• Nem Talált Eredményt

2. Materials and methods 30

2.6. Calculations

2.6.2. Yield

In the most general approach, the yield (Y) of the extract and raffinate can be calculated by comparing them to the total amount of starting material:

Ye= me

mrac+mres (2.6a)

Yr= mr

mrac+mres (2.6b)

In the definitions above,mdenotes mass, the indiceseand r refer to the extract and raffinate and the indices rac and res refer to racemate and resolving agent. Unlike the approaches presented below, this definition can be used regardless of the experi-mental technique. However, a significant disadvantage of this approach is that yields are not normalized, thus for different racemate–resolving agent pairs, the yield val-ues will vary depending on the molar masses and molar ratios (mr) of the compounds involved.

One approach for normalizing yields is to use an ideal resolution as the reference, assuming a complete, irreversible equimolar reaction between the resolving agent and the racemate:

Ye= me

(1−mr)·mrac (2.7a)

Yr= mr

mr·mrac+mres (2.7b)

As with Eq. 2.6a and 2.6b,mdenotes mass, the indiceseandrrefer to the extract and raffinte and the indices rac and res refer to racemate and resolving agent, while mr refers to the molar ratio. The multiplication of the molar ratio (defined in Eq.

1.2 as the ratio of two molar quantities) by a mass, although unusual, does in fact produce correct dimensions. Consider the product mr·mracin the denominator of Eq.

2.7b and substitute the definition of mr from Eq. 1.2:

mr·mrac= nres nrac ·mrac

Using the identity n= Mm, wherendenotes molar quantity, mdenotes mass and M denotes molar mass, yields:

mr·mrac= nres

mrac

Mrac

·mrac

mr·mrac=nres·Mrac (2.8)

The physical interpretation of Eq. 2.8, considering the identityn·M=m(where, as above,n, M andmdenote molar quantity, molar mass and mass, respectively), is a mass of the racemate equal in molar amount to the resolving agent, i.e. the mass of racemate consumed in an equimolar reaction with the resolving agent. A similar derivation starting from the denominator of Eq. 2.7a yields:

(1−mr)·mrac= (nracnresMrac (2.9) .

The physical interpretation of Eq. 2.9 is the mass of the racemate minus the mass equal in molar amount to the resolving agent, i.e. the mass of unreacted racemate left over after an equimolar reaction with the resolving agent.

In principle, values ofY range between 0–1, although bothYeandYrcan exceed 1 under certain conditions. If the extraction of the unreacted enantiomers is incomplete (due to factors such as inadequate stirring or insufficient amount of scCO2used for the extraction), the increased amount of racemate enantiomers in the raffinate can cause Yrto exceed 1. On the other hand, if the diastereomer formation does not proceed to completion or is not irreversible, the increased amount of unreacted enantiomers in the extract can causeYeto exceed 1.

Although the incorporation of mr eliminates the variation of Y with the molar masses of the racemate and resolving agent, it introduces an issue: the values, as defined in Eq. 2.7, are only valid if mr < 1. If mr = 1, the value of Ye is invalid (division by zero), while for mr > 1, Ye is negative and the denominator of Yr is

physically invalid as it exceeds the total mass of material used for the resolution (due to the term mr·mrac where mr > 1). Although the latter issue could be rectified by replacing mr in the denominator Eq. 2.7b with the expression min(1; mr), Eq.

2.7a cannot be augmented to yield correct values for mr≥ 1. If there is significant dissociation of the diastereomeric salts (as seen, for example, in Section 3.1.2), the resolution might be successful at or above mr=1, and therefore Y cannot be used for these systems.

