PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 10 Haloalkanes and Haloarenes Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes

PSEB 12th Class Chemistry Guide Haloalkanes and Haloarenes InText Questions and Answers

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)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
Answer:

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＝CClC₆H₄I-p
Answer:

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)-l-iodooctane
(v) 2 -Bromobutane
(vi) 4-terf-Butyl-3-iodoheptane
(vii) l-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2 -ene
Answer:

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

1. 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.

2. 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, CHC13 has a small dipole moment of 1.08 D.

3. 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, CH2C12 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 5.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Answer:
CO The hydrocarbon with molecular formula C₅H₁₀ can be either a cycloalkane or an alkene.

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

Question 6.
Write the isomers of the compound having formula C₄H₉Br.
Answer:
There are four isomers of the compound having the formula C₄H₉Br.
These isomers are given below:

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

Question 8.
What are ambident nucleophiles? Explain with an example.
Answer:
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 9.
Which compound in each of the following pairs will react faster in Sn₂ reaction with OH^–?
(i) CH₃Br or CH₃I
(ii) (CH₃)₃CCl or CH₃Cl
Answer:
(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 S_N 2 reactions. Hence, CH₃Cl will react at a faster rate than (CH₃)₃ CCl in a S_N2 reaction with OH^– ion.

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-hromopentane.
Answer:
(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 p-hydrogens in 2-chloro-2-methylbutane are not equivalent, hence on treatment with C₂H₅ONa/C₂H₅OH, it gives two alkenes.

(iii) 2, 2, 3-Trimethyl-3-bromopentane has two different sets of p-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.

Question 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 propynt
(vi) Ethanol to ethyl fluoride
(vii) Bromomethane to propanone
(viii) But-l-ene to but-2-ene
(ix) 1-Chlorobutane to n-octane
(x) Benzene to biphenyl.
Answer:

Question 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 ?
Answer:
(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 an 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 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—Cl bond in chlorobenzene acquires some double character while the C—Cl bond in cyclohexyl chloride is a pure single bond. Thus, C—Cl bond in chlorobenzene is shorter than in 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 negative 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 to give hydrocarbons.

Hence, Grignard reagent must be prepared under anhydrous conditions.

Question 13.
Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
Answer:
Uses of Freon-12(CCl₂F₂)

1. It is used as a refrigerant in refrigerators and air conditioners.
2. It is also used in aerosol spray propellants such as body sprays, hair sprays.

Uses of DDT (p, p’-dichlorodiphenyltrichloroethane)

1. It is very effective against mosquitoes and lice.
2. It is also used in many countries as insecticide for sugarcane and fodder crops. (But due to its harmful effects, its use has been banned in many contries including U.S.A.

Uses of Carbontetrachloride (CCl₄)

1. It is used for manufacturing refrigerants and propellants for aerosol cAnswer:
2. It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals.
3. It is used as a solvent in the manufacture of pharmaceutical products. Until the mid 1960’s, carbon tetrachloride was widely used as a cleaning fluid, a degreasing agent in industries, a spot reamer in homes, and a fire extinguisher.

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. The antiseptic property of iodoform is only due to the liberation of free iodine when it comes in contact with the skin.

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


(v) C₆H₅ONa + C₂H₆Cl →
(vi) CH₃CH₂CH₂OH + SOCl₂ →

(viii) CH₃CH = C(CH₃)₂ + HBr →
Answer:

Question 15.
Write the mechanism of the following reaction

Answer:
The given reaction is

The given reaction is an S_N2 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 16.
Arrange the compounds of each set in order of reactivity towards S_N2 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, l-Bromo-2, 2-dimethylpropane, 1-Bromo -2-methylbutane, 1-Bromo-3-methylbutane.
Answer:

Question 17.
Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolysed by aqueous KOH?
Answer:

In S_N1 reaction, reactivity depends upon the stability of carbocations. carbocation is more stable as compared to . Therefore, C₆H₅CHClC₆H₅ gets hydrolysed more easily than C₆H₅CHCl.

Question 18.
p-Dichlorobenzene has higher m.p. and lower solubility than those of o- and m-isomers. Discuss.
Answer:

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 19.
How the following conversions can be carried out?
(i) Propene to propan-l-ol
(ii) Ethanol to but-l-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-l-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
Answer:

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.
Answer:
In an aqueous solution, KOH almost completely ionises 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 p 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 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.
Answer:
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, C₈H₁₈ which is different from the compound formed when n-butyl bromide reacts with Na metal. Hence, compound (A) must be isobutyl bromide.

Question 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.
Answer:
(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.

Chemistry Guide for Class 12 PSEB Haloalkanes and Haloarenes Textbook Questions and Answers

Question 1.
Write structures of the following compounds :
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert-butyl-3-iodoheptane
(iv) 1-4-Dibromobut-2-ene
(v) 1-Bromo-4-sec-butyl-2-methylbenzene
Answer:

Question 2.
Why is sulphuric acid not used during the reaction of alcohols with KI?
Answer:
H₂SO₄ cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding HI and then oxidises it to I₂.

Question 3.
Write structures of different dihalogen derivatives of propane.
Answer:
(i) ClCH₂CH₂CH₂Cl
(ii) ClCH₂CHClCH₃
(iii) Cl₂CHCH₂CH₃
(iv) CH₃CCl₂CH₃

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
Answer:

All the hydrogen atoms are equivalent and replacement of any hydrogen will give the same product.

(ii)
The equivalent hydrogens are grouped as a, b and c. The replacement of equivalent hydrogens will give the same product. Thus, three isomeric products are possible.

(iii)
The equivalent hydrogens are grouped as a, b, c and d. Thus, four isomeric products are possible.

Question 5.
Draw the structures of major monohalo products in each of the following reactions:

Answer:

Question 6.
Arrange each set of compounds in the order of increasing boiling points.
(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(ii) 1-Chloropropane, Isopropyl chloride, 1 -Chlorobutane.
Answer:
(i) Chloromethane < Bromomethane < Dibromomethane < Bromoform. Boiling point increases with increase in molecular mass.

(ii) Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane. Isopropyl chloride being branched has lower boiling point than 1-Chloropropane.

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

Answer:
(i) CH₃CH₂CH₂CH₂Br
Being primary halide, there won’t be any steric hindrance.

(ii)
Being a secondary halide, there will be less crowding around α-carbon than tertiary halide.

(iii)
The presence of methyl group closer to the halide group will increase the steric hindrance and decrease the rate.

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

Answer:
(i)
2-Chloro-2-methylpropane as the tertiary carbocation is more stable than secondary carbocation.

(ii)
2-Chloroheptane as the secondary carbocation is more stable than primary carbocation.

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

Answer: