Q.Write structures and IUPAC names of

The key idea is the Hoffmann bromamide degradation, where a primary amide reacts with bromine and a strong base to give a primary amine with one fewer carbon atom.

Step 1: Identify the amide for propanamine.

Propanamine ($\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$) has 3 carbons. The amide must have one more carbon (4 carbons) because the reaction removes the carbonyl carbon. The amide is butanamide.

Step 2: Write the structure and IUPAC name for (i).

Structure: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CONH}_2$

IUPAC name: Butanamide

Step 3: Identify the amine from benzamide. …

The Concept: Hoffmann Bromamide Degradation

This reaction is a classic method to shorten a carbon chain by one while converting an amide into a primary amine. The key insight: the amide’s carbonyl carbon is removed as carbonate ($\ce{Na2CO3}$), so the amine that forms has one less carbon than the starting amide.

The reaction proceeds through a rearrangement — the alkyl group attached to the carbonyl carbon migrates to the nitrogen atom. This means the R group in the amide ($\ce{R-CONH2}$) becomes the R group in the amine ($\ce{R-NH2}$). The carbonyl carbon itself is ejected.

Step-by-step Solution

Part (i): Amide that gives propanamine

1. Identify the target amine.

Propanamine is $\ce{CH3CH2CH2NH2}$. It has 3 carbon atoms.

2. Work backwards using the Hoffmann rule.

Since the amide loses one carbon during the reaction, the starting amide must have 4 carbon atoms. The amide’s general structure is $\ce{R-CONH2}$, where $\ce{R}$ is the group that will become the amine’s alkyl group.

For propanamine, the alkyl group is propyl ($\ce{-CH2CH2CH3}$). So the amide must be $\ce{CH3CH2CH2-CONH2}$.

3. Name the amide.

The parent chain is butane (4 carbons). Replace the -e with -amide: butanamide.

The IUPAC name is butanamide (common name: butyramide).

Part (ii): Amine from Hoffmann degradation of benzamide

1. Identify the starting amide. …

Here is the clear, concept-first solution for the given problem.

Method: Hoffmann Bromamide Degradation

This is a rearrangement reaction where a primary amide ($\text{RCONH}_2$) is converted into a primary amine ($\text{RNH}_2$) with one less carbon atom in the chain. The reaction uses bromine ($\text{Br}_2$) in the presence of a strong base (like $\text{NaOH}$).

Key Concept: The carbonyl carbon ($\text{C=O}$) of the amide is lost as $\text{CO}_2$ gas. Therefore, the amine produced has one carbon fewer than the starting amide.

(i) Amide which gives propanamine

Step 1: Identify the target amine.

• Propanamine is a primary amine with 3 carbon atoms.
• Structure: $\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$

Step 2: Work backwards (retrosynthetic analysis).

• Since Hoffmann degradation removes one carbon, the starting amide must have 4 carbon atoms.
• The amide group ($\text{-CONH}_2$) is at the end of the chain. The alkyl group attached to the carbonyl must be the same as the alkyl group in the amine.
• For propanamine ($\text{C}_3\text{H}_7\text{NH}_2$), the alkyl group is propyl ($\text{C}_3\text{H}_7-$).
• Therefore, the amide is Butanamide.

Step 3: Write the structure and IUPAC name.

• Structure: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CONH}_2$
• IUPAC Name: Butanamide

Reaction check:

(ii) Amine from Hoffmann degradation of benzamide

Step 1: Identify the starting amide.

• Benzamide is an aromatic amide.
• Structure: $\text{C}_6\text{H}_5\text{CONH}_2$ (benzene ring attached to $\text{-CONH}_2$)

Step 2: Apply the Hoffmann degradation rule.

• The amide loses the carbonyl carbon ($\text{C=O}$) as $\text{CO}_2$. …

Here are the common mistakes students make with this specific Hoffmann bromamide reaction question, and how to avoid each.

The Core Concept (Why it matters)

The Hoffmann bromamide reaction is a rearrangement where an amide ($\text{RCONH}_2$) is treated with bromine ($\text{Br}_2$) and a strong base (like $\text{NaOH}$) to give a primary amine with one less carbon atom in the chain.

• Reaction: $\text{RCONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{RNH}_2 + 2\text{NaBr} + \text{Na}_2\text{CO}_3 + 2\text{H}_2\text{O}$
• Key result: The alkyl group ($\text{R}$) attached to the carbonyl ($\text{C=O}$) in the amide becomes the alkyl group attached to the $\text{NH}_2$ in the amine. The carbonyl carbon is lost as $\text{CO}_2$.

Mistake #1: Counting Carbons Incorrectly (The "One Carbon Less" Trap)

The Error:

For part (i), students see "propanamine" (3 carbons) and write the amide as propanamide ($\text{CH}_3\text{CH}_2\text{CONH}_2$). They forget that the product has one less carbon than the starting amide.

Why it happens:

Students memorize the phrase "amide to amine" but ignore the loss of the carbonyl carbon. They think the carbon skeleton stays the same.

How to Avoid:

• Work backwards. The product is propanamine ($\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$). This has 3 carbons.
• The amide must have one more carbon than the amine. So the amide must have 4 carbons.
• The amide is butanamide ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CONH}_2$).

Correct Answer (i):

• Structure: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CONH}_2$
• IUPAC Name: Butanamide

Mistake #2: Confusing the Starting Material with the Product (Benzamide vs. Aniline)

The Error:

For part (ii), students see "benzamide" and think the product is benzylamine ($\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$) or benzamide itself. They forget that the benzene ring is directly attached to the carbonyl in benzamide.

Why it happens:

Students confuse the structure of benzamide ($\text{C}_6\text{H}_5\text{CONH}_2$) with other aromatic amides. They also forget that the Hoffmann reaction removes the carbonyl carbon.

How to Avoid:

• Draw the structure of benzamide: $\text{C}_6\text{H}_5\text{CONH}_2$. The benzene ring is the $\text{R}$ group.
• Apply the reaction: $\text{RCONH}_2 \rightarrow \text{RNH}_2$. So $\text{C}_6\text{H}_5\text{CONH}_2 \rightarrow \text{C}_6\text{H}_5\text{NH}_2$.
• The product is aniline (aminobenzene), not benzylamine.

Correct Answer (ii):

• Structure: $\text{C}_6\text{H}_5\text{NH}_2$
• IUPAC Name: Aniline (or Benzenamine)

Mistake #3: Writing the Wrong IUPAC Name (Common vs. IUPAC)

The Error:

Students write "propanamine" as "propylamine" or "n-propylamine". For aniline, they write "phenylamine" (which is accepted in some contexts but not the strict IUPAC name). …