Question 3

**Analyzing each product** **P:** 3‑Phenylpropanoic acid undergoes Hell–Volhard–Zelinsky bromination (Br₂/red P, then H₂O) → 2‑bromo‑3‑phenylpropanoic acid, Ph–CH₂–CH(Br)–COOH. Unsaturated carbons: 6 (aromatic) + 1 (C=O) = **7** **Q:** Benzaldehyde + acetaldehyde, aq. NaOH, 293 K → Claisen–Schmidt condensation → cinnamaldehyde, Ph–CH=CH–CHO. Unsaturated carbons: 6 + 2 (C=C) + 1 (C=O) = **9** **R:** Phenylacetylene + NaNH₂ → acetylide; alkylation with allyl bromide → Ph–C≡C–CH₂–CH=CH₂. Hg²⁺/H₃O⁺ hydrates the alkyne (Markovnikov, ketone conjugated with ring) → Ph–CO–CH₂–CH₂–CH=CH₂. Unsaturated carbons: 6 + 1 (C=O) + 2 (C=C) = **9** **S:** 2‑Methylindene: O₃/Zn–H₂O cleaves the ring C=C → 2‑(2‑oxopropyl)benzaldehyde (ring opened, monocyclic). 2 equiv. CH₃MgBr adds to both carbonyls → Ar–CH(OH)CH₃ and Ar–CH₂–C(OH)(CH₃)₂. H⁺/Δ dehydrates both alcohols (toward conjugated/more‑substituted alkenes) → 1‑vinyl‑2‑(2‑methylprop‑1‑enyl)benzene: Ar–CH=CH₂ and Ar–CH=C(CH₃)₂. Unsaturated carbons: 6 + 2 + 2 = **10** S has the highest number of unsaturated carbon atoms.

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