UPSC Combined Geo-Scientist, Mains · 2026 · Mains Paper III
Q10(c). Reimer–Tiemann reaction; bromination of 4-tert-butylcyclohexene
Question
(i) Predict the major product in the given reaction with possible mechanism. Also mention the fast and slow steps in the mechanism: phenol + CHCl₃ + 3 KOH → ?. (ii) Identify the major product formed in the following reaction and explain the stereochemical outcome: 4-tert-butylcyclohexene + Br₂ → ?.
Answer in brief
(i) Salicylaldehyde (2-hydroxybenzaldehyde) by the Reimer–Tiemann reaction via dichlorocarbene; the slow step is the loss of Cl⁻ from CCl₃⁻ to give CCl₂. (ii) trans-1,2-dibromo-4-tert-butylcyclohexane, formed by anti (diaxial) addition through a bromonium ion, with t-Bu equatorial.
Key points
- HO⁻ + CHCl₃ ⇌ H₂O + CCl₃⁻ (reversible, fast pre-equilibrium); phenol is converted to the phenoxide.
- Slow (rate-determining): α-elimination: CCl₃⁻ → CCl₂ + Cl⁻.
- the electron-rich phenoxide (ortho carbon) attacks the electrophilic CCl₂ to give the dichloromethyl-cyclohexadienone, which tautomerises to the aryl–CHCl₂ phenoxide.
- Hydrolysis by HO⁻ (loss of two Cl⁻, aldehyde hydrate) gives, after acid work-up, salicylaldehyde (ortho product; some para).
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
(i) Reimer–Tiemann (5 marks)
- Fast: HO⁻ + CHCl₃ ⇌ H₂O + CCl₃⁻ (reversible, fast pre-equilibrium); phenol is converted to the phenoxide.
- Slow (rate-determining): α-elimination: CCl₃⁻ → CCl₂ + Cl⁻.
- Fast: the electron-rich phenoxide (ortho carbon) attacks the electrophilic CCl₂ to give the dichloromethyl-cyclohexadienone, which tautomerises to the aryl–CHCl₂ phenoxide.
- Hydrolysis by HO⁻ (loss of two Cl⁻, aldehyde hydrate) gives, after acid work-up, salicylaldehyde (ortho product; some para).
(ii) Bromination (5 marks)
- Br₂ polarises and attacks the C=C to form a cyclic bromonium ion; the faces are not equivalent, because the t-Bu group at C-4 (equatorial, locked) makes one face more hindered.
- Br⁻ opens the bromonium from the side opposite to the bromine (anti addition). In a cyclohexane ring the opening that gives a chair with both new C–Br bonds axial (trans-diaxial) is preferred (Fürst–Plattner rule); the t-Bu group stays equatorial.
- Product: trans-1,2-dibromo-4-tert-butylcyclohexane (racemic), with 1,2-diaxial Br: one Br cis and the other trans to the t-Bu group. It can ring-flip to the diequatorial form only by putting t-Bu axial, which is prevented.
CoversReimer–Tiemann reactionbromination of 4-tert-butylcyclohexene
All parts of Question 10
- 10(a) Structure from IR, ¹H-NMR, ¹³C-NMR and MS10 marks
- 10(b) Norbornenone from cyclopentadiene; hydroboration with 9-BBN10 marks
- 10(c) Reimer–Tiemann reaction; bromination of 4-tert-butylcyclohexene10 marks