UPSC Civil Services (Mains), Chemistry optional · 2021 · Paper II
Q1(c). 2-Cyclohexylpropan-2-ol with BF₃·OEt₂: products A and B
Question
2-Cyclohexylpropan-2-ol is treated with BF₃·OEt₂ to give A + B. Write the products A and B in the above reaction. Also give the mechanism of their formation. Which one of these is the major product and why? [structure of the alcohol given in the paper]
Answer in brief
A = (propan-2-ylidene)cyclohexane, the major tetrasubstituted (Saytzeff) alkene; B = (prop-1-en-2-yl)cyclohexane; both form by E1 dehydration through the tertiary carbocation.
Key points
- BF₃ (Lewis acid) binds hydroxyl oxygen: OH becomes a good leaving group.
- Ionisation gives a tertiary carbocation: slow step, E1.
- Proton loss: ring β-H gives A; a methyl H gives B.
- A is tetrasubstituted: more hyperconjugation, more stable; B is only disubstituted.
- Statistics favour B (six methyl H, one ring H); A still predominates.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
CoversE1 dehydrationBF3 etherateSaytzeff ruletertiary carbocationLewis acid
All parts of Question 1
- 1(a) Origin of the CN group in RCOO⁻ + BrCN10 marks
- 1(b) Benzyltrimethylammonium salt with sodium amide at low and at high temperature10 marks
- 1(c) 2-Cyclohexylpropan-2-ol with BF₃·OEt₂: products A and B10 marks
- 1(d) Two aldehydes towards a nucleophile in the presence of BF₃·OEt₂10 marks
- 1(e) Two transformations by electrocyclic relays10 marks