Abstract
The ion PhCH2O– undergoes competitive losses of H˙, H2, CH2O, and C6H6 upon collisional activation. The loss of H2 occurs mainly to form (C6H4)–CHO, and ab initio calculations suggest the reaction proceeds by the stepwise mechanism PhCH2O–→[H–(PhCHO)]→(C6H4)–CHO + H2. The losses of CH2O and C6H6 are accompanied (or preceded) by partial phenyl H–benzyl H interchange. The ion Ph(CH2)3O– undergoes many fragmentations including the losses of H2O and, CH2O and loss of H2. The loss of H2 occurs by both 1,2- and 1,3-eliminations. A number of minor fragmentations occur after partial interchange of phenyl hydrogens and hydrogens at position 2. The first example of a specific double proton transfer is noted, viz. Ph(CH2)3O–→ C6H7–+ CH2CH–CHO. Ions Ph(CH2)nO–(n= 2–5) all decompose to produce PhCH2– ions: when n= 3–5 it is proposed that the reactions may involve Smiles intermediates, i.e. reaction (a). [graphic omitted]
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Schedule a callMore From: Journal of the Chemical Society, Perkin Transactions 2
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