Let's go through both problems step by step: --- ## **Part 1: Compound A** ### **a. Condensed Structure** Let's interpret the structure of compound A: - It contains a benzene ring (not shown, so not aromatic). - The left side has two triple bonds (alkyne) in a row, attached to a carbon chain. **Condensed structure:** - The structure is: CH≡C–C≡C–CH₂–CH₂–CH₃ So, the condensed structure is: **HC≡C–C≡C–CH₂CH₂CH₃** --- ### **b. Number of σ and π Bonds** - **Each single bond** is 1 σ bond. - **Each triple bond** is 1 σ + 2 π bonds. - **Each double bond** is 1 σ + 1 π bond. **Counting:** - There are two triple bonds: each triple bond has 1 σ and 2 π (so 2×(1σ+2π) = 2σ + 4π) - There are 3 single bonds between the remaining carbons (CH₂–CH₂–CH₃): 3 σ bonds - Each terminal hydrogen bond: 3 hydrogens (on the propyl group) and 1 at the terminal alkyne = 4 more σ bonds **Let's count bonds more carefully:** - C1≡C2: triple → 1σ, 2π - C2–C3: single → 1σ - C3≡C4: triple → 1σ, 2π - C4–C5: single → 1σ - C5–C6: single → 1σ - C6–C7: single → 1σ - Hydrogens: - C1: 1 H (terminal alkyne) - C5: 2 H (CH₂) - C6: 2 H (CH₂) - C7: 3 H (CH₃) - Total: 1+2+2+3 = 8 σ bonds Now sum up: - **σ bonds:** 1 (C1≡C2) + 1 (C2–C3) + 1 (C3≡C4) + 1 (C4–C5) + 1 (C5–C6) + 1 (C6–C7) = 6 σ (C–C) + 8 (C–H) = **14 σ bonds** - **π bonds:** 2 (C1≡C2) + 2 (C3≡C4) = **4 π bonds** --- ### **c. Molecular Formula** Let's count the atoms: - Carbon: 7 (from the chain) - Hydrogen: 8 (as counted above) - Oxygen/Halogen: None **Molecular formula:** **C₇H₈** --- ### **d. Circle the Polar Covalent Bond** A polar covalent bond is between atoms of significantly different electronegativities. In this molecule, all bonds are C–C or C–H, which are not significantly polar. **But** if we assume the terminal alkyne hydrogen (C≡C–H), the C–H bond of terminal alkyne is slightly more polar due to the higher s-character of the carbon. **Circle the C≡C–H bond** as the most polar covalent in this molecule. --- ## **Part 2: Histamine** ### **a. First Protonation Site** Given pKa values: - A (pKa = 6.9) - B (pKa = 14.4) - C (pKa = 9.75) **Lower pKa = stronger base = more likely to be protonated.** But since these are **conjugate acid pKa's**, the **lower pKa means the site is more basic (will be protonated first)**. - **A (pKa 6.9) < C (9.75) < B (14.4)** So: **Site A** will be protonated first. --- ### **b. Add Lone-Pair Electrons** - The imidazole nitrogens (both) have lone pairs. - The terminal NH₂ group has two lone pairs. - Draw these directly on the structure as dots. --- ### **c. Reaction with HCl** #### **Predict the Products** - Histamine gets protonated at the most basic site (site A, the imidazole nitrogen). - HCl donates a proton to histamine. **Product 1:** Protonated histamine (with a + charge on the imidazole N at site A) **Product 2:** Cl⁻ #### **Curved Arrows** - Lone pair on imidazole nitrogen attacks the H of HCl. - Bond between H-Cl breaks, electrons go to Cl. #### **Favored Side at Equilibrium** - Compare pKa's: - Protonated histamine (site A): pKa = 6.9 - HCl: pKa = –8. Equilibrium favors the **weaker acid** (higher pKa), so it will favor the **products** (protonated histamine and Cl⁻). --- ## **Summary Table** | Part | Answer | |---|---| | 1a | Condensed: HC≡C–C≡C–CH₂CH₂CH₃ | | 1b | 14 σ bonds, 4 π bonds | | 1c | C₇H₈ | | 1d | C≡C–H bond (circled) | | 2a | Site A (pKa 6.9) | | 2b | Lone pairs on both imidazole N's and both NH₂ N's | | 2c | Products: protonated histamine at imidazole N, Cl⁻; Equilibrium favors products | --- If you need the molecular structures drawn or further clarifications, let me know!