NCERT Solutions for Class 12 Chemistry Chapter 6 – Haloalkanes and Haloarenes

Intext Questions with Solutions of Class 12 Chemistry Chapter 6 – Haloalkanes and Haloarenes

6.1

Ans – (i)

(ii)

(iii)

(iv)

(v)

6.2

Ans – When KI reacts with H₂SO₄, it is expected that HI will be produced, converting alcohols (R-OH) to alkyl iodides (R-I). HI created during the reaction is oxidised to I₂ by the strong oxidising agent H₂SO₄, which does not react with alcohol.

Phosphoric acid (H₃PO₄), that provides HI for the reaction instead of I₂, as H₂SO₄ does, is used to solve this problem.

6.3

Ans – There are four possible isomers. They are,

6.4

Ans – (i) Neopentane is the isomer of pentane that yields a single monochloride.

The replacement of any one of the H-atoms will result in the same outcome because they are all identical.

(ii) n-pentane is the isomer of pentane that yields a three isomeric monochlorides.

Three sets of equivalent hydrogen atoms are present in a, b, and c.

(iii) iso-pentane is the isomer of pentane that yields a four isomeric monochlorides

Four sets of equivalent hydrogen atoms are present in a, b, c and d.

6.5

Ans – (i)

(ii)

(iii)

(iv)

(v)

CH₃CH₂Br + NaI —–> CH₃CH₂I + NaBr

(vi)

6.6

Ans – (i) The Van der Waals principles of attraction, which rely on molecule dimensions, are connected to the simmering temperatures of organic substances. In this instance, there is merely a single carbon atom in each of the composites. The measurement of the halogen isotope and the quantity of halogen atoms in various compounds determine the molecular structure. The boiling temperatures are arranged in ascending sequence: CH₃Cl(chloromethane) < CH₃Br (bromomethane) < CH₂Br₂ (dibromomethane) < CHBr₃ (bromoform)

(ii) In this instance, identical standards are applied. As is well known, the isomer’s diameter is reduced by the bifurcation of the carbon atom series of events, which lowers its boiling point in comparison to the directly link isomer. The boiling point’s ascending sequence consists of the following: (CH₃)₂CHCl (isopropylchloride or 2-chloropropane) < ClCH₂CH₂CH₃ (1-chloropropane) < ClCH₂CH₂CH₂CH₃ (1-chlorobutane)

6.7

Ans – Since the sequence of responsiveness to supply the reaction that occurs with S_N2 is 1^∘>2^∘>3^∘, we may answer the questions using the following sequence.

(i) The remaining mixture is a secondary halide, whereas the alkyl halide is the main halide in the initial group. The first ingredient will thus respond faster.

(ii) The alkyl halide is a subsidiary halide in the initial composition and a third halide in the subsequent compound. The first substance is therefore going to respond faster.

(iii) Although the two compounds are supplementary halides, the first compound is going to respond more quickly because the CH₃ category’s steric barrier can exert a greater influence on the remainder of the molecule.

6.8

Ans – Since the sequence of responsiveness to supply the reaction that occurs with S_N1 is 3^∘>2^∘>1^∘, we may answer the question using the following sequence.

(i) The first compound will be tertiary compound while second compound is secondary compound. Thus, a faster S_N1 reaction will occur in the first compound.

(ii) The first compound will be secondary compound while second compound is primary compound. Thus, a faster S_N1 reaction will occur in the second compound.

6.9

Ans –

Exercise Questions with Solutions of Class 12 Chemistry Chapter 6 – Haloalkanes and Haloarenes

6.1

Ans – (i) Name: – 2-Chloro-3methylbutane
Classification: – 2° alkyl halide

(ii) Name: – 3-Chloro-4methyl hexane
Classification: – 2° alkyl halide

(iii) Name: – 1 -Iodo-2,2-dimethylbutane
Classification: – 1 ° alkyl halide

(iv) Name: – l-Bromo-3, 3-dimethyl -1-phenylbutane
Classification: – 2° benzylic halide

