Reduction of benzophenone by SmI2: The role of proton donors in determining product distribution

Article abstract of DOI:10.1021/ol051623q

(Graph Presented) Reduction of benzophenone by SmI₂ yields benzopinacol. Addition of proton donors results in an initial increase in the amount of the benzhydrol formed. However, the ratio benzhydrol/benzopinacol reaches a maximum, decreases, and then levels off as the proton donor concentration is further increased. The position of the maximum and its height depend on the proton donor concentration and its kinetic acidity. The momentary concentration of the intermediate radicals governs the product distribution.

Full text of DOI:10.1021/ol051623q

ORGANIC
LETTERS
2005
Vol. 7, No. 19
4197-4200
Reduction of Benzophenone by SmI2:
The Role of Proton Donors in
Determining Product Distribution
Gregory Kleiner, Alexander Tarnopolsky, and Shmaryahu Hoz*
Department of Chemistry, Bar-Ilan UniVersity, Ramat-Gan 52900, Israel
shoz@mail.biu.ac.il
Received July 11, 2005
ABSTRACT
Reduction of benzophenone by SmI2 yields benzopinacol. Addition of proton donors results in an initial increase in the amount of the benzhydrol
formed. However, the ratio benzhydrol/benzopinacol reaches a maximum, decreases, and then levels off as the proton donor concentration is
further increased. The position of the maximum and its height depend on the proton donor concentration and its kinetic acidity. The momentary
concentration of the intermediate radicals governs the product distribution.
Reduction of carbonyl compounds is one of the focal points
in the chemistry of SmI2. It may yield two products, the
corresponding alcohol and a coupling productspinacol (for
reviews, see refs 1-3).
Besides numerous synthetic applications, several mecha-
nistic aspects of this reaction have also been investigated.
These include the effects of additives such as poly(ethylene
glycol) ethers on the diastereoselectivity of the coupling
reaction,4 the effect of solvents on the latter,5 and the rate
acceleration by amines,6 diols,7 and â-complexation.8,9 Daas-
bjerg and Skrydstrup determined the rate constant for the
electron-transfer step between SmI2 and acetophenone to be
7 M-1 s-1 and concluded that this was an innersphere
process.10 The same conclusion was reached by Flowers and
co-workers who also observed that LiCl or LiBr enhance
the coupling process.11 Recently, Hilmerson and Flowers
reported that reduction and coupling of carbonyl compounds
can be mediated by microwave heating.12 Pertinent to the
(6) Dahlen, A.; Hilmersson, G. Tetrahedron Lett. 2002, 43, 7197-7200.
(7) Dahlen, A.; Hilmersson, G. Tetrahedron Lett. 2001, 42, 5565-5569.
(8) Prasad, E.; Flowers, R. A. J. Am. Chem. Soc. 2002, 124, 6357-
6361.
(9) Prasad, E.; Flowers, R. A. J. Am. Chem. Soc. 2002, 124, 6895-
6899.
(1) Molander, G. A. Org. React. 1994, 46, 211-367 and references
therein.
(2) Molander, G. A. Chem. ReV. 1996, 96, 307-338.
(3) Kagan, H. B. Tetrahedron 2003, 59, 10351-10372 and references
therein.
(10) Enemaerke, R. J.; Daasbjerg, K.; Skrydstrup, T. Chem. Commun.
1999, 343-344.
(4) Pedersen, H. L.; Christensen, T. B.; Enemaerke, R. J.; Daasbjerg,
K.; Skrydstrup, T. Eur. J. Org. Chem. 1999, 565-572.
(5) Chopade, P. R.; Davis, T. A.; Prasad, E.; Flowers, R. A. Org. Lett.
2004, 6, 2685-2688.
(11) Miller, R. S.; Sealy, J. M.; Shabangi, M.; Kuhlman, M. L.; Fuchs,
J. R.; Flowers, R. A. J. Am. Chem. Soc. 2000, 122, 7718-7722.
(12) Dahlen, A.; Prasad, E. Flowers, R. A.; Hilmerson, G. Chem. Eur.
J. 2005, 11, 3279-3284.
10.1021/ol051623q CCC: $30.25
© 2005 American Chemical Society
Published on Web 08/25/2005

