DE2350633C3 - Process for extracting hops - Google Patents

Description

25th

The present invention relates to a new method for the extraction of neutral substances, bitter substances and tannins from hop extracts.

Processes for obtaining hop extracts are known (Steiner Hopfen GmbH, hops and hop refinement products, Oelde 1970, pages 414-421). So far, however, was a production of those contained in hops Ingredients in three separate extracts, each of which contains the essential neutral, Contain bitter or tannins, in an economical manner and by a method that provided extracts, whose use appeared harmless to health, not possible.

Under the term »neutral substances« the group of substances are understood that influence the typical smell and taste of the hops prepared beer and essentially contain the essential oils contained in hops. Links, which fall into this group are essentially terpenes and sesquiterpenes, such as e.g. B. Humulene, j9-caryophyllene. Farnese, Myrcesi furthermore oxy compounds, such as B. Undecanone-2, Geranyl Acetate, Methyldec-4-enoate and Methyldeca-4,6-dienoaL Furthermore The concept of neutral substances includes the group of so-called that has not yet been clarified in terms of its structure "Non-characterized soft resins" (Monthly for Brewery, 22, p. 209 [1969]).

The term "bitter substances" should be understood to mean the group of substances that characterize beer give pleasant bitterness and mainly from humulon, cohumulon, adhumulon, Lupulonen, Colupulon and Adlupulon as well as the oxidative degradation products of the mentioned substances exist.

The term "tannins" is to be understood as meaning the group of substances that essentially consist of tannins, flavonoids and catechins (M. Dadic et al,]. Inst. Brew 76, 276-280 [1970]).

There has now been a process for obtaining the neutral, bitter and tannins by extraction of Hop extracts or hop extract solutions found, which is characterized in that one Hop extract solution containing these substances (primary extract solution) at least two consecutive Subjects liquid-liquid extractions, being different with exclusively organic solvents Polarity, which may contain water, and / or different volume ratios the polar dissimilar solvent is used, so that either in the first extraction stage the tannins or the neutral substances are separated and in the subsequent extraction stage the neutral and bitter substances or tanning and bitter substances from each other remaining in the primary extract solution be separated.

According to the state of the art, only one separation of the bitter substances from the hop extracts has hitherto been carried out or the solutions of the hop extracts in organic solvents and this by treating the extracts or extract solutions with aqueous alkalis (DE-OS 17 92 509), by treatment the extracts or extract solutions first with aqueous organic solvents and then aqueous alkalis (DT-OS 19 57 730) or by treating the extracts or extract solutions with solutions which affect the pH value of compounds (DE-OS 6 19 563).

With the method according to the invention, a method for separating hop extracts is provided for the first time described in bitter, neutral and tannins, which affects the use of the pH value Compounds dispensed with, and exclusively with organic, optionally containing water Solvent works. It was surprising that by using organic solvents alone a separation of the ingredients of the hops into neutral, bitter and tannins is achieved. It it was to be expected that the neutral substances would differ from the bitter substances or the bitter substance / tannin mixtures could not be separated by extraction with organic solvents because the neutral substances were made a multitude of compounds exist, which belong to the most varied of compound classes and should therefore have very different distribution properties. Furthermore, it was not to be expected that the bitter substances were separated from the tannin mixtures with the aid of organic solvents permit; although one can see the adverse effects of alkalis and alkaline compounds has known for a long time, they are always available until the filing date of the present application Separation of the bitter substances applied, evidently in the opinion that alkalis are the only effective Extracting agents for bitter substances from hop extracts are. With the present application, for the first time shown that a separation of the bitter substances also by liquid-liquid extraction with organic alone Solvents is possible. The method according to the invention is therefore not only new, but also surprising compared to the prior art.

The progress that the process according to the invention brings lies in the fact that it succeeds in all of the ingredients of the brewing industry that are important Hops chemically unchanged, i.e. also unchanged in taste, to win. Another advantage lies in the fact that the ingredients of the hops are obtained so separately that they can be used during of the brewing process in each case at the right moment in the desired to avoid losses Amount can use.

