Q10(a). Perkin reaction, then three steps to 2-methylindane
Complete the following reaction and identify the products B, C, D and E. Sketch a plausible mechanism for conversion of A → B: A (benzaldehyde) with (CH₃CH₂CO)₂O, CH₃CH₂COONa, Δ gives B; B with Na–Hg gives C; C with (i) SOCl₂, (ii) AlCl₃, Δ gives D; D with NH₂–NH₂/KOH, Δ gives E. [Structure of A given in the paper]
Answer in brief
B = (E)-2-methyl-3-phenylprop-2-enoic acid; C = 2-methyl-3-phenylpropanoic acid; D = 2-methylindan-1-one; E = 2-methylindane. A → B: Perkin reaction.
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CoversPerkin reactionα-methylcinnamic acidintramolecular Friedel–Crafts acylationWolff–Kishner reduction2-methylindane
All parts of Question 10
- 10(a) Perkin reaction, then three steps to 2-methylindane10 marks
- 10(b) 4π ring opening, then 6π ring closure: the intermediate II5 marks
- 10(c) Cope rearrangement of cis-2,3-divinyloxirane5 marks
- 10(d) Structure from mass, IR and NMR spectra: a bromo acid10 marks