chapter 6 - Haloalkanes and Haloarenes

Chapter 6 (Haloalkanes and Haloarenes) of NCERT Solutions for Class 12 Chemistry provides an important basis for students to achieve excellence on the CBSE board examination as well as in competitive examinations such as JEE Main, JEE Advanced and NEET. It is the first chapter introducing students to the foundations of Organic Chemistry and teaches them the basic principles of mathematical and structural composition that apply to nucleophilic substitution reactions and elimination reactions. These concepts will provide students with a stronger understanding of the principles they are being taught and give them the ability to apply this technology to real-life applications, including the manufacture of pharmaceuticals, polymers, and specialty solvents

NCERT Solutions for Class 12 Chemistry Chapter 6 has been designed to assist students in utilizing reaction mechanisms and stereochemical principles when making predictions about the products of chemical transformations. The detailed explanations of NCERT Solutions provide students with opportunities to establish strong foundations, practice major conversion questions, and enhance their performance in term examinations, board examinations, and national-level competitive (JEE & NEET level) examinations.

1.0Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes: Key Concepts

Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes covers the preparation, properties, nomenclature, and chemical reactions of compounds containing halogens. The chapter focuses on important reaction mechanisms, stereochemistry, and the uses and environmental effects of polyhalogen compounds.

Key Concepts

Classification and Nomenclature: Learn the classification of haloalkanes and haloarenes and their IUPAC nomenclature.
Nature of the C–X Bond: Understand the polarity, bond length, and bond strength of carbon-halogen bonds with different halogens.
Methods of Preparation: Study the preparation of haloalkanes and haloarenes from alcohols and hydrocarbons, along with halogen exchange reactions such as the Finkelstein and Swarts reactions.
Nucleophilic Substitution Reactions: Understand the SN1 and SN2 mechanisms, including their reaction steps, conditions, and stereochemical outcomes.
Stereochemistry: Learn about chirality, enantiomers, optical activity, and the effect of substitution reactions on the configuration of chiral compounds.
Elimination Reactions: Understand dehydrohalogenation and apply Saytzeff’s Rule to identify the major alkene product.
Reactions with Metals: Study the formation and reactions of Grignard reagents and understand the Wurtz-Fittig reaction.
Nucleophilic Substitution in Haloarenes: Learn why haloarenes are less reactive towards nucleophilic substitution and how electron-withdrawing groups affect their reactivity.
Polyhalogen Compounds: Study important compounds such as chloroform, DDT, and freons, including their properties, uses, and environmental effects.

2.0NCERT Solutions Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes: Detailed Solutions

INTEXT QUESTIONS

Write the structure of the following compounds : (i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert-Butyl-3-iodoheptane
(iv) 1,4-Dibromobut-2-ene
(v) 1-Bromo-4-sec-butyl-2-methylbenzene Sol.

2. Why is sulphuric acid not used during the reaction of alcohols with KI? Sol. In the presence of sulphuric acid $(H_{2} SO_{4})$ , KI produces HI

$2 KI + H_{2} SO_{4} ⟶ 2 KHSO_{4} + 2 HI$

Since $H_{2} SO_{4}$ is an oxidizing agent, it oxidizes HI (produced in the reaction to $I_{2}$ ).

$2 HI + H_{2} SO_{4} ⟶ I_{2} + SO_{2} + H_{2} O$

As a result, the reaction between alcohol and HI to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidizing acid such as $H_{3} PO_{4}$ is used.

Write structures of different dihalogen derivatives of propane.

Sol.

Among the isomeric alkanes of molecular formula $C_{5} H_{12}$ , identify the one that on photochemical chlorination yields. (i) A single monochloride. (ii) Three isomeric monochlorides. (iii) Four isomeric monochlorides. Sol:

Arrange each set of compounds in order of increasing boiling points. (i) Bromomethane, Bromoform, Chloromethane, Dibromomethane. (ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane. Sol.

Which alkyl halide from the following pairs would you expect to react more rapidly by an $S_{N} 2$ mechanism? Explain your answer.

