Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes Textbook Questions and Answers, Additional Important Questions, Notes.

BSEB Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Bihar Board Class 12 Chemistry Haloalkanes and Haloarenes Intext Questions and Answers

Question 1.
Write structures of the following compounds:
(i) 2-Chloro-3-methyl pentane
(ii) l-Chloro-4-ethyl cyclohexane
(iii) 4-tert-Butyl-3-iodoheptane
(iv) 1, 4-Dibromobut-2-ene
(v) l-Bromo-4-sec-butyl-2-methylbenzene.
Answer:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 1

Question 2.
Why is sulphuric acid not used during the reaction of alcohols with KI.
Answer:
Sulphuric acid (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 oxidizes it to I2.
(i) KI + H2SO4 → KHSO4 + HI
(ii) H2SO4 → H2O + SO2 + O.
2HI + O → H2O +I2

Question 3.
Write structures of different dihalogen derivatives of propane.
Answer:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 2
(ii)ClCH2CH2CH2Cl
(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:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 3
All the hydrogen atoms of 4 methyl groups are equivalent and replacement of any hydrogen will give the same product.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 4
The equivalent hydrogens are grouped as a, b and c. The replacement of equivalent hydrogens will give the same product. Thus, three isomeric products possible.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 5
Similarly, the equivalent hydrogens are grouped as a, b, c and d. Thus, four isomeric products are possible.

Question 5.
Draw the structures of major moon halo products in each of the following reactions:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 6
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 7
Answer:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 8

Question 6.
Arrange each set of compounds in order of increasing boiling points.
(i) Bromomet-hane, Bromoform, Chloromethane, Dibromo- methane.
(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Answer:
(i) Chloromethane, Bromomethane, Dibromomethane Bromoform. Boiling point increases with increase in molecular mass.
(ii) Isopropylchloride < 1-Chloropropane < 1-Chlorbutane. Isopropyl chloride is a branched-chain compound and hence has lowered. p. than l-Chloropropane.

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by SN2 mechanism? Explain your Answer.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 9
Answer:
(i) CH3CH2CH2CH2Br. Being primary halide, there would not be any steric hindrance.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 10
Secondary alkyl halide would react faster than tertiary halide,
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 11 The presence of methyl group close to the halide group will increase the steric hindrance and decrease the rate of the reaction.

Question 8.
In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 12
Answer:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 13
Tertiary halides react faster than secondary halides because of the greater stability of test-carbocation.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 14
Because of the greater stability of the secondary carbocation than primary.

Question 9.
Identify A, B, C, D, E, R and R’ in the following: –
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 15
Answer:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 16
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 17

Bihar Board Class 12 Chemistry Haloalkanes and Haloarenes Text Book 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)2 CHCH(Cl)CH3,
(ii) CH3CH2CH (CH3)CH(C2H5)Cl
(iii) CH3CH2C (CH3)2CH2I
(iv) (CH3)3 CCH2CH(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:
(i) 2-chloro-3-methylbutane, 2° alkyl halide.
(ii) 3-Chloro-4-methyl hexane, 2° alkyl halide.
(iii) i-lodo-2,2-dimethylbutane, 1 alkyl halide.
(iv) l-Bromo-2,2-dimethyl-l-phenylbutane,2° benzylic halide.
(v) 2-Bromo-3-methylbutane/ 2° alkyl halide.
(vi) t-Bromo-2-ethyl-2-methyl butane, 1° alkyl halide.
(vii) 3-chloro-3-methylpentane, 3° alkyl halide.
(viii) 3-chloro-5-methyl-hex-2-ene, vinylic halide
(ix) 4-Bromo-4-methylpent-2-ene, allylic halide.
(x) l-(4-Chlorophenyl)-2-methylpropane, aryl halide.
(xi) l-chloromethyl-3-(2,2-dimethyl propyl) benzene, 1° benzylic halide.
(xii) l-Bromo-2-(l-methyl propyl) benzene, aryl halide.

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 = ClC6H4I-p
Answer:
(i) 2-Bromo-3-dilorobutane.
(ii) 1-Bromo-l-Chloro-l, 2,2-trifluoroethane.
(iii) l-Bromo-4-chlorobut-2-yne.
(iv) 2-Trichloromethyl-l, 1,1,2,3,3,3-heptachloropropane-
(v) 2,2-Bis (4-Chlorophenyl) butane.
(vi) l-Chloro-l-(4-iodophenyl)-3,3-dimethylbut-l-ene.

