Question

The reaction(s) that produce(s) 2-methylindole as the major product is(are)

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Count the ring size each route builds first (six-membered = quinoline, not indole), then check what group actually ends up at indole C2.

Answer 1

The Correct Option is B, D

A good way through this question is to first sort the four routes by WHICH ring size they build, since that alone eliminates one option before even worrying about the methyl group position.

1. (A): aldol condensation of 2-nitrobenzaldehyde with acetone, then reduction of the nitro group, is the standard Friedlander route to quinaldine (2-methylquinoline). Counting the chain atoms between the new amino nitrogen and the ketone carbon it condenses with shows a six-membered ring forms, not the five-membered pyrrole ring of an indole. Wrong ring size, so this is rejected outright.
2. (B): an ortho-nitrostyrene bearing a $\beta$-methyl group, heated with $\mathrm{P(OEt)_3}$, is the classic Cadogan-Sundberg reductive cyclization: the phosphite deoxygenates the nitro group to a nitrene, which closes directly onto the alkene to build the five-membered ring in one step, with the methyl substituent landing at C2. This is a textbook one-pot synthesis of 2-methylindole.
3. (C): 2-nitrotoluene condensed with diethyl oxalate, then reduced, is the original 1897 Reissert indole synthesis, but it installs an ester group ($\mathrm{-COOEt}$) at C2, not a methyl group, giving ethyl indole-2-carboxylate. Right ring, wrong substituent, so rejected.
4. (D): N-chloroaniline plus a $\beta$-keto sulfide is the Gassman indole synthesis: a $[2,3]$-sigmatropic rearrangement of the intermediate sulfonium species builds the pyrrole ring with the ketone's methyl group at C2 and the sulfide's methylthio group temporarily at C3. Raney nickel then strips out that C3 sulfur, leaving 2-methylindole as the final product.

Let's summarize:

• Option (A) actually builds a six-membered ring (a quinoline), so it can never give an indole regardless of substituents.
• Option (C) builds the right five-membered ring but puts an ester, not a methyl group, at C2.
• Options (B) and (D) both build the indole ring with a methyl group correctly placed at C2, one directly via nitrene cyclization, the other via a sigmatropic rearrangement followed by desulfurization.

So the reactions that give 2-methylindole are (B) and (D).