Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes ncert solutions

Haloalkanes and Haloarenes Class 12 ncert solutions: Chemistry Class 12 Chapter 6 exercise solutions

Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 6
Chapter Name	Haloalkanes and Haloarenes ncert solutions
Category	Ncert Solutions
Medium	English

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Question 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)Cl
• (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 6.1: (i)

2−Chloro−3−methylbutane

(Secondary alkyl halide)

(ii)

3−Chloro−4−methyhexane

(Secondary alkyl halide)

(iii)

1−Iodo−2, 2 −dimethylbutane

(Primary alkyl halide)

(iv)

1−Bromo−3, 3−dimethyl−1−phenylbutane

(Secondary benzyl halide)

(v)

2−Bromo−3−methylbutane

(Secondary alkyl halide)

(vi)

1−Bromo−2−ethyl−2−methylbutane

(Primary alkyl halide)

(vii)

3−Chloro−3−methylpentane

(Tertiary alkyl halide)

(viii)

3−Chloro−5−methylhex−2−ene

(Vinyl halide)

(ix)

4−Bromo−4−methylpent−2−ene

(Allyl halide)

(x)

1−Chloro−4−(2−methylpropyl) benzene

(Aryl halide)

(xi)

1−Chloromethyl−3−(2, 2−dimethylpropyl) benzene

(Primary benzyl halide)

(xii)

1−Bromo−2−(1−methylpropyl) benzene

(Aryl halide)

Question 6.2: Give the IUPAC names of the following compounds:

• (i) CH₃ CH(Cl) CH(Br) CH₃
• (ii) CH F₂ CBr ClF
• (iii) Cl CH₂ C ≡ C CH₂ Br
• (iv) ( CCl ₃) ₃ CCl
• (v) CH₃ C (p – Cl C₆ H₄ )₂ CH(Br) CH₃
• (vi) (CH₃ )₃ C CH = C Cl C₆ H₄ I – p

Solution 6.2: (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

Question 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 6.3: (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

Question 6.4: Which one of the following has the highest dipole moment? (i) CH₂Cl₂ (ii) CHCl₃ (iii) CCl₄

Solution 6.4: (i) Dichlormethane (CH2Cl2), μ = 1.60D

(ii) 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.

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

Question 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 6.5: A hydrocarbon with the molecular formula, C₅H₁₀ belongs to the group with a general molecular formula CnH2n. Therefore, it may either be an alkene or a cycloalkane.

Since hydrocarbon does not react with chlorine in the dark, it cannot be an alkene. Thus, it should be a cycloalkane.

Further, the hydrocarbon gives a single monochloro compound, C₅H₉Cl by reacting with chlorine in bright sunlight. Since a single monochloro compound is formed, the hydrocarbon must contain H−atoms that are all equivalent. Also, as all H−atoms of a cycloalkane are equivalent, the hydrocarbon must be a cycloalkane. Hence, the said compound is cyclopentane.

Cyclopentane (C₅H₁₀)

The reactions involved in the question are:

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

Solution 6.6: The molecular formula C₄H₉Br corresponds to butyl bromide isomers, which are structural isomers where the arrangement of carbon atoms varies. For this formula, there are four possible isomers, classified as primary, secondary, and tertiary alkyl halides. Here are the isomers:

(a) 1−Bromobutane

(b) 2−Bromobutane

(c) 1−Bromo−2−methylpropane

(d) 2−Bromo−2−methylpropane

Question 6.7: Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol (ii) 1-chlorobutane (iii) but-1-ene.

Solution 6.7:

(i)

(ii)

(iii)

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

Solution 6.8: 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.

Question 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 6.9: (i) In SN2 mechanism, the order in which the halides react to some alkyl group is constant. This is because the halide ion becomes a better leaving group when the size of the ion increases.

R-F << R-Cl << R-Br < R-I

Hence, the reactivity of CH₃I is faster as compared to CH₃Br in S_N2 reactions with OH^–.

(ii)

In S_N2 mechanism, the nucleophile attacks at the atom bearing the leaving group. The attack of nucleophile in (CH₃)₃CCl at the carbon atom is hindered as the carbon atoms (bearing the leaving group) contains bulky substituents.   

However, CH₃Cl does not consist of bulky substituents on the carbon atom bearing the leaving group.

Hence, the reactivity of CH₃Cl is faster as compared to (CH₃)₃CCl in S_N2 reaction with OH^–.

Question 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 6.10: (i)

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

(ii)

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.

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

Hence, alkene (I) i.e., 3,4,4-trimethylpent-2-ene is the major product in this reaction.

Question 6.11: 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.

Solution 6.11: (i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

Question 6.12: 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?

Solution 6.12: (i)

The Cl- atom in chlorobenzene is linked to sp² hybridized carbon atom, whereas in cyclohexyl chloride it is linked to sp³ hybridized carbon atom. Now, as sp² hybridized carbon has more s-character, it is more electronegative than sp³ hybridized carbon atom. Hence, the density of electrons of C-Cl bond near the Cl atom is less in chlorobenzene than in cyclohexyl chloride.

Moreover, the electron density is reduced by the –R effect of the benzene ring of chlorobenzene. Hence, chlorobenze has a decreased polarity of C-Cl bond, also the dipole moment is lower than that of cyclohexyl chloride.

(ii) When the solute water force o attraction is stronger than solute-solute and water-water force of attraction, detmines the miscibility with water. Dipole – dipole interactions hold the polar molecules of alkyl halides together. Now, the alkyl halide- alkyl halide and water-water forces of attraction are stronger than the new force of attraction between alkyl halides and water molecules. As a result, alkyl halides (though polar) are immiscible with water.

