NCERT Solutions For Class 12 Chemistry Chapter 10

haloalkanes and haloarenes ncert solutions

NCERT Solutions for Class 12 Chemistry Chapter 10 – Haloalkanes and Haloarenes have been prepared with a detailed explanation of all the questions that would help the students to understand the concepts better. The First 10 NCERT Exercise Solutions of the Chapter Haloalkanes and Haloarenes has been done in this unit. The other solutions will be provided in the next unit.

NCERT Exercise Solutions | Haloalkanes and Haloarenes | Page 310

Solution 1:

(i) 2-Chloro-3-methylbutane, 2° alkyl halide,

(ii) 3-Chloro-4-methylhexane, 2° alkyl halide

(iii) 1-Iodo-2, 2-dimethylbutane, 1° alkyl halide

(iv) 1-Bromo-3,3-dimethy1-1-phenylbutane, 2° benzylic halide

(v) 2-Bromo-3-methylbutane, 2°alkyl halide

(vi) 1-Bromo-2-ethyl-2-methylbutane, 1° alkyl halide

(vii) 3-Chloro-5-methylpentane, 3° alkyl halide

(viii) 3-Chloro-5-methylhex-2-ene, vinylic halide

(ix) 4-Bromo-4-methylpent-2-ene, allylic halide

(x) 1-Chloro-4-(2-methylpropyl) benzene, aryl halide

(xi) 1-Chloromethy1-3-(2, 2-dimethylpropyl) benzene, 1° benzylic halide

(xii) 1-Bromo-2-(1-methylpropyl) benzene, aryl halide.


Solution 2:

(i) 2-Bromo-3-chlorobutane

(ii) 1-Bromo-l-chloro-1, 2, 2-trifluoroethane

(iii) 1-Bromo-4-chlorobut-2-yne

(iv) 2-(Trichloromethyl)-1, 1, 1, 2, 3, 3, 3-heptachloropropane

(v) 2-Bromo-3, 3- Bis (4-chlorophenyl) butane

(vi) 1-Chloro- 1-(4-iodopheny1)-3, 3-dimethylbut- 1 -ene


Solution 3:

haloalkanes and haloarenes ncert solutions


NCERT Exercise Solutions Page 311

Solution 4:

The three-dimensional structures of the three compounds along with the direction of the dipole moment in each of their bonds are given as follows:

haloalkanes and haloarenes ncert solutions

CCl4 being symmetrical has zero dipole moments. In CHCl3, the resultant of two C—Cl dipoles is opposed by the result of the C—H and C—Cl bond. As the latter result is expected to be smaller than the former, CHCl3 has a finite dipole (1.03 D) moment. In CH2Cl2, the resultant of two C—Cl dipole moments is reinforced by the resultant of two C—H dipoles, hence, CH2Cl2 (1.62 D) has a dipole moment higher than that of CHCl3.

Therefore. CH2Cl2 has the highest dipole moment.


Solution 5:

(i) The hydrocarbon with M.F. C5H10 can be either a cycloalkane or an alkene.

(ii) Since the hydrocarbon does not react with Cl2 in the dark, it cannot be an alkene but must be a cycloalkane.

(iii) As the cycloalkane reacts with Cl2 in the presence of bright sunlight, to give a single monochloro compound, C5H9Cl, therefore all the ten hydrogen atoms of the cycloalkane must be equivalent. Therefore, the cycloalkane is cyclopentane.

haloalkanes and haloarenes ncert solutions


Solution 6:

There are four isomers of the compound having formula C4H9Br.

isomers of C4H9Br


Solution 7:

(i) 3CH3CH2CH2CH2OH + PBr3→→ from (P & Br2)→ 3CH3CH2CH2CH2Br + H3PO3

(ii) CH3CH2CH2CH2Cl + NaBr — acetone, heat —-> CH3CH2CH2CH2Br + NaCl (Finkelstein reaction)

(iii) CH3CH2CH=CH2 + HBr —-peroxide —> CH3CH2CH2CH2Br (Anti -Mark. addition)


Solution 8:

The nucleophiles which can attack through two different sites are called ambident nucleophiles. For example, cyanide ion is a resonance hybrid of the following two structures:

ambident nucleophile

It can attack through carbon to form cyanides and through N to form isocyanides or carbylamines.


