Question #787db

1 Answer
Nov 20, 2016

Answer:

Well, for #"ethane"#, carbon has an oxidation state of #-III#; in ethylamine the ipso carbon also has an oxidation state of #-I#.

Explanation:

For #"ethylamine"#, #H_2NCH_2CH_3# , the methyl carbon again has an oxidation state of #-III#, but since the ipso carbon is bound to nitrogen, an atom with greater electronegativity than carbon, the nitrogen gets the electron, i.e. #H_2N^(-) +""^+CH_2-R#, and so this centre has a formal oxidation state of #-I#:

#-CH_2NH_2rarr H_2C^(+) + ""^(-)NH_2rarr2xxH^(+) rarr C^(-I)#

To assign oxidation number, I break all the element-element bonds, and the bonding electron pair is retained by the most electronegative atom. Remember that this exercise is a formalism, as are all designations of oxidation state.