Identify the reactions which give aniline as the major product.
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- Hoffmann Bromamide Degradation (\(\text{RCONH}_2 \xrightarrow{\text{X}_2/\text{OH}^-} \text{RNH}_2\)) cuts out the \(\text{C}=\text{O}\) group entirely.
- \(\text{NaBH}_4\) cannot reduce aromatic \(-\text{NO}_2\) groups; you need active metals in acid (\(\text{Fe/HCl}, \text{Sn/HCl}\)) or catalytic hydrogenation (\(\text{H}_2/\text{Pd}\)) to do that.
Step 1: Know what aniline needs. Aniline is \(C_6H_5NH_2\): an \(-NH_2\) group joined directly to the benzene ring. The correct route must put nitrogen straight on the ring, not on a side chain. Step 2: Test reaction A. Benzonitrile \(C_6H_5CN\) reduced by \(LiAlH_4\) gives benzylamine \(C_6H_5CH_2NH_2\). The \(-NH_2\) ends up on a side carbon, not on the ring, so this is not aniline. Step 3: Test reaction B. A typical side-chain or wrong-position route in option B again fails to give the amino group directly on the ring, so it does not give aniline as the major product. Step 4: Test reaction C. Reducing nitrobenzene \(C_6H_5NO_2\) (for example with Sn/HCl or \(H_2\) with a catalyst) places \(-NH_2\) right on the ring: \[ C_6H_5NO_2 \rightarrow C_6H_5NH_2 \] This gives aniline as the major product. Step 5: Test reaction D. Reaction D does not deliver the ring-bound \(-NH_2\) cleanly, so aniline is not its major product. Step 6: Choose the option. Only reaction C gives aniline, which is option 3 (FigC). \[ \boxed{\text{Reaction C (reduction of nitrobenzene) gives aniline}} \]
