The complete set of important questions for CBSE class 12th chemistry prepared by our subject experts contains all the explanations and step by step solutions for the chapters and topics in chemistry. The more you practice and prepare well in the concepts learned, the better you will score in your exams. So, download now and start preparing!
Important questions based on NCERT syllabus for Chapter 10 - Haloalkanes and Haloarenes:
Question-1: p-Dichlorobenzene has higher m.p. and lower solubility than those of o-and m-isomers. Discuss.
Solution: The p-isomer being more symmetrical fits closely in the crystal lattice and thus has stronger inter-molecular forces of attraction than o- and m-isomers. Since during melting or dissolution, the crystal lattice breaks, therefore, a large amount of energy is needed to melt or dissolve the p-isomer than the corresponding o-and m-isomers. In other words, the melting point of the p-isomer is higher and its solubility lower than the corresponding o-and m-isomers.
Question-2: 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.
Solution: If aqueous solution, KOH is almost completely ionized to give OH– ions which being a strong nucleophile brings about a substitution reaction on alkyl halides to form alcohols. Further in the aqueous solution, OH– ions are highly solvated (hydrated). This solvation reduces the basic character of OH– ions which, therefore, fails to abstract a hydrogen from the P-carbon of the alkyl chloride to form alkenes. In contrast, an alcoholic solution of KOH contains alkoxide (RO–) ion which being a much stronger base than OH– ions perferentially eliminates a molecule of HCl from an alkyl chloride to form alkenes.
Question-3: Arrange each set of compounds in order of increasing boiling points.
(i)Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(ii)1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Solution: (i) Chloromethane < Bromomethane <
Dibromomethane < Bromoform
The reason is:
(a)for same alkyl group, B.Pt increases with size of halogen atom.
(b)B.Pt increases as number of halogen atoms increase.
(ii)Isopropyl chloride < 1 – Chloropropane < 1 – Chlorobutane
(a)For same halogen, B.Pt. increases as size of alkyl group increases.
(b)B.Pt. decreases as branching increases.