See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Bond length is influenced by bond order and resonance. In methyl acetate (CH3-C(=O)-O-CH3), the C-O single bond labeled 'a' (between the carbonyl carbon and the ester oxygen) has partial double bond character due to resonance. The lone pairs on the oxygen donate into the carbonyl system, creating a resonance structure where the C-O (ester) bond has some double bond character: CH3-C(=O-)-O+=CH3 <-> CH3-C(-O-)=O+ ... more precisely, the resonance involves delocalization of the oxygen lone pair into the C=O pi system, giving the C-O(a) bond partial double bond character. Reasoning: Bond 'a' is the C(carbonyl)-O bond in the ester linkage. Due to resonance delocalization (the lone pair on the ester oxygen overlaps with the pi system of the carbonyl), bond 'a' has partial double bond character, making it shorter than a typical C-O single bond. Bond 'b' is the O-CH3 bond, which is a pure single bond with no resonance contribution (the methyl group has no pi system to accept electron density). Therefore, bond 'b' is a normal C-O single bond, which is longer than bond 'a' which has partial double bond character. This means b > a. Why other options fail: (a) a = b is wrong because the two bonds have different bond orders due to resonance. (c) b < a is wrong because it reverses the correct relationship. (d) Impossible to predict is wrong because resonance theory clearly predicts the relationship. Therefore, the correct answer is B.