Question

Question #787db

Answer 1

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.