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NCERT Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes
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Haloalkanes and Haloarenes
Chapter: 10
| Part – II |
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
Ans: 3-Chloro-2-methylbutane.
(ii) CH3CH2CH(CH3) CH(C2H5)Cl
Ans: 1-Chloro-3-ethyl-2-methylbutane.
(iii) CH3CH2C(CH3)2 CH2I
Ans: 1-Iodo-2, 2-dimethylpropane.
(iv) (CH3)3 CCH2CH(Br)C6H5
Ans: 1-Bromo-3-phenyl-2, 2-dimethylpropane.
(v) CH3CH(CH3)CH(Br)CH3
Ans: 2-Bromo-3-methylbutane.
(vi) CH3C(C2H5)2 CH2Br
Ans: 1-Bromo-2,2-dimethylpropane.
(vii) CH3C(Cl)(C2H5)CH2CH3
Ans: 2-Chloro-3-ethylpentane
(viii) CH3CH = C(Cl)CH2CH(CH3)2
Ans: 3-Chloro-4-methyl-2-pentene.
(ix) CH3CH = CHC(Br)(CH3)2
Ans: 3-Bromo-3-methyl-1-butene.
(x) p-ClC6H4CH2CH(CH3)2
Ans: 1-(4-Chlorophenyl)-2-methylpropane.
(xi) m-ClCH2C6H4CH2C(CH3)3
Ans: 1-(3-Chloromethyl Phenyl)-2,2-dimethylpropane.
(xii) o-Br-C6H4CH(CH3 )CH2CH3
Ans: 1-(2-Bromophenyl)-1-methylpropane.
2. Give the IUPAC names of the following compounds:
(i) CH3CH(Cl)CH(Br)CH3
Ans: 2-Bromo-3-chlorobutane.
(ii) CHF2CBrClF.
Ans: 1-Bromo-1-chloro-2,2-difluoroethane.
(iii) ClCH2C ≡ CCH2Br.
Ans: 1-Bromo-3-chloro-1-butyne.
(iv) (CCl3) 3CCl.
Ans: 2-(Trichloromethyl) – 1,1,1,2,3,3,3-heptafluoropropane.
(v) CH3C(p-ClC6H4 ) 2CH(Br)CH3
Ans: 2-Bromo-3,3-bis (4-chlorophenyl) butane
(vi) (CH3 ) 3CCH = CClC6H4 I-p.
Ans: 4-Iodophenyl – 2 – chloro-3,3-dimethylbut-1-ene
3. Write the structures of the following organic halogen compounds.
(i) 2-Chloro-3-methylpentane.
Ans:
(ii) p-Bromochlorobenzene.
Ans:
(iii) 1-Chloro-4-ethylcyclohexane.
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(iv) 2-(2-Chlorophenyl)-1-iodooctane.
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(v) 2-Bromobutane.
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(vi) 4-tert-Butyl-3-iodoheptane.
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(vii) 1-Bromo-4-sec-butyl-2-methylbenzene.
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(viii) 1, 4-Dibromobut-2-ene.
Ans:
4. Which one of the following has the highest dipole moment?
(i) CH2Cl2
(ii) CHCl3
(iii) CCl4
Ans: CH2Cl2 has highest dipole moment (1.60 D). The dipole moment of CCl4 is zero whereas that of CHCl3 is 1.03 D because third C-Cl bond dipole opposes the resultant of two other C-Cl bond dipoles.
Therefore, the compounds can be arranged in increasing order of their dipole moments as CCl4 < CHCl3 < CH2Cl2
In CHCl₃, the dipole moments of two C-Cl bonds are partially opposed by the dipole moments of one C-H and one C-Cl bond. Since the opposition is partial, CHCl₃ has a small net dipole moment.
CCl₄ is a symmetrical molecule, so the dipole moments of all four C-Cl bonds cancel each other out. As a result, the overall dipole moment of CCl₄ is zero.
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: Higher alkanes generally do not react in the dark, but when exposed to ultraviolet light or high temperatures, they produce a monochloro derivative as the main product. Similarly, pentene undergoes chlorination in the presence of sunlight to give a monohalogen derivative.
