NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes

Haloalkanes And Haloarenes Class 12 NCERT Solutions includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 12 exams must go through NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

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Haloalkanes And Haloarenes Class 12 NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.

Haloalkanes And Haloarenes Class 12 NCERT Solutions – INTEXT Questions

Question 1.
Write the structures of the following compounds :

(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcydohexane
(iii) 4-terf-Butyl-3-iodoheptane
(iv) 1, 4-Dibromobut-2-ene
(v) 1 -Bromo-4-sec-butyl-2-methylbenzene

Solution:

Haloalkanes And Haloarenes Class 12 NCERT Solutions Question 2.
Why is sulphuric acid not used during the reaction of alcohols with Kl ?
Solution:
H₂SO₄ is a strong oxidising agent. Therefore, when it is used in presence of KI, it tends to convert KI to HI and finally oxidises it to I₂.

Question 3.
Write structures of different dihalogen derivatives of propane.
Solution:
The structures of all possible dihalogen derivatives of propane are

Haloalkanes And Haloarenes Class 12 NCERT Solutions Question 4.
Among the isomeric alkanes of molecular formula C₅H₁₂, identify the one that on photochemical chlorination yields
(i) A single monochloride
(ii) Three isomeric monochlorides
(iii) Four isomeric monochlorides.
Solution:
An alkane with molecular formula C₅H₁₂ can exist in the following isomeric forms :

Haloalkanes And Haloarenes Class 12 NCERT Solutions Question 5.
Draw the structures of major monohalo products in each of the following reactions :


Solution:
The major haloderivatives formed in the given reactions are

Question 6.
Arrange each set of compounds in order of increasing boiling points.
(i) Bromomethane, bromoform, chloromethane, dibromomethane.
(ii) 1 – Chloropropane, isopropyl chloride, 1 – chlorobutane.
Solution:
The boiling points of organic compounds depend on the strength of the intermolecular forces in them. These forces are :
(a) van der Waals forces and
(b) dipole-dipole interactions These forces are dependent on the
(i) molecular mass and
(ii) surface area of the molecules

(i) As the molecular mass of the compound increases, the boiling point also increases. Therefore the correct order is
chloromethane < bromomethane < dibromomethane < bromoform

(ii) Amongst molecules with same mass, it is the size of the molecule that determines the boiling point. Branched compounds are more compact and therefore have less surface area as compared to their straight chain counterparts and therefore lower boiling point. The order of boiling point is
iso-propyl chloride < 1-chloropropane < 1-chlorobutane

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by an S_N² mechanism? Explain your answer.

Haloalkanes And Haloarenes Class 12 NCERT Solutions:
We know that S_N² mechanism involves a transition state wherein both, the incoming nucleophile as well as the leaving group are present around the carbon atom. There are 5 atoms simultaneously bonded to it.

Thus, for such a transition state to be possible, there should be minimum steric hindrance. Hence, 1° alkyl halides are most reactive towards S_N² followed by 2° and finally 3°.
1° RX > 2° RX > 3° RX

Question 8.
In the following pairs of halogen compounds, which compound undergoes faster S_N¹ reaction?

Haloalkanes And Haloarenes Class 12 NCERT Solutions:
S_N¹ reaction proceeds via the formation of a carbocation intermediate. This intermediate is formed by the cleavage of the C — X bond. More stable is the resultant carbocation faster is the S_N¹ reaction.
Order of stability of carbocation is
3° carbocation > 2° carbocation >1° carbocation

Question 9.
Identify A, B, C, D, E, R and R’ in the following.

Solution:

10. 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 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.

NCERT Exercises

Question 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)₂CH₂l
(iv) (CH₃)₃CCH₂CH(Br)C₆H₅
(v) CH₃CH(CH₃)CH(Br)CH₃
(vi) CH₃C(C₂H₅)₂CH₂Br
(vii) CH₃C(CI)(C₂H₅)CH₂CH₃
(viii) CH₃CH = C(Cl)CH₂CH(CH₃)₂
(ix) CH₃CH = CHC(Br)(CH₃)₂
(x) p-ClC₆H₄CH₂CH(CH₃)₂
(xi) m-CICH₂C₆H₄CH₂C(CH₃)₃
(xii) o-BrC₆H₄CH(CH₃)CH₂CH₃

Haloalkanes And Haloarenes Class 12 NCERT Solutions:

Question 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 = C(CI) C₆H₄ l-p
Solution:

Question 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) Perfluorobenzene
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1 -Bromo-4-sec-butyl-2-methylbenzene
(viii) 1, 4-Dibromobut-2-ene
Solution:
Structures of the given compounds are :

Question 4.
Which one of the following has the highest dipole moment?
(i) CH₂Cl₂
(ii) CHCl₃
(iii) CCl₄
Solution:

Question 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:
A number of structural isomers are possible for molecular formula C₅H₁₀. But, the given compound gives a single monochloro derivative when reacted with Cl₂ in sunlight suggests that, all the H-atoms in the compound are equivalent. This is possible only if the compound is a cyclic alkane.

