Indian Forest Service (Mains), Chemistry optional · 2016 · Paper II
Q1(f). Saytzeff and Hofmann eliminations
Question
Write the products formed in the following reactions:
(i) [structure] → (KOH, 4 m, EtOH, Δ) ?(ii) [structure] → (KOH, Δ) ?
Answer in brief
(i) 1-methylcyclopentene (Saytzeff, major); (ii) methylenecyclopentane (Hofmann, major) + Me₃N + H₂O.
Key points
- (i) Br⁻ is a good leaving group: C–Br breaks well in the TS, which has much double-bond character. The more stable, trisubstituted alkene forms faster. A little 1-ethoxy-1-methylcyclopentane (SN1 in EtOH) may also form.
- (ii) -N+Me₃ is a poor leaving group: C–H breaks ahead of C–N, giving an E1cb-like TS with negative charge on the β-carbon. The most acidic, least hindered β-H (CH₃, three H) is removed. The bulky NMe₃⁺ also shields the ring…
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.
CoversSaytzeffHofmannleaving groupgood leaving groupdouble-bond character
All parts of Question 1
- 1(a) Aromatic, antiaromatic or non-aromatic5 marks
- 1(b) Crossover experiments5 marks
- 1(c) Quinuclidine against triethylamine with ethyl iodide5 marks
- 1(d) Order of carbocation stability5 marks
- 1(e) Basicity of pyrrole, triethylamine and pyridine5 marks
- 1(f) Saytzeff and Hofmann eliminations5 marks
- 1(g) 6,6-Dimethylfulvene, phenyllithium and acetaldehyde5 marks
- 1(h) Pinacol rearrangement with ring expansion5 marks