In the Hofmann-bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of

Q: Do I need to memorize the balanced equation for this reaction?

While memorizing the balanced equation is helpful for quick recall, understanding the reaction mechanism allows you to logically deduce the stoichiometry without rote memorization.

Q: Why are exactly 4 moles of NaOH required?

In the mechanism, one mole of NaOH deprotonates the amide, a second mole deprotonates the bromoamide intermediate, and two more moles are consumed during the alkaline hydrolysis of the isocyanate intermediate to form the amine and carbonate.

Q: Can KOH be used instead of NaOH?

Yes, KOH can be used interchangeably with NaOH in this reaction. The stoichiometry remains exactly the same: 4 moles of KOH per mole of $\text{Br}_2$.

Q: What happens to the carbonyl carbon of the amide?

The carbonyl carbon is lost as a carbonate ion ($\text{CO}_3^{2-}$), which is why the resulting primary amine has one less carbon atom than the starting amide.

Q: Is this reaction applicable to secondary or tertiary amides?

No, the Hofmann bromamide degradation is strictly applicable only to primary amides ($\text{RCONH}_2$). Secondary and tertiary amides do not have the necessary two protons on the nitrogen atom to proceed through the mechanism.

Question

In the Hofmann-bromamide degradation reaction, the number of moles of NaOH and $	ext{Br}_2$ used per mole of amine produced are

Options:

(A) four moles of NaOH and two moles of $	ext{Br}_2$

(B) two moles of NaOH and two moles of $	ext{Br}_2$

(C) four moles of NaOH and one mole of $	ext{Br}_2$

(D) one mole of NaOH and one mole of $	ext{Br}_2$

Sigma Percentile · Accepted Answer

### The Magic of Hofmann Bromamide Degradation

The Hofmann bromamide degradation is one of the most elegant and frequently tested reactions in organic chemistry. It serves as a powerful synthetic tool to convert a primary amide into a primary amine, effectively "stepping down" the carbon chain by removing the carbonyl carbon as a carbonate ion. But beyond just knowing the reactants and products, mastering the stoichiometry of this reaction is crucial for competitive exams.

### The Master Equation

To answer the question of how many moles of sodium hydroxide and bromine are required, we must look at the overall balanced chemical equation for the process:

$$ 	ext{RCONH}_2 + 4	ext{NaOH} + 	ext{Br}_2 ightarrow 	ext{RNH}_2 + 	ext{Na}_2	ext{CO}_3 + 2	ext{NaBr} + 2	ext{H}_2	ext{O} $$

By simply inspecting this balanced equation, the answer becomes immediately clear. For every **one mole** of the primary amide ($	ext{RCONH}_2$) that is degraded, the reaction consumes exactly **four moles** of sodium hydroxide ($	ext{NaOH}$) and **one mole** of bromine gas ($	ext{Br}_2$).

### Why 4 Moles of NaOH? A Mechanistic Insight

While memorizing the balanced equation is a quick way to get the answer, understanding *why* four moles of base are needed will solidify the concept in your mind forever. Let's break down where each mole of hydroxide goes during the mechanism:

1. **The First Mole:** The reaction begins with the base deprotonating the slightly acidic nitrogen of the primary amide, forming an amide anion which then attacks the bromine molecule to form an N-bromoamide.
2. **The Second Mole:** The N-bromoamide is more acidic than the starting amide. A second mole of base removes the remaining proton on the nitrogen, creating an unstable intermediate that undergoes an alkyl shift to form an alkyl isocyanate ($	ext{R}-	ext{N}=	ext{C}=	ext{O}$).
3. **The Third and Fourth Moles:** The alkyl isocyanate is highly reactive and undergoes alkaline hydrolysis. Two moles of hydroxide are consumed here to attack the carbonyl carbon of the isocyanate, ultimately cleaving the molecule to release the primary amine ($	ext{RNH}_2$) and a carbonate ion ($	ext{CO}_3^{2-}$).

By tracing the journey of the hydroxide ions, the stoichiometry of **4 moles of NaOH** to **1 mole of $	ext{Br}_2$** makes perfect logical sense. This deep understanding ensures you will never have to blindly guess the coefficients again!

Animated Solution for Chemistry - Organic Chemistry: In the Hofmann-bromamide degradation reaction, the number of moles of NaOH and $Br_{2}$ used per mole of amine produced are

Select Answer:

Visualized Solution

$Hofmann Bromamide Degradation$

$Balanced Chemical Equation$

$Moles of NaOH$

$Moles of Br_{2}$

$Final Conclusion$

$The Way Forward$

The Sigma Insight: Amines

The Magic of Hofmann Bromamide Degradation

The Master Equation

Why 4 Moles of NaOH? A Mechanistic Insight

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Animated Solution for Chemistry - Organic Chemistry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the Hofmann-bromamide degradation reaction, the number of moles of NaOH and Br2​ used per mole of amine produced are