See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Relative stability of resonance structures is determined by applying the rules of resonance stability: 1. Structures with more covalent bonds (complete octets, no charge separation) are more stable. 2. If charge separation is present, structures where negative charge resides on more electronegative atoms are more stable. 3. Structures with like charges on adjacent atoms or positive charge on electronegative atoms are least stable. Step 1 - Analyze structure (i): Structure (i) has no formal charges on any atom. All atoms have complete octets and normal bonding. The CN group is a triple bond (C≡N), the carbonyl is C=O, and there is a conjugated alkene. This is the most stable resonance contributor because there is no charge separation. Step 2 - Analyze structure (ii): Structure (ii) has charge separation: the carbonyl carbon carries a positive charge and the oxygen carries a negative charge (C(+)-O(-)). The negative charge is on oxygen, which is highly electronegative — this is favorable. The positive charge is on carbon adjacent to oxygen. While charge separation is present (making it less stable than (i)), the charges are placed on appropriate atoms (negative on O, the most electronegative). This makes (ii) the second most stable among charge-separated structures. Step 3 - Analyze structure (iii): Structure (iii) has a positive charge on a carbon in the chain (on the carbon bearing only alkyl/H substituents, less stabilized) and a negative charge on the nitrogen of the CN group (C≡N(-)). Although nitrogen is electronegative and can bear negative charge, the positive charge in (iii) is on a carbon that is less stabilized compared to structure (ii) where positive charge is on a carbon directly bonded to oxygen (which can stabilize it through induction/resonance less effectively). Additionally, in (iii) the positive charge is further from the electronegative nitrogen relative to the negative charge location, making it less favorable than (ii). Thus (iii) is less stable than (ii) but the key distinction: in (ii) the negative charge is on oxygen (more electronegative than nitrogen), making (ii) more stable than (iii). Step 4 - Order: (i) no charge separation — most stable (ii) charge separation with negative on O (most electronegative) — second (iii) charge separation with positive on carbon (poorly stabilized) and negative on N — least stable Order: (i) > (ii) > (iii), which corresponds to option (a). Why other options fail: - (b) places (ii) above (i): impossible since charge separation always destabilizes relative to no charge separation. - (c) places (iii) above (ii): wrong because O is more electronegative than N, so negative on O is more stable. - (d) places (ii) first but also above (i): same error as (b). Therefore, the correct answer is A.