Question 3.
Write the structure 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) Perfluorobenene, (vi) 4-tert-Butyl-3-iodoheptane,
(vii) l-Bromo-4- sec-butyl-2-methylbenene,
(viii) 1,4-Dibromobut-2-ene.
Answer:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 18
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 19
(viii) Br-CH2CH=CH-CH2Br.

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 4.
Which one of the following has highest dipole moment?
(a) CH2Cl2,
(b) CHCl3,
(C) CCl4
Answer:
(b) CHCl3.

Question 5.
A hydrocarbon C5H10 does not react with chlorine in dark but gives a single mono chloro compound C5H9 Cl in bright sunlight. Identify the hydrocarbon.
Answer:
C5H10 cannot be an alkene since it does not give addition reaction with Cl2 in dark. Therefore it must be a cyclic hydrocarbon. Moreover, it gives a single mono chloro compound C5H9 Cl with Cl2 in bright sunlight, therefore, it must be symmetrical, i.e., Cyclopentane Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 20

Question 6.
Write the isomers of the compound having formula C4H9Br.
Answer:
(i) CH3-CH2-CH2-CH2Br:1- Bromobutane
2- Bromobutane
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 21
2-Bromobutane shows optical isomerism. One of them is dextro or d-form and the other is lalvo or 1-form.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 22
Non-Superimposible image of the object,
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 23

Question 7.
Write the equations for the preparation of 1-iodobutane from (a) 1-butanol, (b) l-chlorobutanol, (c) but-l-ene.
Answer:
(a) Preparation of 1-iodobutane from 1-butanol
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 24
(b) From 1-chloro butane
CH3 -CH2-CH2,-CH2Cl + KI → CH3-CH2-CH2-CH2I + KCl
(c) From but-l-ene
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 25

Question 8.
What are ambident nucleophiles? Explain with an example.
Answer:
Ambident nucleophiles are the reagents which have two nucleophilic centres. Groups like Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 26 which can attack either from carbon or nitrogen. Actually, cyanide group is a hybrid of two contributing structures Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 27 and can act as a nucleophile in two different ways, i.e., linking through carbon atom resulting in the formation of alkyl cyanides and through nitrogen atom leading to the formation of isocyanide.

Bihar Board Class 12 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)3 CCl or CH3Cl.
Answer:
(i) CH3I
(ii) CH3Cl.

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-l-methylcyclohexane,
(ii) 2-Chloro-2-methylbutane,
(iii) 2,2,3-Trimethyl-3-bromopentane.
Answer:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 28
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 29

Question 11.
How will you bring the following conversions?
(i) Ethanol to but-l-yne,
(ii) Ethane to bromoethene,
(iii) Propene to 1-nitro-propane,
(iv) Toluene to benzyl alcohol,
(v) Propene to propyne,
(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:
(i) Ethanol to but-l-yne
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 30
(ii) Ethane to bromoethene
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 31
(iii) Propene to 1-nitro-propane
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 32
(iv) Toluene to benzyl alcohol
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 33
(v) Propene to propyne
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 34
(vi) Ethanol to ethyl fluoride
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 35
(vii) Bromomethane topropanone
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 36
(viii)But-l-enetobut-2-ene
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 37
(ix) 1-Chlorobutane to n-octane
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 38
(x) Benzene to biphenyl
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 39

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’s reagents should be prepared under anhydrous conditions?
Answer:
(i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 40 because C attached to Cl in Chlorobenzene is sp2 hybridized and so more electronegative than C attached to Cl is cyclohexyl chloride which is sp3 hybridized and so less electronegative. Therefore C of C—Cl bond in cyclohexyl chloride is more willing to release electrons to chlorine and thus is more polar.

(ii) Alkyl halides though polar (2.05-2.15 D) are insoluble in water because they can neither form hydrogen bonds with water nor can they break the hydrogen bonds already existing between water molecules.

(iii) Grignard reagents are highly reactive and react with any source of proton to give hydrocarbons. Even water is sufficiently acidic to convert them to corresponding hydrocarbons.
R-Mg-X + H2O → R-H + Mg (OH) X
Therefore, Grignard reagents should be prepared under only anhydrous conditions like in ether.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 41

Bihar Board Class 12 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-It is widely used as:(i) a refrigerant (cooling agent) in refrigerators and air conditions.
(ii) a propellant in aerosols and foams (i.e, hair sprays, deodorants, shaving creams, cleansers, insecticides etc.)
Uses of DDT-It is a cheap, but powerful insecticide. It is widely used for sugarcane and fodder crops and to kill mosquitoes and other insects. Through its use malaria has virtually been minimised in India. However, its use has been stopped in several advanced countries of the world.

