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 Bihar Board Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

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:

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:

(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:

All the hydrogen atoms of 4 methyl groups are equivalent and replacement of any hydrogen will give the same product.

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:

Answer:

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.

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by SN2 mechanism? Explain your Answer.

 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?

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 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:

(viii) Br-CH2CH=CH-CH2Br.

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 

Question 6.
Write the isomers of the compound having formula C4H9Br.
Answer:
(i) CH3-CH2-CH2-CH2Br:1- Bromobutane
2- Bromobutane

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

Question 8.
What are ambident nucleophiles? Explain with an example.
Answer:
Ambident nucleophiles are the reagents which have two nucleophilic centres. Groups like  which can attack either from carbon or nitrogen. Actually, cyanide group is a hybrid of two contributing structures  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.

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:

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

(vi) Ethanol to ethyl fluoride

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  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.

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:

(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.

Question 15.
Write the mechanism of the following reaction:

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

(v) Benzene to 4-bromonitrobenzene

(ix) 2-Chlorobutane to 3,4-dimethyl hexane

(xiii) 2-chloropropane to 1-propanol

(xvii) Chloroethane to butane

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  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].

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:

n-Butyl bromide gives a different product (C8H18) on treatment with sodium than that given by isobutyl bromide (C8H18).

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)

(ii) When bromobenzene is treated with Mg in the presence of dry ether, phenyl magnesium bromide (Grignard’s reagent) is formed.