6. Write the isomers of the compound having formula C4H9Br.
Ans: There are four distinct isomers of the compound with the formula C₄H₉Br.
(i)
(ii)
(iii)
(iv) CH3CH2CHBrCH3
2-bromobutane
7. Write the equations for the preparation of 1-iodobutane from:
(i) 1-butanol.
Ans:
(ii) 1-chlorobutane.
Ans:
(iii) but-1-ene.
Ans:
8. What are ambident nucleophiles? Explain with an example.
Ans: Nucleophiles that can attack through two different sites are referred to as ambident nucleophiles. For instance, the cyanide ion is an ambident nucleophile. It can attack through either the carbon atom or the nitrogen atom, resulting in the formation of either an alkyl cyanide or an alkyl isocyanide. React through either of these centres. Depending on the reagent and the reaction conditions, the reaction may primarily occur at one of these centres.
Example are C = N: (reagent KCN,NaCN)
9. Which compound in each of the following pairs will react faster in SN2 reaction with –OH?
(i) CH3Br or CH3I.
Ans: CH3 – 1 will react faster because bond dissociation enthalpy of C-I band is less than that of C-Br bond.
(ii) (CH3 ) 3CCl or CH3Cl.
Ans: CH₃Cl will react faster due to less steric hindrance compared to (CH₃)₃CCl.
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.
Ans:
So, two products will be formed.
(ii) 2-Chloro-2-methylbutane.
Ans:
(iii) 2,2,3-Trimethyl-3-bromopentane.
Ans:
11. How will you bring about the following conversions?
(i) Ethanol to but-1-yne.
Ans:
(ii) Ethane to bromoethene.
Ans:
(iii) Propene to 1-nitropropane.
Ans:
(iv) Toluene to benzyl alcohol.
Ans:
(v) Propene to propyne.
Ans:
(vi) Ethanol to ethyl fluoride.
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(vii) Bromomethane to propanone.
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(viii) But-1-ene to but-2-ene.
Ans:
(ix) 1-Chlorobutane to n-octane.
Ans:
(x) Benzene to biphenyl.
Ans:
12. Explain why:
(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
Ans: In chlorobenzene, the C atom of the C-Cl bond is sp2 hybridised and has more s character and electronegativity. In cyclohexyl chloride, the C atom of the C-Cl bond is sp3 hybridised, has less s character and less electronegative. Hence, the C-Cl bond in chlorobenzene is less polar than the C-Cl bond in cyclohexyl chloride.
(ii) alkyl halides, though polar, are immiscible with water?
Ans: All the halogen derivatives of hydrocarbon are polar in nature but they are insoluble in water, because they are unable to form hydrogen bond with water or to break the hydrogen bond already present in water. They are soluble in organic solvents.
(iii) Grignard reagents should be prepared under anhydrous conditions?
Ans: Grignard reagents (R-Mg-X) are highly reactive and can easily decompose in the presence of water, leading to the formation of alkanes. This is why it’s crucial to prepare them under anhydrous conditions to prevent any moisture from interfering. Instead of water, ether is used as a solvent during the preparation of Grignard reagents because it provides a dry environment that keeps the reagent stable and functional.
13. Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
Ans: The uses of various compounds ar given below:
(i) Freon 12
(a) It is used as a refrigerant (cooling agent) in refrigerators and air conditioners.
(b) It is also used as a propellant in aerosols and foams to spray out deodorants, cleaners, hair sprays shaving creams.
(c) It is used in aerosol spray propellants such as body sprays, hair sprays etc.
(ii) DDT(p,p Dichloro diphenyl trichloro ethane)
(a) It is a powerful insecticide. It is highly stable and not easily decomposed.
(b) It is used for killing insects and mosquitoes.
(iii) Carbon Tetrachloride (CCI)
(a) It is used as an antiseptic because of the iodine it releases. However, due to its very unpleasant odour, it has been replaced by more effective antiseptics.
(b) It is effective against mosquitoes and lice.
