See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

We analyze each sub-question by considering factors that affect acidity (pKa): resonance stabilization of conjugate base, inductive effects, electronegativity, and aromaticity of conjugate base. **A. Allylic/vinylic alcohols with varying degrees of conjugation:** - (a) has a conjugated diene ending in CH2OH — the conjugate base (alkoxide) is stabilized by extended conjugation through the diene system. - (b) has OH directly on the conjugated diene system (vinylic/allylic type with more direct resonance) — maximum resonance stabilization. - (c) has only one double bond (less conjugation, less stabilization of alkoxide). Acidity order: c < a < b (less conjugation = less acidic; more conjugation/direct attachment = more acidic). **B. Substituted phenols:** - (a) 2-aminophenol: -NH2 is electron-donating by resonance and induction, destabilizes phenoxide, least acidic. - (b) 2-methoxyphenol: -OCH3 is electron-donating by resonance but withdrawing by induction; net electron-donating, less acidic than unsubstituted phenol. - (c) 3-(trifluoromethyl)phenol: -CF3 is strongly electron-withdrawing by induction (meta position, no resonance donation to ring), stabilizes phenoxide, most acidic. Acidity order: a < b < c. **C. Halohydrins — inductive effect of halogens on adjacent carbon bearing OH:** The acidity depends on how close the electronegative halogens are to the OH group (distance and number of EWG near the acidic proton). - (a) has Br and F on the carbon bearing OH, and Cl on adjacent carbon — two halogens directly on the carbinol carbon (Br + F alpha) plus Cl beta; strong inductive withdrawal. - (b) has F on the carbinol carbon and Br on adjacent carbon, Cl on far carbon — F alpha (strongest EWG), Br beta, Cl gamma. - (c) has Cl on the carbinol carbon and Br on adjacent carbon, F on far carbon — Cl alpha (weaker than F), Br beta, F gamma (weakest position contribution). F is most electronegative, so having F directly on the carbinol carbon (b) is very effective, but (a) has both Br and F on the carbinol carbon making it most acidic. (c) has only Cl on the carbinol carbon (weakest). Acidity order: c < b < a. **D. Cyclic diketones and monoketones — acidity of alpha-H in 1,3-dicarbonyl vs monoketone:** - (a) monoketone (cyclohexanone derivative): alpha-H flanked by only one C=O, least acidic. - (b) 1,3-diketone with methyl groups: alpha-H between two carbonyls, much more acidic; gem-dimethyl adjacent. - (c) and (d) are also 1,3-diketones but differ in substitution pattern affecting enolization and steric/electronic effects. For 1,3-diketones, more substitution on the alpha carbon (steric) reduces acidity, and electron-donating alkyl groups on carbons adjacent to carbonyls slightly reduce acidity. - (d) has gem-dimethyl directly on one carbonyl carbon — most alkyl substitution near the acidic H, but the acidic CH is between two C=O groups; the gem-dimethyl is on the same carbon as a carbonyl reducing the effective withdrawal slightly compared to (b) or (c). - (b) has gem-dimethyl at C5 (away from the 1,3-dione system) and methyl at C3 — acidic CH at C2 between two carbonyls, moderately substituted. - (c) has gem-dimethyl and arrangement making the alpha CH most activated. Overall order considering alkyl substitution reducing acidity in 1,3-diketones: d < c < a is wrong for monoketone being less acidic... Correct order given is d < c < a < b, meaning (b) is most acidic (1,3-diketone more activated), (a) is less acidic than (b) but more than (c), suggesting (a) might be a 1,3-diketone too with specific substitution. Re-examining: (a) monoketone least acidic would be < all diketones, but order given places a between c and b. All four likely involve different enolizable systems. The given answer d < c < a < b reflects the relative stabilization of enolate in each case. **E. Lactone-ketone bicyclic isomers:** The acidity of the alpha-H depends on position of the acidic CH relative to both carbonyl groups. - (a) arrow points to the lactone oxygen position — alpha-H activated by both lactone C=O and ketone C=O. - (b) and (c) differ in which C=O is the lactone vs ketone and position of acidic H. Order: c < a < b — in (b), the alpha-H is most activated by both carbonyls in the most favorable geometry; (c) least activated. **F. C-H acidity based on hybridization and stability of carbanion:** - (a) cyclopentadienyl with exo-methylene type: the sp3 C-H on cyclopentadiene ring — deprotonation gives aromatic cyclopentadienyl anion (6 pi electrons, aromatic), very acidic (pKa ~16). - (b) cyclopentadiene sp3 C-H: same aromatic anion formed, pKa ~16 (essentially same as a, but without additional substituent). - (c) benzyl/styryl C-H: deprotonation gives a carbanion stabilized only by conjugation with benzene ring, not aromatic gain; pKa ~43. Acidity order: a > b > c (a and b both form aromatic anions upon deprotonation, a is slightly more acidic than b due to additional conjugation from the exo double bond; c is least acidic as no aromaticity gain). Therefore, the correct answer is {"A": "c<a<b", "B": "a<b<c", "C": "c<b<a", "D": "d<c<a<b", "E": "c<a<b", "F": "a>b>c"}.