NCERT Exercise Solutions – Chemistry Chapter 10 Haloalkanes and Haloarenes

10.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
(i) (CH3 ) 2CHCH(Cl)CH3
(ii) CH3CH2CH(CH3 )CH(C2H5 )Cl
(iii) CH3CH2C(CH3 ) 2CH2 I
(iv) (CH3 ) 3CCH2CH(Br)C6H5
(v) CH3CH(CH3 )CH(Br)CH3
(vi) CH3C(C2H5 ) 2CH2Br
(vii) CH3C(Cl)(C2H5 )CH2CH3
(viii) CH3CH=C(Cl)CH2CH(CH3 ) 2
(ix) CH3CH=CHC(Br)(CH3 ) 2
(x) p-ClC6H4CH2CH(CH3 ) 2
(xi) m-ClCH2C6H4CH2C(CH3 ) 3
(xii) o-Br-C6H4CH(CH3 )CH2CH3

Ans –

10.2 Give the IUPAC names of the following compounds:
(i) CH3CH(Cl)CH(Br)CH3
(ii) CHF2CBrClF
(iii) ClCH2C≡CCH2Br
(iv) (CCl3 ) 3CCl
(v) CH3C(p-ClC6H4 ) 2CH(Br)CH3
(vi) (CH3 ) 3CCH=CClC6H4 I-p

Ans –

(i) 2−Bromo−3−chlorobutane

(ii) 1−Bromo−1−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−iodophenyl)−3, 3−dimethylbut−1−ene

10.3 Write the structures of the following organic halogen compounds.

(i) 2-Chloro-3-methylpentane
(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-methylbenzene
(viii) 1,4-Dibromobut-2-ene

Ans – (i)

Secondary prefix = −Cl (Chloro)
Substitute = −CH3 (methyl)
IUPAC name = Secondary prefix + Primary prefix + Word root + primary suffix + secondary suffix
2 − chloro − 3 − methylpentane

(ii)

Secondary prefix = −Cl (Chloro), −Br (Bromo)
p-Bromochlorobenzne

(iii)

Secondary prefix = −Cl (Chloro)
Substitute = −C2H5 (ethyl)
IUPAC name = Secondary prefix + Primary prefix + Word root + primary suffix + secondary suffix
1 − Chloro − 4 − ethylcyclohexane

(iv)

Secondary prefix = −I (iodo)
Substitute = (Chlorophenyl)
IUPAC name = Secondary prefix + Primary prefix + Word root + primary suffix + secondary suffix
2 − (2 − Chlorophenyl) − 1 − iodooctane
(v)

2 − Bromobutane
(vi)

Secondary prefix = −I (iodo)
Substitute = −C(CH3 )3 (tert-Butyl)
4 − tert − Butyl − 3 − iodoheptane
(vii)

Secondary prefix = −Br (Bromo)
Substitute = (sec-butyl), −CH3(methyl)
1 − Bromo − 4 − sec − butyl − 2 − methylbenzene
(viii)

IUPAC name = Secondary prefix + Primary prefix + Word root + primary suffix + secondary suffix
1,4 − Dibromobut − 2 − ene

10.4 Which one of the following has the highest dipole moment?
(i) CH2Cl2 (ii) CHCl3 (iii) CCl4

Ans – Dipole moment = Charge × Distance

μ = q × d

It is a vector quantity.

CCl₄ is a symmetrical molecule. Therefore, the dipole moments of all four C−Cl bonds cancel each other. Hence, its resultant dipole moment is zero.

As , in CHCl₃, the resultant of dipole moments of two C−Cl bonds is opposed by the resultant of dipole moments of one C−H bond and one C−Cl bond. Since the resultant of one C−H bond and one C−Cl bond dipole moments is smaller than two C−Cl bonds, the opposition is to a small extent. As a result, CHCl₃ has a small dipole moment of 1.08 D.

On the other hand, in case of CH₂Cl₂, the resultant of the dipole moments of two C−Cl bonds is strengthened by the resultant of the dipole moments of two C−H bonds. As a result, CH₂Cl₂ has a higher dipole moment of 1.60 D than CHCl₃ i.e., CH₂Cl₂ has the highest dipole moment.

Hence, the given compounds can be arranged in the increasing order of their dipole moments as:

CCl₄ < CHCl₃ < CH₂Cl₂

10.5 A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.

Ans – The group of hydrocarbons having the general molecular form CnH2n includes the hydrocarbon with the chemical formula C5H10. Since hydrocarbon can not react with chlorine in the dark, it can only be one of two things: an alkene or a cycloalkane. Additionally, the hydrocarbon produces a single monochloro molecule, C5H9Cl, when it reacts with chlorine in bright sunlight since monochloro compounds always have comparable C-H bonds. Consequently, a cycloalkane should be the chemical. The molecule is therefore C5H10 (cyclopentane).

