### 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}$)
**Answer:** Aniline (primary aromatic amine with one fewer carbon than benzamide).$
Part of OC-08 — Amines & Diazonium Salts
Worked Problems
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