NCERT Solution Chemistry Class 12 Chapter 10 Haloalkanes and Haloarenes
Haloalkanes and haloarenes class 12 notes
NCERT Solution Chemistry Class 12 Chapter 10 Haloalkanes and Haloarenes all questions and answers. Chemistry Class 12 10th Chapter Haloalkanes and Haloarenes exercise solution and experts answer. As one of online learning platforms, we (netex.) are excited to offer the NCERT Solution Chemistry Class 12 Chapter 10. This solution is designed to help students who are looking to brush up on their Chemistry concepts on Chapter 10 Haloalkanes and Haloarenes.
Questions And Answers
(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
Ans- The IUPAC names and its classification of given compounds are given below.
Sr.
No |
Name Of Compound |
Classification |
(i) | 2-Chloro-3-methylbutane | Secondary alkyl halide |
(ii) | 3-Chloro-4-methylhexane | secondary alkyl halide |
(iii) | 1-Iodo-2,2-dimethylbutane | primary alkyl halide |
(iv) | 1-Bromo-3,3-dimethyl-1-phenylbutane | secondary benzyl
halide |
(v) | 2-Bromo-3-methylbutane | (secondary alkyl halide |
(vi) | 1-Bromo-2-ethyl-2-methylbutane | primary alkyl halide |
(vii) | 3-Chloro-3-methylpentane | tertiary alkyl halide |
(viii) | 3-Chloro-5-methylhex-2-ene | vinyl halide |
(ix) | 4-Bromo-4-methylpent-2-ene | allyl halide |
(x) | 1-Chloro-4-(2-methylpropyl) benzene | aryl halide |
(xi) | 1-Chloromethyl-3-(2,2-dimethylpropyl)
benzene |
primary benzyl halide |
(xii) | 1-Bromo-2-(1-methylpropyl) benzene | aryl halide |
(2) Give the IUPAC names of the following compounds:
(i) CH3CH(Cl)CH(Br)CH3
(ii) CHF2CBrClF
(iii) ClCH2C≡CCH 2Br
(iv) (CCl3)3CCl
(v) CH3C(p-ClC6H 4) 2CH(Br)CH3
(vi) (CH3)3CCH=CClC6H4I-p
Ans- The IUPAC names of given compounds are given below in order
(i) 2-Bromo-3-chlorobutane
(ii) 1-Bromo-1-chloro-1,2,2-trifluoroethane
(iii) 1-Bromo-4-chlorobut-2-yne
(iv) 2-(Trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane
(v) 2-Bromo-3,3-bis (4-chlororphenyl) butane
(vi) 1-Chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene
(3) Write the structures of the following organic halogen compounds.
(i) 2-Chloro-3-methylpentane
(ii) p-Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-1-iodooctane
(v) 2-Bromobutane
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2-ene
Ans-
(i)
2-chloro-3-nethylpentane
(4) Which one of the following has the highest dipole moment ?
(i) CH2Cl2
(ii) CHCl3
(iii) CCl4
Ans- CCl4 is symmetrical therefore it has zero dipole moment. dipole moment of CHCl3 is 1.03D. dipole moment of CH2Cl2 is 1.62 D which is a higher than CHCl3. Thus , among the given compounds CH2Cl2 has the highest dipole moment.
(5) A hydrocarbon C5H10 does not react with chlorine in dark but gives a single Monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.
Ans- The general formula of alkene is CnH2n . A hydrocarbon have molecular formula C5H10 . It means the hydrocarbon is may be alkene or cyclloalkane. As per given condition it does not react with chlorine in dark. Therefore this hydrocarbon cannot be alkene. Therefore, it must be cycloalkane. If it gives a single monochloro compound C5H9Cl in bright sunlight. Therefore , the hydrocarbon is cyclopentane. The reaction is as follow :
(6) Write the isomers of the compound having formula C4H9Br.
Ans- There are four possible structural isomers for C4H9Br. They are as follows –
(i) 2-Bromobutane (CH3CH2CHBrCH3)
(ii) 1-Bromo-2-methylpropane ((CH3)2CHCH2CH2Br)
(iii) 1-Bromobutane (CH3CH2CH2CH2Br)
(iv) tert-Butyl bromide or 2-Bromo-2-methylpropane ((CH3)3CBr)
(7) Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-1-ene.
Ans- The preparations of 1-iodobutane from the given organic compounds are as follows –
(8) What are ambident nucleophiles? Explain with an example.
Ans- The nucleophiles that can attack through two different sites is called as ambident nucleophiles. Groups like cyanides and nitrites possess two nucleophilic centres and are called ambident nucleophiles. cyanide group is a hybrid of two structures and hence can act as a nucleophile in two different ways as ,
VC≡N ⇔ :C=NV
linking through carbon atom resulting in alkyl cyanides and through nitrogen atom leading to isocyanides.
(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
Ans- (i) methyl iodide will react faster than methyl bromide in SN2 reaction with –OH ion because Iodide ion is better leaving group than bromide ion.
(ii) methyl chloride is more reactive than tert butyl chloride in SN2 reaction with –OH ion. Because of steric hinderance in tert butyl chloride .
(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-3bromopentane.
Ans- (i)
When 1-Bromo-1-methylcyclohexane is dehydrohalogenated with sodium ethoxide in ethanol, 1-methylcyclohexene is the major product.
(ii) When 2-Chloro-2-methylbutane is dehydrohalogenated with sodium ethoxide in ethanol, Two alkenes are produced i.e. 2-methyl-2-butene and 2-methyl-1-butene according to Saytzeff’s rule , More substituted alkene is the major product as it is more stable. Hence , 2-methyl-2-butene is the major product.
(iii) When 2,2,3-Trimethyl-3bromopentane is dehydrohalogenated with sodium ethoxide in ethanol, Two alkenes are produced i.e. 3,4,4-Trimethylpent-2-ene and 2,2- dimethyl-3-methylenepentane according to Saytzeff’s rule , More substituted alkene is the major product as it is more stable. Hence , 3,4,4-Trimethylpent-2-ene is the major product.
11) How will you bring about the following conversions?
(i.)Ethanol to but-1-yne
(ii.) Ethane to bromoethene
(iii.) Propene to 1-nitropropane
(iv.) Toluene to benzyl alcohol
(v.) Propene to propyne
(vi.) Ethanol to ethyl fluoride
(vii.) Bromomethane to propanone
(viii.) But-1-ene to but-2-ene
(ix.) 1-Chlorobutane to n-octane
(x.) Benzene to biphenyl.
Ans-
(i) Ethanol reqcts woth SOCl2 forms ethyl chloride .
pyridine
CH3 −CH2 −OH +SOCl2 →−−−−→ CH3 −CH2 −Cl + SO2 ↑+ HCl↑
Ethyl alcohol Ethyl chloride
Acetylene reacts with NaNH2 in excess of liq. Ammonia gives sodium acetylide.
(12) Explain why
(1) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
(2) Alkyl halides, though polar, are immiscible with water?
(3) Grignard reagents should be prepared under anhydrous conditions?
Ans- (i) In The Carbon atom of the C-Cl bond is sp3 hybridized in cyclohexyl chloride. the s character is less and hence electronegativity. Therefore C-Cl bond is more polar. On the other hand, The Carbon atom of the C-Cl bond is sp2 hybridized in chlorobenzene . the s character is more and hence electronegativity. Hence, C-Cl bond is less polar. There is partial double bond character to C-Cl bond . therefore partial negative charge decreases. It affects dipole moment of chlorobenzene and its magnitude decreases . therefore chlorobenzene has lower dipole moment than cyclohexyl chloride.
(ii) There are dipole dipole interactions in alkyl halides . intermolecular hydrogen bonds are present in water molecules. The forces already present between CH3-X molecules and water molecules is greater than the intermolecular attractive force between an alkyl halide and water molecules when CH3-X are added to H2O . Hence, alkyl halides are water-immiscible.
(iii) We should prepare Grignard reagents in the anhydrous condition because grignard reagents are highly reactive with moisture.
