Problem 1: Identify the product of Hoffmann bromamide degradation starting from benzamide ($C_{6}H_{5}CONH_{2}$)

Step 1: Identify the starting material.

• Benzamide = $C_{6}H_{5}CONH_{2}$ (7 carbons: phenyl ring + carbonyl carbon)

Step 2: Apply Hoffmann bromamide equation. $\text{C}_6\text{H}_5\text{CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}$

Step 3: Count carbons in product.

• Benzamide: 7 carbons ($C_{6}H_{5}CO$–)
• Aniline product: 6 carbons ($C_{6}H_{5}$–), because the carbonyl carbon (–CO–) is lost as $Na_{2}CO_{3}$
• Product = Aniline (C_{6}H_{5}$$NH_{2})

Answer: Aniline (primary aromatic amine with one fewer carbon than benzamide).