I don't really understand why E is the correct answer. Could someone explain the logic behind the resonance stability concept?
2007 DAT Sample Test question on OC
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I don't really understand why E is the correct answer. Could someone explain the logic behind the resonance stability concept?
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If I recall, because oxygen has a negative charge it has a pair of electrons that you can utilize/contribute to the conjugated bonds on the ring. Anytime you can increase the number of resonance structures, you increase stability.
HTH
HTH
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The lone pair from the O- can drop down so the electrons in the pi bonds are delocalized. All of the other options have a double bond that's too far away from the oxygen to resonate.
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Rule of thumb with resonance is you'll always move from a location of electron density (electron rich) to an electron poor location via the p-orbitals which freely move around.
Thus the electron will either start from a negative charged atom or a double bond to a location that can be reached via adjacent pi bonds.
Choice E is the only possibility for this to occur.
Thus the electron will either start from a negative charged atom or a double bond to a location that can be reached via adjacent pi bonds.
Choice E is the only possibility for this to occur.