Sol

In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?

Sol:

Identify A, B, C, D, E, R and $R^{1}$ in the following

Sol:

NCERT EXERCISE

Name of the following halides according to IUPAC system and classify them as alkyl,allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides :

$(i) : (C H_{3})_{2} C H C H (Cl) C H_{3}$

$(ii) : C H_{3} C H_{2} C H (C H_{3}) C H (C_{2} H_{5}) Cl$

$(iii) : C H_{3} C H_{2} C (C H_{3})_{2} C H_{2} I$

$(i v) : (C H_{3})_{3} CC H_{2} C H (B r) C_{6} H_{5}$

$(v) : C H_{3} C H (C H_{3}) C H (B r) C H_{3}$

$(v i) : C H_{3} C (C_{2} H_{5})_{2} C H_{2} B r$

$(v ii) : C H_{3} C (Cl) (C_{2} H_{5}) C H_{2} C H_{3}$

$(v iii) : C H_{3} C H = C (Cl) C H_{2} C H (C H_{3})_{2}$

$(i x) : C H_{3} C H = C H C (B r) (C H_{3})_{2}$

$(x) : p - Cl C_{6} H_{4} C H_{2} C H (C H_{3})_{2}$

$(x i) : m - ClC H_{2} C_{6} H_{4} C H_{2} C (C H_{3})_{3}$

$(x ii) : o - B r - C_{6} H_{4} C H (C H_{3}) C H_{2} C H_{3}$

Sol:

ncert-exer-1-sol-class-12-chap-6-chemistry

2. Give the IUPAC names of the following compounds

$(i) : C H_{3} C H (Cl) C H (B r) C H_{3}$

$(ii) : C H F_{2} CB r ClF$

$(iii) : ClC H_{2} C \equiv CC H_{2} B r$

$(i v) : (CC l_{3})_{3} CCl$

$(v) : C H_{3} C (p - Cl C_{6} H_{4})_{2} C H (B r) C H_{3}$

$(v i) : (C H_{3})_{3} CC H = CCl C_{6} H_{4} I - p$

Sol:

chap-6-class-12-ncert-exer-ques-2-sol-chemistry

Write the structures of the following organic halogen compounds.
(i) 2-Chloro-3-methyl pentane
(ii) p-Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-1-iodooctane
(v) 2-Bromobutane
(vi) 4-tert-butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methyl benzene
(viii) 1,4-Dibromobut-2-ene

Sol:

ncert-exer-ques-3-sol-class-12-chap-6-chemistry

Which of the following has the highest dipole moment?

$(i) : C H_{2} C l_{2}$

$(ii) : C H Cl - 3$

$(iii) : CC l_{4}$

Sol.

ncert-exer-ques-4-sol-class-12-chap-6-chemistry

A hydrocarbon $C_{5} H_{10}$ does not react with chlorine in dark but gives a single monochloro compound $C_{5} H_{9} C I$ in bright sunlight. Identify the hydrocarbon.

Sol.

class-12-ncert-exer-ques-5-sol-chap-6-chemistry

Write the isomers of the compound having formula $C_{4} H_{9} B r .$

Sol. There are four isomers of the compound having the formula $C_{4} H_{9} B r$ . These isomers are given below.

chemistry-class-12-chap-6-ncert-exer-ques-6-sol

Write the equations for the preparation of 1-iodobutane from

(i) 1-butanol

(ii) 1-chlorobutane

(iii) but-1-ene

Sol:

chap-6-ncert-exer-ques-7-sol-class-12-chemistry

What are ambident nucleophiles? Explain with an example.

Sol. Ambident nucleophiles are nucleophiles having two nucleophilic sites. For example, nitrite ion is an ambident nucleophile.

$[\overset{ˉ}{O} - \ddot{N} = O]$

Which compound in each of the following pairs will react faster in $S_{N} 2$ reaction with OH– ?