10.6 Write the isomers of the compound having formula C4H9Br.
Ans – (a) 1-Bromobutane

(b) 2-Bromobutane

(c) 1-Bromo-2-methylpropane

(d) 2-Bromo-2-methylpropane

10.7 Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol (ii) 1-chlorobutane (iii) but-1-ene.

Ans –

10.8 What are ambident nucleophiles? Explain with an example.

Ans – Ambident nucleophiles are nucleophiles with two nucleophilic sites. Ambident nucleophiles can thus attack through two different locations.

An example of an ambident nucleophile is the nitrite ion.

Alkyl nitrites can be produced when the nitrite ion attacks through oxygen, while nitroalkanes can be produced when it attacks through nitrogen.

10.9 Which compound in each of the following pairs will react faster in SN 2 reaction with –OH?
(i) CH3Br or CH3 I (ii) (CH3 ) 3CCl or CH3Cl

Ans – (i) The reactivity of halides for the same alkyl group rises in order in the SN2 mechanism. This occurs as a result of the halide ion’s improved leaving group as the size grows.

R−F << R−Cl < R−Br < R−I

(ii) The nucleophile attacks the atom containing the leaving group as part of the SN2 process. However, in the instance of (CH3)3CCl, the presence of bulky substituents on the carbon atom carrying the leaving group prevents the nucleophile from attacking that carbon atom. On the other hand, the carbon atom in CH3Cl that is carrying the leaving group does not have any bulky substituents. Therefore, in the SN2 reaction with OH, CH3Cl reacts more quickly than (CH3)3CCl.

10.10 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) 2,2,3-Trimethyl-3-bromopentane.

Ans – (i) There are two -hydrogen atoms in 1-bromol-l-methylcyclohexane. The dehydrohalogenation of this will result in a combination of two alkenes. According to Saytzeff’s rule, alkene (B) is more substituted, making it more stable and the main product. The other alkyl halides are similarly subject to the same rule.

(ii) According to Saytzeff’s rule, the alkene with a greater number of alkyl groups connected to a doubly bonded carbon atom is preferable formed in dehydrohalogenation reactions. Alkene (I), namely 2- methylbut-2-ene, is the main byproduct of this process.

(iii) According to Saytzeff’s rule, in dehydrohalogenation reactions, the alkenes having a greater number of alkyl groups attached to the doubly bonded carbon atom is preferably formed

10.11 How will you bring about the following conversions?
(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.

Ans –

10.12 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 reagents should be prepared under anhydrous conditions?

Ans – (i) Dipole moment is a quantity of vectors.

The Cl-atom is fused to a sp2 hybridised carbon atom in chlorobenzene. The Cl-atom is fused to an sp3 hybridised carbon atom in cyclohexyl chloride.

The s-character of sp2 hybridised carbon is now greater than that of sp3 hybridised carbon. The former is therefore more electronegative than the latter.

Therefore, compared to cydohexyl chloride, chlorobenzene has a lower density of C–Cl bond electrons close to the Cl–atom.

Additionally, the electron density of the C Cl bond close to the Cl-atom lowers due to the benzene ring of chlorobenzene’s -R effect.

(ii) Intermolecular hydrogen bonds hold the molecules of H2O together in water. Alkyl halides have polar C-X bonds as well, however they are unable to sever the hydrogen bonds in H20 molecules. This indicates that there isn’t much room for alkyl halide and water molecules to connect. As a result, they are immiscible with one another and exist as independent layers.

(iii) Grignard chemicals are very reactive. Because R acts as a base and water contains acidic hydrogen, they react in the presence of water and produce alkanes.

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

Ans – Uses of Freon –12

CFC is the popular name for Freon-12 (dichlorodifluoromethane, CF2CI2).

It serves as a refrigerant in air conditioners and freezers. Additionally, it serves as a propellant in aerosol sprays like body mists and hairsprays. But it harms the ozone layer. As a result, its production was outlawed in the US and many other nations in 1994.

Uses of DDT

The IUPAC name of D.D.T., which is the abbreviation for p, p’-dichlorodiphenyltrichloroethane, has been provided above. Chiorobenzene and chlorai (trichioroacetaldehyde) are heated together to create it while H2S04 is present.

Uses of carbontetrachloride (CCl₄)

(i) It is employed in the production of aerosol can propellants and refrigerants.

(ii) It serves as a feedstock for the synthesis of compounds such as chlorofluorocarbons.

(iii) It is a solvent used in the production of medicinal goods.

(iv) Up to the middle of the 1960s, carbon tetrachloride was frequently utilised as a solvent for cleaning, a degreasing agent in businesses, a spot reamer in residences, and a fire extinguisher.

