opsomath

The old-school method is to use iodometric titration.

http://pubs.acs.org/doi/abs/10.1021/ac60071a041?journalCode=ancham&quickLinkVolume=24&quickLinkPage=1843&volume=24

This is for NaBH4, but it can be used for the cyano version as well although the stoichiometry is different.

Otherwise, I would hydrolyze in highly acidic solution and use various cyanide assays which are available.

Something seems very odd here. The starting material is C₁₄H₁₁O₂N, and the "rearrangement" product is C₁₄H₁₁ON. In other words, this reaction represents the loss of O, and is therefore a reduction. You do not have any reducing agents listed, just sulfuric acid: did you leave anything out?

Also, I infer you add nitrosobenzene in the second step. This reaction is simple enough; ethoxide deprotonates the N-oxide, the double bond on the indole adds at the 3-position to N=O, and you get a bond. Proton shuffle to protonate the resulting N-O^- compound, then another equiv. of EtO- takes off the 3-indole proton and eliminates OH^- in an E2 reaction between the 3-indole carbon and the nitrosobenzene nitrogen, and you get a double bond between the two atoms. The net result is the formation of a double bond and water as a byproduct. Very similar to the condensation of an arylamine and a nitrosobenzene to form an azobenzene and water.

Hi, I'm new around here. I'm a chemist, I work for a gov't research lab in the US and teach organic chemistry at a community college.