Chemistry Class 12 NCERT Solutions: Chapter 10 Haloalkanes and Haloarenes Part 10

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Q: 16. Arrange the compounds of each set in order of reactivity towards displacement:

(i)

(ii)

(iii)

Answer:

(i)

Q 16 i 2-Bromo-2-methylbutane,1-Bromopentane,2-Bromopentane

Q 16 I 2-Bromo-2-Methylbutane,1-Bromopentane,2-Bromopentane

Q 16 i 2-Bromo-2-methylbutane,1-Bromopentane,2-Bromopentane

An 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 displacement decreases. Due to the presence of substituents, hindrance to the approaching nucleophile increases in the following order.

Hence, the increasing order of reactivity towards displacement is:

(ii)

Q 16 ii 1-Bromo-3-methylbutane,2-Bromo-2-methylbutane,3-Bromo-2- methylbutane

1-Bromo-3-Methylbutane,2-Bromo-2-Methylbutane

Q 16 ii 1-Bromo-3-methylbutane,2-Bromo-2-methylbutane,3-Bromo-2- methylbutane

Since steric hindrance in alkyl halides increases in the order of , the

increasing order of reactivity towards displacement is .

Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards displacement as:

(iii)

The steric hindrance to the nucleophile in the 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:

Hence, the increasing order of reactivity of the given compounds towards displacement is:

Q: 17. Out of and , which is more easily hydrolysed by aqueous ?

Answer:

Q 17 Benzyl chloride and 1degree carbocation

Q 17 Benzyl Chloride and 1degree Carbocation

Q 17 Benzyl chloride and 1degree carbocation

Q 17 Chlorodiphenylmethane 2degree carbocation

Q 17 Chlorodiphenylmethane 2degree Carbocation

Q 17 Chlorodiphenylmethane 2degree carbocation

Hydrolysis by aqueous proceeds through the formation of carbocation. If carbocation is stable, then the compound is easily hydrolyzed by aqueous . Now, forms carbocation, while forms carbocation, which is more stable than carbocation. Hence, is hydrolyzed more easily than by aqueous

Q: 18. p-Dichlorobenzene has higher m.p. and lower solubility than those of o- and m-isomers. Discuss.

Answer:

Q 18 p-Dichlorobenzene, o-Dichlorobenzene, m-Dichlorobenzene

Q 18 P-Dichlorobenzene, O-Dichlorobenzene, M-Dichlorobenzene

Q 18 p-Dichlorobenzene, o-Dichlorobenzene, m-Dichlorobenzene

p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, it fits more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. As a result, p-dichlorobenzene has a higher melting point and lower solubility than o-and m-isomers.