(iii) Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes.

Therefore, Grignard reagents should be prepared under anhydrous conditions.

Question 6.13: Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

Solution 6.13: Uses of Freon -12

 Common name given to Freon-12 (dichlorodifluoromethane, CF₂Cl₂) is CFC. It is used in aerosol spray propellants such as deodorants, hair sprays, etc. It is also used in refrigerators and air-conditioners as a refrigerant. It’s manufacturing was banned in United States and many other countries in 1994 as it was severely damaging the ozone layer.

Uses of DDT

The most commonly used insecticide is DDT (p, p−dichlorodiphenyltrichloroethane). It has various harmful effects due to which it was banned in United States in 1973, though it is very effective aganst mosquitoes and lice.

Uses of carbontetrachloride (CCl₄ )

(i) Refrigerants and propellant for aerosol cans are manufactured using CCl₄.

(ii) Chloroflourocarbons and other chemicals are synthesized.

(iii) Used as a solvent in the manufacture of pharmaceutical products.

(iv) Used as cleaning fluid, a spot reamer in homes and a fire extinguisher during the mid 1960s.

Uses of Iodoform (CHI₃)

Iodoform was used earlier as an antiseptic, but now it has been replaced by other formulations-containing iodine-due to its objectionable smell.

Question 6.14: Write the structure of the major organic product in each of the following reactions:

Solution 6.14:

Question 6.15: Write the mechanism of the following reaction:

n-BuBr + KCN \(\underrightarrow { { EtOH-H }_{ 2 }O } \) nBuCN

Solution 6.15: The given reaction is:

n-BuBr + KCN \(\underrightarrow { { EtOH-H }_{ 2 }O } \) nBuCN

The given reaction is an SN2 reaction. In this reaction, CN^– acts as the nucleophile and attacks the carbon atom to which Br is attached. CN^– ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.

Question 6.16: 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, 3-Bromo-2- methylbutane
• (iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1Bromo-3- methylbutane.

Solution 6.16: (i)

An SN2 reaction involves the approaching of the nucleophile to the carbon atom to which the leaving group is attached. When the nucleophile is sterically hindered, then the reactivity towards S_N2 displacement decreases. Due to the presence of substituents, hindrance to the approaching nucleophile increases in the following order.

1-Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane

Hence, the increasing order of reactivity towards SN2 displacement is:

2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane

(ii)

Since steric hindrance in alkyl halides increases in the order of 1° < 2° < 3°, the increasing order of reactivity towards SN2 displacement is 3° < 2° < 1°.

Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards S_N2 displacement as:

2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane

[2-Bromo-3-methylbutane is incorrectly given in NCERT]

(iii)

The steric hindrance to the nucleophile in the SN2 mechanism increases with a decrease in the distance of the substituents from the atom containing the leaving group. Further, the steric hindrance increases with an increase in the number of substituents. Therefore, the increasing order of steric hindrances in the given compounds is as below:

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

Hence, the increasing order of reactivity of the given compounds towards SN2 displacement is:

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

Question 6.17: Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolysed by aqueous KOH?

Solution 6.17:

Hydrolysis by aqueous KOH proceeds through the formation of carbocation. If carbocation is stable, then the compound is easily hydrolyzed by aqueous KOH. Now,

C₆H₅CH₂Cl forms 1^o– carbocation, while C₆H₅CHClC₆H₅ forms 2^o– carbocation, which is more stable than 1^o– carbcation. Hence, C₆H₅CHClC₆H₅ is hydrolyzed more easily than C₆H₅CH₂Cl by aqueous KOH.

Question 6.18: p-Dichlorobenzene has higher m.p. and lower solubility than those of o- and m-isomers. Discuss.

Solution 6.18:

p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, it fits more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. As a result, p-dichlorobenzene has a higher melting point and lower solubility than o-and m-isomers.

Question 6.19: How can the following conversions 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

Solution 6.19:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(xiv)

(xv)

(xvi)

(xvii)

(xviii)

(xix)

(xx)

Question 6.20: 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 6.20: In an aqueous solution, KOH almost completely ionizes to give OH− ions. OH− ion is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol.

On the other hand, an alcoholic solution of KOH contains alkoxide (RO−) ion, which is a strong base. Thus, it can abstract a hydrogen from the β-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.

OH− ion is a much weaker base than RO− ion. Also, OH− ion is highly solvated in an aqueous solution and as a result, the basic character of OH− ion decreases. Therefore, it cannot abstract a hydrogen from the β-carbon.

Question 6.21: 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 6.21: There are two primary alkyl halides having the formula, C4H9Br. They are n − bulyl 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, C8H18, which is different from the compound formed when n−butyl bromide reacts with Na metal. Hence, compound (a) must be isobutyl bromide.

Thus, compound (d) is 2, 5−dimethylhexane.

It is given that compound (a) reacts with alcoholic KOH to give compound (b). Hence, compound (b) is 2−methylpropene.

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

Question 6.22: 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) methylchloride is treated with KCN.

Solution 6.22: (i) When n−butyl chloride is treated with alcoholic KOH, the formation of but−l−ene takes place. This reaction is a dehydrohalogenation reaction.

(ii) When bromobenzene is treated with Mg in the presence of dry ether, phenylmagnesium bromide is formed.

(iii) Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm to form phenol.

(iv) When ethyl chloride is treated with aqueous KOH, it undergoes hydrolysis to form ethanol.

(v) When methyl bromide is treated with sodium in the presence of dry ether, ethane is formed. This reaction is known as the Wurtz reaction.

(vi) When methyl chloride is treated with KCN, it undergoes a substitution reaction to give methyl cyanide.