See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify criteria for aromaticity, non-aromaticity, and anti-aromaticity. - Aromatic: cyclic, planar, fully conjugated, follows Huckel rule (4n+2 pi electrons) - Anti-aromatic: cyclic, planar, fully conjugated, follows 4n pi electrons - Non-aromatic: does not meet all criteria for aromatic or anti-aromatic (not planar, not fully conjugated, or not cyclic) Step 2: Analyze each compound. (a) Naphthalene: bicyclic, planar, fully conjugated, 10 pi electrons (4n+2, n=2). AROMATIC. (b) C8H8^{2-}: This is the cyclooctatetraene dianion. COT is tub-shaped non-aromatic, but adding 2 electrons gives 10 pi electrons (4n+2, n=2), planar. AROMATIC. (c) Cyclobutadiene dianion (square with inner square, drawn as cyclobutadiene^{2-}): C4H4^{2-} has 6 pi electrons (4n+2, n=1), planar, cyclic, fully conjugated. AROMATIC. (d) Cyclooctatetraene (COT): tub-shaped, NOT planar, not fully conjugated. NON-AROMATIC. (e) Imidazoline (4,5-dihydroimidazole): five-membered ring, has one C=N double bond but C4-C5 bond is saturated (sp3 carbons), not fully conjugated. NON-AROMATIC. (f) Benzene: 6 pi electrons (4n+2, n=1), planar, cyclic, fully conjugated. AROMATIC. (g) Cyclohexene: six-membered ring with one double bond, not fully conjugated. NON-AROMATIC. (h) 1,3,5-triazine with lone pairs: six-membered ring with 3 N and 3 C, alternating double bonds, 6 pi electrons, planar. AROMATIC. (The lone pairs shown are in sp2 orbitals, not contributing to pi system in the way that makes it non-aromatic; the ring pi system has 6 electrons.) (i) C3H3^{+1} (cyclopropenyl cation): three-membered ring, 2 pi electrons (4n+2, n=0), planar. AROMATIC. (j) Cyclopropenyl cation with OH: essentially a hydroxycyclopropenyl cation. The three-membered ring is aromatic (2 pi electrons). AROMATIC. (k) Four-membered ring with N (azete or 1-azetine type shown with two double bonds in a four-membered ring containing N): If fully conjugated 4-membered ring with 4 pi electrons (4n, n=1), it is ANTI-AROMATIC. This appears to be azete (1H-azete) or similar with 4 pi electrons in a planar ring. ANTI-AROMATIC. (l) 1,2-dihydronaphthalene: two fused six-membered rings, one ring is aromatic (benzene), the other has one double bond and two sp3 carbons. The molecule is not fully aromatic overall; it contains one non-aromatic ring. NON-AROMATIC (as a whole molecule). Step 3: Count w (aromatic), x (non-aromatic), y (anti-aromatic). Aromatic (w): (a) naphthalene, (b) C8H8^{2-}, (c) C4H4^{2-} (cyclobutadiene dianion), (f) benzene, (h) triazine, (i) C3H3^{+}, (j) hydroxycyclopropenyl cation = 7 compounds. w = 7. Anti-aromatic (y): (c) if cyclobutadiene neutral... Re-examining: compound (c) is drawn as a square with inner square suggesting cyclobutadiene itself (neutral), which has 4 pi electrons -> ANTI-AROMATIC. Then C8H8^{2-} (b) is the dianion of COT = 10 pi electrons, AROMATIC. Revised analysis: (c) Cyclobutadiene (neutral, square): 4 pi electrons, ANTI-AROMATIC. (k) Four-membered ring with N: 4 pi electrons (azete), ANTI-AROMATIC. Anti-aromatic (y): (c) cyclobutadiene, (k) azete = 2 compounds. y = 2. Aromatic (w): (a) naphthalene, (b) C8H8^{2-}, (f) benzene, (h) triazine, (i) C3H3^{+}, (j) hydroxycyclopropenyl cation = 6 compounds. w = 6. Non-aromatic (x): (d) COT, (e) imidazoline, (g) cyclohexene, (l) 1,2-dihydronaphthalene = 4 compounds. x = 4. Check: 6 + 4 + 2 = 12. Total compounds = 12. Correct. Step 4: Determine z (compounds that readily undergo dimerization at room temperature). Compounds that dimerize readily are those that are anti-aromatic and highly reactive (cyclobutadiene dimerizes via Diels-Alder at room temperature) and other strained/reactive species. - Cyclobutadiene (c): famously dimerizes at room temperature. YES. - C3H3^{+} (i) and (j): cationic species, not known for dimerization. - Azete (k): also very reactive and can dimerize. YES. - Cyclohexadiene or similar: 1,3-cyclohexadiene can dimerize but compound (g) is cyclohexene, less reactive. For z = 2: compounds (c) cyclobutadiene and (k) azete undergo ready dimerization. z = 2. Step 5: Sum. w + x + y + z = 6 + 4 + 2 + 2 = 14. Therefore, the correct answer is w+x+y+z=14.