In a set of reactions, ethylbenzene yielded a product D. [figure] D would be
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Answer
(4) [figure]
Solution
KMnO4/KOH oxidises the whole side chain of an alkylbenzene to –COOH (as benzoate; benzoic acid after acidification).
B = benzoic acid, C6H5COOH
–COOH is a deactivating, meta-directing group, Br2/FeCl3 brominates the meta position.
C = m-bromobenzoic acid
C2H5OH/H+ esterifies the –COOH group.
D = ethyl m-bromobenzoate (option 4)
Answer: (4)