Discussion
The suitable starting material for the following reaction is :
- A.
- B.
- C.
- D.
Answer: A
P is (A), acetophenone phenylhydrazone.
This is the Fischer indole synthesis. With ZnCl2 the phenylhydrazone tautomerises to an ene-hydrazine, which undergoes a [3,3]-sigmatropic rearrangement forming a new C-C bond to the ortho position of the aryl ring. Cyclisation and loss of NH3 then give the aromatic indole. The methyl of acetophenone becomes C-3 (as CH) and the phenyl-bearing carbon becomes C-2, so the product is 2-phenylindole.
Propiophenone phenylhydrazone (D) would give 3-methyl-2-phenylindole. An azo compound (B) and an anil without the N-N bond (C) cannot lose NH3 in this way.
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