Phenols undergo electrophilic substitution reactions readily due to the strong
activating effect of the OH group attached to the benzene ring. Since the OH group
increases the electron density more to the ortho- and para- positions, one of the
examples of the aldehyde group being introduced on the aromatic ring of phenol, ortho
to the hydroxyl group. This is a general method used for the ortho-formylation of
phenols.
Answer the following questions:
Question: 1
Br\(_2\)/CS\(_2\)
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The presence of the hydroxyl group makes phenol highly reactive to electrophilic substitution, resulting in tribromination in the ortho and para positions.
When phenol reacts with bromine in carbon disulfide (CS\(_2\)), an electrophilic substitution reaction occurs, where the hydroxyl group (OH) activates the ring towards substitution at the ortho and para positions. The major product is 2,4,6-Tribromophenol, as bromine atoms are added at the ortho and para positions relative to the hydroxyl group.
\[
\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \xrightarrow{\text{CS}_2} \text{C}_6\text{H}_2\text{Br}_3\text{OH}
\]
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Question: 2
Conc. HNO\(_3\)
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Concentrated nitric acid causes electrophilic nitration of phenol, predominantly at the ortho and para positions, leading to trinitrophenol (picric acid).
When phenol reacts with concentrated nitric acid (HNO\(_3\)), it undergoes nitration at the ortho and para positions relative to the hydroxyl group. The major product is 2,4,6-Trinitrophenol (picric acid), which is highly reactive due to the electron-donating effect of the hydroxyl group, activating the ring for nitration.
\[
\text{C}_6\text{H}_5\text{OH} + 3\text{HNO}_3 \rightarrow \text{C}_6\text{H}_2\text{NO}_3\text{OH}
\]
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Question: 3
Why phenol does not undergo protonation readily?
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The resonance of the lone pair on the oxygen in phenol with the benzene ring decreases its availability to accept a proton, making protonation less likely.
Phenol does not undergo protonation readily because the lone pair of electrons on the oxygen of the hydroxyl group is delocalized into the benzene ring through resonance. This reduces the electron density on the oxygen atom, making it less available to accept a proton. As a result, phenol is less basic than alcohols and does not readily undergo protonation.
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Question: 4
Which is a stronger acid – phenol or cresol? Give reason.
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Electron-donating groups like the methyl group in cresol reduce the acidity of phenols by stabilizing the negative charge on the oxygen less effectively.
Phenol is a stronger acid than cresol because cresol has a methyl group (\(-CH_3\)) attached to the aromatic ring. The methyl group is an electron-donating group, which increases the electron density on the ring and on the oxygen atom of the hydroxyl group. This reduces the ability of the oxygen atom to lose a proton, making cresol less acidic than phenol. In contrast, phenol’s hydroxyl group is less stabilized and more likely to lose a proton, making it more acidic.
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Question: 5
Write the IUPAC name of the product formed in the Reimer-Tiemann reaction.
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The Reimer-Tiemann reaction introduces an aldehyde group ortho to the hydroxyl group on the benzene ring of phenol.
The Reimer-Tiemann reaction involves the formylation of phenol, where the hydroxyl group is activated to form an ortho-formylated product. In the presence of chloroform (CHCl\(_3\)) and a base like NaOH, the major product is 2-formylphenol (also known as ortho-hydroxybenzaldehyde).
The IUPAC name of the product is 2-formylphenol.