Question 6.
Write the isomers of the compound having formula C₄H₉Br.
Solution:
The possible isomers of C₄H₉Br are

Question 7.
Write the equations for the preparation of 1 -iodobutane from
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-1-ene
Solution:

Question 8.
What are ambident nucleophiles? Explain with an example.
Solution:
Ambident nucleophiles are nucleophiles that are capable of attacking the substrate (alkyl halide) through two different atoms.

It so ‘happens due to the presence of two nucleophilic centres which arise from the contributing (resonance) structures that are possible for the ion.

e.g., In NO^–₂ ion, there is a lone pair of electrons on N and therefore makes it nucleophilic while oxygen by virtue of the negative charge acts as a nucleophile. Thus, NO^–₂ can attack via O or N atom thereby making it ambidentate.

Question 9.
Which compound in each of the following pairs will react faster in SN2 reaction with OH^– ?
(i) CH₃Br or CH₃l
(ii) (CH₃)₃CCI or CH₃Cl
Solution:
(i) Between CH₃Br and CH₃I, CH₃I will react faster via the S_N² mechanism. In S_N² mechanism, C – X bond breaks and the faster it breaks faster is the reaction.

I- is a better leaving group. Owing to its large size, the C – I bond breaks faster than the C – Br bond and reaction proceeds further at a greater rate.

(ii) The order of reactivity in an S_N² reaction depends on minimal steric hindrance around the carbon involved in the C – X bond. Lesser the steric hindrance felt by the incoming nucleophile, more reactive will be the alkyl halide towards S_N² reaction.
Based on this, CH₃Cl will react faster than (CH₃)₃CCl.

Question 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:

Question 11.
How will you bring about the following conversions?

1. Ethanol to but-1 -yne
2. Ethane to bromoethene
3. Propeneto 1-nitropropane
4. Toluene to benzyl alcohol
5. Propene to propyne
6. Ethanol to ethyl fluoride
7. Bromomethane to propanone
8. But-1 -ene to but-2-ene
9. 1-Chlorobutane ton-octane
10. Benzene to biphenyl

Solution:

Question 12.
Explain why

1. the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
2. alkyl halides, though polar, are immiscible with water?
3. Grignard reagents should be prepared under anhydrous conditions?

Solution:
(i) (a) In order to understand the lower dipole moment of chlorobenzene we need to look into the contributing structures of the molecules.

(b) From the above structures we find that the C – Cl bond in chlorobenzene has a partial double bond character (structure II, III and IV). As a result, the C – Cl bond length here is shorter than the C – Cl single bond but longer than the C – Cl double bond.

(c) Also evident is the positive charge on Cl atom which reduces the partial negative (δ-) charge which it is expected to carry by the virtue of its electronegativity.

(d) Consequently, the dipole moment, which is a product of bond length and partial negative charge on Cl atom, reduces. However, in cyclohexyl chloride this does not happen. It is an alkyl halide and carbon is purely sp3 hybridised and C – Cl bond has the bond length of a single bond and 8“ appearing on Cl is also higher, thus, the greater dipole moment.

(ii) Only those compounds which can form hydrogen bonds with water are miscible with it. Alkyl halides, though polar due to the presence of electronegative halogen atom, are immiscible since they cannot form hydrogen bonds.

(iii) Grignard reagents R – Mg – X is a class of highly reactive compounds which can extract a proton even from water molecule. They thus, turn into the corresponding alkanes and render any other desired reaction ineffective.

This is why Grignard reagents are prepared in the absolute absence of water (anhydrous conditions), (e.g.,)

Question 13.
Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
Solution:
(i) Freon 12 (CCl₂F₂) is

• used in aerosol propellants
• refrigeration
• air-conditioning.

(ii) DDT (p, p’- dichlorodiphenyltrichloroethane) is

• used as an insecticide,
• mainly used against mosquitoes.

