Q6(a). Chlorocarbene from CH₂Cl₂; ring expansion of 3-methylindole
Predict the product(s) A and B in the following reaction and explain the mechanism involved: CH₂Cl₂ with n-BuLi/THF at −78 °C gives [A], which reacts with 3-methylindole to give [B]. [Reaction scheme with the structure of 3-methylindole given in the paper]
Answer in brief
A = chlorocarbene, :CHCl (formed through the carbenoid LiCHCl₂). B = 4-methylquinoline (lepidine), C₁₀H₉N.
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Coverschlorocarbeneα-eliminationCiamician–Dennstedt rearrangementindole ring expansionquinolinecarbenoid
All parts of Question 6
- 6(a) Chlorocarbene from CH₂Cl₂; ring expansion of 3-methylindole4 marks
- 6(b) Ozonolysis of camphene, then Wolff–Kishner reduction4 marks
- 6(c) Reductive amination, then intramolecular aza-Michael addition4 marks
- 6(d) Benzaldehyde and ethane-1,2-dithiol: a 1,3-dithiolane4 marks
- 6(e) Amidomercuration–demercuration of hex-1-ene in acetonitrile4 marks
- 6(f) Dimerisation of cyclopentadiene: endo rule by frontier orbitals4 marks
- 6(g) Two isomeric esters C₇H₁₄O₂ from ¹H NMR4 marks
- 6(h) Mass spectrum of 3-butyn-2-ol: the ion of m/z 554 marks
- 6(i) Acetone: solvent shift of the n → π* band4 marks
- 6(j) Butyrophenone: McLafferty rearrangement and α-cleavage4 marks