$(i) : C H_{3} B ror C H_{3} I$

$(ii) : (C H_{3})_{3} CCl or C H_{3} Cl$

Sol. (i) $C H_{3} - I$ is more reactive because IΘ is better leaving group than BrΘ

(ii) In case of $(C H_{3})_{3} CC I$ , the attack of the nucleophile at the carbon atom is hindered of the presence of bulky substituents on that carbon atom bearing the leaving group.

Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene. (i) 1-Bromo-1-methylcyclohexane (ii) 2-Chloro-2-methylbutane (iii) 3-Bromo-2,2,3-trimethyl pentane Sol.

ncert-exer-ques-10-sol-class-12-chemistry-chap-6

How will you bring about the following conversion ?

(i) Ethanol to but-1-yne
(ii) Ethane to bromoethene
(iii) Propene to 1-nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl fluoride
(vii) Bromomethane to propanone
(viii) But-1-ene to but-2-ene
(ix) 1-Chlorobutane to n-octane
(x) Benzene to biphenyl

Sol.

ncert-exer-ques-11-sol-class-12-chap-6-chemistry

12. Explain why (i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride? (ii) Alkyl halides, though polar, are immiscible with water ? (iii) Grignard reagent should be prepared under anhydrous conditions ?

Sol.

(i)

ncert-exer-12-(i)-sol-chemistry-class-12-chap-6

(ii) Less energy is released when new attractions are set up between the haloalkanes and the water molecules as these are not as strong as the original hydrogen bonds in water.

(iii) Grignard reagents in the presence of moisture, they react with H2O to give alkanes.

ncert-exer-ques-12-(iii)-sol-class-12-chap-6-chemistry

Give the uses of Freon 12, DDT, carbon tetrachloride and iodoform.

Sol. Uses of Freon - 12 : Freon-12 (dichlorodifluoromethane, $CF_{2} Cl_{2}$ ) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners.

Uses of DDT : ( $p - p -$ dichlorodiphenyltrichloroethane) is one of the best known insecticides.

Uses of carbontetrachloride $(CCl_{4})$

(i) It is used for manufacturing refrigerants and propellants for aerosol cans. (ii) It is used as a solvent in the manufacture of pharmaceutical products. (iii) It is used as a fire extinguisher. Uses of iodoform $(CHI_{3})$ Iodoform was used earlier as an antiseptic. The antiseptic property of iodoform is only due to the liberation of free iodine when it comes in contact with the skin.

Write the structure of the major organic product in each of the following reactions. (i) $CH_{3} CH_{2} CH_{2} Cl + Nal acetone heat$ (ii) $(CH_{3})_{3} CBr + KOH ethanol heat$ (iii) $CH_{3} CH (Br) CH_{2} CH_{3} + NaOH water$ (iv) $CH_{3} CH_{2} Br + KCN aq.acetone$ (v) $C_{6} H_{5} ONa + C_{2} H_{5} Cl ⟶$ (vi) $CH_{3} CH_{2} CH_{2} OH + SOCl_{2} ⟶$ (vii) $CH_{3} CH_{2} CH = CH_{2} + HBr Peroxide$ (viii) $CH_{3} CH = C (CH_{3})_{2} + HBr ⟶$

Sol.

class-12-chemistry-chap-6-ncert-exer-ques-14-sol

15. Write the mechanism of the following reaction:

ncert-exer-ques-15-class-12-chap-6-chemistry

Sol. The given reaction is : The given reaction is an $S_{N} 2$ reaction. In this reaction, $CN^{-}$ acts as the nucleophile and attacks the carbon atom to which $Br^{-}$ in attached. $CN^{-}$ ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C- atom.

class-12-chemistry-chap-6-ncert-exer-ques-15-sol

Arrange the compounds of each set in order of reactivity towards $S_{N} 2$ displacement : (i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane (ii) 1-Bromo-3-methylbutane,2-Bromo-2-methylbutane,3-Bromo-2-methylbutane (iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane,1-Bromo-3-methylbutane.

