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