Aromatic Substitution Reactions

Step 1: Reaction with hydroxylamine (NH OH·HCl).
Benzil reacts with hydroxylamine to form the benzil monoxime. The oxime formation occurs at one of the carbonyl groups, giving a C=NOH functionality. This step does not alter aromatic rings.
Step 2: Beckmann rearrangement (H SO , heat).
The oxime undergoes Beckmann rearrangement.
In benzil monoxime, migration occurs of the phenyl group trans to the oxime OH. This produces a benzil-derived amide in which one phenyl migrates, forming: Ph–CO–NH–Ph (benzamide derivative).
Step 3: Bromination using Br /FeBr .
Electrophilic aromatic substitution occurs on the ring attached to nitrogen.
Anilide (Ar–NH–CO–Ph) strongly directs electrophiles to the para position because –NHCO– is an activating, ortho/para directing group.
Hence bromine enters para to the –NHCO– group on the anilide ring.
Step 4: Identify the correct structure.
Option (B) shows the amide with bromine at the para position relative to the –NHCO– group, which matches the expected regioselectivity.
Other options show ortho substitution or bromination on the wrong ring, which is not favored.
Step 5: Conclusion.
Thus the major product is structure (B).

Step 1: Identify required transformation.
The product is p-cyano anisole, meaning CN must replace the bromine atom. The only reliable method for aryl-CN introduction is via diazonium salt (Sandmeyer reaction).
Step 2: Steps needed.
We must convert Ar–Br → Ar–NH → Ar–N → Ar–CN.
Thus the reagents must do the following:
– Nitration to form p-nitro anisole.
– Reduction of –NO to –NH .
– Diazotization with NaNO /HCl.
– Replacement of N with CN using CuCN + heat.
Step 3: Matching with options.
Option (B) exactly matches:
(i) Nitration, (ii) Reduction, (iii) Diazotization, (iv) Sandmeyer cyanation.
Step 4: Conclusion.
Thus the correct sequence is option (B).

Step 1: Apply Huckel’s rule (4n+2 π electrons).
A compound is aromatic if it is:
• cyclic
• planar
• fully conjugated
• has (4n + 2) π electrons
Step 2: Examine each given compound.
1. **Phthalic anhydride derivative** → conjugation is broken in carbonyl groups → non-aromatic.
2. **Imide ring (cyclic imide)** → not fully conjugated → non-aromatic.
3. **Borole derivative** → 4π electrons → antiaromatic → not aromatic.
4. **1,3,4-oxadiazole derivative** → 6π system → aromatic → yes.
5. **Furan derivative** → 6π electrons → aromatic → yes.
Therefore, only three are aromatic:
• the 1,3,4-oxadiazole derivative
• the N-methyl imidazole-like structure
• the furan derivative
Step 3: Conclusion.
Thus, the number of aromatic compounds is 3.