Sol.

chemistry-ncert-exer-ques-16-sol-class-12-chap-6

(more branching at nearer distance)

Hence, the increasing order of reactivity of the given compounds towards SN2 is : 1-Bromo-2, 2-dimethylpropane < 1-Bromo-2- methylbutane < 1-Bromo-3-methylbutane < 1-Bromobutane

Out of $C_{6} H_{5} CH_{2} Cl$ and $C_{6} H_{5} CHClC_{6} H_{5}$ , which is more easily hydrolysed by aqueous KOH ?

Sol :

chap-6-ncert-exer-ques-17-sol-class-12-chemistry

p-Dichlorobenzene has higher melting point than those of o-and m-isomers. Discuss. Sol.

ques-18-sol-ncert-exer-class-12-chap-6-chemsitry

p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, molecules of p-dichlorobenzene are more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene.

How the following conversions can be carried out?

(i) Propene to propan-1-ol

(ii) 1-Bromopropane to 2-bromopropane

(iii) Benzene to 4-bromonitrobenzene

(iv) Benzyl alcohol to 2-phenylethanoic acid

(v) Ethanol to propanenitrile

(vi) Aniline to chloro benzene

(vii) 2- Chlorobutane to 3, 4-dimethylhexane

(viii) 2-Methyl-1-propene to 2-chloro-2-methyl propane

(ix) Ethyl chloride to propanoic acid

(x) But-1-ene to n-butyliodide

(xi) 2-Chloropropane to 1- propanol

(xii) Isopropyl alcohol to iodoform

(xiii) Chlorobenzene to p-nitrophenol

(xiv) 2-Bromopropane to 1-bromopropane

(xv) Chloroethane to butane

(xvi) tert-Butyl bromide to isobutyl bromide

(xvii) Aniline to phenylisocyanide

Sol.

chemsitry-class-12-chap-6-ques-19-sol-ncert-exer

The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH , alkenes are major products. Explain. Sol. $OH^{-}$ ion is highly solvated in an aqueous solution and as a result, the basic character of $OH^{-}$ ion decreases. Therefore, it cannot abstract a hydrogen from the $β$ -carbon.

$R - Cl + Aq. KOH ⟶ R - OH + KCl$

So on the other hand, an alcoholic solution of KOH contains alkoxide $(RO^{-})$ ion, which is a strong base. Thus, it can abstract hydrogen from the $β$ -carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.

$R - CH_{2} - CH_{2} - Cl + KOH (alc) Alkyl chloride \to R - CH = CH_{2} + KCl + H_{2} O Alkene$

Primary alkyl halide $C_{4} H_{9} Br (a)$ reacted with alcoholic KOH to given compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), $C_{8} H_{18}$ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions. Sol. There are two primary alkyl halides having the formula, $C_{4} H_{9} Br$ . They are n-butyl bromide and isobutyl bromide. Therefore, compound (a) is either n-butyl bromide or isobutyl bromide.

ncert-exer-ques-21-sol-(a)-class-12-chemistry-chap-6

Therefore, compound (a) is either n-butyl bromide or isobutyl bromide.

Now, compound (a) reacts with Na metal to give compound (b) of molecular formula, $C_{8} H_{18}$ . Which is different from the compound formed when n-butyl bromide reacts with Na metal.

Hence, compound (a) must be isobutyl bromide.

chemistry-necrt-exer-ques-21-(ii)-sol-class-12-chap-6

Thus, compound (d) is 2, 5–Dimethylhexane, It is given that compound (a) reacts with alcoholic KOH to give compound (b). Hence, compound (b) is 2–Methylpropene.

class-12-chap-6-ncert-exer-ques-21-(c)-sol-chemistry

Also, compound (b) reacts with HBr to given compound (c) which is an isomer of (a). Hence compound (c) is -2-Bromo-2-methylpropane.

chap-6-class-12-chemistry-ncert-exer-ques-21-(d)-sol

What happens when

(i) n-butyl chloride is treated with alcoholic KOH , (ii) bromobenzene is treated with Mg in the presence of dry ether, (iii) chlorobenzene is subjected to hydrolysis, (iv) ethyl chloride is treated with aqueous KOH , (v) methyl bromide is treated with sodium in the presence of dry ether, (vi) methyl chloride is treated with KCN.

Sol.

ncert-exer-ques-22-sol-chemistry-class-12-chap-6

3.0NCERT Solutions Class 12 Chemistry | Chapter-wise Links

Find chapter wise NCERT Solutions for Class 12 Chemistry with simple explanations, key formulae, important concepts and step wise numerical solutions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Solutions
Chapter 2	Electrochemistry
Chapter 3	Chemical Kinetics
Chapter 4	d- and f-Block Elements
Chapter 5	Coordination Compounds
Chapter 7	Alcohols, Phenols and Ethers
Chapter 8	Aldehydes, Ketones and Carboxylic Acids
Chapter 9	Amines
Chapter 10	Biomolecules

4.0Key Features of NCERT Solutions for Class 12 Chemistry Chapter 6

Clear Reaction Mechanisms: Step-by-step explanations of $S_{N 1}$ and $S_{N 2}$ reactions to make each reaction step easy to understand.
Detailed Stereochemistry: Simple explanations of Walden inversion and racemization, with suitable structural representations.
Important Name Reactions: Focused explanations of important reactions such as Sandmeyer, Fittig, and Friedel-Crafts reactions.
Easy Conversion Methods: Step-by-step approaches for solving conversion-based questions from haloalkanes and haloarenes.

NCERT Solutions Class 12 Chemistry Chapter 6 - Haloalkanes and Haloarenes

1.0Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes: Key Concepts

Key Concepts

Classification and Nomenclature: Learn the classification of haloalkanes and haloarenes and their IUPAC nomenclature.
Nature of the C–X Bond: Understand the polarity, bond length, and bond strength of carbon-halogen bonds with different halogens.
Methods of Preparation: Study the preparation of haloalkanes and haloarenes from alcohols and hydrocarbons, along with halogen exchange reactions such as the Finkelstein and Swarts reactions.
Nucleophilic Substitution Reactions: Understand the SN1 and SN2 mechanisms, including their reaction steps, conditions, and stereochemical outcomes.
Stereochemistry: Learn about chirality, enantiomers, optical activity, and the effect of substitution reactions on the configuration of chiral compounds.
Elimination Reactions: Understand dehydrohalogenation and apply Saytzeff’s Rule to identify the major alkene product.
Reactions with Metals: Study the formation and reactions of Grignard reagents and understand the Wurtz-Fittig reaction.
Nucleophilic Substitution in Haloarenes: Learn why haloarenes are less reactive towards nucleophilic substitution and how electron-withdrawing groups affect their reactivity.
Polyhalogen Compounds: Study important compounds such as chloroform, DDT, and freons, including their properties, uses, and environmental effects.

2.0NCERT Solutions Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes: Detailed Solutions

INTEXT QUESTIONS

Write the structure of the following compounds : (i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert-Butyl-3-iodoheptane
(iv) 1,4-Dibromobut-2-ene
(v) 1-Bromo-4-sec-butyl-2-methylbenzene Sol.

2. Why is sulphuric acid not used during the reaction of alcohols with KI? Sol. In the presence of sulphuric acid $(H_{2} SO_{4})$ , KI produces HI

$2 KI + H_{2} SO_{4} ⟶ 2 KHSO_{4} + 2 HI$

Since $H_{2} SO_{4}$ is an oxidizing agent, it oxidizes HI (produced in the reaction to $I_{2}$ ).

$2 HI + H_{2} SO_{4} ⟶ I_{2} + SO_{2} + H_{2} O$

Write structures of different dihalogen derivatives of propane.

Sol.