For experiments carried out in the batch reactor, the extraction efficiency can be taken into consideration by treating the reactor as a continuously stirred tank reac-tor (CSTR). For a dissolved compound, let ˙m(t) and m(t) refer, respectively, to its effluent mass flow rate and dissolved mass in the reactor, at time t. According to the conservation of mass, the following are related by the following equation:

˙

m(t) =−dm(t) dt

Substituting ˙m(t) =V˙(tc(t), where ˙V andc(t)denote the effluent volumetric flow rate and the effluent concentration, respectively, at time t, yields:

V˙(tc(t) =−dm(t) dt

The shorthand notation ˙V(t)is expanded to dVdt(t), whereV(t)denotes the cumu-lative effluent volume at time t, and the resulting equation is simplified:

dV(t)

dt ·c(t) =−dm(t) dt dV(tc(t) =−dm(t)

The effluent concentration at time t is given by the mass of the dissolved com-pound in the reactorm(t)and the reactor volumeVreactorbyc(t) = Vmreactor(t) . Substituting this into the equation above and separating the variables yields:

dV(tm(t)

Vreactor =−dm(t) dV(t)

Vreactor =−dm(t)

m(t) (2.10)

Let t=0 denote the start of the extraction, letV(0) =0 and letVextractiondenote the volume of CO2 used for the extraction (at the pressure and temperature of the reactor). Furthermore, letm0andmdenote the mass of the material at the start and the end of extraction, respectively. Integrating Eq. 2.10 gives:

1 Vreactor

Z Vextraction

0

dV =− Z m

m0

dm(t) m(t)

Vextraction

Vreactor =−lnm−lnm0

Combining the logarithms, the mass of material remaining in the reactor is ob-tained as a function of the volume of CO2used for extraction:

m=m0·e

Vextraction Vreactor

From the equation above, the mass of material extracted by a given volume of CO2 is obtained:

m0m=m0m0·e

Vextraction Vreactor

m0m=m0·

1−e

Vextraction Vreactor

 (2.11)

Dividing Eq. 2.11 bym0, the amount of material extracted relative to the amount of starting material can be calculated:

m0m

m0 =1−e

Vextraction

Vreactor (2.12)

An ideal resolution (as defined on p. 47), i.e. a complete, irreversible, equimolar reaction between the racemate and resolving agent is assumed. In this case, the initial massm0of the unreacted enantiomeric mixture, as shown in Eq. 2.9, ismrac·(1−mr).

Substituting this into Eq. 2.12 and rearranging yields:

m0m

(1−mr)·mrac =1−e

Vextraction Vreactor

Let ˆme denote the expected theoretical mass of the extract m0m. It can be expressed by rearranging the above equation:

ˆ

me= (1−mr)·mrac·

1−e

Vextraction Vreactor

 (2.13)

The mass of the raffinate can be expressed as the sum of three terms: the mass of racemate consumed in an equimolar reaction with the resolving agent mr·mrac, the mass of the resolving agent mres, and the mass of the unreacted enantiomers (1−mr)·mrac minus the mass of the extract ˆme:

ˆ

mr=mr·mrac+mres

(1−mr)·mracmˆe—

Expanding the term (1−mr) in the above equation, simplifying and grouping terms containingmrac together yields:

ˆ

mr=mr·mrac+mrac−mr·mracmˆe+mres ˆ

mr=mracmˆe+mres (2.14)

The extract and raffinate yields are calculated simply based on the theoretical masses defined in Eqs. 2.13 and 2.14:

Yˆe= me ˆ

me (2.15a)

Yˆr= mr ˆ

mr (2.15b)

Values of ˆY are expected to range from 0–1, but both ˆYe and ˆYr can exceed 1.

The conditions under which this occurs are similar to those described on p. 48 forY: values of ˆYe can exceed 1 if the diastereomer formation is incomplete or if the

dia-stereomers dissociate, ˆYrcan exceed 1 if the extraction is less efficient than calculated by the idealized CSTR model.

In a further similarity toY, values of ˆY become invalid if mr≥1. Due to the term (1−mr), the value of ˆmebecomes zero if mr=1, causing the value of ˆYeto become invalid due to a division by zero. At this point, ˆYr is still valid, although ˆmr is now equal to the entire mass of materials used for the resolution. If mr>1, both ˆmeand mˆrbecome physically invalid by decreasing below zero and by exceeding the mass of available material, respectively. Therefore, as withY, if diastereomer dissociation in a system causes the resolution to proceed atmr ≥ 1, ˆY cannot be used to describe the yields for the system.