In the Hofmann bromamide degradation reaction, the number of moles of $\mathrm{NaOH}$ and $\mathrm{Br}_{2}$…
In the Hofmann bromamide degradation reaction, the number of moles of $\mathrm{NaOH}$ and $\mathrm{Br}_{2}$ used per mole of amine produced are
Four moles of $\mathrm{NaOH}$ and two moles of $\mathrm{Br}_{2}$
Two moles of $\mathrm{NaOH}$ and two moles of $\mathrm{Br}_{2}$
Four moles of $\mathrm{NaOH}$ and one mole of $\mathrm{Br}_{2}$
One moles of $\mathrm{NaOH}$ and one mole of $\mathrm{Br}_{2}$
Solution
The net reactions is $\mathrm{RCONH}_{2}+\mathrm{Br}_{2}+4 \mathrm{NaOH} ightarrow \mathrm{RNH}_{2}+2 \mathrm{NaBr}+\mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{H}_{2} \mathrm{O}$