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CBSE Class 12 Chapter Test: HALOALKANES & HALOARENES
General Instructions:
- This question paper consists of 16 questions divided into 5 sections: A, B, C, D, and E.
- Section A comprises 6 MCQs and 2 Assertion-Reasoning questions of 1 mark each.
- Section B comprises 2 Very Short Answer questions of 2 marks each.
- Section C comprises 3 Short Answer questions of 3 marks each.
- Section D comprises 1 Case-Based question of 4 marks.
- Section E comprises 2 Long Answer questions of 5 marks each.
Directions for Q7 & Q8: These questions consist of two statements, each printed as Assertion (A) and Reason (R). Select the correct answer from the codes (a), (b), (c), and (d) as given below:
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Reason (R): Branching leads to a more spherical shape, decreasing the surface area and resulting in weaker van der Waals forces of attraction.
Reason (R): The C-X bond in haloarenes acquires partial double bond character due to resonance.
(i) Chirality
(ii) Enantiomers
(i) Alkyl halides, though polar, are immiscible with water.
(ii) Sulphuric acid is not used during the reaction of alcohols with KI.
(iii) Allyl chloride is highly reactive towards the SN1 reaction.
(i) Sandmeyer's Reaction
(ii) Finkelstein Reaction
(iii) Wurtz-Fittig Reaction
(i) CH3-CH=CH2 + HBr → ?
(ii) CH3-CH=CH2 + HBr →(peroxide) ?
(iii) CH3CH2Br + AgCN → ?
Halogen atoms attached to an aromatic ring are highly electronegative, hence they withdraw electrons from the benzene ring through the Inductive effect (-I effect). Because of this, haloarenes are less reactive than benzene towards electrophilic substitution reactions. However, halogens also possess lone pairs of electrons which can be donated to the benzene ring via the Resonance effect (+R effect). This creates a clash of effects. Ultimately, electrophilic substitution in haloarenes happens slowly and requires drastic conditions, but the incoming electrophile is directed to specific positions on the ring.
(a) Is the halogen atom an activating or deactivating group? Give a reason. [1 Mark]
(b) What are the specific positions to which the halogen directs the incoming electrophile? [1 Mark]
(c) Why does the incoming electrophile go to those specific positions despite the halogen being deactivating? Explain using the concept of resonance. [2 Marks]
(a) Which compound in each of the following pairs will react faster in an SN2 reaction and why?
(i) CH3Br or CH3I
(ii) (CH3)3CCl or CH3Cl [2 Marks]
(b) Haloarenes are extremely unreactive towards Nucleophilic Substitution reactions. State any three distinct reasons for this unreactivity. [3 Marks]
(i) Propene to Propan-1-ol
(ii) Ethanol to But-1-yne
(iii) 1-Bromopropane to 2-Bromopropane
(iv) Toluene to Benzyl alcohol
(v) Aniline to Chlorobenzene
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SOLUTIONS & MARKING SCHEME
1. Answer: (b) CH3Cl
Reasoning: Dipole moment depends on both charge and bond distance (μ = q × d). Even though Fluorine is the most electronegative, its extremely small size makes the C-F bond distance very short. The product of charge and distance is maximum for the C-Cl bond, making CH3Cl (1.86 D) higher than CH3F (1.84 D).
2. Answer: (b) Ethanol
Reasoning: Aqueous KOH provides hydroxide ions (OH-) which act as strong nucleophiles. It brings about the nucleophilic substitution of the bromide ion, yielding the corresponding alcohol (Ethanol). (Alcoholic KOH would cause elimination to yield Ethene).
3. Answer: (c) (CH3)3C-Br
Reasoning: SN1 reactions proceed via a carbocation intermediate. The reactivity order depends on the stability of the carbocation formed (3° > 2° > 1°). (CH3)3C-Br forms a highly stable tertiary (3°) carbocation, so it reacts fastest via SN1.
4. Answer: (c) Swarts reaction
Reasoning: The synthesis of alkyl fluorides by heating an alkyl chloride/bromide with metallic fluorides (like AgF, Hg2F2, CoF2) is known as the Swarts reaction.
5. Answer: (b) Saytzeff's rule
Reasoning: Saytzeff's (Zaitsev) rule states that in dehydrohalogenation reactions, the preferred alkene is the one which is more highly substituted (has a greater number of alkyl groups attached to the double-bonded carbons). But-2-ene is a disubstituted alkene, while but-1-ene is monosubstituted.
6. Answer: (c) Enantiomers in equal proportions
Reasoning: A racemic mixture (or racemate) is defined as a mixture containing two enantiomers in equal (50:50) proportions. Because they rotate plane-polarized light in opposite directions by the same magnitude, the net optical rotation is zero.
7. Answer: (a) Both A and R are true and R is the correct explanation of A.
Reasoning: As branching increases in isomeric haloalkanes, the molecule approaches a spherical shape. This reduces the surface area of contact, weakening the intermolecular van der Waals forces, which directly leads to a decrease in the boiling point.
8. Answer: (d) A is false but R is true.
Reasoning: Assertion is totally false; haloarenes are extremely unreactive (less reactive) towards nucleophilic substitution compared to haloalkanes. The Reason is true, as the partial double bond character due to resonance makes the C-X bond very strong and difficult to break, which is exactly why they are unreactive.
9. Solution:
(i) Chirality: The property of an object or molecule to be non-superimposable on its mirror image is called chirality. A chiral molecule lacks a plane of symmetry and is optically active.
(ii) Enantiomers: The stereoisomers that are non-superimposable mirror images of each other are called enantiomers. They have identical physical properties but rotate plane-polarized light in opposite directions.
10. Solution:
Grignard reagents (R-Mg-X) are highly reactive and act as strong bases. If even a trace amount of moisture (water) is present, the Grignard reagent will immediately react with the acidic hydrogen of water to form an alkane, destroying the reagent.
Equation: R-Mg-X + H2O → R-H (Alkane) + Mg(OH)X
11. Solution:
(i) To dissolve in water, energy is needed to break H-bonds between water molecules. Alkyl halides cannot form hydrogen bonds with water. The energy released by weak dipole-dipole attractions between water and the alkyl halide is not enough to break the strong H-bonds in water, making them immiscible.
(ii) Concentrated H2SO4 is a strong oxidizing agent. It oxidizes the HI formed during the reaction into Iodine gas (I2) preventing the substitution of the -OH group by an iodine atom.
(iii) Allyl chloride reacts fast via SN1 because the intermediate allyl carbocation (CH2=CH-CH2+) formed after the cleavage of the C-Cl bond is highly stabilized by resonance.
12. Solution:
(i) Sandmeyer's Reaction:
C6H5N2+Cl- (Benzene diazonium chloride) + Cu2Cl2 / HCl → C6H5-Cl (Chlorobenzene) + N2
(ii) Finkelstein Reaction:
R-X (X = Cl, Br) + NaI →(Dry Acetone) R-I (Alkyl iodide) + NaX↓
(iii) Wurtz-Fittig Reaction:
C6H5-X (Haloarene) + 2Na + R-X (Haloalkane) →(Dry Ether) C6H5-R (Alkylarene) + 2NaX
13. Solution:
(i) CH3-CH=CH2 + HBr → CH3-CH(Br)-CH3 (2-Bromopropane). [Markovnikov addition]
(ii) CH3-CH=CH2 + HBr →(peroxide) CH3-CH2-CH2Br (1-Bromopropane). [Anti-Markovnikov / Kharasch effect]
(iii) CH3CH2Br + AgCN → CH3CH2-NC (Ethyl isocyanide) + AgBr. [AgCN is covalent, forces attack through Nitrogen lone pair].
14. Solution (Case-Based):
(a) The halogen atom is a deactivating group. Because of its high electronegativity, the Inductive (-I) effect is stronger than the Resonance (+R) effect, causing a net withdrawal of electron density from the ring, making it less reactive towards electrophiles.
(b) The halogen directs the incoming electrophile to the ortho and para positions.
(c) Although the overall ring is deactivated due to the dominant -I effect, the +R effect (donation of the halogen's lone pair) specifically increases the electron density at the ortho and para positions relative to the meta position. Resonance structures show a negative charge appearing only at the ortho and para positions, stabilizing the intermediate carbocation when attack happens there.
15. Solution:
(a) (i) CH3I will react faster. Iodine is larger, making the C-I bond longer and weaker than the C-Br bond. Iodide is a better leaving group.
(a) (ii) CH3Cl will react faster. SN2 reactions are extremely sensitive to steric hindrance. CH3Cl has no bulky alkyl groups, allowing easy backside attack. (CH3)3CCl is a 3° halide, heavily hindering the nucleophile.
(b) Three reasons for unreactivity of Haloarenes:
1. Resonance Effect: The lone pair on halogen is in conjugation with the ring, giving the C-X bond partial double bond character. It becomes stronger and harder to cleave.
2. Hybridization: The carbon attached to halogen is sp2 hybridized (more s-character, more electronegative), holding the electron pair more tightly than the sp3 carbon in haloalkanes.
3. Instability of Phenyl Cation: SN1 is ruled out because the phenyl cation is highly unstable and cannot be stabilized by resonance.
16. Solution:
(i) Propene to Propan-1-ol:
CH3-CH=CH2 + HBr(peroxide) → CH3CH2CH2Br
CH3CH2CH2Br + aq. KOH → CH3CH2CH2OH
(ii) Ethanol to But-1-yne:
CH3CH2OH + SOCl2 → CH3CH2Cl + SO2 + HCl
CH3CH2Cl + HC≡C-Na+ (Sodium acetylide) → CH3CH2C≡CH + NaCl
(iii) 1-Bromopropane to 2-Bromopropane:
CH3CH2CH2Br + alc. KOH → CH3-CH=CH2 (Propene) + KBr + H2O
CH3-CH=CH2 + HBr → CH3-CH(Br)-CH3
(iv) Toluene to Benzyl alcohol:
C6H5CH3 + Cl2(hν) → C6H5CH2Cl (Benzyl chloride)
C6H5CH2Cl + aq. KOH → C6H5CH2OH + KCl
(v) Aniline to Chlorobenzene:
C6H5NH2 + NaNO2 + 2HCl (0-5°C) → C6H5N2+Cl- (Diazonium salt)
C6H5N2+Cl- + Cu2Cl2/HCl → C6H5Cl + N2
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