Solution 9:

( i) Since I ion is a better leaving group than Br ion, hence, CH3I reacts faster than CH3Br in SN2 reaction with OH ion.

(ii) On steric grounds, 1° alkyl halides are more reactive than tert-alkyl halides in SN2 reactions. Hence, CH3Cl will react at a faster rate than (CH3)3CCl in a SN2 reaction with OH ion.


Solution 10:

(i) In 1-Bromo- 1 -methylcyclohexane, the β-hydrogens on either side of the Br atom are equivalent, therefore, only 1-alkene is formed.


(ii) All the nine β-hydrogens in 2-chloro-2-methylbutane are equivalent, hence on treatment with C2H5ONa/

C2H5OH, it gives a single alkene.


(iii) 3-Bromo-2, 2, 3-trimethylpentane has two different sets of β-hydrogen and therefore, in principle, can give two alkenes (I and II). But according to Saytzeff rule, more highly substituted alkene (II), being more stable is the major product.


Solution 11:

(i) Ethanol to But-1-yne

CH3CH2OH + SOCl2pyridine –> CH3CH2Cl + SO2

HC ≡ CH + NaNH2Liq. NH3, 196 K –> HC ≡ CNa+

CH3CH2Cl + HC ≡ CNa+ —-> CH3CH2C ≡ CH + NaCl

(ii) Ethane to Bromoethane

CH3CH3 + Br2hv, 520-670 K —> CH3CH2Br + HBr

(iii) Propene to 1- Nitropropane

CH3 —CH = CH2 + HBr —-peroxide —> CH3CH2CH2Br — AgNO2, C2H5OH/H2O —> CH3CH2CH2NO2

(iv) Toluene to Benzyl Alcohol

toluene to benzyl alcohol


(v) Propene to Propyne

CH3 —CH = CH2 + Br2 –CCl4 —> CH3CH(Br)CH2(Br) — KOH (alc), Δ –> CH3C ≡ CH

(vi) Ethanol to Ethyl fluoride

CH3CH2OH — SOCl2, pyridine –> CH3CH2Cl —Hg2F2 –> CH3CH2F

(vii) Bromomethane to Propanone

bromoethane to propanone

(viii) But-1-ene to But-2-ene

CH3CH2CH = CH2 + HBr – Mark, addn –> CH3 —CH2 —CH(Br)—CH3KOH (alc), Δ –> CH3 —CH = CH—CH3

(ix) 1-Chlorobutane to n-Octane

2 CH3CH2CH2CH2Cl + 2 Na — Dry Ether —>CH3CH2CH2CH2 —CH2CH2CH2CH3 + 2NaCl (x)

(x) Benzene to Biphenyl

benzene to biphenyl

Solution 12:

(i) Because of greater s-character, an sp2 -hybrid carbon is more electronegative than an sp3 -hybrid carbon. Thus, the sp2-hybrid carbon of C—Cl bond in chlorobenzene has less tendency to release electrons to Cl than a sp3 -hybrid carbon of cyclohexyl chloride

Hence, the C—Cl bond in chlorobenzene is less polar than that in cyclohexyl chloride. In other words, the magnitude of negative charge, i.e., δ is less on Cl atom of chlorobenzene than in cyclohexyl chloride. Now, due to delocalisation of lone pairs of electrons of the Cl atom over the benzene ring, C—CI bond in chlorobenzene acquires some double character while the C—CI bond in cyclohexyl chloride is a pure single bond. Thus, the C—Cl bond in chlorobenzene is shorter than in cyclohexyl chloride.

chlorobenzene and cyclohexyl chloride

As dipole moment is a product of charge and distance, chlorobenzene has lower dipole moment than cyclohexyl chloride due to lower magnitude of -ve charge on the Cl atom and shorter C—Cl distance.

