Chapter 6 Haloalkanes and Halorenes NCERT Solutions Class 12 Chemistry- PDF Download
Page No. 189
6.1. 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)CI
(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
Solution
(i) 2-Chloro-3methylbutane, 2° alkyl halide
(ii) 3-Chloro-4methyl hexane, 2° alkyl halide
(iii) 1-Iodo-2,2-dimethylbutane, 1° alkyl halide
(iv) l-Bromo-3, 3-dimethyl-1-phenylbutane, 2° benzylic halide
(v) 2-Bromo-3-methylbutane, 2° alkyl halide
(vi) 1-Bromo-2-ethyI-2-methylbutane, 1° alkyl halide
(vii) 3-Chloro-3-methylpentane, 3° alkyl halide
(viii) 3-Chloro-5-methylhex-2-ene, vinylic halide
(ix) 4-Bromo-4-methylpent-2-ene, allylic halide
(x) 1-Chloro-4-(2-methylpropyl) benzene, aryl halide
(xi) 1-Chloromethyl-3-(2,2-dimethylpropyl) benzene, 1° benzylic halide.
(xii) 1-Bromo-2-(l-methylpropyl) benzene, aryl halide.
6.2. Give the IUPAC names of the following compounds:
(i) CH3CH(Cl)CH (Br)CH3
(ii) CHF2CBrCIF
(iii) ClCH2C≡CCH2Br
(iv) (CCl3)3CCl
(v) CH3C(p-ClC6H4)2CH(Br)CH3
(vi) (CH3)3CCH = C(Cl)C6H4I-p
Solution
(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-chlorophenyl) butane
(vi) 1-Chloro-1-(4-iodophenyl)-3,3- dimethylbut-1-ene.
6.3. 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
Solution
(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
6.4. Which one of the following has the highest dipole moment?
(i) CH2Cl2
(ii) CHCl3
(iii) CCl4
Solution
1) Dichlormethane (CH2Cl2)
μ = 1.60D
2) Chloroform (CHCl3)
μ = 1.08D
(iii) Carbon tetrachloride (CCl4)
μ = 0D
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.
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, CHCl3 has a small dipole moment of 1.08 D.
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, CH2Cl2 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
6.5. A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.
Solution
The hydrocarbon with molecular formula C5H10 can either a cycloalkane or an alkene.
Since, the compound does not react with Cl2 in the dark, therefore it cannot be an alkene but must be a cycloalkane. Since 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 cycloalkanes must be equivalent. Thus, the cycloalkane is cyclopentane.
6.6. Write the isomers of the compound having formula C4H9Br.
Solution
There are 4 isomers of the compound, C4H9Br which are given below:
(i) 1-Bromobutane
(ii) 2-Bromobutane
(iii) 1-Bromo-2-methylpropane
(iv) 2-Bromo-2-methylpropane
6.7. Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol
(ii) 1-chlorobutane
(iii) But-1-ene
Solution
(i) 1-iodobutane from 1-butanol
(ii) 1-iodobutane from 1-chlorobutane
(iii) 1-iodobutane from But-1-ene
6.8. What are ambident nucleophiles? Explain with an example.
Solution
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.
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.
6.9. 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
Solution
(i) In the SN2 mechanism, the reactivity of halides for the same alkyl group increases in the order. This happens because as the size increases, the halide ion becomes a better leaving group.
R−F << R−Cl < R−Br < R−I
Therefore, CH3I will react faster than CH3Br in SN2 reactions with OH−.
(ii) The SN2 mechanism involves the attack of the nucleophile at the atom bearing the leaving group. But, in case of (CH3)3CCl, the attack of the nucleophile at the carbon atom is hindered because of the presence of bulky substituents on that carbon atom bearing the leaving group. On the other hand, there are no bulky substituents on the carbon atom bearing the leaving group in CH3Cl. Hence, CH3Cl reacts faster than (CH3)3CCl in SN2 reaction with OH−.
6.10. 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-bromopentane.
Solution
(i) 1-Bromo-1-methylcyclohexane
(ii) 2-Chloro-2-methylbutane
In the given compound, there are two different sets of equivalent β-hydrogen atoms labelled as a and b. Thus, dehydrohalogenation of the compound yields two alkenes.