(13) Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
Ans-Freon 12-
(1) It is used in aerosol spray propellants e.g. body sprays, hair sprays etc.
(2) It is used as a refrigerant in electronic devices e.g. refrigerators and air conditioners.
DDT –
(1) It is used in insecticide.
(2) It is effective against insects like lice , mosquitoes and. Carbon tetrachloride –
(1) It is used in manufacturing refrigerants
(2) It is used as a solvent in pharmaceutical products.
(3) It is used in manufacturing propellants for aerosol cans.
(4) It is used for the synthesis of chlorofluorocarbons.
(5) It was used as a fire extinguisher. Iodoform –
(1) It is used as an antiseptic .
(2) It is used for synthesis of many chemical products .
(14) Write the structure of the major organic product in each of the following reactions:
acetone
(i) CH3CH2CH2Cl + NaI → −−−−−→
heat
Ans-
(15) Write the mechanism of the following reaction: nBuBr + KCN → nBuCN
Ans- When potassium cynide reacts with n-butyl bromide, bromine is removed and substitution of cyanide occurs to yeild butyl cyanide whereas potassium bromide is byproduct . this reaction is SN2 reaction The mechanism is shown as-
CN– ion can attack through both carbon and nitrogen since it is ambient nucleophile . here it attacks with Carbon atoms . CN– ion acts strong nucleophile and attacks on CH2– Br bond due which Br is replaced by CN– ion .
(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,2-Bromo-3methylbutane
(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2 methylbutane, 1- Bromo-3-methylbutane.
Ans- Steric factors is one of the factor that affect the reactivity during SN2 displacements. In case of alkyl halides , Reactivity is inversely proportional to steric hinderance. therefore, the decreasing order of the reactivity of alkyl halides is as follows
Primary > secondary > tertiary
(1) 1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylbutane
(2) 1-Bromo-3-methylbutane > 2-Bromo-3-methylbutane > 2-Bromo- methylbutane
(3) 1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane
> 1-Bromo-2, 2-dimethylpropane
(17) Out of C6H5CH2Cl and C6H5CHClC6H5 , which is more easily hydrolysed by aq. KOH ?
Ans- In case of C6H5CHClC6H5 there is formation of a secondary carbocation . it is stabilized by resonance with two phenyl groups.
In case of C6H5CH2Cl, there is formation of a primary carbocation during hydrolysis . It is stabilized by resonance with only one phenyl group hence it is less stable therefore possibility of formation is less .
Therefore , C6H5CHClC6H5 is more easily hydrolysed by aqueous KOH.
(18) p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss. Ans-
(1) p-Dichlorobenzene has a symmetrical structure.
(2) Its molecules fits closely in the crystal lattice. As a result , intermolecular forces of attraction are stronger .
(3) Energy is required to break its structure is maximum . hence, it melts at higher temperature. This is the reason that p-Dichlorobenzene has higher m.p. than those of o- and m-isomers.
(19) How the following conversions can be carried out?
(i) Propene to propan-1-ol
(ii) Ethanol to but-1-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3, 4-dimethylhexane
(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) Tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide
Ans-
(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. R−Cl+KOH(aq)→R−OH+KCl
Ans- KOH produces -OH ions and -OH ions are strong nucleophiles. Hence, alkyl chlorides to form alcohol. This is substitution reaction .
R−Cl + KOH(aq) → R−OH + KCl
R− CH2− CH2−Cl+KOH(alc)→R−CH=CH2+KCl+H2O
alkoxide ion is a strong base. Alcoholic KOH solution gives alkoxide ion. β hydrogen atom of alkyl chloride removed and forms water molecule and potassium . hence alkene is formed as major product .
(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.
Ans- Given primary alkyl halide having molecular formula C4H9Br . There are two possibilities for compounds i.e. n-butyl bromide and isobutyl bromide. 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. therefore , compound a is isobutyl bromide and compound d is 2,5-dimethylhexane.
(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?
Ans-
(i) n-butyl chloride is treated with alcoholic KOH, gives But-1-ene with KCl and water molecules.
(iii) There is no reaction if chlorobenzene is subjected to hydrolysis.
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