Uses of Carbon tetrachloride-

  • It is used as an industrial solvent for oils, fats, resins, lacquers etc. and also in dry-cleaning. —
  • as a fire extinguisher under the name of Pyrene.
  • in the industrial preparation of chloroform.
  • as a medicine for hookworms.

Uses of iodoform-

  1. It is used as an antiseptic for dressing wounds.
  2. It is used in the preparation of certain pharmaceuticals.

Question 14.
Write the structure of the major organic product in each of the following reactions:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 42
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 43
(v) C6H5ONa+C2H5Cl →
(vi) CH3CH2CH2OH+SOCl2
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 44
(viii) CH3CH = C(CH3)2+HBr →
Answer:
(i) CH3 CH2CH2 I,
(ii) (CH3)2 C = CH2,
(iii) CH3 CH (OH) CH2CH3
(iv) CH3CH2CN,
(v) C6H5OC2H5,
(vi) CH3CH2CH2Cl,
(vii) CH3 CH2 CH2 CH2 CH2 Br,
(viii) CH3 CH2 CBr (CH3)2.

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 15.
Write the mechanism of the following reaction:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 45
Answer:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 46

Question 16.
Arrange the compounds of each set in order of reactivity towards SN2 displacement:
(i) 2-Bromo-2-methyl butane, 1-Bromopentane, 2-Bromopentane
(ii) l-Bromo-3-methylbutane, 2-Bromo-2-methylb utane, 3-Bromo-2-methylbutane
(iii) 1-Bromobutane, l-Bromo-2,2-dimethylpropane l-Bromo-2- methylbutane < l-Bromo-3-methyl butane.
Answer:
(i) 2-Bromo-2-methylbutane < 2-Bromopentane < 1- Bromopentane.
(ii) 2-Bromo-2-methyl butane < 3-Bromo-2-methylbutane < 1-Bromo-3-methyl butane.
(iii) l-Bromo-2,2-dimethylpropane < l-Bromo-2-methylbutane < l-Bromo-3-methyl butane.

Question 18.
p-Dichldrobenzene has higher m.p. andsolubility toll those of o- and m-isomers. Discuss.
Answer:
p-Dichlorobehzene has higher m.p. and solubility than those of o- and m-isomers because p-isomer is more symmetrical and hence its molecular forces of attraction are stroriger and hence the p-isomer melts at a higher temperature and has comparatively higher solubility.

Question 19.
How the following conversions can be carried out?
(i) Propene to propah-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 propanenitrate
(viii) Aniline to chlorobenzene
(ix) 2-chlorobutane to 3,4-dimethyl hexane
(x) 2-Methyl-l-propene to 2-chloro-2-methyl propane
(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-Bfombpropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to Diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenyl isocyanide.
Answer:
(i) Propene to propan-l-ol
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 47
(ii) Ethanol to but-l-yne
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 48
(v) Benzene to 4-bromonitrobenzene
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 50
(vi) Benzyl alcohol to 2-phenylethanoic acid
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 51
(vii) Ethanol to propane nitrile
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 52
(viii) Aniline to chlorobenzene
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 53
(ix) 2-Chlorobutane to 3,4-dimethyl hexane
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 54
(x) 2-Methyl-l-propene to 2-chloro-2-methyl-propane
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 55
(xi) Ethyl chloride to propanoic acid
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 56
(xii) But-l-ene ton-butyl iodide ‘
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 57
(xiii) 2-chloropropane to 1-propanol
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 58
(xiv) Isopropyl alcohol to iodoform
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 59
(xv) Chlorbenzene to p-nitrophenol
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 60
(xvi) 2-Bfombpropane to 1-bromopropane
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 61
(xvii) Chloroethane to butane
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 62
(xviii) Benzene to Diphenyl
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 63
(xix) tert-Butyl bromide to isobutyl bromide
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 64
(xx) Aniline to phenyl isocyanide.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 65

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 aqueous solution KOH is almost completely ionised to give OH ions which being a strong nucleophile brings about a; substitution reaction on alkyl chloride to form alcohols. Further in the aq. solution OH ions are highly solvated (hydrated). This solvation reduces the basic character of OH ions which, therefore, fail to abstract a hydrogen from the β-carbon of the alkyl chloride to form an alkene.

