Question

What is the theoretical yield in a reaction using 0.50g of N-tert-butoxycarbonyl-L-alanine, 0.63g methyl-L-phenylalaninate hydrochlroide, 0.4mL isobutyl chloroformate, and 0.3mL N-methylmorpholine to yield methyl N-tert-butoxycarbonyl-alanyl-phenylalaninate (Boc-Ala-Phe-OMe)?

Answer 1

This reaction basically involves the formation of an amide linkage (peptide bond) between the carboxylic group of N-butoxycarbonyl-L-alanine and the amine group of methyl-L-phemylalalinate hydrochloride (as shown). Chloroformate and N-methylmorpholine are coupling agents that facilitate the reaction and are not incorporated the the product.

We first need to determine the number of moles of each reactant using their respective molecular weights,

Number of moles (n)= weight used/molecular weight

Number of moles of N-butoxycarbonyl-L-alanine= 0.5g/189.21 = .00264

Number of moles of methyl-L-phemylalalinate hydrochloride= 0.63/215.676 = .0029

Since the reaction is a 1:1 reaction, number of moles of product formed= .0026 moles (moles og N-butoxycarbonyl-L-alanine are limiting).

Thus the expected theoretical yield = .0026 x molecular weight of product= .0026 x 350.38 = 0.911 g