Angewandte
Chemie
naphthalenide and the resulting boryl radical 3 again reacts
with a second equivalent of sodium naphthalenide. However,
we suggest that radical–radical anion coupling occurs in this
second encounter rather than electron transfer. This provides
anion 6, which undergoes intramolecular substitution to give
the products 2. Recent studies on diverse substitution
reactions of NHC-boryl halides and triflates provide prece-
dent for this last step.[10]
The reaction between 3 and sodium naphthalenide is
effectively a radical–radical combination whose rate could
approach the diffusion-controlled limit. This explains why
external traps cannot “outcompete” the reagent for trapping.
Radical–molecule reactions cannot compete with radical–
radical reactions when one of the radicals (here NaN) is
present in high concentration. The negligible barrier of this
reaction also accounts for the lack of stereoselectivity in
formation of the two diastereomers of 6 and hence 2.
This mechanism has precedent in the reactions of sodium
naphthalenide with alkyl halides summarized in Scheme 3.
These reactions have low preparative value because they give
complex product mixtures.[11] For example, treatment of alkyl
À37.4 ppm that corresponds to the doubly reduced, proton-
ated product 7 (diMe-Imd-BH3, < 5%) and an intriguing
sharp triplet at À24.6 ppm (15%).
The formation of 7 suggested that our reaction medium
had traces of a proton source (water?) that was not present in
Braunschweigꢀs experiments. This in turn suggested that some
diMe-Imd-BH2Cl (8) was formed during our reaction. Could
the triplet at À24.6 ppm be due to a coupling product derived
from 8 and NaN?
To answer this question, we prepared 8 (see Supporting
Information) and reduced it with sodium naphthalenide. The
results of two key experiments are shown in Scheme 4. The
Scheme 3. Reductions of alkyl halides with NaN provide complex
À
mixtures of aliphatic and naphthalene-derived products. (R( H) is an
alkene formed from disproportionation.)
Scheme 4. Reduction of mono-chloroborane 8 gives stable coupled
product 10.
halides with sodium naphthalenide provides aliphatic prod-
ucts of reduction (RH), coupling (RR) and disproportiona-
tion (R(-H)) along with assorted naphthalene coupling
products.[12] All the aliphatic products are said to arise directly
or indirectly from alkylsodium intermediates. The ratio of
11B NMR spectrum of the reaction product resulting from the
addition of 2.5 equiv NaN in THF to 8 showed no resonances
for 2. There were three resonances: the sharp triplet at
À24.6 ppm (about 25%), the quartet for 7 (about 25%), and
in between a new, very broad triplet at À33.0 ppm (about
50%). This last product might be diborane 9,[15] but it did not
survive flash chromatography.
Inverse addition of 8 to excess NaN (7.5 equiv) provided a
cleaner crude product; the broad triplet resonance was absent
and the sharp triplet resonance predominated over the
quartet of diMe-Imd-BH3 (7) (about 65% to 35%). Flash
chromatography provided the new product 10 in 24%
isolated yield. The initial structure assignment by an HH
COSY experiment was confirmed by X-ray crystallography
(Figure 1).
naphthalene-derived products to aliphatic products increases
[12a,13]
À
in order of R X primary < secondary < tertiary.
This trend originates from a competition in the reactions
of alkyl radicals with sodium naphthalenide between electron
transfer (reduction) and radical–radical anion coupling. As
the radical RC becomes more difficult to reduce to an anion
(primary < secondary < tertiary), the radical–radical anion
coupling takes over and the amounts of naphthalene-derived
products increase. Tellingly, chemical and electrochemical
reduction studies have shown that NHC-boryl radicals are
easy to oxidize[5a,14] but difficult to reduce.[4b] So boryl radicals
are prime candidates for coupling.
We suggest that 10 arises from the coupling of the boryl
radical derived from 8 with NaN followed by protonation.[16]
Although this radical (diMe-Imd-BH2C) is not the same as the
previous radical 3 (diMe-Imd-BHClC), it is similar. So this
result provides additional evidence that the radical–radical
anion coupling mechanism for formation of 2 is viable.
To start, we repeated the experiment to prepare 2.[7]
Looking for minor side products, we recorded an 11B NMR
spectrum of the crude product mixture. As expected, the
major peaks at À33.2 and À36.9 ppm are from the two
diastereomers of 2.[7] We also observed a small quartet at
Angew. Chem. Int. Ed. 2012, 51, 1602 –1605
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1603