In contrast, an alcoholic solution of KOH contains alkoxide Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 66 ions which being a much stronger base than OH ions preferentially eliminates a molecule of HC1 from an alkyl halide to form alkenes.
R-Cl + KOH (aq) → R-OH + KCl [Substitution]
CH3CH2CH2Cl + KOH (ale.) → CH3CH = CH2 [Elimination].

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 21.
Primary alkyl halide C4H9Br (a) reacted with alcoholic, KOH to give compound (b). Compound (b) reacted with HBr to give (c) which is an isomer of (a). When (a) was reacted with sodium metal it gives a compound (d), C8H18 which is different from the compound I when n-butyl bromide was reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions. s
Answer:
C4H9Br has two primary alkyl halides:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 67
n-Butyl bromide gives a different product (C8H18) on treatment with sodium than that given by isobutyl bromide (C8H18).
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 68
Therefore (a) is isobutyl bromide. It is confirmed from the following reactions:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 69
Evidently (c) is an isomer of (a).
Therefore compound (a) is isobutyl bromide and not n-Butyl 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 (aq.) KOH,
(v) methyl bromide is treated with sodium in the presence of dry ether,
(vi) methyl chloride is treated with KCN?
Answer:
(i)
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 70
(ii) When bromobenzene is treated with Mg in the presence of dry ether, phenyl magnesium bromide (Grignard’s reagent) is formed.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 71
(iii) When chlorobenzene is subject to hydrolysis with NaOH at 300°C and 200 atm. pressure, phenol is formed.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 72
(iv) When ethyl chloride is treated with aq. KOH, ethyl alcohol is formed.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 73
(v) When methyl bromide is treated with sodium in the presence of dry ether, ethane is formed. [Wurtz reaction]
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 74
(vi) When methyl chloride is treated with KCN, ethanenitrile is formed.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 75

Bihar Board Class 12 Chemistry Haloalkanes and Haloarenes Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
Name the alkyl halide which can be used to prepare methane and ethane in single steps.
Answer:
Methyl iodide (CH3I)
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 76

Question 2.
What happens when chlorobenzene is treated with soda mide in liquid ammonia. (W.B. JEE 2001)
Answer:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 77

Question 3.
Iodoform gives precipitate with silver nitrate, but chloroform does not. Explain.
Answer:
C-I bond is much weaker than C-Cl bond. Therefore when CHI, is heated, it gives T ions which give a yellow ppt of Agl on reaction with AgNO3 solution. On the other hand C-Cl bond of CHCl3 does not break on heating to give Cl ions and hence ppt. of AgCl is not formed.

Question 4.
Which alkyl halide has the highest density and why?
Answer:
CH3I. Because of its smallest carbon content and heaviest halogen, i.e., I.

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 5.
Arrange the following in order of their increasing reactivity towards sulphonation with fuming sulphuric acid:
Benzene, toluene, methoxybenzene, chlorobenzene.
Answer:
Chlorobenzene < benzeoe < toluene < methoxy benzene.

Question 6.
Which of the following compounds would show optical isomerism and why?
(i) 1-Bromobutane,
(ii) 2-Bromobutane.
Answer:
2-Bromobutane Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 78 because it contains a chiral carbon.

Question 7.
Arrange the following in order of their increasing reactivity in nucleophilic substitution reactions: CH3F, CH3I, CH3Br, CH3Cl.
Answer:
CH3F < CH3Cl < CH3Br < CH3I.

Question 8.
Under what conditions, 2-methyl propene can be converted into isobutyl bromide by hydrogen bromide?
Answer:
In the presence of peroxides.

Question 9.
Explain why thionyl chloride method is preferred for preparing alkyl chlorides from alcohols?
Answer:
Because the by-products of the reaction i.e,, SO2 and HCl being gassed escape into the atmosphere leaving alkyl chlorides in almost pure state.

Question 10.
Which compounds responds to iodoform test?
Answer:
Compounds containing CH3 CH(OH)- or CH3-C = O groups.

Question 11.
Name an aldehyde which will respond to iodoform test.
Answer:
Acetaldehyde, (CH3-CHO).

Question 12.
Give one chemical test to distinguish between C2H5Br and C6H5Br.
Answer:
Hydrolysis of C2H5Br with KOH followed by acidification with dil. HNO3 and subsequent treatment with AgNO3 gives light yellow, ppt. of AgBr whereas C2H5Br does not give this test.