Saytzeff’s rule implies that in dehydrohalogenation reactions, the alkene having a greater number of alkyl groups attached to a doubly bonded carbon atoms is preferably produced.
Therefore, alkene (I) i.e., 2-methylbut-2-ene is the major product in this reaction.
(iii) 2,2,3-Trimethyl-3-bromopentane.
In the given compound, there are two different sets of equivalent β-hydrogen atoms labelled as a and b. Thus, dehydrohalogenation of the compound yields two alkenes.
6.11. 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
Solution
(i) Converting Ethanol to but-1-yne:
(ii) Converting Ethane to bromoethene:
(iii) Converting Propene to 1-nitropropane:
(iv) Converting Toluene to benzyl alcohol:
(v) Converting Propene to propyne:
(vi) Converting Ethanol to ethyl fluoride:
(vii) Converting Bromomethane to propanone:
(viii) Converting But-1-ene to but-2-ene:
(ix) Converting 1-Chlorobutane to n-octane:
(x) Converting Benzene to biphenyl:
6.12. Explain
(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
(ii) why alkyl halides, though polar, are immiscible with water?
(iii) grignard reagents should be prepared under anhydrous conditions?
Solution
(i) Due to sp2 hybridisation of C-atom in chlorobenezens, C-atom is more electronegative (greater s-character), whereas in cyclohexyl chloride, C-atom is in sp3 hybridisation i.e., less electronegative (lesser s-chararter). So, polarity of C—Cl bond in chlorobenzene is less than the C—Cl bond in cyclohexyl chloride. Therefore, the dipole moments of chlorobenzene is lower than that of cyclohexyl chloride.
(ii) To be miscible with water, the solute-water force of attraction must be stronger than the solute-solute and water-water forces of attraction. Alkyl halides are polar molecules and so held together by dipole-dipole interactions. Similarly, strong H-bonds exist between the water molecules. The new force of attraction between the alkyl halides and water molecules is weaker than the alkyl halide-alkyl halide and water-water forces of attraction. Hence, alkyl halides (though polar) are immiscible with water.
Or,
Alkyl halides are polar molecules, therefore, their molecules are held together by dipole-dipole attraction. The molecules of H2O are hold together by H-bonds. Since the new forces of attraction between water and alkyl halide molecules are weaker than the forces of attraction already existing between alkyl halide – alkyl halide molecules and water-water molecules, therefore, alkyl halides are immiscible (not soluble) in water. Alkyl halide are neither able to form H- bonds with water nor are able to break the H-bounding network of water.
(iii) Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes.
Therefore, Grignard reagents should be prepared under anhydrous conditions.
6.13. Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
Solution
Uses of Freon-12
Freon-12 (dichlorodifluoromethane, CF2Cl2) is commonly known as CFC.
- It is used as a refrigerant in refrigerators and air conditioners.
- It is also used in aerosol spray propellants such as body sprays, hair sprays, etc.
However, it damages the ozone layer. Hence, its manufacture was banned in the United States and many other countries in 1994.
Uses of DDT
- DDT (p, p−dichlorodiphenyltrichloroethane) is one of the best known insecticides.
- It is very effective against mosquitoes and lice.
But due its harmful effects, it was banned in the United States in 1973.
Uses of carbontetrachloride (CCl4)
- It is used for manufacturing refrigerants and propellants for aerosol cans.
- It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals.
- 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.
The antiseptic property of iodoform is only due to the liberation of free iodine when it comes in contact with the skin.
6.14. Write the structure of the major organic product in each of the following reactions:
Solution
6.15. Write the mechanism of the following reaction:
Solution
KCN is a resonance hybrid of the following two contributing structures:
Thus, CN– ion is an ambident nucleophile. Therefore, it can attack the “carbon atom of C—Br bond in n-BuBr either through C or N. Since C—C bond is stronger than C—N bond, therefore, attack occurs through C to form n-butyl cyanide.
6.16. 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, 1-Bromo-2, 2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methyl butane.
Solution
(i) An SN2 reaction involves the approaching of the nucleophile to the carbon atom to which the leaving group is attached. When the nucleophile is sterically hindered, then the reactivity towards SN2 displacement decreases.
Due to the presence of substituents, hindrance to the approaching nucleophile increases in the following order.
