3 METHODS
3.1 Breaking N-C21 Bond and
C14-C15 Bond
Upon considering the procedure to connect C-21 and
nitrogen in the pentagon, and the connection of C-14
and C-15, the basic logic was to make form the easier
bond and then the other bond. This logic allows us to
avoid the influence of the problem of choosing the
point to connect. This means that with the application
of a sequential procedure rather than a spontaneous
reaction, we can form the desired bond first.
In this case, the bond between C-21 and that
particular nitrogen is considered to be formed through
formation of a trisubstituted amine. The reaction
starts with the central nitrogen in a substituted indole.
We will use a base to deprotonate the nitrogen,
producing a nitrogen anion. In the lower part, there
are initially two carbons connect with halogens, that
are C-21 and C-15. In this case, with the contribution
of the lone pair electron from the nearby nitrogen of
C-21, the amine formation of C-21 is more likely to
perform rather than C-15. The reason is that the lone
pair nitrogen is likely to attack C-21 due to the
inductive effect of bromine’s negativity, which makes
C-21 more electropositive. The negatively charged N
will then perform a SN2 nucleophilic substitution to
connect to C-21, forming the target bond between C-
21 and nitrogen in the indole (ABU-SHANAB, 2010).
Attempt to make the C-14, C-15 bond:
With a treatment of a base, a carbon anion could
form at C-14 on the indole structure of the upper
section. The base deprotonates the methyl group of C-
14 and forms a carbon anion. This anion again
performs an SN2 nucleophilic substitution on the
electropositive C-15 that is bonded to a Cl. Upon
attacking C-15 with the carbon anion and ejecting Cl-,
the desired C15-C14 bond is formed, so is the desired
product (Figure 3).
It is worthy noticing that benefiting from the
amine alkylation of the previous step, the strong base
added to the reaction in this stage only deprotonates
C-14 instead of forming anion on the nitrogen that is
to connect with C-21 (Hunt, 2021).
3.2 Breaking the Cyclohexene
To form the cyclohexene, use 2 smaller molecules:
Figure 4: The starting materials to make the cyclohexene.
With the starting ingredients shown in Figure 4,
the initializing step here uses a double-bonded O
instead of C on molecule 1 because the O can
mesmerically withdraw electrons, which makes the
left most C more electropositive; the reason why not
separate the ester out from part 2 and name it as part
3 is that the ester can also draw electrons toward the
O in the ester through resonance, which also helps
make C bonded to Cl more electropositive; and
connect N to a R-group because it can help stabilize
the electrons on N. All of these helps the Dials-Alder
reaction proceed (Wilson, 2001; Juhl M, 2009).
Figure 3: The mechanism for making C-14, C-15 bond.