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In the following reaction sequence starting from o-ethyltoluene: o-ethyltoluene is oxidised by KMnO4/H+ to give X; X reacts with excess NH3 to give Y; Y reacts with Br2/KOH to give Z1; Y heated gives Z2; Z2 reacts with Br2/KOH to give Z3. The final product in the sequence is called phthalimide. Which statement about the products in each step is correct?

1. Y is a compound with two -CONH2 groups on adjacent positions of a benzene ring (phthalamide)
2. Z1 is o-phenylenediamine
3. Z2 is phthalimide
4. Z3 is potassium 2-aminobenzoate (potassium anthranilate)

Correct answer: Z2 is phthalimide

Solution

X = phthalic acid; Y = phthalamide (two -CONH2); heating Y loses H2O to give phthalimide (Z2). Z1 from Hofmann on Y = o-phenylenediamine; Z3 from Hofmann on Z2 = potassium anthranilate.

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