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For the molecule CH3-O-CH=CH-CH=CH2, consider these canonical (resonance) structures: (I) CH3-O-CH=CH-CH=CH2 (original, no charge separation) (II) CH3-O(+)=CH-CH=CH-CH2(-) (oxygen double bond, terminal carbanion) (III) CH3-O(+)=CH-CH=CH2-CH2(-) (oxygen double bond, internal carbanion) What is the relative order of their contribution to the resonance hybrid?

1. I > II > III
2. III > II > I
3. I > III > II
4. III > I > II

Correct answer: I > II > III

Solution

In resonance, structures with (a) no formal charges, (b) charges on electronegative atoms, (c) more bonds, and (d) no atom with incomplete octet contribute more. Structure I has no charge separation -> greatest contribution. Structures II and III both have charge separation (+O, -C). In II, negative is on terminal carbon (primary carbanion, less stable). In III, negative is on a different carbon position. Since negative charge on carbon adjacent to another carbon with double bond provides slightly more delocalization, but in general the structure with more bonds and less charge separation wins. I > II > III is the standard answer.

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