(v) Name: – 2-Bromo-3-methylbutane
Classification: – 2° alkyl halide

(vi) Name: – 1-Bromo-2-ethyI-2-methylbutane
Classification: – 1° alkyl halide

(vii) Name: – 3-Chloro-3-methylpentane
Classification: – 3° alkyl halide

(viii) Name: – 3-Chloro-5-methylhex-2-ene
Classification: – vinylic halide

(ix) Name: – 4-Bromo-4-methylpent-2-ene
Classification: – allylic halide

(x) Name: – 1-Chloro-4-(2-methylpropyl) benzene
Classification: – aryl halide

(xi) Name: – 1-Chloromethyl-3- (2,2-dimethylpropyl) benzene
Classification: – 1° benzylic halide.

(xii) Name: – 1-Bromo-2-(l-methylpropyl) benzene
Classification: – aryl halide.

6.2

Ans – (i) 2-Bromo-3-chlorobutane
(ii) 1 JBromo-1 -chloro-1,2,2-trifluoroethane
(iii) l-Bromo-4-chlorobut-2-yne
(iv) 2-(Trichloromethyl)-l, 1,1,2,3,3,3- heptachloropropane
(v) 2-Bromo-3,3-bis-(4-chlorophenyl) butane
(vi) l-Chloro-l-(4-iodophenyl)-3,3- dimethylbut-l-ene.

6.3

Ans – (i) The structure of the compound,

(ii) The structure of the compound,

(iii) The structure of the compound,

(iv) The structure of the compound,

(v) The structure of the compound,

(vi) The structure of the compound,

(vii) The structure of the compound,

(viii) The structure of the compound,

6.4

Ans – The 3 combinations’ 3D configurations and the dipole period that indicates the orientation of each bonding are shown in the one that follows:

CCl₄‘s symmetry prevents it from having a dipole moment. The addition of 2 C-Cl dipole energies to CHCl₃ causes the C-H and C-Cl links to clash with one another. Since the second resultant’s dipole angle is expected to be smaller than the first’s, CHCl₃ has a restricted dipole value (1.03 D). This indicates that compared to CHCl3, the resultant dipole moment of C-Cl couples is larger in CH₂Cl₂. CH₂Cl₂ is stronger because of its dipole moment attraction.

6.5

Ans – The molecules of hydrocarbons may have an alkene or cycloalkane structure in its molecules. The molecular structure must be a cycloalkane because it fails to interact with Cl₂ in the absence of light. Each of the 10 hydrogen atoms that make up cycloalkanes must be exactly equal because, when exposed to intense sunlight, the cycloalkane reacts with Cl₂ to generate an individual monochloride molecule, C₅H₉Cl. Therefore, the cycloalkane becomes cyclopentane.

6.6

Ans – In the beginning, we must determine C₄H₉Br’s dual-bond equivalent (DBE). It is provided the one that follows:

Since the correct reply will be zero, every single component of the isomers is molecules of the provided chemical are going to have either an extended or linear cord, indicating that the substance has neither a circle nor an unsaturated state in them. Additionally, there will consequently be 4 isomeric compounds, which are listed as follows: 
(a) 1- Bromobutane – CH₃CH₂CH₂CH₂Br
(b) 1-Bromo-2-Methylpropane

(c) 2-Bromo Butane

(d) 2-Bromo-2-Methylpropane

6.7

Ans – (i) Phosphoric acid and potassium iodide are going to interact with 1-butanol to produce 1-iodobutane. The following makes for the final reaction:

CH₃CH₂CH₂CH₂OH + KI + H₃PO₄ → CH₃CH₂CH₂CH₂I + H₂O + KH₂PO₄

(ii) In an environment consisting of acetone, 1-chlorobutane is going to combine with potassium iodide to produce 1-iodobutane. The following equation serves as the reaction:

(iii) When peroxide is present, but-1-ene might come in contact with hydrogen bromide to produce 1-bromobutane. When acetone is present, the compound is going to interacts with sodium iodide to produce 1-iodobutane. The subsequent reaction is mentioned by the following equation:

6.8

Ans – The ability to contribute two separate electrons coming from two distinct atoms enables ambident nucleophiles to assault a potential electrophile from any location. When two distinct atoms containing individual pairs of electrons are present, usually underneath an identical functional classification, this simultaneous reactivity develops.
For instance, the two types of nitrogen plus carbon elements in the cyanide ion (CN⁻) have the ability to function as nucleophilic sites. The kind of electrically conductive compounds, the solvent being used, and the resultant circumstances all affect the precise location of the assault. Based on which particular element takes responsibility for the chemical process, ambient nucleophiles may culminate in the creation of regioisomers.
Nucleophiles that are capable of assault at 2 different sites are known as ambident nucleophiles. It constitutes a resonant hybridization of the cyanide ion and the subsequent pair of molecules. The cyanide ion is a resonant hybrid comprising the two distinct compounds shown below:

Based on the chemical substance created in nature, cyanide could target nitrogen to make isocyanide or carbon compounds to generate the chemical cyanide.

6.9

Ans – (i) Although the ion that makes up iodide is a bigger atom as opposed to the bromide ion, both substances comprise alkyl halides. Hence, I^– ion is more adept at quitting a network compared to Br^– ion. As a result, CH₃I is going to interact with the hydroxyl ion present in the reaction caused by S_N2 more quickly rather than CH₃Br.

(ii) The steric barrier is expected to be significantly lower in the S_N2 process. CH₃Cl possesses a lower steric hindrance than (CH₃)₃CCl, which exhibits comparatively greater steric hindrance. Therefore, CH₃Cl is likely to interact with the hydroxyl ion in the S_N2 transition more quickly.

6.10

Ans – (i) 1-Methylcyclohexane is the main result of the reaction between 1-Bromo-1-methylcyclohexane and the presence of sodium ethoxide in ethanol as the solvent. The following constitutes the reaction:

(ii) 2-Chloro-2-methyl butane could theoretically yield 2 alkaline compounds since it has 2 separate kinds of comparable beta-hydrogen. Nevertheless, according to Satzeff’s principle, the principal result is an extremely stronger exchanged alkene, which is greater in a stabilized property. The following serves as the reaction:

(iii) Theoretically, 2, 2, 3-triethyl-3-bromopentane could produce two alkenes since it has 2 separate collections of identical beta-hydrogen atoms. Meanwhile, Satzeff’s rule states that the principal result is an improved approach that strongly replaces alkene, which is of greater stability. The one that follows pertains to the response:

6.11

Ans – (i) SOCl₂ and pyridine are going to combine along with ethanol to produce chloroethane. Sodium acetylide is created when acetylene and NaNH₂ interact. But-1-yne will subsequently be created by the reaction of sodium acetylide and chloroethane. Those responses have been listed as follows:

(ii) Ethane will come into contact alongside bromine to generate bromoethane. Ethene is going to be generated by this bromoethane reaction with some KOH, followed by the creation of 1, 2-Dibromoethane. Bromoethene is currently being created when 1, 2-Dibromoethane, and alcoholic KOH contact. The responses have been as follows:

(iii) During a peroxide reaction, propene and hydrogen bromide will come together to generate 1-bromopropane. Silver nitrate and 1-bromopropane shall subsequently bond to produce 1-nitropropane. The one that follows represents my response:

(iv) In an environment full of illumination, toluene along with chlorine will blend together to form benzyl chloride, which will subsequently combine using aqueous KOH to form benzyl alcohol. What follows constitutes the response:

(v) In an atmosphere of the element carbon tetrachloride, propene and bromine can combine to form 1, 2-Dibromopropane. Propyne is supposed to be produced by the reaction of 1, 2-Dibromopropane together with NaNH₂ plus liquid ammonia. The one that follows serves as the response:

(vi) Ethyl chloride is created when ethanol reacts alongside pyridine & SOCl₂. Ethyl fluoride they’ll soon be created when ethyl chloride and Hg₂F₂ combine. The information that follows describes the response:

(vii) When ethanol is present, bromomethane will come into contact with KCN to produce acetonitrile. Propanone is produced via the reaction of acetonitrile using CH₃MgBr in an environment composed of ether plus water. That which follows makes up the response:

(viii) Hydrogen bromide while participating but-1-ene is going to combine producing 2-bromobutane, which will subsequently combine alongside alcoholic KOH to transform into but-2-ene. The one that follows serves as the response:

(ix) 1-N-octane is created when chlorobutane reacts alongside salt and dehydrated ether. What follows constitutes the response:

(x) When FeBr₃ is present, benzene and bromine will combine to form bromobenzene. Biphenyl is created when sodium and bromobenzene react. The following is the reaction: 

6.12

Ans – (i) Because of its higher s-character, the sp²-hybrid carbon surrounding chlorobenzene ends up being higher electronegative compared to the sp³-hybrid carbon within cyclohexyl chloride. Therefore, relative to the carbon atom with cyclohexyl chloride, the C atomic structure of chlorobenzene possesses a lower propensity to transfer particles of electrons toward Cl.
Consequently, compared to cyclohexyl chloride, the bond formed by C-C in chlorobenzene becomes relatively low polar. Furthermore, the bond between the chlorine atoms in chlorobenzene takes on certain double-binding characteristics as a result because of the Cl atom’s independent pairs of ions delocalizing around the band of benzene. Whereas, the C-Cl link in cyclohexyl chloride is a genuine independent bond.
To put it another way, compared to cyclohexyl chloride, the bond between chlorine and chlorine in chlorobenzene is considerably smaller. Because of the reduced C-Cl separation & smaller negative electrical ions attached to the Cl atomic unit, chlorobenzene exhibits a weaker dipole force over cyclohexyl chloride.

(ii) The chemical connection between alkyl halides is maintained via dipole-dipole magnetism. The individual molecules containing H₂O are held attached through hydrogen bonds. Due to the fact that the additional pulling forces of adhesion among water & alkyl halide particles remain weaker compared to the ones that already occur around alkyl halide particles & water particles, alkyl halides are insoluble in liquid. Furthermore, alkyl halides cannot form hydrogen-bonded structures when mixed with liquid substances or interfere by disrupting water’s H-bonded structure.

(iii) The reagent used by Grignard has highly elevated reactive properties. Compounds easily respond and generate alkanes immediately after they develop into touch alongside water. Following is the common response:
R-Mg-X + H-OH → R-H + Mg(OH)X
As a result, the Grignard reagent needs to be prepared anhydrously.

6.13

Ans – Iodoform: This element has been commonly used as an antibacterial agent in the beginning. Yet rather than the material itself, the properties are due to the liberated iodine that has been emitted. Because of its unpleasant smell, it was eventually replaced by substitute products that include iodine.
Carbon Tetrachloride: For use in garment drying along with the manufacturing industry to obtain oil, fats, and resins. Additionally, the company claims that CCl₄ gases are very combustible. CCl₄ is thus marketed as pyrene, a form of fire extinguisher substance. utilized in the manufacturing of chemical propellants & aerosol is employed as refrigerants. 
Freons: The term “Freons” encompasses a class comprising halogenated hydrocarbons, mainly CFCs (chlorofluorocarbons), or similar substances with carbon, fluorine, and chlorine atoms within them. Because these substances are chemically resistant, incombustible, and non-hazardous compounds they’re frequently employed as chemical solvents, propeller fluid for aerosol sprays, and refrigerant gases. The frequently encountered type of coolant used in the business comprises Freon-12 (CCl₂F₂). Aerosol propellants, refrigeration fluids, or air conditioning additives.
DDT: The success it had over crop-damaging pests and malaria-carrying mosquitoes caused their use to skyrocket globally after the Second World War. Conversely, DDT has already been in widespread usage throughout the 1940s. Several bug species developed resistance to DDT, which is hazardous to aquatic organisms. DDT cannot be easily metabolized in mammals. Rather, it builds up and stores away in adipose cells. DDT accumulates inside the body of living creatures throughout their lives as they consume it starting at an identical pace.

6.14

Ans –

6.15

Ans – KCN represents a resonating hybridization of two distinct fundamental components that are demonstrated below:

CN^– atom is hence an ambident protein nucleophile. Consequently, the enzyme may use whether C or N to make contact with the “carbon molecule that makes up the C-Br connection within n-BuBr.” Assault happens via C to generate n-butyl cyanide because the link between C and C has become more potent compared to the C-N connection.