(ii) Alkyl halides, though polar, are immiscible in water because they are unable to form hydrogen bonds with water molecules.

(iii) Grignard reagents are very reactive. They react with moisture present in the apparatus or the starting materials.

R—MgX + H—OH –> R—H + Mg(OH)X

Hence, Grignard reagents must be prepared under anhydrous conditions.


Solution 13:

Freon – 12: They have been used extensively as propellants, for aerosols and foams to spray out deodorants, cleansers, hair spray, shaving creams.

Carbon Tetrachloride: It is widely used as an industrial solvent for oils, fats, resins, lacquers, etc. It is also used as fire extinguisher. It is also used in the manufacture of propellants and refrigerants for aerosol cans, as feedstock in the synthesis of CFCs and other chemicals.

DDT: It is used as an insecticide againsgt mosquitoes and lice.


Solution 14:

Solution 15:

The reactivity in SN2 reactions depends upon steric hindrance; more the steric hindrance, slower the reaction.

(i) chapter 10 class 12 question 15 (i)

Due to steric reasons, the order of reactivity in SN2 reactions follows the order: 1° > 2° > 3°, therefore, order of reactivity of the given alkyl bromides is as follows:

1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylpentane.

(ii) chapter 10 class 12 question 15 (ii)

Due to steric reasons, the order of reactivity of alkyl halides in SN2 reactions follows the order: 1° > 2° > 3°, therefore, the order of reactivity of the given alkyl bromides is as follows:

1-Bromo-3-methylbutane > 2-Bromo-3-methylbutane > 2-Bromo-2-methylbutane.

iii) chapter 10 class 12 question 15 (iii)

As in case of 1° alkyl halides steric hindrance increases in the order: n-alkyl halides, alkyl halide with a substituent at any position other than the β-position, one substituent at the β -position, two substituents at the β -position, therefore the reactivity decreases in same order. Hence, the reactivity of the given alkyl bromides decreases in the order:

1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 2-dimethylpropane.

Solution 16:

C6H5CH2Cl is a 1° aralkyl halide while C6H5CH(Cl)C6H5 is a 2° aralkyl halide. In SN1 reactions, the

haloalkanes and haloarenes ncert solutions

reactivity depends upon the stability of carbocations. Since the carbocation C6H5CH+C6H5 (where

the +ve charge is delocalised over two C6H5 rings) derived from C6H5CH(Cl)C6H5 is more stable than the carbocation C6H5CH2+ (where +ve charge is delocalised over one C6H5 ring) derived from C6H5CH2Cl, therefore, C6H5CH(Cl)C6H5 gets hydrolysed more easily than C6H5CH2Cl under SN1 conditions.

Although, under SN2 conditions, the reactivity depends on steric hindrance, therefore, under SN2 conditions, C6H5CH2Cl get hydrolysed more easily than C6H5CH(Cl)C6H5.

NCERT Exercise Solutions Page 312

Solution 17:

The p-isomer being more symmetrical fits closely in the crystal lattice and thus has stronger intermolecular forces of attraction than o-and m-isomers. As during melting or dissolution, the crystal lattice breaks, therefore, a larger amount of energy is needed to melt or dissolve the p-isomer than the corresponding o-and m-isomers. In other words, the melting point of the p-isomer is higher and its solubility lower than the corresponding o-and m-isomers.


Solution 18:

chapter 10 class 12 question 18


Solution 19:

In aqueous solution, KOH is almost completely ionized to give OH ions which being a strong nucleophile brings about a substitution reaction on alkyl halides to form alcohols. In the aqueous solution, OHions are highly hydrated. This reduces the basic character of OHions which fail to abstract a hydrogen from the β-carbon of the alkyl chloride to form an alkene.

On the other hand, an alcoholic solution of KOH contains alkoxide (OR) ions which being a much stronger base than OH ions preferentially eliminates a molecule of HCl from an alkyl chloride to form alkenes.


Solution 20:

haloalkanes and haloarenes ncert solutions

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top
Share via
Copy link