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 13.
What happens when chlorine is passed through boiling toluene in the presence of sunlight?
Answer:
Benzyl chloride Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 79 is formed.

Question 14.
For a given alkyl group, what is the trend in the b.pts of the following halides RBr, RC1, RF, RI?
Answer:
B. Pts. of RI > RBr > RC1 > RF.

Question 15.
Among the isomeric dichlorobenzenes, p-isomer has the highest melting point. Why?
Answer:
Because of the symmetry of its structures Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 80.

Question 16.
Which of the SN1 and SN2 reactions proceeds with complete stereochemical inversion of configuration and which proceeds with racemisation?
Answer:
SN2 reactions proceed with complete stereochemical inversion of configuration and SN1 reaction proceed with racemisation.

Question 17.
Write a chemical reaction to illustrate Saytzeff’s rule.
Answer:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 81

Question 18.
Arrange the following in order of increasing ease towards nucleophilic substitution. 4-nitrochlorobenzene, Chlorobenzene, 2,4, 6-trinitrochlorobenzene, 2,4-dinitrochlorobenzene.
Answer:
Chlorobenzene < 4-nitrochlorobenzene < 2, 4, dinitrochlorobenzene < 2,4 6-trinitrochlorobenzene.

Question 19.
A hydrocarbon C5H12 gives only one monochlorination product. Identify the hydrocarbon.
Answer:
It must be neopentane (CH3)4C in which all the hydrogen atoms attached to 4 methyl groups are identical.

Question 20.
An alkyl halide having molecular formula C4H9Cl is optically active. What is its structure?
Answer:
2-Chlorobutane. CH3-CHCl-CH2-CH3.

Question 21.
What type of isomerism is shown by 1,2-dichloroethene?
Answer:
Cis-trans or geometrical isomerism.

Question 22.
Out of chloromethane and chlorobenzene which is more reactive towards nucleophilic substitution reactions.
Answer:
Chloromethane (CH3Cl) because it is an alkyl halide.

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 23.
Which is a better nucleophile, a bromide ion or an iodine ion?
Answer:
Iodide ion (I ).

Question 24.
Write the structure of the main product obtained by the action of cone. H2SO4 on 2-methylbutan-l-ol.
Answer:
2-Methylbut-2-ene. The 1° carbocation initially formed img being less stable rearranges to more stable 2° carbocation img (CH3)2 through 1, 2-hydride shift during dehydration to give more stable alkene in keeping with Saytzeff rule.

Question 25.
A hydrocarbon C5H10 does not react with chlorine in dark but gives a single mono chloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.
Answer:
C5H10 is a saturated compound and all hydrogen atoms are same. Therefore C5H10 is a symmetrical molecule having cyclic structure: Cyclopentane.

Question 26.
What is the name of polyhaloalkaneTesponsible for depletion ozone layer?
Answer:
It is Freon-12 (dichlorodifluoro methane (CCl2F2).

Question 27.
Wurtz reaction fails in the case of tert-alkyl halides. Explain.
Answer:
Only 1°, 2° alkyl halides respond to Wurtz reaction to form higher alkanes, while 3° alkyl halides prefer to undergo dehydrohalogenation to form alkenes.

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 28.
Why alkyl halides are insoluble in wafers?
Answer:
Because they do not form hydrogen bonds with water.

Question 29.
What name would you give to the formation of phenol from chlorobenzene reaction?
Answer:
Nucleophilic Substitution reaction.

Question 30.
Which insecticide is used to eradicate malaria?
Answer:
DDT.

Short-Answer Type Questions

Question 1.
p-Dichlorobenzene has higher melting point and lower solubility than those of o- and m-isomers. Explain why?
Answer:
Tire para isomer is more symmetrical and fits closely in the crystal lattice and thus has stronger intermolecular forces of attraction than o- and m- isomers. Since during melting or dissolution, the crystal lattice breaks, therefore larger amount of heat is required to melt or dissolve the Midsomer In other words, its melting point is higher and solubility lower.

Question 2.
Haloarenes are insoluble in water but soluble in benzene.
Answer:
Haloarenes are insoluble in water because neither they can form H-bonds with water nor can they break the H-bonds already existing in water. (Like dissolves like). Due to larger hydrocarbon part in haloarenes, they are soluble in organic solvents like benzene, petroleum ether etc.