The inventive method can, for. B. be carried out in such a way that the primary extract solution is subjected to two successive liquid-liquid extractions, the neutral and bitter substances are separated from the tannins in an extraction stage A and the neutral substances and the bitter substances are separated in an extraction stage B .

One can proceed in such a way that the extractant 1 used for extraction stage A is matched in its polarity to the solvent composition of the primary extract solution so that in extraction stage A a practically complete separation of the bitter and neutral substances from the tannins is achieved and furthermore The polarity of the extractant 2 used for extraction stage B is matched to the solvent composition of the solution of bitter and neutral substances remaining in extraction stage A , so that in extraction stage B a practically complete separation of the neutral substances and the bitter substances is achieved.

For the method according to the invention it has proven advantageous, for example, as a primary extract solution to use an alcoholic or alcoholic aqueous solution, preferably mixtures lower aliphatic alcohols with water. Furthermore, it is also possible as a solvent for the primary extract solution Halogenated hydrocarbons with 1 to 3 carbon atoms and up to 3 halogen atoms, preferably chlorine atoms and benzene or .alkylbenzenes with a total carbon number of up to Use 8 carbon atoms. Primary extracts are preferably used which act as solvents Have alcohol or alcohol-water mixtures, methylene chloride or benzene. Particularly preferred are methanol and ethanol or mixtures of these alcohols with water.

For example, when using alcoholic or alcoholic-aqueous primary extract solutions as extraction agent 1 for extraction stage A, aliphatic hydrocarbons with 5 to 10 carbon atoms and / or cycloaliphatic hydrocarbons with 5 to 8 carbon atoms and / or aromatic hydrocarbons from the group of benzene and alkyl-substituted benzene derivatives Find use, the neutral and bitter substances being extracted into the hydrocarbon phase and the tannins remaining in the alcoholic-aqueous phase. The extract from extraction stage A is then subjected to extraction stage B in accordance with the process according to the invention, for which, for example, mixtures of lower aliphatic alcohols and water can be used as extraction agent 2, the bitter substances being extracted into the alcoholic-aqueous phase and the neutral substances being extracted from the Hydrocarbon phase remain.

In this embodiment, the polarity of the coal is f according to the process of this invention, ₀ lenwasserstoffphase with respect to the alcoholic-aqueous phase for the extraction stage A sufficiently large that, in the neutral and bitter substances are extracted into the hydrocarbon phase and relatively to the extraction stage B for Extraction stage A lowered so far that the bitter substances are extracted from the hydrocarbon phase in extraction stage B.

Instead of hydrocarbons as extraction agents for extraction stage A , it is also possible to use halogenated hydrocarbons with 1 to 3 hydrocarbon atoms and up to 3 halogen atoms, preferably methylene chloride or chloroform. The extraction stage A takes place in a completely analogous manner with the formation of a halogenated hydrocarbon extract which contains the neutral and bitter substances and which is extracted for further separation in extraction stage B with mixtures of lower aliphatic alcohols and water, preferably with aqueous methanol.

When using primary extract solutions in benzene or alkylbenzenes as well as in halogenated hydrocarbons, aliphatic alcohols with 1 to 3 carbon atoms or mixtures of these alcohols with water, preferably methanol and ethanol, particularly preferably methanol, are used as extraction stage A, the tannins being used in the alcoholic or alcoholic-aqueous phase are extracted and the neutral and bitter substances remain in the hydrocarbon or halogenated hydrocarbon phase. The raffinate containing neutral and bitter substances from extraction stage A is then subjected to extraction with mixtures of lower aliphatic alcohols and water in accordance with the process according to the invention in extraction stage B ^{, wherein;} In the case of the presence of a halogenated hydrocarbon raffinate, aqueous methanol is preferably used as the extractant for extraction stage B.

