In the Hoffmann bromamide degradation reaction, the number of moles of NaOH and $$Br_2$$ used per mole of amine produced are:

In the Hoffmann bromamide degradation, an unsubstituted amide $$RCONH_2$$ is converted into an amine with one carbon atom less. The general reaction in basic medium is

$$RCONH_2 + Br_2 + NaOH \;\rightarrow\; RNH_2 + \dots$$

To find the exact stoichiometry, we write the complete ionic reaction and then balance every atom and charge. Let us begin by recalling that:

1. Each molecule of $$Br_2$$ is reduced to two bromide ions $$2\,Br^-$$.
2. The amide carbonyl carbon finally appears in the carbonate ion $$CO_3^{2-}$$ produced from hydroxide.

First, write the skeletal ionic equation showing the essential species:

$$RCONH_2 + Br_2 + OH^- \;\rightarrow\; RNH_2 + Br^- + CO_3^{2-} + H_2O$$

Now we balance step by step.

Balancing bromine. One mole of $$Br_2$$ gives two bromide ions, so we need the coefficient 2 in front of $$Br^-$$ on the product side:

$$RCONH_2 + Br_2 + OH^- \;\rightarrow\; RNH_2 + 2\,Br^- + CO_3^{2-} + H_2O$$

Balancing nitrogen and the organic part. Nitrogen already balances: one $$NH_2$$ in the amide and one $$NH_2$$ in the amine.

Balancing carbon. The carbonyl carbon of the amide ends up in the carbonate ion. One $$RCONH_2$$ gives one $$CO_3^{2-}$$, so the carbon atoms are balanced.

Balancing oxygen and hydrogen. Count oxygen atoms on each side:

Left side: from one $$OH^-$$ there is 1 oxygen; from $$RCONH_2$$ there is 1 oxygen. Total = 2.

Right side: in $$CO_3^{2-}$$ there are 3 oxygens; in $$H_2O$$ there is 1 oxygen. Total = 4.

We are short by 2 oxygens on the left. Each additional $$OH^-$$ contributes one oxygen. Therefore we add 2 more $$OH^-$$ to the left:

$$RCONH_2 + Br_2 + 3\,OH^- \;\rightarrow\; RNH_2 + 2\,Br^- + CO_3^{2-} + H_2O$$

Now recount oxygens:

Left: 3 oxygens from $$3\,OH^-$$ + 1 oxygen from amide = 4.
Right: 3 in $$CO_3^{2-}$$ + 1 in $$H_2O$$ = 4.

Oxygens balance. Next, balance hydrogen atoms. Hydrogen count:

Left: in $$RCONH_2$$ there are 2 hydrogens; in $$3\,OH^-$$ there are 3 hydrogens. Total = 5.

Right: in $$RNH_2$$ there are 2 hydrogens; in $$H_2O$$ there are 2 hydrogens. Total = 4.

We are short by 1 hydrogen on the right. To supply one more hydrogen (and one more oxygen so that oxygen balance stays intact) we must place another $$H_2O$$ on the right and correspondingly add another $$OH^-$$ on the left. Therefore we write:

$$RCONH_2 + Br_2 + 4\,OH^- \;\rightarrow\; RNH_2 + 2\,Br^- + CO_3^{2-} + 2\,H_2O$$

Now, hydrogen count:

Left: 2 (from amide) + 4 (from $$4\,OH^-$$) = 6.
Right: 2 (in amine) + 4 (in $$2\,H_2O$$) = 6.

Oxygen count:

Left: 4 oxygens in $$4\,OH^-$$ + 1 in amide = 5.
Right: 3 in carbonate + 2 in $$2\,H_2O$$ = 5.

Everything balances, and the charges also balance: left side charge $$= 4(-1) = -4$$; right side charge $$= 2(-1) + (-2) = -4$$.

Thus the balanced ionic equation is confirmed. Converting the ions back to the molecular form using sodium ions as counter-ions, we multiply every ionic species by $$Na^+$$ as required:

$$RCONH_2 + Br_2 + 4\,NaOH \;\rightarrow\; RNH_2 + 2\,NaBr + Na_2CO_3 + 2\,H_2O$$

From this balanced molecular equation we read directly:

Per mole of amine $$RNH_2$$ produced,
moles of $$NaOH$$ required $$= 4$$,
moles of $$Br_2$$ required $$= 1$$.

This matches Option B, which states “Four moles of NaOH and one mole of $$Br_2$$.”

Hence, the correct answer is Option B.