Question #787db

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Question #787db

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Answer 1 · 2016-11-20T08:13:54

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

Explanation:

For $ethylamine$ $"ethylamine"$ , $H_{2} N C H_{2} C H_{3}$ $H_2NCH_2CH_3$ , the methyl carbon again has an oxidation state of $- I I I$ $-III$ , but since the ipso carbon is bound to nitrogen, an atom with greater electronegativity than carbon, the nitrogen gets the electron, i.e. $H_{2} N^{-} +^{+} C H_{2} - R$ $H_2N^(-) +""^+CH_2-R$ , and so this centre has a formal oxidation state of $- I$ $-I$ :

$- C H_{2} N H_{2} \to H_{2} C^{+} +^{-} N H_{2} \to 2 \times H^{+} \to C^{- I}$ $-CH_2NH_2rarr H_2C^(+) + ""^(-)NH_2rarr2xxH^(+) rarr C^(-I)$

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.

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Question #787db

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