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:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 1
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 2

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:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 3

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:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 4
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 5

PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 4.
Which one of the following has the highest dipole moment?
(i) CH2Cl2
(ii) CHCl3
(iii) CCl4
Answer:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 6
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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 7

PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

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:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 8

Question 7.
Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-l-ene.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 9

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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 10
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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 11

PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

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 SN 2 reactions. Hence, CH3Cl will react at a faster rate than (CH3)3 CCl in a SN2 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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 12

(ii) All p-hydrogens in 2-chloro-2-methylbutane are not equivalent, hence on treatment with C2H5ONa/C2H5OH, it gives two alkenes.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 13

(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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 14

PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

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:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 15
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 16
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 17

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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 18
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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 19
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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 20
Hence, Grignard reagent must be prepared under anhydrous conditions.

PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

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
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 21
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 22
(v) C6H5ONa + C2H6Cl →
(vi) CH3CH2CH2OH + SOCl2
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 23
(viii) CH3CH = C(CH3)2 + HBr →
Answer:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 24

Question 15.
Write the mechanism of the following reaction
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 25
Answer:
The given reaction is
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 26
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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 27

PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 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, l-Bromo-2, 2-dimethylpropane, 1-Bromo -2-methylbutane, 1-Bromo-3-methylbutane.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 28

Question 17.
Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH?
Answer:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 29
In SN1 reaction, reactivity depends upon the stability of carbocations. PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 30 carbocation is more stable as compared to PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 31. 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:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 32
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.

PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

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:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 33
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 34
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 35
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 36
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 37
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 38
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 39

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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 40
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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 41
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.

PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 42
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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 43

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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 44

(ii) When bromobenzene is treated with Mg in the presence of dry ether, phenylmagnesium bromide is formed.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 45

(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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 59

PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 46

(iv) When ethyl chloride is treated with aqueous KOH, it undergoes hydrolysis to form ethanol.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 47

(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.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 48

(vi) When methyl chloride is treated with KCN, it undergoes a substitution reaction to give methyl cyanide.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 49

PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

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:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 50

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

PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

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:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 51
All the hydrogen atoms are equivalent and replacement of any hydrogen will give the same product.

(ii) PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 52
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) PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 53
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:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 54
Answer:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 55

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.

PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism ? Explain your answer.
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 56
Answer:
(i) CH3CH2CH2CH2Br
Being primary halide, there won’t be any steric hindrance.

(ii) PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 57
Being a secondary halide, there will be less crowding around α-carbon than tertiary halide.

(iii) PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 58
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 SN1 reaction ?
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 59
Answer:
(i) PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 60
2-Chloro-2-methylpropane as the tertiary carbocation is more stable than secondary carbocation.

(ii) PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 61
2-Chloroheptane as the secondary carbocation is more stable than primary carbocation.

PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 9.
Identify A, B, C, D, E, R and R’ in the following:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 62
Answer:
PSEB 12th Class Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 63

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