Question:
Which of the following is NOT a phenol ?
Which of the following is NOT a phenol ?
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A phenol has an OH group bonded directly to a carbon of an aromatic ring.
Updated On: Oct 1, 2026
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The Correct Option is D
Solution and Explanation
Step 1: Understand the concept
A phenol is a compound in which a hydroxyl group is attached directly to a carbon of an aromatic ring. Substituents on the ring do not change this.
Step 2: Check options (A), (B) and (C)
Option (A) is a benzene ring with NO2 and OH on neighbouring carbons (2-nitrophenol), so OH is on the ring. Option (B) is naphthalene with OH on a ring carbon (2-naphthol), which is also a phenol. Option (C) is a benzene ring with Br and OH on adjacent carbons (2-bromophenol).
Step 3: Check option (D)
Option (D) is a benzene ring joined to a \(\text{CH}_2\) group that carries the OH. The OH sits on a side-chain \(sp^3\) carbon, not on the ring. This is benzyl alcohol, an alcohol.
Step 4: Conclusion
Benzyl alcohol lacks a ring-bonded OH group, so it is not a phenol.
Final Answer:
In benzyl alcohol the OH is on a side-chain carbon, not on the ring. This is option (D). \[ \boxed{\text{(D) }\text{Benzyl alcohol}} \]
A phenol is a compound in which a hydroxyl group is attached directly to a carbon of an aromatic ring. Substituents on the ring do not change this.
Step 2: Check options (A), (B) and (C)
Option (A) is a benzene ring with NO2 and OH on neighbouring carbons (2-nitrophenol), so OH is on the ring. Option (B) is naphthalene with OH on a ring carbon (2-naphthol), which is also a phenol. Option (C) is a benzene ring with Br and OH on adjacent carbons (2-bromophenol).
Step 3: Check option (D)
Option (D) is a benzene ring joined to a \(\text{CH}_2\) group that carries the OH. The OH sits on a side-chain \(sp^3\) carbon, not on the ring. This is benzyl alcohol, an alcohol.
Step 4: Conclusion
Benzyl alcohol lacks a ring-bonded OH group, so it is not a phenol.
Final Answer:
In benzyl alcohol the OH is on a side-chain carbon, not on the ring. This is option (D). \[ \boxed{\text{(D) }\text{Benzyl alcohol}} \]
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