6.16

Ans – The steric barrier determines the susceptibility of the S_N2 process. The more gradual the response is, the greater the steric barrier. Consequently, 1° > 2° > 3° shall represent the sequence of reaction.

Hence, the rate of reaction declines in the equivalent sequence as the steric hindrance grows in the scenario of 1° alkyl halides, n-alkyl halides, alkyl halides that have a replacement during any point apart from the β-position, a single substitution at the β-position, and two substituted molecules at the β-position. Accordingly, 1-Bromobutane > l-Bromo-3-methylbutane > l-Bromo-2-methyjbutane > 1-Bromo-2,2-dimethyl propane is the sequence in which the reacting capacity of the specified alkyl bromides diminishes.

6.17

Ans – While C₆H₅CHClC₆H₅ is an intermediate aryl halide, C₆H₅CHCl₂ is the principal aryl halide. The long-term stability that exists in the carbocation determines whether the reaction that occurs with water-soluble KOH is a reaction classified as S_N1. The section that follows describes how a carbocation forms:

Because the additional carbocation is highly resilient compared to the main one, the C₆H₅CHClC₆H₅ will respond quicker than usual. However, since the mechanism of action happens in the reverse sequence from C₆H₅CHCl₂ is going to respond more quickly if its reaction mediator is S_N2.

6.18

Ans – Because p-isomers appear more symmetric and therefore integrate within the crystalline lattice with greater precision than o-isomers, there are significantly stronger atomic forces of adhesion connecting species. The p-isomer requires higher amounts of potential to undergo meltdown or disintegrate compared to the o- & m-isomers since the crystal lattice fractures throughout these processes. Accordingly, p-isomers are less soluble and possess greater degrees of melting compared to their o- and m-isomer counterparts.

6.19

Ans – (i) Propene will react with hydrogen bromide in the presence of peroxide to produce 1-bromopropane. The reaction between 1-Bromopropane and aqueous KOH produces Propan-1-ol.

(ii) Iodoethane will be produced from ethanol’s reaction with iodine and phosphorous; this will then convert to ethene. Ethene reacts with bromine in the presence of carbon tetrachloride to generate 1,2-Dibromoethane, which upon dehydrohalogenation and interaction with NaNH₂ creates disodium acetylide. The reaction with methyl iodide will then produce But-2-yne.

(iii) Alcoholic KOH reacts with 1-bromopropane to form propene. The reaction between propene and HBr produces 2-Bromopropane.

(iv) Toluene and chlorine will react to create benzyl chloride in the presence of light; this will subsequently react with aqueous KOH to produce benzyl alcohol.

(v) Benzene and bromine react in presence of FeBr₃ to form bromobenzene. 4-Bromonitrobenzene is produced by bromobenzene reacting with concentrated nitric acid and concentrated sulphuric acid.

(vi) Benzyl alcohol produces benzyl chloride when reacting with SOCl₂. Benzyl chloride produces benzyl cyanide when it reacts with KCN. Benzyl cyanide produces 2-Phenylethanoic acid when hydrolysed.

(vii) Ethanol will react with iodine when phosphorous is present to produce iodoethane. Iodoethane and KCN reaction in aqueous ethanol produces propanenitrile.

(viii) Diazotising aniline results in benzene diazonium chloride. Benzene diazonium chloride will react with copper chloride when hydrochloric acid is present to generate chlorobenzene.

(ix) When dry ether is present, 2-chlorobutane reacts with sodium to produce 3, 4-dimethylhexane.

(x) Hydrogen chloride reacts with 2-methyl-1-propene to generate 2-chloro-2-methylpropane.

(xi) Propanenitrile is produced by ethyl chloride reacting with KCN. Hydrolysis of propanenitrile produces propanoic acid.

(xii) But-1-ene reacts with HBr in peroxide to generate 1-Bromobutane. When acetone is present, bromobutane reacts with NaI to generate n-butyl iodide.

(xiii) 2-Chloropropane generates propene by dehydrohalogenation. Propene will react with HBr in peroxide’s presence to produce 1-bromopropane. 1-propanol results from the reaction of 1-bromopropane with KOH.