Among the isomeric alkanes of molecular formula $C_{5} H_{12}$ , identify the one that on photochemical chlorination yields. (i) A single monochloride. (ii) Three isomeric monochlorides. (iii) Four isomeric monochlorides. Sol:

Arrange each set of compounds in order of increasing boiling points. (i) Bromomethane, Bromoform, Chloromethane, Dibromomethane. (ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane. Sol.

Which alkyl halide from the following pairs would you expect to react more rapidly by an $S_{N} 2$ mechanism? Explain your answer.

Sol

In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?

Sol:

Identify A, B, C, D, E, R and $R^{1}$ in the following

Sol:

NCERT EXERCISE

Name of the following halides according to IUPAC system and classify them as alkyl,allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides :

$(i) : (C H_{3})_{2} C H C H (Cl) C H_{3}$

$(ii) : C H_{3} C H_{2} C H (C H_{3}) C H (C_{2} H_{5}) Cl$

$(iii) : C H_{3} C H_{2} C (C H_{3})_{2} C H_{2} I$

$(i v) : (C H_{3})_{3} CC H_{2} C H (B r) C_{6} H_{5}$

$(v) : C H_{3} C H (C H_{3}) C H (B r) C H_{3}$

$(v i) : C H_{3} C (C_{2} H_{5})_{2} C H_{2} B r$

$(v ii) : C H_{3} C (Cl) (C_{2} H_{5}) C H_{2} C H_{3}$

$(v iii) : C H_{3} C H = C (Cl) C H_{2} C H (C H_{3})_{2}$

$(i x) : C H_{3} C H = C H C (B r) (C H_{3})_{2}$

$(x) : p - Cl C_{6} H_{4} C H_{2} C H (C H_{3})_{2}$

$(x i) : m - ClC H_{2} C_{6} H_{4} C H_{2} C (C H_{3})_{3}$

$(x ii) : o - B r - C_{6} H_{4} C H (C H_{3}) C H_{2} C H_{3}$

Sol:

2. Give the IUPAC names of the following compounds

$(i) : C H_{3} C H (Cl) C H (B r) C H_{3}$

$(ii) : C H F_{2} CB r ClF$

$(iii) : ClC H_{2} C \equiv CC H_{2} B r$

$(i v) : (CC l_{3})_{3} CCl$

$(v) : C H_{3} C (p - Cl C_{6} H_{4})_{2} C H (B r) C H_{3}$

$(v i) : (C H_{3})_{3} CC H = CCl C_{6} H_{4} I - p$

Sol:

Write the structures of the following organic halogen compounds.
(i) 2-Chloro-3-methyl pentane
(ii) p-Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-1-iodooctane
(v) 2-Bromobutane
(vi) 4-tert-butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methyl benzene
(viii) 1,4-Dibromobut-2-ene

Sol:

Which of the following has the highest dipole moment?

$(i) : C H_{2} C l_{2}$

$(ii) : C H Cl - 3$

$(iii) : CC l_{4}$

Sol.

A hydrocarbon $C_{5} H_{10}$ does not react with chlorine in dark but gives a single monochloro compound $C_{5} H_{9} C I$ in bright sunlight. Identify the hydrocarbon.

Sol.

Write the isomers of the compound having formula $C_{4} H_{9} B r .$

Sol. There are four isomers of the compound having the formula $C_{4} H_{9} B r$ . These isomers are given below.

Write the equations for the preparation of 1-iodobutane from

(i) 1-butanol

(ii) 1-chlorobutane

(iii) but-1-ene

Sol:

What are ambident nucleophiles? Explain with an example.

Sol. Ambident nucleophiles are nucleophiles having two nucleophilic sites. For example, nitrite ion is an ambident nucleophile.

$[\overset{ˉ}{O} - \ddot{N} = O]$

Which compound in each of the following pairs will react faster in $S_{N} 2$ reaction with OH– ?

$(i) : C H_{3} B ror C H_{3} I$

$(ii) : (C H_{3})_{3} CCl or C H_{3} Cl$

Sol. (i) $C H_{3} - I$ is more reactive because IΘ is better leaving group than BrΘ

(ii) In case of $(C H_{3})_{3} CC I$ , the attack of the nucleophile at the carbon atom is hindered of the presence of bulky substituents on that carbon atom bearing the leaving group.

Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene. (i) 1-Bromo-1-methylcyclohexane (ii) 2-Chloro-2-methylbutane (iii) 3-Bromo-2,2,3-trimethyl pentane Sol.

How will you bring about the following conversion ?

Sol.

Sol.

(i)

(ii) Less energy is released when new attractions are set up between the haloalkanes and the water molecules as these are not as strong as the original hydrogen bonds in water.

(iii) Grignard reagents in the presence of moisture, they react with H2O to give alkanes.

Give the uses of Freon 12, DDT, carbon tetrachloride and iodoform.

Sol. Uses of Freon - 12 : Freon-12 (dichlorodifluoromethane, $CF_{2} Cl_{2}$ ) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners.

Uses of DDT : ( $p - p -$ dichlorodiphenyltrichloroethane) is one of the best known insecticides.

Uses of carbontetrachloride $(CCl_{4})$

Write the structure of the major organic product in each of the following reactions. (i) $CH_{3} CH_{2} CH_{2} Cl + Nal acetone heat$ (ii) $(CH_{3})_{3} CBr + KOH ethanol heat$ (iii) $CH_{3} CH (Br) CH_{2} CH_{3} + NaOH water$ (iv) $CH_{3} CH_{2} Br + KCN aq.acetone$ (v) $C_{6} H_{5} ONa + C_{2} H_{5} Cl ⟶$ (vi) $CH_{3} CH_{2} CH_{2} OH + SOCl_{2} ⟶$ (vii) $CH_{3} CH_{2} CH = CH_{2} + HBr Peroxide$ (viii) $CH_{3} CH = C (CH_{3})_{2} + HBr ⟶$

Sol.

15. Write the mechanism of the following reaction:

Arrange the compounds of each set in order of reactivity towards $S_{N} 2$ displacement : (i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane (ii) 1-Bromo-3-methylbutane,2-Bromo-2-methylbutane,3-Bromo-2-methylbutane (iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane,1-Bromo-3-methylbutane.

Sol.

(more branching at nearer distance)

Hence, the increasing order of reactivity of the given compounds towards SN2 is : 1-Bromo-2, 2-dimethylpropane < 1-Bromo-2- methylbutane < 1-Bromo-3-methylbutane < 1-Bromobutane

Out of $C_{6} H_{5} CH_{2} Cl$ and $C_{6} H_{5} CHClC_{6} H_{5}$ , which is more easily hydrolysed by aqueous KOH ?

Sol :

p-Dichlorobenzene has higher melting point than those of o-and m-isomers. Discuss. Sol.

How the following conversions can be carried out?

(i) Propene to propan-1-ol

(ii) 1-Bromopropane to 2-bromopropane

(iii) Benzene to 4-bromonitrobenzene

(iv) Benzyl alcohol to 2-phenylethanoic acid

(v) Ethanol to propanenitrile

(vi) Aniline to chloro benzene

(vii) 2- Chlorobutane to 3, 4-dimethylhexane

(viii) 2-Methyl-1-propene to 2-chloro-2-methyl propane

(ix) Ethyl chloride to propanoic acid

(x) But-1-ene to n-butyliodide

(xi) 2-Chloropropane to 1- propanol

(xii) Isopropyl alcohol to iodoform

(xiii) Chlorobenzene to p-nitrophenol

(xiv) 2-Bromopropane to 1-bromopropane

(xv) Chloroethane to butane

(xvi) tert-Butyl bromide to isobutyl bromide

(xvii) Aniline to phenylisocyanide

Sol.

The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH , alkenes are major products. Explain. Sol. $OH^{-}$ ion is highly solvated in an aqueous solution and as a result, the basic character of $OH^{-}$ ion decreases. Therefore, it cannot abstract a hydrogen from the $β$ -carbon.