The polarity of the hydrocarbon or halogenated hydrocarbon phase in relation to the alcoholic-aqueous phase, on the one hand, by changing the water content of the alcoholic-aqueous phase as well as the choice of alcohol. For example, the polarity of the alcoholic-aqueous Phase with increasing water content and when changing the alcohol in the sequence Propanol, ethanol, methanol increased, on the other hand by changing the solvent composition the hydrocarbon phase, e.g. B. by changing the proportion of cycloaliphatic hydrocarbons or, in particular, of benzene hydrocarbons. For example, the polarity is the Hydrocarbon phase with an increasing proportion of cycloaliphatic and especially aromatic Hydrocarbons larger.

Instead of changing the polarity of the place for the extraction stage B using extraction means 2 can in the extraction stages A l and at respective same solvent systems; nd B, the volume ratio of the extract to the extraction stage Λ to the extraction medium 2 against the corresponding volume ratio which is in the extraction stage A application , are reduced to such an extent that in extraction stage B a practically complete separation of the neutral substances and the bitter substances is achieved. In general, it has proven to be particularly expedient to reduce this volume ratio in extraction stage B to at least about half, preferably to at least one third, of the corresponding value in extraction stage A.

The method according to the invention can also be carried out in such a way that from the primary extract solution in an extraction stage C first the neutral substances from the bitter and tannins

5 6

are separated and in the extraction stage D the In this embodiment is corresponding to the

Tannins are separated from the bitter substances. The methods of the invention the polarity of the extraction stages mentioned Kohneu C and D correspond lenwasserstoffphase with respect to the hydroalcoholic the already mentioned extraction stages A and B. phase for the extraction stage C just chooses as large give characterized 5 in that in the extraction stage C, only the neutral substances from extracted from the alcoholic-aqueous phase

Extraction stage A will continue to be so for extraction stage D.

the first extraction stage, if in this the much increased, that then the bitter substances from the alcoholic tannins are extracted from the neutral and bitter substances holic-aqueous phase,
be separated, io When using primary extract solutions in Ben

Find zol or alkylbenzenes as extractants

Extraction stage C for the extraction stage C with aliphatic alcohols

the erite extraction stage, if in this the 1 to 3 carbon atoms or mixtures of these alcohol-neutral substances get from the tannins and bitter substances with water, preferably methanol and ethanol, be separated, 15 particularly preferably methanol, use, wherein

the tannins and bitter substances in the alcoholic or

Extraction stage B alcoholic-aqueous phase are extracted and the

the extrak neutral substances following the first extraction substance in the hydrocarbon phase, if the neutral phase is in this phase. The tannin and bitter substances containing raffinate bitter substances are separated, 20 from the extraction stage C is then accordingly

the process of the extraction extraction stage D _stu f _e D subjected to the invention.

the extraction following the first extraction stage, if primary extract solutions are used in the cation stage, if in this the tanning hydrocarbons are found as extraction agents Bitter substances are separated. 25 for extraction stage C methanol or mixtures

of methanol with water, using the

The extraction agents 3 and 4 correspond to the ex-bitter and tannins in which the alcoholic or alcoholic traction agents 1 and 2 are extracted in the same way as the holic-aqueous phase and the new extraction stages C and D correspond to the extraction stages A tral substances in the halogenated hydrocarbon phase. and B. 30 remain. The raspberry containing tannins and bitter substances

One can proceed in such a way that the extract used for finate from extraction stage C is then subjected to the extraction stage C in its polarity to solution stage D in the process according to the invention.

