Q.Write resonance structures of CH₃COO⁻ and show the movement of electrons by curved arrows.
The acetate ion () has two equivalent resonance structures where the negative charge is delocalised equally over both oxygen atoms. The actual ion is a hybrid of these two forms, with each C–O bond having a bond order of 1.5.
Why Resonance Matters Here
The acetate ion is a classic example of resonance stabilisation. If you try to draw a single Lewis structure for , you'll face a dilemma: which oxygen gets the negative charge? The answer is that neither oxygen "owns" the charge permanently — instead, the charge is shared between them through a delocalised π-electron system.
This delocalisation makes the acetate ion much more stable than a simple Lewis structure would suggest. The two C–O bonds become identical, with a bond length somewhere between a single and a double bond.
Drawing the Resonance Structures
Step 1: Draw the basic skeleton
The acetate ion has a methyl group () attached to a carboxylate group (). The carbon of the carboxylate is hybridized, forming a planar structure.
O
||
H₃C — C
|
O⁻
This is one possible Lewis structure — but it's not the whole story.
Step 2: Identify the delocalisable electrons
The key players are:
- The π-bond (double bond) between carbon and the top oxygen
- The lone pair on the negatively charged bottom oxygen
These four electrons (2 from the π-bond, 2 from the lone pair) can move because they're in a conjugated system — the orbitals on all three atoms (C, O₁, O₂) overlap.
Step 3: Push the electrons with curved arrows
From the structure above, the lone pair on the bottom oxygen moves to form a π-bond between that oxygen and carbon. Simultaneously, the existing π-bond between carbon and the top oxygen breaks, and those electrons move onto the top oxygen as a lone pair.
The curved arrow starts at the electron source (the lone pair) and points to where the bond forms. A second arrow starts at the π-bond and points to the oxygen that receives the electrons.
This gives the second resonance structure:
O⁻
|
H₃C — C
||
O
Step 4: Recognise the equivalence
These two structures are identical in energy — they're just mirror images. The methyl group doesn't participate in the resonance, so it stays unchanged.
A quick way to check: count the total number of electrons in each structure. Both have 24 valence electrons (4 from C, 1 from each H, 6 from each O, plus 1 for the negative charge). The connectivity is the same; only the π-bond location changes.
Step 5: Draw the resonance hybrid
The actual acetate ion is not flipping between these two forms — it's a single, stable hybrid. The two C–O bonds are identical, each with a bond order of 1.5. The negative charge is spread equally over both oxygen atoms.
You can represent this with a dashed line between the two oxygens and the carbon, or with a δ⁻ on each oxygen.
A common mistake is to draw the curved arrow from the π-bond to the oxygen without also showing the lone pair moving to form the new π-bond. Both arrows are essential — you're moving four electrons total, not just two. If you only show one arrow, you'll end up with an impossible structure (a carbon with only 6 electrons).
The Final Structures
Here are the two resonance structures with curved arrows:
Structure 1 (left form):
O O⁻
|| |
H₃C — C → H₃C — C
| ||
O⁻ O
The arrow from the lone pair on points toward the C–O bond region. The arrow from the C=O π-bond points toward the top oxygen.
Structure 2 (right form):
O⁻ O
| ||
H₃C — C → H₃C — C
|| |
O O⁻
The arrows are reversed — the lone pair on the new moves to form the π-bond, and the old π-bond breaks to give a lone pair on the other oxygen.
The resonance hybrid is often written as:
with a dashed line or a on both oxygens to show equal charge distribution.
The acetate ion has two equivalent resonance structures, with the negative charge delocalised equally over both oxygen atoms, represented by curved arrows showing the movement of a lone pair to form a π-bond and the simultaneous breaking of the existing π-bond.
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