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Discussion

Complete the following reaction by suggesting the appropriate substrates (I & II) :

L-alanine on protection gives I and II, which couple with DCC and HOBt to give Boc-Ala-Val-OMe; L-valine is drawn below the arrow

  • A.Pair I and II: Boc-protected alanine acid with hashed methyl and valine methyl ester with hashed isopropyl
  • B.Pair I and II: Boc-protected alanine acid drawn inverted with hashed methyl and valine methyl ester drawn inverted with bold isopropyl
  • C.Pair I and II: Boc-protected valine acid with bold isopropyl and alanine methyl ester with bold methyl
  • D.Pair I and II: Boc-protected alanine acid with hashed methyl and valine methyl ester with hashed isopropyl, drawn inverted

Answer: A

The dipeptide Boc-Ala-Val-OMe is built from Boc-L-alanine (I) and L-valine methyl ester (II), option A.

To join alanine to valine with DCC/HOBt, the amino group of alanine and the carboxyl group of valine must be blocked: alanine is protected as N-Boc, leaving its COOH free to be activated, and valine as its methyl ester, leaving its NH2 free to attack. The starting amino acids are drawn as the L forms (substituent hashed on a lower carbon, wedged on an upper carbon), and the product keeps the same drawing, so I and II must keep it too. In A both stereocentres match the product and the starting materials.

In B and D the alanine carbon is drawn with the methyl hashed on an upper carbon, which is the mirror image (D-alanine). C carries the residues the wrong way round, Boc-valine and alanine methyl ester, which would give Boc-Val-Ala-OMe.

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