Q6(c). Reductive amination, then intramolecular aza-Michael addition
In the following reaction sequence, reaction of aldehyde A with benzylamine followed by sodium borohydride (NaBH₄) reduction results in formation of compound B which upon heating gives compound C (tertiary amine). Identify A, B and C. Which name reaction is involved in conversion of B to C? [Structure of A given in the paper]
Answer in brief
A = methyl (E)-7-oxohept-2-enoate; B = methyl (E)-7-(benzylamino)hept-2-enoate; C = methyl 2-(1-benzylpiperidin-2-yl)acetate. B → C: intramolecular aza-Michael (conjugate) addition.
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Coversreductive aminationaza-Michael additionconjugate additionpiperidine synthesisNaBH4
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