Medium composition of the primary extract - The polarity of the hydrocarbon or halogen is correct that in the extraction stage C a practically 35 hydrocarbon phase in relation to the alcoholic complete separation of the neutral substances from the tanning aqueous phase can be achieved in the same way as with and bitter substances and furthermore that for the extraction stages A and B are influenced, the subsequent extraction stage D use fin- Instead of changing the polarity of the

Dende extraction means 4 may in Ex 40 tion means 4 also at each same solution traction stage C remaining solution of bitter and medium systems in the extraction stages C and D, the tannins tunes in polarity so on extraction step D using place Extrakdie solvent composition that in the extraction stage D volume ratio of the extractant 4 to Ex a practically complete separation of the tannins tract of the extraction stage C compared to the corresponding and bitter substances is achieved. the corresponding volume ratio, which is

For example, it has been found in application of a aiko- ₄₅ tion step c application, be increased so holisch-aqueous primary extract solution that a virtually full instructed stage to be advantageous in the extraction stage D as extractant 3 constant for extraction separation of the bitter principles and tanning C aliphatic hydrocarbons with 5 to is achieved. In general, it has proven particularly expedient to use 10 carbon atoms and / or cycloaliphatic hydrocarbons to increase this volume of hydrocarbons with 5 to 8 carbon atoms and / 50 times, preferably to one or more aromatic hydrocarbons from the group of twenty times the corresponding value Extract of benzene and alkyl-substituted benzene derivatives to increase ion level C.

to use, whereby the neutral substances in the coal The preparation of the primary extract solution is carried out

Hydrogen phase are extracted and the bitter. Usually by extraction of hops with and tanning agents in the alcoholic-aqueous phase reduce lower aliphatic alcohols, in particular with stay. The methanol, ethanol, propanol and isopropanol remaining after extraction stage C. Further alcoholic-aqueous The phase is then correspondingly towards the preparation of the primary extract the inventive method of extraction-extraction of hops with halogenated hydrocarbons D subjected, with the extraction agent 4 ali- substances with 1 to 3 carbon atoms and up to Phatic hydrocarbons with 5 to 10 carbon 60 3 halogen atoms, preferably chlorine atoms and with atoms and / or cycloaliphatic hydrocarbons- benzene and alkylbenzenes with a total of carbons with 5 to 8 carbon atoms and / or an aromatic number of up to 8 carbon atoms. as Matic hydrocarbons from the group of ben-halogenated hydrocarbons are, for example, gezols and alkyl-substituted benzene derivatives used: methylene chloride, chloroform, 1,1-dichloroethane, can find fertilization, whereby the bitter substances in the 65 l ^ -Dichlorätnanundl-Chlorpropane.bevorzugt methyl-coblene hydrogen phase are extracted and the lene chloride, _and as alkylbenzenes are toluene, Tannins in the alcoholic-aqueous phase, xylene and ethylbenzoL are preferably used as extracts. tion medium for the preparation of the primary extract solution

Use of methanol or ethanol. The extraction is carried out in a manner known per se by simple Extraction (see Steiner Hopfen GmbH, Hopfen und Hopfenveredlungsprodukte, Oelde 1970, pages 414-421) or e.g. according to the principle of carousel extraction (see Chemiker Zeitung, 94 [1970], pages 56-62).

The preparation of the primary extract solution for the process according to the invention can, for. B. with commercially available, dried raw hops are used as the starting material. Of course, you can use it as a Starting material also hop powder, concentrated hop powder or lupulin as well as water-containing ones Process fresh hops accordingly.

Furthermore, commercially available hop extracts, which contain neutral, bitter and tannins, can be added the method according to the invention are separated. The commercial extract is used for this purpose before separation according to the invention in alcohol, benzene or alkylbenzenes or halogenated hydrocarbons as Added solvent and this solution used as the primary extract solution.

For the practical implementation of the method according to the invention, there are correspondingly large number of possible variations of the solvent systems, the volume ratios of the extractant phase to be extracted and the engineering design of the extractors very many possible variations.

As aliphatic hydrocarbons, which are used in extractants 1, 3 and 4 can include: pentane, hexane, heptane, octane, nonane, decane and the isomers of these compounds. It is of course also possible to use mixtures of these compounds, for example a fraction of aliphatic hydrocarbons in the boiling range from approx. 34 to 40 C (pentane fraction), the boiling range from approx. 65 to 70 C (hexane fraction), the boiling range from approx. 95 to 100 C (heptane fraction), the boiling range from approx. 125 to 130 C (octane fraction), the boiling range from approx. 145 to 155 C (nonane fraction) or the boiling range from approx. 170 up to 180 C (decane fraction), preference is given to hexane, heptane and octane or the corresponding hexane, Heptane and octane fractions use.

