Chapter 10 Haloalkanes and Haloarenes Solutions
Question - 21 : - Explain why
(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
Answer - 21 : -
(ii) alkyl halides, though polar, are immiscible with water?
(iii) Grignard reagents should be prepared under anhydrous conditions?
Answer
In chlorobenzene, the Cl-atom is linked toa sp2 hybridized carbon atom. In cyclohexyl chloride, theCl-atom is linked to a sp3 hybridized carbonatom. Now, sp2 hybridized carbon has more s-character than sp3 hybridizedcarbon atom. Therefore, the former is more electronegative than the latter.Therefore, the density of electrons of C−Cl bond near the Cl-atom is less in chlorobenzenethan in cydohexyl chloride.
Moreover, the −R effect of the benzene ring ofchlorobenzene decreases the electron density of the C−Cl bond near the Cl-atom.As a result, the polarity of the C−Cl bond in chlorobenzene decreases. Hence,the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(ii) To be miscible with water, the solute-water force ofattraction must be stronger than the solute-solute and water-water forces ofattraction. Alkyl halides are polar molecules and so held together bydipole-dipole interactions. Similarly, strong H-bonds exist between the watermolecules. The new force of attraction between the alkyl halides and watermolecules is weaker than the alkyl halide-alkyl halide and water-water forcesof attraction. Hence, alkyl halides (though polar) are immiscible with water.
(iii) Grignard reagents are very reactive. In the presence ofmoisture, they react to give alkanes.
Therefore, Grignard reagents should be prepared underanhydrous conditions.
Question - 22 : - Give the uses of freon 12, DDT, carbon tetrachloride andiodoform.
Answer - 22 : -
Uses of Freon − 12
Freon-12 (dichlorodifluoromethane, CF2Cl2) iscommonly known as CFC. It is used as a refrigerant in refrigerators and airconditioners. It is also used in aerosol spray propellants such as body sprays,hair sprays, etc. However, it damages the ozone layer. Hence, its manufacturewas 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 mosquitoesand lice. But due its harmful effects, it was banned in the United States in1973.
Uses ofcarbontetrachloride (CCl4)
(i) It is used for manufacturing refrigerants andpropellants for aerosol cans.
(ii) It is used as feedstock in the synthesis ofchlorofluorocarbons and other chemicals.
(iii) It is used as a solvent in the manufacture ofpharmaceutical products.
(iv) Until the mid 1960’s, carbon tetrachloride waswidely used as a cleaning fluid, a degreasing agent in industries, a spotreamer in homes, and a fire extinguisher.
Uses of iodoform(CHI3)
Iodoform was used earlier as an antiseptic, but now ithas been replaced by other formulations-containing iodine-due to itsobjectionable smell. The antiseptic property of iodoform is only due to theliberation of free iodine when it comes in contact with the skin.
Question - 23 : - Write the structure of the major organic product in eachof the following reactions:
Answer - 23 : - 
Answer
Question - 24 : - Write the mechanism of the following reaction:
Answer - 24 : - 
Answer
The given reaction is:
The given reaction is an SN2reaction. In this reaction, CN− acts as thenucleophile and attacks the carbon atom to which Br is attached. CN− ionis an ambident nucleophile and can attack through both C and N. In this case,it attacks through the C-atom.
Answer - 25 : -
(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-methylbutane.
Answer
(i)
An SN2 reaction involves theapproaching of the nucleophile to the carbon atom to which the leaving group isattached. When the nucleophile is sterically hindered, then the reactivitytowards SN2 displacement decreases. Due to the presence ofsubstituents, hindrance to the approaching nucleophile increases in thefollowing order.
1-Bromopentane < 2-bromopentane <2-Bromo-2-methylbutane
Hence, the increasing order of reactivitytowards SN2 displacementis:
2-Bromo-2-methylbutane < 2-Bromopentane <1-Bromopentane
(ii)
Since steric hindrance in alkyl halidesincreases in the order of 1° < 2° < 3°, the increasing order ofreactivity towards SN2 displacement is
3° < 2° < 1°.
Hence, the given set of compounds can bearranged in the increasing order of their reactivity towards SN2 displacementas:
2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane <1-Bromo-3-methylbutane
[2-Bromo-3-methylbutane is incorrectly given in NCERT]
(iii)
The steric hindrance to the nucleophile inthe SN2 mechanism increases with a decrease in the distance ofthe substituents from the atom containing the leaving group. Further, thesteric hindrance increases with an increase in the number of substituents.Therefore, the increasing order of steric hindrances in the given compounds isas below:
1-Bromobutane < 1-Bromo-3-methylbutane <1-Bromo-2-methylbutane
< 1-Bromo-2, 2-dimethylpropane
Hence, the increasing order of reactivity ofthe given compounds towards SN2 displacement is:
1-Bromo-2, 2-dimethylpropane <1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane < 1-Bromobutane
Question - 26 : - Out of C6H5CH2Cland C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH?
Answer - 26 : - 
Hydrolysis by aqueous KOHproceeds through the formation of carbocation. If carbocation is stable, thenthe compound is easily hydrolyzed by aqueous KOH. Now,
forms 1°-carbocation, while C6H5CHClC6H5forms 2°-carbocation, which is more stablethan 1°-carbocation. Hence,
is hydrolyzed more easilythan by aqueous KOH.
Question - 27 : - p-Dichlorobenzenehas higher m.p. and lower solubility than those of o– and
m-isomers.Discuss.
Answer - 27 : -
p-Dichlorobenzeneis more symmetrical than o-and m-isomers. For thisreason, it fits more closely than o-and m-isomers inthe crystal lattice. Therefore, more energy is required to break the crystallattice of p-dichlorobenzene. As a result, p-dichlorobenzenehas a higher melting point and lower solubility than o-and m-isomers.
Question - 28 : - How the following conversions can be carried out?
Answer - 28 : -
(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
Answer
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
(xiii)
(xiv)
(xv)
(xvi)
(xvii)
(xviii)
(xix)
(xx)
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 - 29 : -
In an aqueous solution, KOH almostcompletely ionizes to give OH− ions. OH− ionis a strong nucleophile, which leads the alkyl chloride to undergo asubstitution reaction to form alcohol.
On the other hand, an alcoholic solution ofKOH contains alkoxide (RO−) ion, which is a strong base. Thus, it can abstract ahydrogen from the β-carbon of the alkyl chloride and form an alkene byeliminating a molecule of HCl.
OH− ion is a muchweaker base than RO− ion. Also, OH− ion is highlysolvated in an aqueous solution and as a result, the basic character of OH− iondecreases. Therefore, it cannot abstract a hydrogen from the β-carbon.
Question - 30 : - Primary alkyl halide C_4H_9Br (a) reacted with alcoholic KOH to give compound (b).Compound (b) is reacted with HBr to give
Answer - 30 : -
(c) which is an isomer of (a). When (a) is reacted with sodiummetal it gives compound (d), C8H18 whichis different from the compound formed when n-butyl bromide is reacted withsodium. Give the structural formula of (a) and write the equations for all thereactions.
Answer
Therefore, compound (a) is either n−butylbromide or isobutyl bromide.
Now, compound (a) reacts with Na metal togive compound (b) of molecular formula, C8H18, which is different from the compound formedwhen n−butyl bromide reacts with Na metal. Hence, compound (a) mustbe isobutyl bromide.
Thus, compound (d) is 2, 5−dimethylhexane.
It is given that compound (a) reacts with alcoholic KOH togive compound (b). Hence, compound (b) is 2−methylpropene.
Also, compound (b) reacts with HBr to give compound (c)which is an isomer of (a). Hence, compound (c) is 2−bromo−2−methylpropane.
