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

Step 1: Recognize this as a "spot the named indole synthesis" question.
Each option draws a classic named synthesis. We need the one(s) that specifically build the INDOLE ring (a five-membered pyrrole fused to benzene) with a methyl group landing at C2, not some other heterocycle or a differently substituted indole.

Step 2: Check option (A), actually a quinoline synthesis, not an indole synthesis.
2-Nitrobenzaldehyde and acetone under \(\mathrm{NaOH}\) undergo an aldol (Claisen-Schmidt) condensation to give the enone \(\mathrm{Ar\text{-}CH{=}CH\text{-}CO\text{-}CH_3}\) (\(\mathrm{Ar}=\) 2-nitrophenyl). Reducing the nitro group with \(\mathrm{Sn/HCl}\) generates an ortho-amino group, which then condenses intramolecularly with the ketone carbon three atoms further down the chain. Counting atoms around that ring closure gives a SIX-membered ring fused to benzene, i.e. this is the Friedlander-type synthesis of quinaldine (2-methylquinoline), not an indole. So (A) does not give 2-methylindole.

Step 3: Check option (B), the Cadogan (Sundberg) reductive cyclization.
The substrate is an ortho-nitro-\(\beta\)-methylstyrene, \(\mathrm{Ar(NO_2)\text{-}CH{=}CH\text{-}CH_3}\). Heating with the trivalent phosphite \(\mathrm{P(OEt)_3}\) deoxygenates the nitro group to a nitrene at nitrogen; this nitrene undergoes intramolecular cyclization onto the adjacent alkene (with loss of \(\mathrm{P(OEt)_3{=}O}\)), closing the five-membered pyrrole ring directly. The carbon that carried the methyl group becomes indole C2 (next to nitrogen), so this gives 2-methylindole directly, in one pot. So (B) works.

Step 4: Check option (C), the classical Reissert synthesis of an indole-2-CARBOXYLATE, not 2-methylindole.
2-Nitrotoluene's benzylic \(\mathrm{CH_3}\) (activated by the ortho nitro group) is deprotonated by \(\mathrm{NaOEt}\) and undergoes a Claisen condensation with diethyl oxalate, installing a \(\mathrm{-CH_2COCOOEt}\) side chain. Reducing the nitro group with \(\mathrm{Zn/AcOH}\) triggers reductive cyclization to give ethyl indole-2-carboxylate, an ester at C2, not a methyl group. So (C) gives the wrong substituent at C2 and is not the answer.

Step 5: Check option (D), the Gassman indole synthesis.
N-Chloroaniline reacts with the \(\beta\)-keto sulfide \(\mathrm{MeS\text{-}CH_2\text{-}CO\text{-}CH_3}\): the sulfide attacks the electrophilic N-Cl nitrogen, and base (\(\mathrm{Et_3N}\), heat) promotes a \([2,3]\)-sigmatropic rearrangement onto the ortho position of the ring, followed by cyclization, to give a 3-(methylthio)-2-methylindole intermediate (the "\(\mathrm{Me}\)" from the ketone lands at C2, and the sulfide sulfur ends up at C3). \(\mathrm{Raney\text{-}Ni}\) then desulfurizes, removing the C3 \(\mathrm{-SMe}\) group and leaving plain 2-methylindole. So (D) works.

Final Answer:
Only the Cadogan-type reductive cyclization (B) and the Gassman synthesis (D) deliver 2-methylindole; (A) gives a quinoline and (C) gives an indole-2-carboxylate. \[ \boxed{\text{(B) and (D)}} \]