3
20
H. Debbeche et al. / Journal of Molecular Catalysis B: Enzymatic 66 (2010) 319–324
esters [11]. The product obtained from acylation is a hemiester
which is easily separated from the unreacted alcohol by extraction
with an aqueous sodium bicarbonate solution, avoiding the use of a
chromatographic separation. Furthermore the hemiester liberates
the enantiomeric alcohol through saponification. The enantiomeric
purities of both the unreacted alcohol and the produced one were
measured by chiral HPLC. Conditions for HPLC analysis were set
up to record baseline separations of enantiomers of the mixture
of alcohols. Conversions were calculated from these values. Evalu-
ation of the E-values for each individual substrate would then be
compared with the values recorded for reactions performed onto
two- and four-substrate mixtures. This would indicate whether
the substrates behave independently, and, accordingly, which sub-
strates show high reactivity and selectivity.
(3) 1-(6-Methoxy-2-naphthyl)ethanol (3): hexane/i-PrOH, 90/10;
ꢀ = 254 nm; 1 ml/min; tR(3) (S) = 8.70 min, tR(3) (R) = 11.85 min.
(4) 1-(2-Naphthyl)ethanol (4): hexane/EtOH, 98/2; ꢀ = 254 nm;
0.8 ml/min; tR(4) (S) = 28.2 min, tR(4) (R) = 30.3 min. [(S,S)-ulmo-
column: hexane/i-PrOH, 95/5; ꢀ = 254 nm; 0.5 ml/min; tR(4)
(S) = 12.6 min, tR(4) (R) = 14.5 min].
(5) 1-Acenaphthenol (5): hexane/i-PrOH, 90/10; ꢀ = 254 nm;
0.5 ml/min; tR(5) (S) = 16.0 min, tR(5) (R) = 17.0 min.
(6) 1-(1-Naphthyl)ethanol (6): hexane/i-PrOH, 90/10; ꢀ = 254 nm;
0.9 ml/min; tR(6) (S) = 10.5 min, tR(6) (R) = 16.6 min.
(7) 1-Phenylethanol (7): hexane/i-PrOH, 95/5; ꢀ = 254 nm;
1 ml/min; tR(7) (S) = 8.37 min, tR(7) (R) = 9.67 min.
(8) 1-(4-Methoxyphenyl)ethanol (8): hexane/i-PrOH, 95/5;
ꢀ = 254 nm; 0.8 ml/min; tR(8) (S) = 13.5 min, tR(8) (R) = 15.1 min.
The use of cyclic anhydride in this multi-substrate screening
methodology is of interest: (i) the work-up is easily carried out,
since it does not require a chromatographic separation; (ii) a single
set up ofanalysishas to be done, since boththe unreacted substrates
and the products have the same structure (arylalkylethanols); (iii)
the accurate measurement (chiral HPLC) of ee’s of these alcohols
allows to calculate the conversion, and hence the E factor for each
substrate. We made the assumption that the reactions are irre-
versible and hence the E-values remained constant throughout the
reaction [12].
2
.4. HPLC analysis of the alcohols in two-substrate mixture
(
(
(
(
(
(
(
(
(
1) Alcohols [(1) + (6)]: hexane/EtOH, 98/2; ꢀ = 254 nm;
.8 ml/min; tR(1) (S) = 15.3 min, tR(1) (R) = 17.0 min; tR(6)
S) = 26.9 min, tR(6) (R) = 46.3 min.
2) Alcohols [(6) + (8)]: hexane/EtOH, 98/2; ꢀ = 254 nm;
.8 ml/min; tR(6) (S) = 27.3 min, tR(6) (R) = 46.7 min; tR(8)
S) = 20.0 min, tR(8) (R) = 23.8 min.
3) Alcohols [(1) + (8)]: hexane/EtOH, 98/2; ꢀ = 254 nm;
.8 ml/min; tR(1) (S) = 15.7 min, tR(1) (R) = 17.7 min; tR(8)
S) = 20.7 min, tR(8) (R) = 24.5 min.
4) Alcohols [(2) + (3)]: hexane/EtOH, 98/2; ꢀ = 254 nm;
.8 ml/min; tR(2) (S) = 13.5 min, tR(2) (R) = 15.0 min; tR(3)
S) = 31.1 min, tR(3) (R) = 52.2 min.
5) Alcohols [(2) + (5)]: hexane/EtOH, 98/2; ꢀ = 254 nm;
.8 ml/min; tR(2) (S) = 13.7 min, tR(2) (R) = 15.2 min; tR(5)
S) = 28.7 min, tR(5) (R) = 32.9 min.
6) Alcohols [(7) + (8)]: hexane/EtOH, 98/2; ꢀ = 254 nm;
.8 ml/min; tR(7) (S) = 13.7 min, tR(7) (R) = 17.1 min; tR(8)
S) = 19.9 min, tR(8) (R) = 21.9 min.