(xiv) Iodoform is produced when sodium hydroxide and iodine react with isopropyl alcohol.

(xv) P-Chloronitrobenzene is produced when chlorine benzenes reacts with concentrated nitric acid and concentrated sulphuric acid. P-nitrophenol will be produced from 15% sodium hydroxide and hydrochloric acid reacting with p-chloronitrobenzene.

(xvi) Propene is produced when 2-bromopropane reacts with alcoholic KOH. Propene combines with HBr in the presence of peroxide to produce 1-bromopropane.

(xvii) The reaction between chloroethane and sodium in dry ether produces butane.

(xviii) Benzene and bromine will react to form bromobenzene when FeBr₃ is present. Biphenyl is formed when sodium and bromobenzene react.

(xix) 2-methyl-1-propene is produced from tert-butyl bromide reacting with alcoholic KOH. In presence of peroxide, 2-methyl-1-propene reacts with HBr to produce isobutyl bromide.

(xx) Phenylisocyanide is produced from aniline reacting with chloroform and alcoholic KOH.

6.20

Ans – To produce OH^– ions, KOH would fully disintegrate in aqueous environments such as water. These ions may trigger the replacement of alkyl halides to produce alcoholic substances since they are a potent nucleophile. The OH^– ions are going to be very hydrated at the exact same period. In order to create alkenes, agents would be unable to remove a proton particle (H⁺) from their p-carbon atomic structure.
Stated differently, OH^– ions will react as weakened bases in aqueous media, making it unable to eliminate them and produce alkenes. In addition to OH^– ions, ethoxide-charged particles (C₂H₅O^–) are also likely to be present in the resulting solution of alcoholic KOH. Since they come from much more powerful bases than OH^– ions, protons are going to remove an H⁺ ion from every β-carbon atom, resulting in the production of alkene whenever the end byproduct of dehydrohalogenation happens.

6.21

Ans – (i) Two fundamental alkyl halides consist of the molecular combination denoted by C₄H₉Br.

(ii) Because the substance (d) that has the chemical structure C₈H₁₈ was produced whenever product (a) interacted alongside Na metallic material, it differed in composition from the die component that was produced while n-butyl bromide interacted with Na metals. Consequently, chemical (d) needs to be 2,3-dimethylhexane, while (a) needs to contain isobutyl bromide.

(iii) The chemical (b) that is produced upon reaction with alcoholic potassium hydroxide is required to be 2-methyl-1-propane when the substance (a) remains isobutyl bromide.

(iv) According to the Markownikoff procedure, component (b) yields component (c) after being treated along with HBr. Tert-butyl bromide, the isomeric form of chemical (a) is hence a component of (c) which is denoted by isobutyl bromide.

Hence, the following compounds are formed as a result of the following equation.
(a) becomes isobutyl bromide,
(b) becomes 2-methyl-1 -propane,
(c) becomes tert-butylbromide &
(d) becomes 2,5-dimethylhexane.

6.22

Ans –

Related Study Resources of Chapter 6 – Haloalkanes and Haloarenes

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1	Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes – Important Questions

NCERT Solutions for Class 12 Haloalkanes and Haloarenes

An essential chapter in Class 12 Chemistry, Haloalkanes and Haloarenes addresses the structure, classification, synthesis, and characteristics of organic halogen compounds. Our NCERT Solutions for Class 12 Chemistry Chapter 6 help students understand fundamental ideas such nucleophilic substitution reactions, elimination reactions, and reaction processes with thorough, step-by-step explanations for all intext and exercise questions. These answers are required for competitive tests like JEE and NEET as well as for board exam preparation and match the most recent NCERT syllabus.

Through working with our Class 12 Chemistry Chapter 6 NCERT Solutions, students may deepen their knowledge of the physical and chemical characteristics of haloalkanes and haloarenes, their applications, and their effects on the environment. These guarantees conceptual clarity by streamlining difficult reaction processes and stereochemical elements. Having a good basis in this chapter, students will be able to boldly answer theoretical questions and numerical problems in tests.