Q.E. Identify more favourable resonance structure from the following. Justify.
a. CH3-C(=O)-OH and CH3-C(-OH)=O, one neutral and one bearing a positive charge on oxygen (protonated carboxylic-acid-type resonance contributors)
b. CH2=CH-CH=C(H)-O and CH2-CH=CH-CH=O, two charge-bearing resonance contributors of an unsaturated carbonyl-type species, differing in where the positive charge sits

Note on the source text. The precise structural diagrams for 1E are read with some uncertainty from the extracted source, consistent with this chapter's known garbling risk for resonance-arrow diagrams -- but the GOVERNING RULE (section 14.6.5) is stated unambiguously in the chapter's own text and is applied here directly.

Step 1. Recall the rule. Between two candidate resonance structures, the more stable/more important one has (a) more covalent bonds,

(b) more atoms with a complete octet,

(c) less separation of opposite charges,

(d) any negative charge on the more electronegative atom and any positive charge on the more electropositive atom, and (e) more dispersal of charge.

Step 2a. In (a), CH3-C(=O)-OH is the plain, neutral carboxylic-acid structure: every atom (C, both O's) has a complete octet and there is no charge separation at all. The alternative, CH3-C(-OH)=O(⊕) (or an equivalent contributor carrying a positive charge on an oxygen), necessarily has one fewer covalent bond in the resonance sense and places a positive charge on the highly electronegative oxygen -- both unfavourable by rules (a)-(d). So the neutral structure is more favourable. …

Applying section 14.6.5's five resonance-structure-stability rules (covalent bond count, octet completeness, charge separation, charge placement by electronegativity, ch …

• Assuming a charged resonance structure can never be drawn or matters (charged contributors ARE valid and can matter for the hybrid, they are simply LESS important than a neutral, fully-octet-complete one when both exist). …