As cycloaliphatic hydrocarbons which are used in extractants 1, 3 and 4 can be mentioned: cyclopentane, cyclohexane, cycloheptane, cyclooctane, as well as lower alkyl groups substituted derivatives of these compounds; preference is given to cyclopentane and cyclohexane.

As aromatic hydrocarbons which are used in extractants 1, 3 and 4 can, be for example benzene, which is optionally replaced by methyl, ethyl, propyl or isopropyl groups is mono- or polysubstituted, may be mentioned as preferred substituted benzene derivatives Called toluene, ethylbenzene and xylene

Lower aliphatic alcohols which are used in extraction stage A, B or C include, for example, methanol, ethanol, propanol and isopropanol; methanol and ethanol are preferred

For extraction stage A or C according to the process according to the invention, an alcoholic-aqueous, particularly preferably a methanolic or ethanolic-aqueous primary extract solution is used and the water content of this primary extract solution is increased to approx. 5 to 70% by volume, preferably to approx.

5 to 40% by volume, particularly preferably adjusted to about 5 to 30% by volume.

When carrying out the extraction run A , one can expediently proceed in such a way that the S alcoholic-aqueous primary extract solution is extracted with an extractant 1 which consists of a mixture of aliphatic hydrocarbons with cydoaliphatic hydrocarbons or with aromatic hydrocarbons. Preferred

ίο are mixtures with aromatic hydrocarbons, preferably benzene, toluene or xylene being used as aromatic hydrocarbons. For example, it has proven to be advantageous for extraction stage A to adjust the water content of the alcoholic-aqueous phase to about 10 to 40% by volume and the aromatic and cydoaliphatic hydrocarbons content of the extractant 1 preferably to about 10 to 60% by volume .-% to be set. For example, if the alcoholic primary extract solution has a water content of approx. 10% by volume, a hydrocarbon mixture can be used as extraction agent 1, the aromatic hydrocarbon content of which is preferably approx. 20 to 60% by volume, and if the water content is approx. 20% by volume preferably approx. 20 to 40% by volume and, with a water content of approx. 30% by volume, preferably approx. 10 to 30% by volume.

When using a primary extract solution with benzene or alkylbenzenes and halogenated hydrocarbons as solvents, these are extracted in a corresponding manner with alcohol or alcohol-water mixtures, preferably methanol or ethanol, particularly preferably with methanol as the extracting agent, with alcohol-water mixtures preferably having a water content of about 10-40% by volume, particularly preferably about 20 to 40% by volume, are used. For extraction stage A , the volume ratio of the hydrocarbon phase to the alcoholic-aqueous phase can be varied within wide limits and, for economic reasons, is in the range from approx. 1 to 10.

4c Volume ratios of about 1 to 4 are preferred adhered to. When using primary extract solutions with halogenated hydrocarbons, that is Volume ratio of the halogenated hydrocarbon phase to the alcoholic-aqueous phase in general in the range from approx. 0.1 to 2, preferably from approx. 0.2 to 1.5.

To carry out the extraction stage B one can proceed in such a way that the hydrocarbon or halogenated hydrocarbon phase resulting from the extraction stage A is fed to the extraction stage B either unchanged in its solvent composition or changed within the scope of the mixtures used for the Exiraküunssiuic B application proven advantageous to use a hydrocarbon phase for extraction stage B according to the inventive method, which consists of aliphatic and / or cydoaliphatic and / or aromatic hydrocarbons and optionally has a content of cydoaliphatic and / or aromatic hydrocarbons which is below approx. 50 vol. %, preferably below about 40% by volume

The extractants 2 are generally alcohol-water mixtures Application, where the water content of the extraction agent is expediently in The range is from about 10 to 30% by volume

