Chapter 6 Haloalkanes and Halorenes NCERT Solutions Class 12 Chemistry- PDF Download

Page No. 189

6.1. 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)CI

(iii) CH₃CH₂C(CH₃)₂CH₂I

(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

Solution

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

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

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

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

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

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

(vii) 3-Chloro-3-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-Chloromethyl-3-(2,2-dimethylpropyl) benzene, 1° benzylic halide.

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

6.2. Give the IUPAC names of the following compounds:

(i) CH₃CH(Cl)CH (Br)CH₃

(ii) CHF₂CBrCIF

(iii) ClCH₂C≡CCH₂Br

(iv) (CCl₃)₃CCl

(v) CH₃C(p-ClC₆H₄)2CH(Br)CH₃

(vi) (CH₃)₃CCH = C(Cl)C₆H₄I-p

Solution

(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.

6.3. 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

Solution

(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

6.4. Which one of the following has the highest dipole moment?

(i) CH₂Cl₂

(ii) CHCl₃

(iii) CCl₄

Solution

1) Dichlormethane (CH₂Cl₂)

μ = 1.60D

2) Chloroform (CHCl₃)

μ = 1.08D

(iii) Carbon tetrachloride (CCl₄)

μ = 0D

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 shown in the above figure, 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

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

Solution

The hydrocarbon with molecular formula C₅H₁₀ can either a cycloalkane or an alkene.

Since, the compound does not react with Cl₂ in the dark, therefore it cannot be an alkene but must be a cycloalkane. Since 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 cycloalkanes must be equivalent. Thus, the cycloalkane is cyclopentane.

6.6. Write the isomers of the compound having formula C₄H₉Br.

Solution

There are 4 isomers of the compound, C₄H₉Br which are given below:

(i) 1-Bromobutane

(ii) 2-Bromobutane

(iii) 1-Bromo-2-methylpropane

(iv) 2-Bromo-2-methylpropane

6.7. Write the equations for the preparation of 1-iodobutane from

(i) 1-butanol

(ii) 1-chlorobutane

(iii) But-1-ene

Solution

(i) 1-iodobutane from 1-butanol

(ii) 1-iodobutane from 1-chlorobutane

(iii) 1-iodobutane from But-1-ene

6.8. What are ambident nucleophiles? Explain with an example.

Solution

Ambident nucleophiles are nucleophiles having two nucleophilic sites. Thus, ambident nucleophiles have two sites through which they can attack.

For example, nitrite ion is an ambident nucleophile.

Nitrite ion can attack through oxygen resulting in the formation of alkyl nitrites. Also, it can attack through nitrogen resulting in the formation of nitroalkanes.

6.9. Which compound in each of the following-pairs. Will react faster in S_N2 reaction with –OH?

(i) CH₃Br or CH₃I

(ii) (CH₃)₃CCl or CH₃Cl

Solution

(i) In the S_N2 mechanism, the reactivity of halides for the same alkyl group increases in the order. This happens because as the size increases, the halide ion becomes a better leaving group.

R−F << R−Cl < R−Br < R−I

Therefore, CH₃I will react faster than CH₃Br in S_N2 reactions with OH⁻.

(ii) The S_N2 mechanism involves the attack of the nucleophile at the atom bearing the leaving group. But, in case of (CH₃)₃CCl, the attack of the nucleophile at the carbon atom is hindered because of the presence of bulky substituents on that carbon atom bearing the leaving group. On the other hand, there are no bulky substituents on the carbon atom bearing the leaving group in CH₃Cl. Hence, CH₃Cl reacts faster than (CH₃)₃CCl in S_N2 reaction with OH⁻.

6.10. 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.

Solution

(i) 1-Bromo-1-methylcyclohexane

In the given compound, there are two types of β-hydrogen atoms are present. Thus, dehydrohalogenation of this compound gives two alkenes.

(ii) 2-Chloro-2-methylbutane

In the given compound, there are two different sets of equivalent β-hydrogen atoms labelled as a and b. Thus, dehydrohalogenation of the compound yields two alkenes.

Saytzeff’s rule implies that in dehydrohalogenation reactions, the alkene having a greater number of alkyl groups attached to a doubly bonded carbon atoms is preferably produced.

Therefore, alkene (I) i.e., 2-methylbut-2-ene is the major product in this reaction.

(iii) 2,2,3-Trimethyl-3-bromopentane.

In the given compound, there are two different sets of equivalent β-hydrogen atoms labelled as a and b. Thus, dehydrohalogenation of the compound yields two alkenes.