Class 12 Chemistry Important Chapter 10 Haloalkanes and Haloarenes

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Class 12 Chemistry Important Chapter 10 Haloalkanes and Haloarenes

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Haloalkanes and Haloarenes

Chapter: 10

PART – II

IMPORTANT QUESTION AND ANSWER

1. Why are haloalkanes generally less soluble in water?

Ans: Haloalkanes are less soluble in water because the energy required to break the hydrogen bonds between water molecules is greater than the energy released when new attractions between haloalkanes and water molecules form. Also, haloalkanes are non-polar or weakly polar, so they do not interact well with polar water molecules.

2. Arrange the following alkyl halides in order of increasing boiling point: Bromomethane, Bromoform, Chloromethane, Dibromomethane.

Ans: Chloromethane < Bromomethane < Dibromomethane < Bromoform

This order is due to increasing molecular mass and van der Waals forces.

3. What is the difference between SN1 and SN2 reactions in haloalkanes?

Ans: SN1 is a two-step unimolecular nucleophilic substitution where a carbocation intermediate forms, and rate depends only on the alkyl halide concentration. SN2 is a one-step bimolecular nucleophilic substitution where nucleophile attacks simultaneously as the leaving group leaves, and rate depends on both alkyl halide and nucleophile concentrations.

4. Why are tertiary alkyl halides more reactive in SN1 reactions compared to primary ones?

Ans: Because tertiary carbocations formed in SN1 reactions are more stable due to greater alkyl group stabilization, making the rate-determining step (carbocation formation) faster for tertiary alkyl halides.

5. What is the Wurtz reaction?

Ans: The Wurtz reaction is the coupling of two alkyl halides with sodium metal in dry ether to form a hydrocarbon with double the number of carbon atoms.

6. Why are haloarenes less reactive towards nucleophilic substitution than haloalkanes?

Ans: Due to resonance stabilization of the C–X bond (partial double bond character), sp² hybridization of the carbon, shorter C–X bond length, and difficulty in forming phenyl cation intermediates, haloarenes resist nucleophilic substitution.

7. Explain why chlorobenzene undergoes substitution more slowly than benzene undergoes electrophilic substitution.

Ans: Chlorine withdraws electrons via inductive effect (deactivating the ring) but donates electrons via resonance at ortho- and para- positions, making substitution slower and requiring harsher conditions.

8. How can chlorobenzene be converted into phenol?

Ans. By heating chlorobenzene with aqueous sodium hydroxide at 623 K and 300 atmospheres pressure, the chlorine is replaced by hydroxyl group to form phenol.

9. What is an ambident nucleophile? Give examples.

Ans: An ambident nucleophile can attack through two different atoms. Examples include cyanide ion (CN⁻) which can attack through carbon or nitrogen, and nitrite ion (NO₂⁻) which can attack through oxygen or nitrogen.

10. What are Grignard reagents and why must they be prepared under anhydrous conditions?

Ans: Grignard reagents (RMgX) are organomagnesium compounds formed by reacting haloalkanes with magnesium in dry ether. They must be prepared in anhydrous conditions because they react readily with water or alcohols to give hydrocarbons, destroying the reagent.

11. Explain the electronic configurations of the first row transition elements and discuss the exceptions to the general configuration. Illustrate your answer with examples.

Ans:The first row transition elements include scandium (Sc) to zinc (Zn), with atomic numbers 21 to 30. Their general electronic configuration is 

(n−1)d¹⁻¹⁰ns¹⁻². This means the d orbitals of the penultimate shell and the outermost s orbital are progressively filled.

However, there are exceptions due to the small energy difference between the 

(n−1)d and ns orbitals, and the relative stability of half-filled and fully filled orbitals.

For example, chromium (Cr, Z = 24) has the configuration 3d54s13d, instead of the expected 3d44s2. This is because half-filled d5 orbitals are more stable due to exchange energy.

Similarly, copper (Cu, Z = 29) has the configuration 3d104s1 rather than 3d94s2 because a fully filled d10 set is more stable.

Thus, stability factors cause these deviations in the electron configuration of some transition metals.