See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Tautomerism to form an aromatic product means the compound can undergo a proton-transfer tautomerism (keto-enol or analogous) to give a fully aromatic (benzene-like, 4n+2 pi electrons, planar) ring system. Concept: For a six-membered ring compound to tautomerize to an aromatic product, it needs to convert to benzene (all-carbon) or a heteroaromatic compound (like pyridine) by proton transfer. This requires the non-aromatic tautomer to have the right number of degrees of unsaturation and the correct arrangement of heteroatoms or functional groups such that moving a proton yields a fully conjugated aromatic ring. Analysis of each compound: (a) 1,2,3,6-tetrahydropyridin-4(1H)-one (shown with N-H, C=O, and one C=C in the ring): This is the keto tautomer. The enol tautomer would involve the N-H proton migrating: if the N-H proton transfers and the C=O becomes C-OH with full conjugation, the ring can achieve a pyridin-4-ol (= 4-hydroxypyridine, which is aromatic — pyridine ring with OH substituent). The lactam-to-hydroxypyridine tautomerism is well established: 4-pyridinone <-> 4-hydroxypyridine. 4-Hydroxypyridine is an aromatic compound. So (a) YES — tautomerizes to give aromatic 4-hydroxypyridine. (b) Cyclohex-2-en-1-one (2-cyclohexenone): This has one C=O and one C=C in the six-membered ring. Enolization of the alpha position adjacent to C=O can extend conjugation. The full enol tautomer of cyclohex-2-en-1-one would be phenol (C6H5OH) if two proton shifts occur to give complete conjugation — actually cyclohex-2-en-1-one can tautomerize stepwise to phenol (an aromatic compound) via the dienol intermediate. So (b) YES — can tautomerize to phenol (aromatic). (c) Cyclohexa-2,4-dien-1-one (2,4-cyclohexadienone): This compound is already one proton transfer away from phenol. Moving the C1 carbonyl OH (i.e., enolization at C1) directly gives phenol. This is the classic phenol-cyclohexadienone tautomerism. Phenol is aromatic. So (c) YES — directly tautomerizes to phenol. (d) Cyclohexa-2,5-diene-1,4-dione (para-benzoquinone): This compound has two C=O groups. Its enol tautomers would require both carbonyls to enolize to give hydroquinone (1,4-dihydroxybenzene), which is aromatic. However, para-benzoquinone itself is the oxidized form, and its tautomer hydroquinone is aromatic. But wait — the structure shown appears to be cyclohexa-2,5-diene-1,4-dione (p-benzoquinone). For tautomerism to aromatic product, a simple proton shift is needed. p-Benzoquinone does not have an acidic C-H adjacent to the carbonyls in a way that a single tautomeric shift gives a fully aromatic product — it would require two enolizations. Simple keto-enol tautomerism of one C=O does not give a fully aromatic ring. So (d) NO — does not simply tautomerize to an aromatic product by a single tautomeric step. (e) Cyclohex-2-en-1-one (2-cyclohexenone, same connectivity as b but drawn as the saturated ring form — actually this appears to be 2-cyclohexenone again, a six-membered ring with C=O and one double bond, fully saturated otherwise): Similar to (b), but looking more carefully, (e) appears to be cyclohex-2-en-1-one without the extra conjugation needed. A single enolization step gives cyclohexa-2,4-dien-1-ol, not directly phenol. Multiple tautomeric steps would be needed. Strict tautomerism (single step proton transfer) from cyclohex-2-enone does not directly yield an aromatic compound. So (e) NO. (f) Cyclobut-2-en-1-one (cyclobutenone): Four-membered ring — even if enolization occurred, a four-membered ring cannot be aromatic (would need to be cyclobutadiene, antiaromatic). So (f) NO. (g) 4,4-dimethylcyclohexa-2,5-dien-1-one: The gem-dimethyl group at C4 means that enolization at C4 is blocked (no C-H at C4 due to two methyl groups). The compound cannot tautomerize to give a fully aromatic benzene ring because C4 has no hydrogen to lose/gain to complete aromatization. So (g) NO. (h) 4-(Hydroxyimino)cyclohexa-2,5-dien-1-one (para-quinone oxime, shown as six-membered ring with C=O at top and C=N-OH at bottom with conjugated double bonds): This compound can tautomerize via oxime-nitroso or more relevantly via the C=O enolization. The para-quinone oxime can tautomerize to 4-nitrosophenol or to para-aminophenol... more precisely, the N-OH group can shift: the C=N-OH (oxime) with the C=O can tautomerize such that C-OH forms and C=N remains, or the ring fully conjugates. Actually, 4-(hydroxyimino)cyclohexa-2,5-dien-1-one is the tautomer of 4-nitrosophenol: the quinoid oxime tautomerizes to the aromatic nitrosophenol (or via the para-quinone monoxime to 4-aminophenol route). The key point: the quinone oxime (non-aromatic, quinoid form) tautomerizes to the aromatic nitrosophenol form, which has a benzene ring. So (h) YES — tautomerizes to 4-nitrosophenol (aromatic). Summary: (a) YES, (b) YES, (c) YES, (d) NO, (e) NO, (f) NO, (g) NO, (h) YES. Count = 4 compounds: a, b, c, h. Therefore, the correct answer is 4 (a, b, c, h).