(iii) Carbontetrachloride (CCl₄) is used

• in manufacture of refrigerants and propellants for aerosol cans
• in synthesis of chlorofluorocarbons
• as degreasing agent
• as cleansing agent
• as a solvent in laboratories

(iv) Iodoform (CHI₃) is used as an antiseptic.

Question 14.
Write the structure of the major organic product in each of the following reactions :

Solution:

Question 15.
Write the mechanism of the following reaction

Solution:

Question 16.
Arrange the compounds of each set in order of reactivity towards SN2 displacement :

1. 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
2. 1-Bromo-3-methylbutane, 2-Bromo-2- methylbutane, 3-Bromo-2-methylbutane
3. 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methyl butane, 1-Bromo-3- methylbutane.

Solution:
S_N² reaction proceeds via the formation of transition state where the carbon atom is surrounded by 5 other atoms (groups). Thus, for such a transition state to form, the steric interactions have to be minimum. Therefore, the most favourable substrates for S_N² reactions are 1° alkyl halides followed by 2° and 3° alkyl halide. Order of reactivity towards S_N² : 1° > 2° > 3° > aryl halide.

In example (iii), although all the given alkyl halides are 1° but the steric hindrance around the carbon bearing the -Br atom decides the order of reactivity. More the number of bulky groups around this carbon lower will be its reactivity towards S_N².

Question 17.
Out of C₆ H₅ CH₂ Cl and C₆H₅ CHClC₆ H₅, which is more easily hydrolysed by aqueous KOH?
Solution:
C₆ H₅ CHClC₆ H₅ is hydrolysed faster.
(a) Hydrolysis of an alkyl halide is an example of nucleophilic substitution reaction. In case of aryl halides this follows the S_N¹ pathway i.e., via the formation of carbocation.
(b) C₆ H₅ CH₂ Cl or benzyl chloride gives

(c) Out of I & II, carbocation II is more stable. The reason is the presence of two phenyl rings attached to the carbon carrying the positive charge.
(d) As a result, the delocalisation of the +ve charge is greater and the carbocation is more stable. Due to this, (II) is formed faster and the corresponding halide is hydrolysed with greater ease as compared to benzyl chloride.

Question 18.
p-Dichlorobenzene has higher m.p. and solubility than those of o- and m-isomers. Discuss.
Solution:
The para-isomers have high melting points and solubility as compared to their ortho and meta isomers due to symmetry of para-isomers that fits into crystal lattice better than ortho and para isomers.

Question 19.
How can the following conversions be carried out?

1. Propene to propan-1 -ol
2. Ethanol to but-1 -yne
3. 1 -Bromopropane to 2-bromopropane
4. Toluene to benzyl alcohol
5. Benzene to 4-bromonitrobenzene
6. Benzyl alcohol to 2-phenylethanoic acid
7. Ethanol to propanenitrile
8. Aniline to chlorobenzene
9. 2-Chlorobutane to 3,4-dimethylhexane
10. 2-Methyl-1 -propene to 2-chloro-2- methylpropane
11. Ethyl chloride to propanoic acid
12. But-1-ene to n-butyliodide
13. 2-Chloropropaneto 1-propanol
14. Isopropyl alcohol to iodoform
15. Chlorobenzene to p-nitrophenol
16. 2-Bromopropane to 1-bromopropane
17. Chloroethane to butane
18. Benzene to diphenyl
19. ferf-Butyl bromide to isobutyl bromide
20. Aniline to phenylisocyanide

Solution:

Question 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.
Solution:
Formation of alcohols from the reaction between alkyl chlorides and aqueous KOH is an example of simple nucleophilic substitution.

But when aqueous KOH is replaced by alcoholic KOH, alkenes are formed instead of alcohols due to elimination of HCl from an alkyl halide.

Question 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.
Solution:

Question 22.
What happens when

1. n-butyl chloride is treated with alcoholic KOH,
2. bromobenzene is treated with Mg in the presence of dry ether,
3. chlorobenzene is subjected to hydrolysis,
4. ethyl chloride is treated with aqueous KOH,
5. methyl bromide is treated with sodium in the presence of dry ether,
6. methyl chloride is treated with KCN?

Solution:


Chlorobenzene is highly unreactive towards nucleophilic substitution. However, it can be hydrolysed to phenol by heating in aqueous sodium hydroxide solution at a temperature of 623 K and 300 atm pressure. The presence of an electron withdrawing group increases the reactivity of haloarenes.

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NCERT Solutions for Class 12 Chemistry