# Question #76b00

Oct 18, 2015

tert-butoxide (${\left(C {H}_{3}\right)}_{3} C - {O}^{-}$), sodium ion ($N {a}^{+}$) and elemental hydrogen (${H}_{2}$).

#### Explanation:

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

The products of this reaction are: tert-butoxide (${\left(C {H}_{3}\right)}_{3} C - {O}^{-}$), sodium ion ($N {a}^{+}$) and elemental hydrogen (${H}_{2}$).

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

The $\textcolor{red}{red}$ 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 $\textcolor{b l u e}{b l u e}$ arrows show that sodium gives its valence electron to oxygen and then there is a hemolytic cleavage of the $O - H$ bond to form hydrogen radical and tert-butoxide.

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

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