Problem 7. Which is (±)-Trikentrin A?

Although the indole skeleton is ubiquitous in nature, annulated indoles at any of the benzenoid positions are uncommon. The trikentrins and the structurally similar herbindoles represent fascinating such examples of 6,7-annulated indole or polyalkylated cyclopent[g]indole natural products. The trikentrins were isolated from the marine sponge Trikentrion flabelliforme and possess antibacterial activity. Possible structures for trikentrin A are shown in the Figure below. In this problem, we will find out which of these structures is trikentrin A.

There are several ways to synthesize trikentrin A. Two routes below involve aryne-based and hydrovinylation strategies and both finally lead to the formation of trikentrin A. The first step for problems 8.1 and 8.2 is the Bartoli reaction or Bartoli indole synthesis, which is the organic reaction of ortho-substituted nitroarenes with vinyl Grignard reagents to yield substituted indoles. In particular, it is the most efficient route to 7-substituted indoles.

(±)-Trikentrin A: 13C NMR (CDCl3): δ 143.4101.6 (8 signals), 44.815.1 (7 signals).

Aryne-based strategy

7.1. Draw the structures of A–I.

7.2. Draw the structure of the aryne involved as a reaction intermediate in step D → E.

Hydrovinylation strategy

7.3. Chemical transformation of bromo-nitrobenzene to corresponding 7-vinylindole J includes in Bartoli reaction followed by the vinylation step with vinylstannane. Draw the structure of J.

7.4. The second step is the Ni(II)-catalyzed asymmetric hydrovinylation of J. The ligands (K1–K4) used for hydrovinylation are given above.

Note: ee = enantiomeric excess; % ee = % major enantiomer - % minor enantiomer

Choose the correct statement(s):

☐ Ligand 3 gave the best enantioselectivity. ☐ Ligand 4 gave a racemic mixture. ☐ Each of the ligands K1–K4 is chiral. ☐ Each of the ligands K1–K4 gave excellent yield (>95%) of the product.

7.5. For the hydrovinylation step, choose the correct statement(s):

☐ (allyl)2Ni2Br2 or [(allyl)NiBr]2 is a source of vinyl. ☐ In this Ni-allyl complex, each nickel has oxidation number +2. ☐ In this Ni-allyl complex, the electron count of Ni is 18. ☐ This complex has a square planar geometry.

7.6. Draw the structures of L–P. The absolute configuration of the asymmetric center in the hydrovinylation product is S. Hint: In the 13C NMR spectrum of compound M, one carbonyl carbon signal was observed at δ = 178.3 ppm.

Solution:

7.1.

7.2.

7.3.

7.4.

☒ Ligand 3 gave the best enantioselectivity. ☒ Ligand 4 gave a racemic mixture. ☐ Each of the ligands K1–K4 is chiral. ☐ Each of the ligands K1–K4 gave excellent yield (>95%) of the product.

7.5.

☐ (allyl)2Ni2Br2 or [(allyl)NiBr]2 is a source of vinyl. ☒ In this Ni-allyl complex, each nickel has oxidation number +2. ☐ In this Ni-allyl complex, the electron count of Ni is 18. ☒ This complex has a square planar geometry.

7.6.