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प्रश्न
Give reasons for the following:
o-nitrophenol is more acidic than o-methoxyphenol.
Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?
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उत्तर १
Due to the strong –R and –I effect of the NO2 group, the electron density on the O–H bond decreases; hence, the proton is released easily.

After losing the proton, the remaining o-nitrophenoxide ion becomes stable by resonance.

Ortho-nitrophenoxide ion resonance is stable; hence, o-nitrophenol is a strong acid. On the other hand, due to the +R effect of the OCH3 group, the electron density on the O–H bond increases; hence, the removal of the proton becomes difficult.

Now, the o-methoxyphenoxide ion, which remains after the loss of the proton, gets destabilised due to resonance.

The two negative charges repel each other and destabilise the o-methoxyphenoxide ion, so o-nitrophenol is more acidic than o-methoxyphenol.
उत्तर २

The acidity of phenol is due to the dissociation of a hydrogen ion (H+) from the hydroxyl (O–H) group. This loss increases if the O–H bond polarity intensifies and H carries a significant δ+ charge.
The presence of the strongly electron-withdrawing NO2 group at the ortho position in (I) increases the polarity of the O–H bond.

This helps in the release of H+ and increases acidity.
The reaction (II) is less acidic because –OCH3 is an electron-releasing group and increases the electron density on O of OH, and H+ release is suppressed. This can be understood from the resonance structures of II, which are:

संबंधित प्रश्न
Write the mechanism of the following reaction:

Give the equation of the following reaction:
Dilute HNO3 with phenol.
Name the reagent used in the following reaction:
Oxidation of a primary alcohol to aldehyde.
When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:
\[\begin{array}{cc}
\phantom{...................................}\ce{Br}\\
\phantom{..................................}|\\
\ce{CH3 - CH - CH - CH3 ->[HBr] CH3 - C - CH2 - CH3}\\
|\phantom{.........}|\phantom{...................................}|\phantom{...........}\\
\ce{CH3}\phantom{...}\ce{OH}\phantom{...............................}\ce{CH3}\phantom{.......}\\
\end{array}\]
Give a mechanism for this reaction.
(Hint: The secondary carbocation formed in step II rearranges to a more
stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)
Lucas reagent is ____________.
In the reduction \[\ce{R - CHO + H2 -> RCH2OH}\] the catalyst used is:
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Which of the following is not true in case of reaction with heated copper at 300°C?
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The process of converting alkyl halides into alcohols involves ______.
Name the factors responsible for the solubility of alcohols in water.
Suggest a reagent for the following conversion.

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Write the chemical equation in support of your answer.
Write the mechanism of acid dehydration of ethanol to yield ethene.