7) Alcohols [(1) + (7)]: hexane/EtOH, 98/2; ꢀ = 254 nm;
.8 ml/min; tR(1) (S) = 14.8 min, tR(1) (R) = 16.5 min; tR(7)
S) = 13.6 min, tR(7) (R) = 17.7 min.
8) Alcohols [(1) + (3)]: hexane/EtOH, 98/2; ꢀ = 254 nm;
.8 ml/min; tR(1) (S) = 14.9 min, tR(1) (R) = 16.9 min; tR(3)
S) = 28.9 min, tR(3) (R) = 40.6 min.
9) Alcohols [(3) + (8)]: hexane/EtOH, 98/2; ꢀ = 254 nm;
.8 ml/min; tR(3) (S) = 31.3 min, tR(3) (R) = 54.1 min; tR(8)
S) = 21.2 min, tR(8) (R) = 23.4 min.
10) Alcohols [(2) + (7)]: hexane/EtOH, 98/2; ꢀ = 254 nm;
.8 ml/min; tR(2) (S) = 19.2 min, tR(2) (R) = 21.9 min; tR(7)
S) = 16.4 min, tR(7) (R) = 17.1 min.
0
(
0
(
The commercially available Candida antarctica lipase B (CAL B)
was chosen for this study, since it is one of the most used for kinetic
resolution of alcohols. Moreover, we showed that it performed bet-
ter than Pseudomonas fluorescens lipase (PFL) [13].
0
(
0
(
2
. Experimental
2.1. Materials and methods
0
(
Substrates 1–8 and anhydrides 9–12 were commercially avail-
able and purchased from Acros Organics (1, 3, 9), Sigma–Aldrich (2, 5,
1, 12), Fluka (4), Fluka Chemika (6), Janssen (7), Janssen Chemika (10)
0
(
1
and Alfa Aesar (8). They were used as received. Lipase acrylic resin
from C. antarctica (Novozym, L4777, 3 units/mg) was purchased
from Sigma–Aldrich.
Reactions were monitored by TLC and the enantiomeric excesses
of the alcohols were determined by chiral HPLC on Chiralcel® OD-H
column (25 cm × 4.6 mm), with UV-detection at 254 nm.
0
(
0
(
0
(
2.2. General procedure for kinetic resolution
(
The mixture of equimolecular amounts of the racemic alcohol
0
(
(
1 mmol) and anhydride (1 mmol) in diethyl ether (5 ml) and the
lipase (150 mg) was stirred for 24 h at room temperature. After fil-
tration of the enzyme, the ethereal solution was extracted with an
aqueous solution of 2 M Na CO3 (2× 2 ml). The combined aqueous
2
extracts were washed with ethyl ether (5 ml) and the combined
2.5. HPLC analysis of the alcohols in four-substrate mixture
organic phases were dried (MgSO ) and evaporated to recover the
4
unreacted alcohol. To the combined aqueous phases was added an
aqueous NaOH solution (5 ml, 10%). After 2 h the aqueous phase was
extracted with ethyl ether (2× 5 ml), the combined ethereal phases
(
1) Alcohols [(1) + (3) + (6) + (8)]: hexane/EtOH, 98/2; ꢀ = 254 nm;
.8 ml/min; tR(1) (S) = 16.3 min, tR(1) (R) = 18.4 min; tR(3)
0
(
(
S) = 34.6 min, tR(3) (R) = 55.9 min; tR(6) (S) = 31.5 min, tR(6)
R) = 50.4 min; tR(8) (S) = 22.3 min, tR(8) (R) = 24.6 min.
were dried (MgSO ) then evaporated to give the alcohol which has
4
reacted in the acylation process.
(
2) Alcohols [(1) + (4) + (7) + (8)]: hexane/EtOH, 98/2; ꢀ = 254 nm;
.8 ml/min; tR(1) (S) = 15.9 min, tR(1) (R) = 18.0 min; tR(4)
(S) = 29.7 min, tR(4) (R) = 33.4 min; tR(7) (S) = 14.3 min, tR(7)
R) = 19.0 min; tR(8) (S) = 21.0 min, tR(8) (R) = 22.9 min.
0
2.3. HPLC analysis of racemic alcohols 1–8
(
(
1) 1-Indanol (1): hexane/i-PrOH, 98/2; ꢀ = 254 nm; 0.8 ml/min;
tR(1) (S) = 20.7 min, tR(1) (R) = 23.9 min.
2) 1,2,3,4-Tetrahydro-1-naphthol (2): hexane/EtOH, 98/2;
(3) Alcohols [(2) + (3) + (5) + (7)]: hexane/EtOH, 98/2; ꢀ = 254 nm;
0.8 ml/min; tR(2) (S) = 10.5 min, tR(2) (R) = 11.2 min; tR(3)
(S) = 21.0 min, tR(3) (R) = 33.7 min; tR(5) (S) = 23.4 min, tR(5)
(R) = 25.8 min; tR(7) (S) = 11.1 min, tR(7) (R) = 14.0 min.
(
ꢀ
= 254 nm; 0.8 ml/min; tR(2) (S) = 9.45 min, tR(2) (R) = 10.1 min.