Proteins can be cleaved specifically at the amide bond on the carboxyl side of
methionine residues by reaction with cyanogen bromide,
\(\mathrm{BrC}=\mathrm{N}\).
The reaction occurs in several steps:
(a) The first step is a nucleophilic substitution reaction of the sulfur on
the methionine side chain with BrCN to give a cyanosulfonium ion,
\(\left[\mathrm{R}_{2} \mathrm{SCN}\right]^{+} .\) Show the structure of the
product, and propose a mechanism for the reaction.
(b) The second step is an internal \(S_{N}^{2}\) reaction, with the carbonyl
oxygen of the methionine residue displacing the positively charged sulfur
leaving group and forming a five-membered ring product. Show the structure of
the product and the mechanism of its formation.
(c) The third step is a hydrolysis reaction to split the peptide chain. The
carboxyl group of the former methionine residue is now part of a lactone
(cyclic ester) ring. Show the structure of the lactone product and the
mechanism of its formation.
(d) The final step is a hydrolysis of the lactone to give the product shown.
Show the mechanism of the reaction.