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.