Talk:Hell–Volhard–Zelinsky halogenation

• Carl Magnus von Hell 8 September 1849 - 11. December 1926
• Jacob Volhard 4 June 1834 – 14 January 1910
• Nicolai (Nikolay Dmitrievich) Zelinsky. 6 February 1861 - 31 July 1952 doi:10.1039/JR9540004059 doi:10.1007/BF01178843--Stone (talk) 07:24, 9 April 2008 (UTC)

Mechanism

Question: after the carbonyl oxygen attacks the carbonyl carbon in the final part, the oxygen atom is left with three bonds and an electron pair, making it positive. How then can two of the pi electrons transfer to the adjacent hydroxyl group without leaving the former oxygen positive? There is no positive charge drawn on the original carbonyl oxygen. Please correct me if I am wrong on this. Thanks, JC Shenk (talk) 14:56, 14 March 2009 (UTC)

You only transfere one electron of the oxygen, because the other electron in the pi-bond is owned by the carbon of the C=O so you transfere two electrons from the C=O bond to the C-OH bond, but the carbon does not lose its electron but only puts it into another bond. With this the + at the C=O gets a zero again and the C-O-H gets positive charge.--Stone (talk) 15:27, 14 March 2009 (UTC)

The curly arrows describing the delocalisation of electrons (resonance) in the first intermediate are incorrect. Interconversion of the canonical forms can be represented by movement of the non-charged oxygen's lone pair electrons to form a new c=O with breaking of the existing C=O and neutralisation of the charge on the other carbonyl oxygen. AS written the movment of electons away from the oxonium ion would increase the charge on O to 2+ and break the octet rule for the other oxygen. (130.95.202.225 (talk) 04:24, 22 April 2010 (UTC) MJP, UWA)