Is there anyone out there who would make an educated guess regarding the reaction mechanism, and if this procedure would work for benzaldehydes too, and not just infole-3-aldehydes?They said in the preambule:
In order to accumulate basic knowledge, we chose 4-nitroindole-3-carboxaldehyde9 (5a) as a substrate and tested various trials employing cyanating reagents in the presence of reducing agents, such as Me3SiCl-NaI-KCN-Et3SiH, Me3SiCl-NaI-KCN-NaBH4, Me3SiCN-NaBH4, and so on. During these studies,8 we observed that simple treatment of 5a sequentially with NaBH4, and then with NaCN in MeOH, produced 4-nitroindole-3-acetonitrile.Here, KCN reacts with the Ar-CHO forming the alpha-hydroxy cyanide, which is consequently transformed to the Ar-CHOSi(CH
3)
3-CN & the NaBH
4 hydride reduction generates the Ar-CH
2-CN.
Procedure:
NaBH4 (1.3 mol eq.) was added to a solution of indole-3-carboxaldehyde in MeOH and NH2CHO. After stirring at rt for 1 h, NaCN (10 mol eq.) was added to the reaction mixture and the whole was refluxed on oil bath at 100°C for 5 h with stirringIn this procedure, they wait 1 hour before adding the NaCN, so it is reasonable to expect that further reaction occurs on the Ar-CH
2OH. I speculate that the alcohol reacts with the formed B(OMe)
3 (from NaBH
4 & MeOH) to yield Ar-CH-OB(OMe)
2 and that nucleophilic displacement with cyanide or NH
2CHO generates 7a and 9a respectively. I guess the rather large excess of cyanide is used to favor the formation of 7a (instead of the solvolysis --> 9a).
Now, the question that remains is: how is 8a formed?
If this would be the rationale after the reaction, I don't see why Ar should be restricted to indole.
Comments on this proposed mechanism are appreciated
.
By the way: this is a rather nice & ingenious procedure
.