Preparing Ethyl Amine Using Bromine and Sodium Hydroxide

The preparation of Ethyl amine from an organic compound using bromine ($\text{Br}_2$) and sodium hydroxide ($\text{NaOH}$) solution is a characteristic transformation achieved through the Hofmann Bromamide Degradation reaction.

Understanding the Hofmann Bromamide Degradation

The Hofmann Bromamide Degradation is a key chemical reaction that converts a primary amide ($\text{R-CONH}_2$) into a primary amine ($\text{R-NH}_2$). A crucial aspect of this reaction is that the product amine has one less carbon atom than the starting amide. The reagents involved are the amide, bromine, and a strong base like sodium hydroxide.

The general equation for the Hofmann Bromamide Degradation is:

\(\text{R-CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{R-NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}\)

We are aiming to prepare Ethyl amine, which has the structure $\text{CH}_3\text{CH}_2\text{NH}_2$. This means the $\text{R}$ group in our target amine is $\text{CH}_3\text{CH}_2\text{-}$. Following the mechanism of the Hofmann degradation, this $\text{R}$ group comes from the starting amide, which must therefore be $\text{CH}_3\text{CH}_2\text{CONH}_2$.

Analyzing Options for Amide Reaction

Let's look at the provided options and determine which one fits the requirement to produce Ethyl amine via the Hofmann Bromamide Degradation:

• Acetamide: This compound has the formula $\text{CH}_3\text{CONH}_2$. If acetamide undergoes Hofmann degradation, the R group is $\text{CH}_3\text{-}$. The product would be methyl amine ($\text{CH}_3\text{NH}_2$), not Ethyl amine.
• Propionamide: This compound has the formula $\text{CH}_3\text{CH}_2\text{CONH}_2$. The R group is $\text{CH}_3\text{CH}_2\text{-}$. When propionamide reacts with bromine and sodium hydroxide in the Hofmann degradation, it yields the amine with this R group attached to $\text{NH}_2$. This product is $\text{CH}_3\text{CH}_2\text{NH}_2$, which is Ethyl amine. The specific reaction is:

\(\text{CH}_3\text{CH}_2\text{CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}\)

• Formamide: This compound is $\text{HCONH}_2$. The R group is $\text{H-}$. The Hofmann degradation of formamide leads to an amine derived from one carbon, which is not Ethyl amine.
• Methyl cyanide: This compound is $\text{CH}_3\text{CN}$. It is a nitrile, not a primary amide ($\text{R-CONH}_2$). The Hofmann Bromamide Degradation is a specific amide reaction. Methyl cyanide does not contain the amide functional group ($\text{-CONH}_2$) necessary for this reaction. Therefore, methyl cyanide will not yield Ethyl amine under these specific bromine and sodium hydroxide conditions designed for amide reactions.

Conclusion: Propionamide Yields Ethyl Amine

To prepare Ethyl amine ($\text{CH}_3\text{CH}_2\text{NH}_2$) using the Hofmann Bromamide Degradation with bromine and sodium hydroxide, the starting amide must have the $\text{CH}_3\text{CH}_2\text{-}$ group attached to the amide functionality. This corresponds to propionamide ($\text{CH}_3\text{CH}_2\text{CONH}_2$). Reacting propionamide with Br and NaOH effectively removes the carbonyl carbon and forms Ethyl amine.