strongest acid?

[donaldduck]

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which of the following is the strongest acid?

A. para-ClC6H4OH
B. Para-O2NC6H4OH
C. Para-H3COC6H4OH
D. C6H5OH

The correct answer is B, I picked C. When looking at the conjugate bases i saw that choice C had more significant resonance contributors than answer B. I Understand that choice B has a greater inductive effects, but that still doesnt beat out the one more significant resonance contributor that choice C has. Where is the flaw in my reasoning?

 

[Rabolisk]

Sorry I misread. Methoxy group is an electron donor through resonance and the nitro group is an electron withdrawer through resonance.

 

[loveoforganic]

C isn't methyl, it's methoxy. That said, methoxy substituent is a resonance donator, and thus would make the acidic proton less acidic

 

[donaldduck]

C isn't methyl, it's methoxy. That said, methoxy substituent is a resonance donator, and thus would make the acidic proton less acidic

This may sound kinda of stupid but, how does that make the proton less acidic?

 

[Rabolisk]

No such thing as a stupid question! The proton is less acidic because the conjugate base, which is the result of acid dissociation is destabilized by an electron donating group like methoxy. The negative charge on the oxygen of the phenoxide wants to be delocalized as much as possible. The lone pairs want to get away from the oxygen through inductive effects or resonance. Methoxy does the opposite.

 

[vsl5]

which of the following is the strongest acid?

A. para-ClC6H4OH
B. Para-O2NC6H4OH
C. Para-H3COC6H4OH
D. C6H5OH

The correct answer is B, I picked C. When looking at the conjugate bases i saw that choice C had more significant resonance contributors than answer B. I Understand that choice B has a greater inductive effects, but that still doesnt beat out the one more significant resonance contributor that choice C has. Where is the flaw in my reasoning?

The answer is B because N2O is the strongest electron withdrawing group (vs Cl or OCH3 or just H as in choice D). Stronger electron withdrawing groups help stabilize the conjugate base (since the -OH loses a proton and becomes negatively charged) by spreading out the negative charge.

I dont understand what ur saying about choice C, choice D can delocalize its electrons more because it has more equivalent resonance structures (when the double bound moves from the different O atoms of N2O).

 

[donaldduck]

No such thing as a stupid question! The proton is less acidic because the conjugate base, which is the result of acid dissociation is destabilized by an electron donating group like methoxy. The negative charge on the oxygen of the phenoxide wants to be delocalized as much as possible. The lone pairs want to get away from the oxygen through inductive effects or resonance. Methoxy does the opposite.

Thanks! that clears it up!!

 

[ElliottB]

No such thing as a stupid question!

http://answers.yahoo.com/question/index?qid=20091116225017AAANtEF