$R - Cl + Aq. KOH ⟶ R - OH + KCl$

$R - CH_{2} - CH_{2} - Cl + KOH (alc) Alkyl chloride \to R - CH = CH_{2} + KCl + H_{2} O Alkene$

Primary alkyl halide $C_{4} H_{9} Br (a)$ reacted with alcoholic KOH to given compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), $C_{8} H_{18}$ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions. Sol. There are two primary alkyl halides having the formula, $C_{4} H_{9} Br$ . They are n-butyl bromide and isobutyl bromide. Therefore, compound (a) is either n-butyl bromide or isobutyl bromide.

Therefore, compound (a) is either n-butyl bromide or isobutyl bromide.

Now, compound (a) reacts with Na metal to give compound (b) of molecular formula, $C_{8} H_{18}$ . Which is different from the compound formed when n-butyl bromide reacts with Na metal.

Hence, compound (a) must be isobutyl bromide.

Thus, compound (d) is 2, 5–Dimethylhexane, It is given that compound (a) reacts with alcoholic KOH to give compound (b). Hence, compound (b) is 2–Methylpropene.

Also, compound (b) reacts with HBr to given compound (c) which is an isomer of (a). Hence compound (c) is -2-Bromo-2-methylpropane.

What happens when

Sol.

3.0NCERT Solutions Class 12 Chemistry | Chapter-wise Links

Find chapter wise NCERT Solutions for Class 12 Chemistry with simple explanations, key formulae, important concepts and step wise numerical solutions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Solutions
Chapter 2	Electrochemistry
Chapter 3	Chemical Kinetics
Chapter 4	d- and f-Block Elements
Chapter 5	Coordination Compounds
Chapter 7	Alcohols, Phenols and Ethers
Chapter 8	Aldehydes, Ketones and Carboxylic Acids
Chapter 9	Amines
Chapter 10	Biomolecules

4.0Key Features of NCERT Solutions for Class 12 Chemistry Chapter 6

Clear Reaction Mechanisms: Step-by-step explanations of $S_{N 1}$ and $S_{N 2}$ reactions to make each reaction step easy to understand.
Detailed Stereochemistry: Simple explanations of Walden inversion and racemization, with suitable structural representations.
Important Name Reactions: Focused explanations of important reactions such as Sandmeyer, Fittig, and Friedel-Crafts reactions.
Easy Conversion Methods: Step-by-step approaches for solving conversion-based questions from haloalkanes and haloarenes.

Frequently Asked Questions

Why are NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes important?

Why are NCERT Solutions for Class 12 Chemistry Chapter 6 helpful for board and other competitive exams?

Are these NCERT Solutions for Chapter 6 aligned with the latest syllabus and how are they useful?

What are the important topics in Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes?

How can I prepare Haloalkanes and Haloarenes using NCERT Solutions?

Join ALLEN!

NCERT Solutions Class 12 Chemistry Chapter 6 - Haloalkanes and Haloarenes

1.0Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes: Key Concepts

Key Concepts

2.0NCERT Solutions Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes: Detailed Solutions

INTEXT QUESTIONS

NCERT EXERCISE

3.0NCERT Solutions Class 12 Chemistry | Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Chemistry Chapter 6

Table of Contents

Frequently Asked Questions

Why are NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes important?

Why are NCERT Solutions for Class 12 Chemistry Chapter 6 helpful for board and other competitive exams?

Are these NCERT Solutions for Chapter 6 aligned with the latest syllabus and how are they useful?

What are the important topics in Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes?

How can I prepare Haloalkanes and Haloarenes using NCERT Solutions?

Join ALLEN!

NCERT Solutions Class 12 Chemistry Chapter 6 - Haloalkanes and Haloarenes

1.0Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes: Key Concepts

Key Concepts

2.0NCERT Solutions Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes: Detailed Solutions

INTEXT QUESTIONS

NCERT EXERCISE

3.0NCERT Solutions Class 12 Chemistry | Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Chemistry Chapter 6

Table of Contents