See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: When a dianion reacts with only one equivalent of acid (H+), protonation occurs at the more basic (less stable) anion site, leaving the more stable (more stabilized) anion intact. Step 1 - Identify the two anionic sites in the dianion: The dianion is CH2=C(O-)-CH=C(O-)-CH3. It has two oxyanion (O-) sites: - Site 1 (left): O- on the terminal vinylic carbon adjacent to CH2= (this is an enolate where the negative charge is on oxygen of a vinyl system CH2=C(O-)-). - Site 2 (right): O- on the internal carbon adjacent to CH3 (this is an enolate where the negative charge is on oxygen of a system -C(O-)-CH3, which upon protonation at carbon gives a ketone, or protonation at oxygen gives an enol). Step 2 - Assess relative stability of the two anions: The key question is which anion is MORE stable (and thus should remain unprotonated after addition of one equivalent of H+). - The right-side anion (-C(O-)-CH3 with conjugation through the double bond) corresponds to a resonance-stabilized enolate that is part of a conjugated system: CH3-C(=O)-CH= (an alpha,beta-unsaturated ketone enolate, i.e., a cross-conjugated or extended enolate). This anion is stabilized by conjugation with the adjacent C=C and inductive effects. - The left-side anion (CH2=C(O-)-) is a simple vinyl alkoxide/enolate with less extended conjugation. - Alternatively, considering that the right anion (enolate of a methyl ketone, -C(O-)-CH3) when protonated gives a stable conjugated enone (methyl vinyl ketone type), while the left anion when protonated gives a less stable species. - Product stability argument: The remaining monoanion after protonation should be the more stable one. The right-side enolate (-C(O-)-CH3) is stabilized because it is an enolate of a ketone in a conjugated environment. If the LEFT anion is protonated, we get: CH3-C(=O)-CH=C(O-)-CH3, which is answer (d) — an enolate stabilized by conjugation with the adjacent carbonyl (forming an extended conjugated system: ketone-CH=C(O-)-CH3, analogous to a stabilized beta-keto enolate). Step 3 - Confirm choice (d): Protonation of the terminal (left) CH2=C(O-)- site converts it to CH3-C(=O)- (a methyl ketone), giving the product CH3-C(=O)-CH=C(O-)-CH3. This remaining monoanion in (d) is highly stabilized because it is a vinylogous or conjugated enolate: the O- is conjugated through C=C to the adjacent ketone carbonyl, creating an extended delocalized system. This cross-conjugated/extended enolate is more stable than the simple vinyl alkoxide on the left side. Therefore protonation preferentially occurs at the less stable (left) anion, leaving the more stable (right, conjugated) anion intact. Step 4 - Why other options fail: - (a) protonates the right O- leaving the left O- intact — the left vinyl alkoxide is less stable, so this is not preferred. - (b) protonates the left O- but gives an enol (OH) rather than tautomerizing to ketone; under acidic conditions the enol would tautomerize, but the structure shown retains OH — this is less stable than the keto form shown in (d). - (c) involves loss of conjugation (saturated bridge CH2), which would require more than simple protonation. - (e) protonates incorrectly and gives a less stable arrangement. Therefore, the correct answer is D.