Animated Solution for Chemistry - Organic Compounds Containing Nitrogen: Which of the following reaction does not involve Hoffmann bromamide degradation?

Hoffmann Bromamide Degradation

The Hoffmann bromamide degradation is a cornerstone reaction in organic chemistry, renowned for its ability to step down a carbon chain. When a primary amide ($R-CON{H}_2$) is treated with bromine ($B{r}_2$) and a strong base like sodium hydroxide ($NaOH$), it undergoes a fascinating rearrangement. The carbonyl carbon is expelled as a carbonate ion ($C{O}_3^{2-}$), and the alkyl or aryl group migrates directly to the nitrogen atom.

The result? A primary amine ($R-N{H}_2$) that boasts exactly one carbon atom less than its parent amide. This unique feature makes it an invaluable tool for synthetic chemists looking to shorten carbon skeletons.

Let's evaluate the given options to see which ones utilize this powerful degradation.

Option (a) presents us with 2-phenylacetamide ($PhC{H}_{2}CON{H}_2$). This is a classic primary amide. When exposed to $B{r}_2$ and $NaOH$, it undergoes the textbook Hoffmann bromamide degradation, smoothly converting into benzylamine ($PhC{H}_{2}N{H}_2$).

Option (b) takes a slightly longer route. It begins with cyanobenzene ($PhCN$). The first step involves alkaline hydrolysis ($KOH,H_{2}O$), which elegantly transforms the nitrile group into an amide, yielding benzamide ($PhCON{H}_2$). The subsequent step introduces $B{r}_2$ and $NaOH$, triggering the Hoffmann bromamide degradation to produce aniline ($PhN{H}_2$).

Option (d) follows a similar logic. Starting with benzoyl chloride ($PhCOCl$), a nucleophilic substitution with ammonia ($N{H}_3$) quickly forms benzamide. Just like in option (b), the addition of $B{r}_2$ and $NaOH$ initiates the Hoffmann bromamide degradation, once again yielding aniline.

Option (c)

Now, we arrive at the outlier: Option (c). The starting material here is 1-phenylpropan-2-one ($PhC{H}_{2}COC{H}_3$), which is a methyl ketone.

When a methyl ketone encounters $B{r}_2$ and $NaOH$, it does not undergo Hoffmann degradation—because it's not an amide! Instead, these reagents trigger the famous Haloform reaction. The methyl group is exhaustively halogenated and cleaved off as bromoform ($CHB{r}_3$), leaving behind a carboxylic acid salt ($PhC{H}_{2}CO{O}^{-}N{a}^{+}$).

Heating this salt with ammonia ($N{H}_3$) converts it into an amide ($PhC{H}_{2}CON{H}_2$). Finally, the introduction of lithium aluminium hydride ($LiAl{H}_4$)—a robust reducing agent—simply reduces the carbonyl group of the amide to a methylene group ($C{H}_2$).

The final product is 2-phenylethanamine ($PhC{H}_{2}C{H}_{2}N{H}_2$). Notice that throughout this entire sequence, no carbon atoms were lost from the main chain via degradation.

By carefully tracing the reaction mechanisms, it becomes abundantly clear that options (a), (b), and (d) all rely on the Hoffmann bromamide degradation as a critical step. Option (c), however, utilizes the Haloform reaction followed by a standard reduction. Therefore, option (c) is the correct answer.