Figure 2. H/P ratio in the reaction of benzophenone (0.022 M)
and SmI2 (0.044 M) as a function of MeOH.
and the pinacol fraction does not change with time, it is clear
that product distribution is determined at the beginning of
the reaction. The intermediate which is the precursor of the
hydrol regenerates the starting materialsbenzophenones
when it reacts with I2 of the quench solution. The most likely
structure of this intermediate is the dibenzylic carbanion.
Under the reaction conditions it cannot generate pinacol,
whereas protonation of it will furnish the hydrol. When this
intermediate is reacted with the quench I2 solution, it reverts
to the starting materialsbenzophenone (eq 2).
Figure 1. Spectra of SmI2 and of the radical anion of benzophenone
(RA).
present study is another observation by Flowers and co-
workers that the reduction of acetophenone is enhanced by
proton donors according to their acidity.13
In the present work, we will use benzophenone as a model
and focus on the unusual effect alcohols have on the product
distribution (benzhydrol vs benzopinacol) in its reduction of
by SmI2 (eq 1).
The existence of a C-Sm bond has been demonstrated
previously.15 However, the fact that it undergoes a slow
protonation may seem somewhat surprising; nevertheless, we
have previously shown that samarium exchanges ligands
surprisingly slowly.16 As a result of the above observation,
all of the product distribution data below was collected 10
min after the beginning of the reaction.
In the absence of a proton source, hydrol is not obtained,
and the coupling yielding pinacol is the only reaction. One
would therefore expect that the hydrol/pinacol (H/P) ratio
would increase with rising concentration of the proton donor.
Surprisingly, using MeOH as a proton donor, we found that
the ratio indeed increases at the beginning but then reaches
a maximum after which it drops and then nearly leVels off
(Figure 2).
The effect of the proton donor on the product distribution
can be explained by assuming the mechanism shown in
Scheme 1.
This is a general mechanism from which various reaction
steps were omitted for the sake of simplicity. According to
this scheme, we start at the radical anion (RA) formed
quantitatively at the dead time. In the absence of a proton
donor, two radical anions paired to samarium will couple to
produce pinacol (k1). As the concentration of the MeOH is
When equivalent amounts of benzophenone and SmI2 are
mixed in a stopped-flow spectrophotometer, the typical
absorption of the SmI2 vanishes in the dead time and the
absorption of the radical anion of benzophenone is im-
mediately obtained (Figure 1).
For a complete formation of benzhydrol, 2 equiv of SmI2
is required. In the following experiments, we have used the
concentrations of 0.0167 M for benzophenone and 0.0334
M for SmI2 in THF. The reactions were performed in the
presence of 0.0334 M MeOH and quenched by an I2/THF
solution. Product analysis was done, in most cases, by both
NMR and HPLC. Based on the difference between the two,
we estimate the experimental error to be in the range of 5%.
It turns out that while the percent of pinacol14 remains
constant at all times, that of the hydrol increases at the
expense of the benzophenone until it reaches its final value
in 10 min. Since the radical anion is obtained upon mixing,
(13) Chopade, P. R.; Prasad, E.; Flowers, R. A. J. Am. Chem. Soc. 2004,
126, 44-45.
(14) The percent yield for pinacol is based on the number of benzophe-
none molecules used for its production, namely, mol % pinacol × 2.
4198
Org. Lett., Vol. 7, No. 19, 2005

Scheme 1. Reaction Mechanism
Table 1. Product Distribution in the Reaction of
Benzophenone (0.022 M) and SmI2 (0.044 M) as a Function of
[ROH]
[ROH] (M) hydrol (%) pinacol14 (%) ketone (%) H/P
MeOH
i-PrOH
TFE
0
0
47
33
31
21
19
0
67
57
55
50
50
0
100
48
63
66
76
77
100
31
42
45
50
50
100
73
79
83
84
82
0
5
4
3
3
4
0
2
1
0
0
0
0
0
0
0
0
0
0.00
0.98
0.52
0.47
0.28
0.25
0.00
2.23
1.38
1.22
1.00
1.00
0
0.044
0.110
0.165
0.220
0.275
0
0.029
0.055
0.114
0.165
0.220
0
0.028
0.055
0.083
0.138
0.275
increased, protonation (k2) to form the radical (R) will be
enhanced and the importance of k1 will diminish. Since R is
a precursor of the hydrol (H), the H/P ratio will begin to
increase. At a high enough concentration of alcohol, the
initially formed radical anion will be fully and rapidly
converted to the radical via k2. Product distribution will be
determined by the relative rates of the pinacol producing
coupling (k3) and further reduction by the SmI2 (k4). Since
the proton donor concentration does not enter the rate
equations for these two processes, the product ratio depen-
dence on the proton donor concentration (Figure 2) must
evolve from the previous stepsthe protonation of the radical
anion (k2). The molecularities of these reactions (k3 and k4)
are such that the formation of pinacol is second order in the
radical, whereas that of the hydrol is only first order in this
radical. Thus, slow protonation of the radical anion will
produce a momentary low concentration of the radical. This
will reduce the probability of a bimolecular reaction of the
radical (pinacol formation via k3) and, therefore, increase
the fraction of the formed hydrol. As the protonation rate
grows faster due to an increase in the MeOH concentration,
the momentary concentration of the radical (R) increases,
raising the probability for the bimolecular coupling to yield
pinacol. Hence, the H/P ratio will decrease. In the extreme
case where protonation is instantaneous, product distribution
will be independent of the proton donor concentration and a
plateau will be achieved. The plateau level will be determined
by the ratio of the coupling (via k3) and the reduction (via
k4) rates.
27
21
17
16
18
0.37
0.27
0.20
0.19
0.22
2-propanol as an alcohol of a weaker acidity (Table 1 and
Figure 3).
Figure 3. H/P ratio in the reaction of benzophenone (0.022 M)
and SmI2 (0.044 M) as a function of alcohol concentration.
It should be pointed out that at alcohol concentrations
lower than 1 equiv, the results are of low relevance. This is
because the alcohol is completely consumed before the
reaction is completed, and in addition, traces of water may
affect the distribution. As a result, data points are not
available for the lower concentration range.
Using computer simulation, we can analyze the data in a
more quantitative manner. Stopped flow measurements18 in
the absence of a proton donor yield a k1 value of 5 × 102
M-1 s-1. The coupling rate constant k3 is known for
acetonitrile as a solvent.19 We will use this value (2 × 108
M-1 s-1) on the assumption that radical coupling is not much
Based on the above mechanism, one can predict a two-
fold consequence of increasing the protonation rate (k2) by
using a more acidic alcohol: (a) the maximum in the H/P
vs [ROH] plot will be achieved at a lower H/P value, and
(b) the plateau will be achieved earlier.17 To examine these
predictions we have conducted experiments with two alco-
hols, trifluoroethanol (TFE) as the more acidic alcohol and
(15) (a) Curran, D. P.; Fevig, T. L.; Totleben, M. J. Synlett 1990, 773-
774. (b) Curran, D. P.; Totleben, M. J. J Am. Chem. Soc. 1992, 114, 6050-
6058. (c) Molander, G. A.; Kenny, C. J. Org. Chem. 1991, 56, 1439-
1445. (d) Namy, J. L.; Collin, J.; Bied, C.; Kagan, H. B. Synlett 1992,
733-734.
(16) Yacovan, A.; Bilkis, I.; Hoz, S. J. Am. Chem. Soc. 1996, 118, 261-
262. This contradicts somewhat the statement made by Kagan describing
organosamarium compounds as highly reactive intermediates.3 However,
the nature of the two systems is entirely different.
(17) One could intuitively conclude that the maximum in the simulation
line will move towards lower [ROH] with the increase in the alcohol acidity.
However when [ROH] < [RA], stoichiometry controls the H/P value, and
for all the alcohols, the maximum value will be achieved at [ROH] ) [RA].
Org. Lett., Vol. 7, No. 19, 2005
4199

