See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: The starting material is 4-methylacetanilide (p-toluidine acetylated at nitrogen), which has a para-methyl group on the benzene ring and an NHAc group. Step 2 - Identify the reagents and conditions: Br2 in CS2 (carbon disulfide) is a classic electrophilic aromatic substitution (EAS) condition. CS2 is a non-polar solvent that favors electrophilic aromatic substitution without promoting free-radical side reactions. This rules out benzylic or side-chain radical bromination (which would require hv or peroxide). Step 3 - Determine directing effects: The NHAc (acetamido) group is a strong ortho/para director. The methyl group is also an ortho/para director. The dominant director is the NHAc group. The para position relative to NHAc is already occupied by the methyl group. Therefore, electrophilic bromination occurs at the ortho position relative to NHAc (which is also ortho to nothing blocking it), specifically at the position ortho to NHAc and ortho to the methyl group would be position 2 (between NHAc at C1 and Me at C4, the open ortho positions are C2 and C6; C2 is ortho to NHAc and also adjacent to the methyl-bearing carbon chain side). Step 4 - Determine regioselectivity: With NHAc at C1 and Me at C4, the ortho positions to NHAc are C2 and C6. The position C2 is also ortho to the methyl group indirectly (meta to methyl), while C6 is meta to methyl. Actually C2 is meta to C4-methyl and C6 is also meta to C4-methyl. Both ortho positions to NHAc are equivalent by symmetry? No - the molecule is 1-NHAc, 4-Me, so C2 and C6 are not equivalent because C2 is between C1(NHAc) and C3, and C6 is between C1(NHAc) and C5. However, C2 is closer to C4(Me) by one carbon than C6... actually both are equidistant (meta) to C4. By convention, bromination of 4-methylacetanilide with Br2 gives predominantly the 2-bromo product (ortho to NHAc, which is also ortho to methyl group on the same side), i.e., bromine enters at C3 relative to methyl or C2 relative to NHAc. The product shown in option (d) shows Br ortho to NHAc with Me at para - this is 2-bromo-4-methylacetanilide. Step 5 - Eliminate other options: Option (a) shows dibromination or incorrect regiochemistry. Option (b) shows alpha-bromination of the carbonyl methyl, which does not occur under EAS conditions in CS2. Option (c) shows benzylic bromination of the ring methyl group, which requires radical conditions (light/peroxide), not Br2/CS2. Step 6 - Conclusion: Br2/CS2 promotes electrophilic aromatic substitution, and the NH-acetyl group directs bromine to the ortho position (C2), giving 2-bromo-4-methylacetanilide, which corresponds to option (d). Therefore, the correct answer is D.