Q6(c). Phenylglycine and mandelic acid from benzaldehyde
How will you prepare the compounds (I) C₆H₅CH(NH₂)COOH and (II) C₆H₅CH(OH)COOH from benzaldehyde using suitable reagents?
Answer in brief
(I) Strecker synthesis: NH₄Cl + NaCN, then acid hydrolysis of the α-aminonitrile. (II) Cyanohydrin (NaCN, H⁺), then acid hydrolysis of the nitrile.
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CoversStrecker synthesiscyanohydrinphenylglycinemandelic acidnitrile hydrolysis
All parts of Question 6
- 6(a) Curtius rearrangement; the isocyanate closes to a cyclic urea4 marks
- 6(b) HBr and a conjugated diene: 1,2- and 1,4-addition4 marks
- 6(c) Phenylglycine and mandelic acid from benzaldehyde4 marks
- 6(d) Coumarin and 3-acetylcoumarin from salicylaldehyde4 marks
- 6(e) Intramolecular oxymercuration–demercuration: a cyclic ether4 marks
- 6(f) trans- and cis-Cyclohexane-1,2-diol from cyclohexanone4 marks
- 6(g) [1,5]-Carbon shift and Cope rearrangement4 marks
- 6(h) Four C₁₀H₁₄ isomers: the one marked by a McLafferty peak4 marks
- 6(i) C₃H₇NO with two N–methyl singlets: N,N-dimethylformamide4 marks
- 6(j) ν(C=C) of methylidenecycloalkanes and ring size4 marks