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We need to determine the major product of the given reaction sequence involving a free-radical bromination followed by an intramolecular cyclization.
Step 1: Free-Radical Bromination ($$\text{Br}_2 / h\nu$$)
When the starting ethylbenzene derivative is treated with bromine in the presence of light ($$h\nu$$), free-radical substitution occurs selectively at the more stable benzylic position. This replaces one benzylic hydrogen atom with a bromine atom, forming a secondary benzylic bromide intermediate:
$$\text{Ar--CH}_2\text{--CH}_3 \xrightarrow{\text{Br}_2, \, h\nu} \text{Ar--CH(Br)--CH}_3$$Step 2: Deprotonation by Base ($$\text{OH}^\ominus$$)
The addition of hydroxide ion ($$\text{OH}^\ominus$$) acts as a base to abstract the acidic proton from the primary amide nitrogen group ($$\text{--CONH}_2$$). This generates a strongly nucleophilic amidate anion intermediate:
$$\text{Ar--CONH}_2 + \text{OH}^\ominus \rightarrow \text{Ar--CONH}^\ominus + \text{H}_2\text{O}$$Step 3: Intramolecular $$\text{S}_\text{N}2$$ Cyclization (Ring Closure)
The negatively charged nitrogen atom ($$\text{--NH}^\ominus$$) undergoes a rapid, favorable intramolecular nucleophilic substitution ($$\text{S}_\text{N}2$$) by attacking the adjacent benzylic carbon bearing the bromine atom, displacing the bromide leaving group ($$\text{Br}^\ominus$$):
$$\text{Ar--CONH}^\ominus\text{--CH(Br)--CH}_3 \xrightarrow{\text{Intramolecular }\text{S}_\text{N}2} \text{6-membered Lactam (Isoquinolinone derivative)} + \text{Br}^\ominus$$This ring closure constructs a stable, six-membered nitrogen-containing heterocyclic ring (a lactam framework) fused directly to the benzene core with a methyl group remaining at the newly closed carbon junction.
The reaction sequence smoothly converts the ortho-substituted ethylbenzamide into a fused six-membered cyclic amide carrying a methyl substituent via regioselective benzylic bromination and subsequent intramolecular displacement.
Answer: Option B
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