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W. E. Kowtoniuk, D. K. MacFarland / Tetrahedron Letters 46 (2005) 451–453
Our initial hypothesis was that a byproduct of the
carbonyl reduction (AlH3 perhaps?) that was coordin-
atively unsaturated could chelate to the BINOL oxy-
gens, then transfer a hydride to give an aluminum
complex. This is not consistent with the observation of
ether cleavage in the case with no carbonyl, however,
as the byproduct is not generated without a reducible
substrate. We also considered the possibility that an
impurity in the LAH was catalyzing cleavage, but the
absence of cleavage in the naphthalene examples is not
consistent with this possibility, as the impurity would
be present in all cases.
It is possible that aluminum could become six coordi-
nate (four hydrides and the chelating oxygens), which
would circumvent the need for prereaction of the
LAH. Six-coordinate aluminum hydride complexes are
known,4 though for hydride alkoxy complexes, it is the
five-coordinate AlH3(OR)2 complexes that are re-
ported.5 It is also possible that at the elevated reaction
temperature, the AlH4ꢀ degrades into AlH3. Thus, the
geometries required for BINOL–LAH chelation are
known and reasonable for either five or six-coordinate
chelation. In either case, the aluminum could act as a
Lewis acid and lead to nucleophilic hydride attack on
the alkyl.
Figure 2. General reduction/cleavage scheme for BINOL and naphthol
ethers derivatives. Reagents and conditions: (i) LAH, dioxane, reflux
60 h.
Such an unusual reaction required further study. Two
questions were of particular interest: (1) is the reaction
general to reduction of other carbonyls? and (2) is the
reaction general to other, non-BINOL ethers? We de-
signed and synthesized two sets of ether-reduction po-
tential substrates. For both the BINOL and
naphthalene cases, we tested three carbonyls (an amide,
acid, and ester), and a ÔcontrolÕ with no carbonyl (Fig.
2). To test if the BINOL was acting as a chelate, we syn-
thesized ether derivatives of 2-hydroxy-6-naphthoic
acid. These naphthalene analogues have the same elec-
tronic state in the aromatic ring, and the same ether to
be cleaved, but cannot chelate a metal.
In each of the BINOL cases, cleavage of some or all of
the ethers was observed. Thus, the cleavage did not de-
pend on the group being reduced (X = acid, ester, or
amide), or even on the presence of a reducible group
(X = H). This last case was most interesting to us be-
cause no alcohol– or amine–aluminum complex could
be present to facilitate the ether cleavage.
3. Experimental
Each of the BINOL carbonyls was cleanly reduced, and
in each case some ether cleavage was observed (Table 1).
Even in the ÔcontrolÕ no carbonyl case, ether cleavage
was observed. Only the amide gave complete cleavage
of both BINOL ethers. All the other cases (acid, ester,
control) gave monoether–mononaphthol as the major
product, with some binaphthol observed as well. Why
the different functional groups give cleavage to different
extents is unclear.
3.1. Reduction procedure
LAH (6 equiv) per carbonyl to be reduced were placed
in a round bottom flask with a minimal amount of
freshly distilled dioxane (15–20 mL). The dioxane was
allowed to reflux while stirring for approximately an
hour until a homogenous slurry was obtained. The reac-
For each of the naphthoic acid derivative ethers 11–13,
reduction of the carbonyl was clean, but no ether cleav-
age was observed (Table 2). No cleavage was observed
for the naphthalene ÔcontrolÕ 14 as well, with only start-
ing material recovered. Thus, we propose that cleavage
involves a chelate of the BINOL ether oxygens. Chelat-
ing BINOL complexes of aluminum are known, though
they are considered four coordinate.3
Table 2. Results of LAH reduction/cleavage of naphthol derivatives
(determined by GC/MS)
Reaction
X
Y
R1 = H % R1 = decyl %
11!15
12!16
13!17
14!18
CONBn2
COOH
CH2NBn2
CH2OH
0
0
0
0
100
100
100
100
COOndecyl CH2OH
H
H
Table 1. Results of LAH reduction/cleavage of BINOL derivatives (determined by GC/MS)
Reaction
X
Y
R1 = R2 = H %
R1 = H, R2 = hex %
R1 = R2 = hex %
1!7
4!8
5!9
6!10
CON(CH3CHPh)2
CH2N(CH3CHPh)2
100
58
0
0
42
90
53
0
0
COOH
COOiPr
H
CH2OH
CH2OH
H
10
21
26