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Why is p-methoxyphenol less acidic than phenol?
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Marian Doscillo
Why is p-methoxyphenol less acidic than phenol?
Deprotonation creates a negative charge on the phenolic oxygen. The better you can stabilise this charge, the more favored this process is going to be.
You are right in saying that the charge can be stabilised by delocalisation in the aromatic ring. However, the electrons of the methoxy oxygen can also be delocalised in the ring (to a much lesser extend, of course). This has the effect of making the ring richer in electrons. It will thus tend to less receive additional electrons, and thus stabilise less the formal negative charge. (However, keep in mind that the ring does still stabilise the negative charge, making p-methoxyphenol still much more acidic than alcohols with pKa values around 15)
Deprotonation creates a negative charge on the phenolic oxygen. The better you can stabilise this charge, the more favored this process is going to be.
You are right in saying that the charge can be stabilised by delocalisation in the aromatic ring. However, the electrons of the methoxy oxygen can also be delocalised in the ring (to a much lesser extend, of course). This has the effect of making the ring richer in electrons. It will thus tend to less receive additional electrons, and thus stabilise less the formal negative charge. (However, keep in mind that the ring does still stabilise the negative charge, making p-methoxyphenol still much more acidic than alcohols with pKa values around 15)
Deprotonation creates a negative charge on the phenolic oxygen. The better you can stabilise this charge, the more favored this process is going to be.
You are right in saying that the charge can be stabilised by delocalisation in the aromatic ring. However, the electrons of the methoxy oxygen can also be delocalised in the ring (to a much lesser extend, of course). This has the effect of making the ring richer in electrons. It will thus tend to less receive additional electrons, and thus stabilise less the formal negative charge. (However, keep in mind that the ring does still stabilise the negative charge, making p-methoxyphenol still much more acidic than alcohols with pKa values around 15)
Deprotonation creates a negative charge on the phenolic oxygen. The better you can stabilise this charge, the more favored this process is going to be.
You are right in saying that the charge can be stabilised by delocalisation in the aromatic ring. However, the electrons of the methoxy oxygen can also be delocalised in the ring (to a much lesser extend, of course). This has the effect of making the ring richer in electrons. It will thus tend to less receive additional electrons, and thus stabilise less the formal negative charge. (However, keep in mind that the ring does still stabilise the negative charge, making p-methoxyphenol still much more acidic than alcohols with pKa values around 15)
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