Question 3.
Explain why alkyl halides are generally not prepared in the laboratory by free radical halogenation of alkanes?
Answer:
Free-radical halogenation is not a suitable method for laboratory synthesis of alkyl halides because of the following reasons :
(i) It gives a mixture of isomeric mono halogenated products where boiling points are so close that they cannot be easily separated from each other.

(ii) Polyhalogenation may also occur to some extent thereby making the mixture mere complex and hence more difficult to separate.

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 4.
An Alkyl naide, X, of formula C6H13Cl treatment with potassium tert butoxide gives two isomeric alkenes Y and Z (C6H12). Both alkenes on hydrogenation give 2,3 dimethylbutane. Predict the structures of X, Y and Z.
Answer:
Since the two isomeric alkenes Y and Z (C6H12) on catalytical hydrogenation give the same 2, 3 dimethyl butane, therefore, Y and Z must differ in the position of the double bond, i.e.,
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 82
If this is so, then the alkyl halide X (C6H13Cl) must have a H-atom on either side of Cl atom, i.e., X must be 2-chloride-2,3-dimethyl butane
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 83

Question 5.
Optically active 2-iodobutane on treatment with Nal in acetone gives a product which does not show a optical activity. Explain.
Answer:
Optically active 2-iodobutane as treatment with Nal in acetone undergoes racemisation and hence the product does not show optical activity as explained :
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 84
I undergoes Walden Inversion (SN2) to give 2-iodobutane (II) which is the enantiomer of I. No II undergoes Walden Inversion to give enantiomer I. As a result of this two Walden Inversion, a 50:50 mixture of 1 and II is obtained. In other words active 2-iodobutane undergoes racemization.

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 6.
Give the structures of the major organic products from 3- ethyl pent-2-ene under each of the following conditions:
(a) HBr in presence of peroxide
(b) Br2/H2O.
Answer:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 85
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 86

Question 7.
Predict the main product in each of the following reactions:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 87
Answer:
CCl3 is m-directing
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 88

Question 8.
Give the names of the following compounds.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 89
Answer:
(i) 3-Chloro-5-Fluoro-3, 5-dim ethyiheptane.
(ii) 3-Bromo-5-chloro-3, 5-diniethyiheptarie.

Question 9.
Out of the various possible isomer of C7H7Cl containing a benzene ring, suggest the structure with the weakest C—Cl bond.
Answer:
Four isomers are possible
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 90
The C-Cl bond in o, m- and p-chlorotoluene has some double-bond character due to resonance. In contrast, in benzyl chloride, the C- Cl bond is a pure single bond. Hence out of 4 isomers the C-Cl bond in benzyl chloride is the weakest.

Question 10.
Account for the following: Haloalkanes undergo nucleophilic substitution reactions whereas haloarenes undergo electrophilic substitutions.
Answer:
Haloalkanes are more polar than haloarenes. Thus the C atom carrying halogen in haloalkanes is more electrons. Haloalkanes do not possess any benzene ring.

Long Answer Type Questions

Question 1.
(a) Draw the structures of all eight isomers that have the molecular formula C5H11Br. Name each isomer according to IUPAC system and classify them as primary, secondary or tertiary bromide. Point out if anyone is optically active.
(b) Amongst the aromatic compounds having the molecular formula C7H7Cl, how many isomers are possible. Write them structures and IUPAC names.
Answer:
(a) The structures and IUPAC names of all 8 isomers are given.
(i) CH3-CH2-CH2-CH2-CH2Br [1-Bromopentane] (1°)
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 91
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 92
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 93
The 2nd C atom is chiral and will show optical activity.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 98

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 95

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 97
2nd C atom is chiral and this isomer will show optical activity.

(b) Four isomers are possible. Their structures and IUPAC names are given below:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 99
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 100

Question 2.
(a) How will you prepare Bromoethane from (i) ethanol, (ii) ethene. How will you convert it into (i) n-Butane, (ii) ethyl acetate, (iii) ethene, (iv) Diethyl ether
(b) How will you prepare chlorobenzene from (i) Benzene, (ii) benzene diazonium chloride. How will you convert it into (i) Phenol, (ii) Aniline, (iii) Benzonitrile, (iv) o- and p-chlorotoluene.
Answer:
(a)
(i) Bromoethane from ethanol
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 101
(ii) Bromo ethane from ethene:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 102
(i) Conversion of Bromoethane into n-Butane
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 103
(ii) Conversion of Bromoethane into ethyl acetate
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 104
(iii) Conversion of Bromoethane into ethene
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 105
(iv) Conversion of Bromoethane into Diethyl ether

