Class 12 Chemistry Chapter 6: Haloalkanes and Haloarenes case study question with answers
One case-based question of the kind the board asks: a passage and a table, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
| Tube | Alkyl chloride | Time for precipitate |
|---|---|---|
| 1 | 1-chlorobutane | 4 min |
| 2 | 2-chlorobutane | 30 min |
| 3 | 2-chloro-2-methylpropane | none after 60 min |
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Answer: (b) NaCl
R–Cl + NaI → R–I + NaCl. NaI dissolves in dry acetone but NaCl does not, so NaCl precipitates (Finkelstein).
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Answer: (b) increasing steric hindrance to the nucleophile from 1° to 3°
Iodide in acetone reacts by SN2, a back-side attack; alkyl groups round the carbon hinder it, so primary > secondary > tertiary. Carbocation stability would give the reverse order.
Answer and marking scheme
Sooner. The alkyl group is the same, but bromide is a better leaving group than chloride (R–I > R–Br > R–Cl in both SN1 and SN2), the C–Br bond being longer, with a lower bond enthalpy (293 against 351 kJ mol⁻¹ for C–Cl), so 1-bromobutane reacts faster; NaBr is also insoluble in dry acetone and precipitates.
- sooner (1)
- bromide the better leaving group, lower C–Br bond enthalpy (1)
Answer and marking scheme
Rate = k[CH₃CH₂CH₂CH₂Cl][I⁻], second order (SN2). In tube 3 the three methyl groups block the back-side approach of I⁻, so SN2 does not occur; the tertiary halide reacts fast only by SN1, which needs a polar protic solvent such as water to ionise it, not dry acetone.
- second-order rate law (1)
- steric block to SN2; SN1 needs a polar protic solvent (1)
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