10.14 Write the structure of the major organic product in each of the following reactions:
(i) CH3CH2CH2Cl + NaI
(ii) (CH3 ) 3CBr + KOH
(iii) CH3CH(Br)CH2CH3 + NaOH
(iv) CH3CH2Br + KCN
(v) C6H5ONa + C2H5Cl
(vi) CH3CH2CH2OH + SOCl2
(vii) CH3CH2CH = CH2 + HBr
(viii) CH3CH = C(CH3 ) 2 + HBr

10.15 Write the mechanism of the following reaction: nBuBr + KCN nBuCN

Ans – The given reaction is:

nBuBr + KCN EtOH−H2O → nBuCN

An SN 2 reaction is what is being described. As the nucleophile in this reaction, CN approaches the carbon atom that Br is bound to with the least amount of resistance as CN ion is an ambident nucleophile and can attack through both C and N. In this instance, the C-atom is the assault vector.

10.16 Arrange the compounds of each set in order of reactivity towards SN2 displacement:
(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane
(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.

Ans – (i) 1-Bromopentane > 2-brornopentane > 2-Bromo-2-rnethylbutane (nu − approach tendency)

Hence, the increasing order of reactivity towards Sn2 displacement is: 2-Bromo-2-methylbutane < 2-Brornopentane < 1-Bromopentan

(ii) Since steric hindrance in alkyl halides increases in the order of

1° < 2° < 3°, the increasing order of reactivity towards SN 2 displacement is 3° < 2° < 1°,

Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards Sn2 displacement as:

2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < l-Brorno-3-rnethylbutane.

(iii) As the distance between the substituents and the atom containing the leaving group decreases in the Sn2 mechanism, the steric barrier to the nucleophile also increases. Alkyl group substituted compounds exhibit less SN2 reaction because the steric hindrance also increases with the number of substituents. As a result, the increasing order of steric hindrances in the supplied compounds is as follows.

1-Bromobutane < l-Bromo-3-methylbutane < l-Bromo-2-methylbutane < l-Bromo2, 2-dimethylpropane

Hence, the increasing order of reactivity of the given compounds towards Sn2 displacement is:

l-Bromo-2, 2-dimethylpropane < l-Bromo-2-methylbutane < l-Brorno-3- methylbutane < 1-Bromobutane

10.17 Out of C6H5CH2Cl and C6H5CHClC6H5 , which is more easily hydrolysed by aqueous KOH.

Ans – A primary aralkyl halide is C6H5CH2Cl, whereas a secondary aralkyl halide is C6H5CH(Cl)C6H5. The following factors make it likely that both of these compounds will undergo SN1 mechanism during the hydrolysis process with aqueous KOH (polar).

(a) The presence of phenyl groups at the a-position in both situations causes the carbocations produced by ionisation to be resonance stabilised (s).

(b) Given that water is a polar solvent, it is predicted that it will favour the ionisation of the two halogen-substituted chemicals, which will result in the SN1 mechanism.

The following are examples of the carbocations that come from slow ionisation:

The relative stability of the carbocation(s) that are created in two situations determines how easily they can be hydrolyzed. The positive charge on the secondary carbocation is delocalized on two phenyl groups that are present at the a-positions, making the secondary carbocation more stable. However, the main carbocation only has one phenyl group that can be used for charge delocalization.

As a result, we can draw the conclusion that C6H5CHClC6H5 is hydrolyzed by aqueous KOH more readily than C6H5CH2Cl.

10.18 p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.

Ans –

The degree to which the carbocation(s) produced in the two scenarios are hydrolyzable depends on their relative stability. The secondary carbocation is stabilised by the delocalization of the positive charge on two phenyl groups that are present at the a-positions. However, there is only one phenyl group available in the primary carbocation that can be employed for charge delocalization.

As a result, we can infer that aqueous KOH hydrolyzes C6H5CHClC6H5 more easily than it does C6H5CH2Cl.

10.19 How the following conversions can be carried out?
(i) Propene to propan-1-ol
(ii) Ethanol to but-1-yne
(iii) 1-Bromopropane to 2-bromopropane.
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3, 4-dimethylhexane
(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide

Ans –

(ix)

10.20 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.

Ans – KOH almost entirely ionises to generate OH ions in an aqueous solution. Since OH ions are potent nucleophiles, they cause the alkyl chloride to undergo a substitution process, resulting in alcohol.

R − Cl Alkyl chloride + KOH(aq) → R − OH Alcohol + KCl

On the other hand, the alkoxide (RO) ion, a strong base, is present in an alcoholic solution of KOH. As a result, it can remove a molecule of HCl and extract a hydrogen from the alkyl chloride’s -carbon to create an alkene.

In contrast, an alcoholic solution of KOH contains alkoxide (RO^–) ion which being a much stronger base than OH^– ions preferentially eliminates a molecule of HCl from an alkyl chloride to form alkenes.

10.21 Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give 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), C8H18 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.

Ans – There are two primary alkyl halides having the formula, C4H9Br. They are n -butyl bromide and 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, C₈H_18, which is different from the compound formed when n−butyl bromide reacts with Na metal. Hence, compound (a) must be isobutyl bromide.

Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a). Hence, compound (c) is 2−bromo−2−methylpropane.

10.22 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?

Ans – (i) But-l-ene is produced when n-butyl chloride is treated with alcoholic KOH.