Question #76b00

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Question #76b00

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Answer 1 · 2015-10-18T15:19:44

tert-butoxide ( ${(C H_{3})}_{3} C - O^{-}$ $(CH_3)_3C-O^-$ ), sodium ion ( $N a^{+}$ $Na^+$ ) and elemental hydrogen ( $H_{2}$ $H_2$ ).

Explanation:

This is the reduction reaction of a tertiary alcohol by sodium, the proposed mechanism is as depicted in the following scheme.

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The products of this reaction are: tert-butoxide ( ${(C H_{3})}_{3} C - O^{-}$ $(CH_3)_3C-O^-$ ), sodium ion ( $N a^{+}$ $Na^+$ ) and elemental hydrogen ( $H_{2}$ $H_2$ ).

The above scheme shows two possible mechanism for sodium to lose its valence electron to.

The $r e d$ $color(red)(red)$ arrows show that sodium gives its valence electron to hydrogen and the hydrogen loses its electron to oxygen to form hydrogen radical and tert-butoxide.

The $b l u e$ $color(blue)(blue)$ arrows show that sodium gives its valence electron to oxygen and then there is a hemolytic cleavage of the $O - H$ $O-H$ bond to form hydrogen radical and tert-butoxide.

Finally, two hydrogen radicals combine together to form $H_{2}$ $H_2$ ; the elemental hydrogen.

Note that Sodium is oxidized (from $0$ $0$ to $+ 1$ $+1$ ) and hydrogen is reduced (from $+ 1$ $+1$ to $0$ $0$ ).

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Question #76b00

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