Questions: When Benzoic acid (PhCO2H) is deprotonated it generates a resonance stabilized anion at which position within the aromatic ring relative to the carboxyl group? Enter all that apply. ortho meta para it is not resonance stabilized by the ring

When Benzoic acid (PhCO2H) is deprotonated it generates a resonance stabilized anion at which position within the aromatic ring relative to the carboxyl group? Enter all that apply. ortho meta para it is not resonance stabilized by the ring
Transcript text: When Benzoic acid $\left(\mathrm{PhCO}_{2} \mathrm{H}\right)$ is deprotonated it generates a resonance stabilized anion at which position within the aromatic ring relative to the carboxyl group? Enter all that apply. ortho meta para it is not resonance stabilized by the ring
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Solution

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Solution Steps

Step 1: Understanding Resonance Stabilization

When benzoic acid (\(\mathrm{PhCO_2H}\)) is deprotonated, it forms the benzoate anion (\(\mathrm{PhCO_2^-}\)). The negative charge on the oxygen can be delocalized through resonance within the carboxylate group.

Step 2: Analyzing Resonance with the Aromatic Ring

The resonance stabilization of the benzoate anion primarily involves the carboxylate group itself. The negative charge is delocalized between the two oxygen atoms of the carboxylate group. This resonance does not extend into the aromatic ring because the carboxylate group is not conjugated with the π-system of the benzene ring.

Step 3: Conclusion on Resonance Stabilization

Since the resonance stabilization of the benzoate anion does not involve the aromatic ring, the positions within the ring (ortho, meta, para) relative to the carboxyl group are not relevant for the resonance stabilization of the anion.

Final Answer

\[ \boxed{\text{it is not resonance stabilized by the ring}} \]

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