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(D) The reaction involves a radical bromination of the ethyl group at the benzylic position using $Br_2/h
u$,followed by an intramolecular nucleophil

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(D) The reaction involves a radical bromination of the ethyl group at the benzylic position using $Br_2/h
u$,followed by an intramolecular nucleophilic substitution (cyclization) in the presence of $KOH$. $1$. The $Br_2/h
u$ reagent performs a free-radical bromination on the benzylic carbon of the ethyl group attached to the benzene ring. $2$. The resulting intermediate has a bromine atom at the benzylic position. $3$. The lone pair on the nitrogen atom of the amide group attacks the benzylic carbon,displacing the bromide ion. $4$. This cyclization leads to the formation of a cyclic amide (lactam). $5$. Based on the structure,the product is $4-methyl-3,4-dihydroquinolin-2(1H)-one$ or a similar cyclic structure as shown in option $D$.

The major product of the following reaction is:

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