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 106
(b)
(i) chlorobenzene from benzene:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 107
(ii) chlorobenzene from benzene diazonium chloride.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 108
(i) Conversion of chlorobenzene into phenol
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 109
(ii) Conversion of chlorobenzene into aniline
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 110
(iii) Conversion of chiorobenzene into benzonitrile
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 111
(iv) o- and p- Cholorotoluene
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 112

Question 3.
(a) Explain why Grignard reagent cannot be prepared from Br CH2C ≡ CH?
(b) When phenol reacts with phosphorus pentachloride, minor amount of chlorobenzene is formed. What is the major product? Write down its structure.
(c) If carbon has rectangular planar geometry, how many isomers are possible for CH2Cl2, (ii) If it has square planar geometry how many isomers for H2Cl2 are possible.
Answer:
(a) Initially Mg reacts with Br CH2C ≡ CH to produce the corresponding Grignard reagent (I). Since Br CH2C ≡ CH has acidic acetylenic hydrogen, it immediately reacts with the Grignard reagent (I) to produce (II) and propyne.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 113
These two reactions proceed in tandem till the whole Br CH2C ≡ CH is consumed. Therefore, Grignard reagent cannot be prepared from it.

(b) Because of resonance C-OH bond in phenols is much stronger than C-OH bond in alcohols and hence cannot be displaced by Cl. Instead phenol reacts as nucleophile and brings about a nucleophilic hydrolysis gives triphenylphosphate.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 114
(c) If carbon has rectangular planar geometry then CH2Cl2 will have the following three stereoisomers:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 115
And if carbon has square planar geometry, then CH2Cl2 will have the following two stereoisomers:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 116

Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Question 4.
(a) Predict the order of reactivity of the following compounds in SN1 reactions:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 117
(b) Predict the order of reactivity of the four isomeric bromobutanes in SN1and SN2 reactions:

(c) Arrange the following halides in the decreasing order of reactivity. CH3CH2Cl (I), CH2 = CH CHClCH3 (II) and CH3CH2 CH Cl CH3 (III).
Answer:
(a) Order of reactivity of towards SN1 reactions:
The first compound is 2° alkyl halide, while all others are tertiary alkyl halides. Since tertiary alkyl halides are more reactive than 2° alkyl halides in SN1 reactions, therefore, first compound is least reactive. Further reactivity increases in the order : Chlorides < bromides < iodide. Titus the order of reactivity is:
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 118

(b) In SN1 reactions, the order of reactivity depends upon the stability of the intermediate carbocations. Therefore (CH3)3 C Br which gives a 3° to carbocation, i.e., Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 120 is most reactive, CH3 CH2 CH (Br) CH3 which
gives a 2° carbocation, i.e., Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 121 and hence is less reactive than (CH3)3 CBr. Out of the remaining two 1° alkyl halides, the carbocation (CH3)2CH CH2 is more stable than the carbocation
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 122 -group and hence the alkyl bromide (CH3)2 CHCH2Br is more reactive than CH3CH2CH2CH2Br. Thus the overall increasing reactivity of the four isomeric bromobutanes towards SN1 reactions follows the order:
CH3 CH2CH2CH2 Br < (CH3)2 CH CH2 Br < CH3 CH2 CH (Br) CH3 < (CH3)3 CBr
The reactivity in SN2 reactions, however, follows the reverse order. CH3 CH2 CH2 CH2 Br > (CH3)2 CH CH2 Br > CH3 CH2 CH (Br) CH3 > (CH3)3 C Br.
Since the steric hindrance around the electrophilic carbon {i.e. a- carbon) increases in that order.

(c)
(i) In SN1 reactions, carbocations are the intermediates. Obviously more stable the carbocation, more reactive the alkyl halide. Since alkyl halide (II) gives an allylic carbocation which is stabilized by resonance. Therefore the alkyl halide (II) is the most reactive.
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 123

(ii) Out of alkyl halides (I) and (III), (III) gives a more stable 2° carbocation while gives a less stable 1° carbocation, therefore, alkyl halide (III) is more reactive than alkyl halide (I’).
Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 124
Thus from the above discussion, it follows that the overall reactivity 1 reaction follows the sequence II > III > I.