14. Write the structure of the major organic product in each of the following reactions:
(i)
Ans:
(ii)
Ans:
(iii)
Ans:
(iv)
Ans:
(v) C6H5ONa + C2H5Cl➡
Ans:
(vi) CH3CH2CH2OH + SOCl2➡
Ans:
(vii)
Ans:
(viii) CH3CH = C(CH3)2 + HBr➡
Ans:
15. Write the mechanism of the following reaction:
Ans: The nucleophile, Nu, approaches the carbon atom from the side opposite to the halogen and forms a new covalent bond with the carbon atom, replacing the halogen as the leaving group. 
as the displaced base, X, leaves.
This is because KCN is an ionic compound, K+ [C= N]–. Since carbon carrying a lone pair of electrons is more reactive than nitrogen carrying a lone pair, the transition state is formed by the carbon of the cyanide ion forming band with the carbon of the Thus X-R+:C = N– → X… R… C = N → R-CN + X–
16. Arrange the compounds of each set in order of reactivity towards SN2 displacement:
(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
Ans:
(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane
Ans:
(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.
Ans:
17. Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH.
Ans: Is more easily hydrolysed by aqueous. It involves formation of a secondary carbocation which is stabilised by resonance with two phenyl groups. On the other hand, during the hydrolysis of C₆H₅CH₂Cl, a primary carbocation is generated. This carbocation is stabilised through resonance with only one phenyl group, making it less stable and less readily formed.
18. p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.
Ans: p-dichlorobenzene has higher melting point than those of o- and m- isomers. It is due to the greater symmetry of para-isomer that fits in the crystal better as compared to ortho and meta-isomers.
19. How the following conversions can be carried out?
(i) Propene to propan-1-ol.
Ans:
(ii) Ethanol to but-1-yne.
Ans:
(iii) 1-Bromopropane to 2-bromopropan.
Ans:
(iv) Toluene to benzyl alcohol.
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(v) Benzene to 4-bromonitrobenzene.
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(vi) Benzyl alcohol to 2-phenylethanoic acid.
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(vii) Ethanol to propanenitrile.
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(viii) Aniline to chlorobenzene.
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(ix) 2-Chlorobutane to 3, 4-dimethylhexane.
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(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane.
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(xi) Ethyl chloride to propanoic acid (xii) But-1-ene to n-butyl iodide.
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(xii) But-1-ene to n-butyl iodide.
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(xiii) 2-Chloropropane to 1-propanol.
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(xiv) Isopropyl alcohol to iodoform.
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(xv) Chlorobenzene to p-nitrophenol.
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(xvi) 2-Bromopropane to 1-bromopropane.
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(xvii) Chloroethane to butane.
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(xviii) Benzene to diphenyl.
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(xix) tert-Butyl bromide to isobutyl bromide.
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(xx) Aniline to phenyl isocyanide.
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.
Ans: In the presence of water, KOH fully dissociates into hydroxide ions. These are strong nucleophiles and produce alcohols from alkyl halides. Hydration of OH ions also takes place. So they are not able to abstract H+ from ẞ-C-atom. So alkenes are not formed: In alcoholic medium, solution also contains CHO, ethoxide ions in addition to OH ions. Being a stronger base, they abstract H ion from ẞ-C-atom giving rise to alkene (Dehydrohalogenation).This difference arises due to the varying reactivity of the base in each medium.
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: There are only two primary alkyl bromides that have the molecular formula CₓHᵧBr.
According to the problem when alkyl bromide (A) was treated with sodium, it gave a compound which is not a straight chain of hydrocarbons. Therefore (A) cannot be butyl bromide. It may be isobutyl bromide. The complete reactions are
22. What happens when:
(i) n-butyl chloride is treated with alcoholic KOH.
Ans:
(ii) bromobenzene is treated with Mg in the presence of dry ether.
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(iii) chlorobenzene is subjected to hydrolysis.
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(iv) ethyl chloride is treated with aqueous KOH.
Ans:
(v) methyl bromide is treated with sodium in the presence of dry ether.
Ans:
(vi) methyl chloride is treated with KCN?
Ans:
