Write the major product when 3-methylphenol is subjected to mononitration.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3-methyl-4-nitrophenol
In electrophilic substitution on a ring that already has two groups, each group directs the incoming electrophile. The –OH group is a strongly activating ortho/para director; the –\( \ce{CH3} \) group is a weakly activating ortho/para director. Where they point to different places, the stronger activator, –OH, wins.
Number 3-methylphenol with –OH at C-1 and –\( \ce{CH3} \) at C-3. The –OH group directs to C-2, C-4 and C-6; the –\( \ce{CH3} \) group directs to C-2, C-4 (its ortho positions) and C-6 (its para position). Both groups thus favour the same three positions:
C-2 lies between the two groups and is sterically hindered, so little substitution occurs there.
C-4 is para to the strongly activating –OH group and ortho to –\( \ce{CH3} \). Attack para to –OH is the most favoured.
C-6 is ortho to –OH and para to –\( \ce{CH3} \); it gives the minor product.
\( \displaystyle \ce{HO-C6H4-CH3 + HNO3} \xrightarrow{\text{dil., 298 K}} 3\text{-methyl-4-nitrophenol}\,(\text{major}) + \ce{H2O} \)
So the major product is 3-methyl-4-nitrophenol, the structure in Option (4).
Why the other options are incorrect
Option (1) 4-nitrophenol — Incorrect. It has lost the methyl group; nitration only replaces a ring hydrogen.
Option (2) 3-methylphenol — Incorrect. This is the starting compound with no nitro group.
Option (3) 2-nitrotoluene — Incorrect. It has lost the –OH group; nitration does not remove substituents from the ring.