affected by the change of solvent from MeCN to THF.
Product distribution at the plateau of TFE (where k1 is
probably not operative) together with the k3 value, yield, by
computer simulation a k4 value of 2 × 107 M-1 s-1. Manual
radical anions and the corresponding radicalssis low. As a
result, the rate of the bimolecular steps leading to pinacol
formation (steps corresponding to k1 and k3 in Scheme 1)
cannot compete effectively with the step leading to reduction.
On the other hand, the radical anion of an aromatic ketone
such as benzophenone is formed very rapidly. As a result,
the momentary concentration of these intermediates is high,
increasing the probability of the bimolecular coupling
reaction to give pinacol.
In conclusion, the key to a good yield of coupling products
is a high momentary concentration of radical anions and,
more importantly, a high concentration of the derived
radicals. Slow reduction will produce a low momentary
concentration of the radical anions/radicals which will favor
the complete reduction of the carbonyl function. Thus, a
recipe based on the reducing power of the reagent,22 its
concentration, and the characteristics of the proton donor can
be tailored, in principle, for most substrates to channel the
reaction to the desired products.
variation of k2 (until reasonable fit to the experimental data
TFE
was obtained) gave the following rate constants: k2
)
i
108 M-1 s-1, k2MeOH ) 2.5 × 106 M-1 s-1, and k2 PrOH ) 7 ×
105 M-1 s-1. Due to the assumptions used in the mechanistic
scheme,20 these can be considered only approximate values.
It should be pointed out that we have performed the
simulation from different starting points to minimize the
probability of the solution being a local minimum.
This simple model also provides an explanation for the
long puzzling difference observed between aromatic and
aliphatic ketones. While under normal conditions (in the
presence of proton donors), aliphatic ketones yield mainly
the corresponding alcohols, benzophenones yield also the
corresponding pinacols.1,21 The key to the product partition
between hydrol and pinacol lies in the momentary concentra-
tion of the radical. The electron-transfer process in aliphatic
ketones is relatively slow. Therefore, the momentary con-
centration of the candidates for bimolecular couplingsthe
Supporting Information Available: Experimental pro-
cedures. This material is available free of charge via the
Internet at http://pubs.acs.org.
OL051623Q
(18) See the Experimental Section in the Supporting Information.
(19) Shield, S. R.; Harris, J. M. Anal. Chem. 1998, 70, 2576-2583.
(20) A more refined scheme should also include additional factors and
steps such as the complexation of MeOH to samarium at high MeOH
concentrations, possible coupling between a radical and a radical anion,
disproportionation of two samarium paired radical anions to give a bridged
dimer, etc.
(21) Kagan, H. B.; Namy, J. L.; Girard, P. Tetrahedron 1981, 37, 175-
180.
(22) This could be also achieved with SmI2 itself since adding HMPA
increases its reduction potential from -0.89 to -1.79V vs SCE (ref 10
above).
4200
Org. Lett., Vol. 7, No. 19, 2005