1-Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane
Hence, the increasing order of reactivity towards SN2 displacement is:
2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane
(ii) Since steric hindrance in alkyl halides increases in the order of 1° < 2° < 3°, the increasing order of reactivity towards SN2 displacement is 3° < 2° < 1°.
Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards SN2displacement as:
2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane
(iii) The steric hindrance to the nucleophile in the SN2 mechanism increases with a decrease in the distance of the substituents from the atom containing the leaving group. Further, the steric hindrance increases with an increase in the number of substituents.
Therefore, the increasing order of steric hindrances in the given compounds is as below:
1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane < 1-Bromo-2, 2-dimethylpropane
Hence, the increasing order of reactivity of the given compounds towards SN2 displacement is:
1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane < 1-Bromobutane
7.17. Out of C6H5CH2Cl and C6H5CHClC6H5 which is more easily hydrolysed by aqueous KOH.
Solution
Hydrolysis by aqueous KOH proceeds through the formation of carbocation. If carbocation is stable, then the compound is easily hydrolyzed by aqueous KOH. Now, C6H5CH2Cl forms 1°-carbocation, while C6H5CHClC6H5 forms 2°-carbocation, which is more stable than 1°-carbocation. Hence, C6H5CHClC6H5 is hydrolyzed more easily than C6H5CH2Cl by aqueous KOH.
7.18. p-dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.
Solution
The three isomers are position isomers which differ in the relative positions of the chlorine atoms in the ring:
As we know, p-isomer is more symmetrical as compared to the other isomers. This means that in the crystal lattice, molecules of the p-isomers are more closely packed as compared to the other isomers. As a result, it has a higher melting point and lower solubility as compared to ortho and meta isomers.
Haloarenes are less polar than haloalkanes and are insoluble in water. This is because of lack of hydrogen bonding. As a result, the attractive forces in haloarenes water system remain less than the attractive forces in H2O molecules which are hydrogen bonded. Haloarenes are soluble in organic solvents of low polarity such as benzene, ether, chloroform, carbon tetrachioride etc.
7.19. 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-Chlropropane 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
Solution
(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-Chlropropane 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
7.20. The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in presence of alcoholic KOH, alkenes are major products. Explain.
Solution
In aqueous medium KOH dissociates completely into OH– ions. These are (OH– ions) are strong nucleophiles which produces alcohols from alkyl halides by substitution reaction.
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 β-carbon of the alkyl chloride and from an alkene by eliminating a molecule of HCl.
OH– ion is a much weaker base than RO– ion. Also, OH– ions 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.
7.21. 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 give 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.
Solution
(i) There are two primary alkyl halides having the molecular formula, C4H9Br.
(ii) Since compound (a) when reacted with Na metal gave a compound (d) with molecular formula C8H18 which was different from die compound obtained when n-butyl bromide was reacted with Na metal, therefore, (a) must be isobutyl bromide and compound (d) must be 2,3-dimethylhexane.
(iii) If compound (a) is isobutyl bromide, than the compound (b) which it gives on treatment with alcoholic KOH must be 2-methyl-1-propane.
(iv) The compound (b) on treatment with HBr gives compound (c) in accordance with Markownikoff rule. Therefore, compound (c) is tert-butyl bromide which is an isomer of compound (a), i.e., isobutyl bromide.
Thus,
(a) is isobutyl bromide,
(b) is 2-methyl-1-propane,
(c) is tert-butylbromide, and
(d) is 2,5-dimethylhexane.
7.22. What happens when a:
(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.
Solution
(i) When n-butyl chloride is treated with alcoholic KOH, the formation of but-1-ene takes place. This reaction is a dehydrohalogenation reaction.
(ii) When bromobenzene is treated with Mg in the presence of dry ether, phenylmagnesium bromide is formed.
(iii) The hydrolysis of chlorobenzene is not possible under normal conditions. In order to subject chlorobenzene for hydrolysis, we need to heat chlorobenzene in an aqueous medium with NaOH solution at temperature 623K and a pressure of 300 atm to form phenol.
(iv) When ethyl chloride is treated with aqueous KOH, it undergoes hydrolysis to form ethanol.
(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.
(vi) When methyl chloride is treated with KCN, it undergoes a substitution reaction to give methyl cyanide.