It has been the case, for example, for the extraction stage Z? Proven to be particularly beneficial by

230 222/179

to start from a hydrocarbon phase containing about 0 to 30% by volume of cycloaliphatic and / or aromatic Contains hydrocarbons, and the water content of the extractant 2 aui about 10 to 20 vol .-%, especially preferably to set approx. 10% by volume. For example, it has proven expedient to use the hydrocarbon phase with a content of aromatic hydrocarbons of approx. 10 to 30 vol .-% with a To extract extraction agent 2, which has a water content of approx. 10 vol .-% or at a content of aromatic hydrocarbons from 0 to 10 vol .-% with an extractant 2, the one Has a water content of approx. 20% by volume.

The volume ratio of hydrocarbon phase to extractant 2 can be varied within wide limits and is generally about 0.1 to 1.0 for economic reasons. Volume ratios are preferred from about 0.1 to 0.6.

If a halogenated hydrocarbon phase is present for further extraction in extraction stage B , it has proven to be particularly advantageous to extract the halogenated hydrocarbon phase with aqueous methanol, which has a water content of approx. 30 to 40% by volume, and to determine the volume ratio of the halogenated hydrocarbon phase to the alcoholic aqueous phase preferably in the range from about 0.05 to 0.2, preferably from about 0.05 to 0.1.

It has been found that it is advantageous for the separation process of the invention, the volume ratio of hydrocarbon or halogenated hydrocarbon phase to the alcoholic-aqueous phase in the extraction stage B with respect to the extraction step A to at least about half, preferably at least one third Reduce.

The conditions used for extraction stage B , in particular the water content of the alcoholic-aqueous phase or aromatic and cycloaliphatic content of the hydrocarbon phase, are expediently matched to the conditions preselected in extraction stage A so that the combination of extraction stages A and B within the scope of the preferred conditions can be done with the least possible effort.

Accordingly, the practical implementation can the method according to the invention preferably be carried out in such a way that one uses a matched optionally by addition of water to a Wassergehall of about 10 to 30 vol .-% alcoholic-aqueous primary extract solution in the extraction stage A, and as an extracting agent 1 mixtures of a aliphatic hydrocarbon with 10 to 60 vol .-% of an aromatic hydrocarbon is used, with a volume ratio of the hydrocarbon phase to the alcoholic-aqueous phase of about 1 to 10 is extracted. In the subsequent extraction stage B , the hydrocarbon phase resulting from extraction stage A is extracted with an extractant 2 which is a lower aliphatic alcohol containing about 10 to 30% by volume of water. The volume ratio of the hydrocarbon phase to the alcoholic-aqueous phase is preferred extraction stage A is reduced by at least half, preferably to at least about one third, and a volume ratio of about 0.1 to 1 is maintained.

In extraction stage B, with the same solvent systems in extraction stages A and B, only the volume ratio of the hydrocarbon phase to the alcoholic-aqueous phase can be reduced, a reduction to approximately one fifth to one tenth of the corresponding value in extraction stage A preferably taking place.

It has also proven to be particularly advantageous in extraction stage B either

a) with the same solvent composition of the hydrocarbon phase in the extraction stages A and B, to reduce the water content of the extractant 2 compared to the water content of the alcoholic-aqueous phase from the extraction stage A by approx. 10 to 20% by volume, or

b) with the same water content of the primary extract solution or of the extractant 2, the content of the Hydrocarbon phase of aromatic and / or cyclic aliphatic hydrocarbons by adding an aliphatic hydrocarbon containing 5 to 10 carbon atoms to reduce approx. 5 to 20% by volume.

Of course, the aforementioned measures a and b can also be combined.

When carrying out extraction stage C , it is expedient to proceed in such a way that the alcoholic-aqueous primary tract solution is extracted with an extractant which consists of the aliphatic and / or cycloaliphatic and / or aromatic hydrocarbons mentioned.

For example, it has proven advantageous for extraction stage C to adjust the water content of the primary extract solution to approx. 5 to 30% by volume, preferably to approx. 10 to 20% by volume, and use an aliphatic hydrocarbon or a mixture of an aliphatic as the extractant To use hydrocarbons with alicyclic or preferably aromatic hydrocarbons. The content of the extractant 3 of cycloaliphatic or aromatic hydrocarbons can, for. B. in the range from about 0 to 50 vol .-%, preferably in the range from about 0 to 30 vol .-%.

When using primary extract solutions with benzene or alkylbenzenes as well as halogenated hydrocarbons The solvent used is of course an alcohol-water mixture in a corresponding manner as an Extraktionsmiitel application, whereby preferably Methanol or ethanol-water mixtures, particularly preferably methanol-water mixtures a water content of approx. 10 to 40 Vel .-% are used.

The volume ratio of extractant 3 to the alcoholic-aqueous phase is within wide limits variable and is generally in the range from about 0.1 to 1 for economic reasons volume ratios of approx. 0.2 to 0.6 are maintained.

To carry out the extraction stage D one can proceed in such a way that the alcoholic-aqueous phase resulting from the extraction stage C is fed to the extraction stage D either unchanged in its water content or modified within the scope of the mixtures used for the extraction stage D.

For extraction stage D , it has proven advantageous to adjust the water content of the alcoholic-aqueous phase to about 10 to 40% by volume, preferably about 10 to 30% by volume, and to set the water content to be aliphatic and / or cycloaliphatic and / or to use extractant 4 containing aromatic hydrocarbons. For example, when using aliphatic hydrocarbons as

11 12

Extraction agent 4 proved to be useful, which are also used in combination for a and b, extracting alcoholic-aqueous phase to approx. 10 to The liquid-liquid extractions according to the invent-20 % By volume. Mixtures of the process according to the invention are preferably found with the aid of an carried out by aliphatic with cycloaliphatic and / or known extraction methods, aromatic hydrocarbons application, whereby 5. B. by mixtures with aromatic shaking in separating funnels, with the aid of mixer hydrocarbons, are particularly preferred be applied. The content of separator systems, furthermore by countercurrent Mixtures of aromatic hydrocarbon traction in columns without a stirrer, such as B. Extracts can generally be between about 0 and ion spray towers or extraction towers with or without 90% by volume, preferably about 10 to 50% by volume. io packing or by countercurrent extraction in co-

The volume ratio of extractant 4 to the lonnen with stirrers, such as. B. Scheibel columns or alcoholic-aqueous phase is variable within wide limits RDC columns (= rotating disk contactor) or with and for economic reasons it is necessary to use extraction centrifuges common in the range of approx. 1 and 10. Preference is given to sieving for continuous operation Volume ratios of about 1 to 4 in-bottom pulsation columns or RDC columns as th. Proven particularly advantageous.

It has been shown that for the separation after the extraction temperature for the individual process according to the invention it is expedient, the traction levels can be varied within wide limits. From the volume ratio of hydrocarbon phase to practical reasons, is in general alcoholic-aqueous phase in the extraction stage D 20 at temperatures between 0 and 50 ⁰ C, preferably in relation to the extraction stage C by at least at temperatures of 5-30 ⁰ C. The Exca. Five-fold, preferably by at least approx. The traction will increase from practical and economic twenty-fold. Reasons generally carried out at normal pressure

For the practical implementation of the invention leads, however, in principle, a work such. B. according to the method you can therefore preferably 25 when using particularly low-boiling Exso proceed that one in the extraction stage C a traction means, possible at overpressure,
optionally by adding water to an aliphatic hydrocarbon obtained from the water content of approx. 10 to 20% by volume brought by the process according to the invention or alcoholic-aqueous primary extract solution and as extractive hydrocarbon solutions or halogenated carbon detergent 3 ₁₀ hydrogen solutions can be the neutral, bitter and optionally mixed with an aromatic tanning agent in the usual way, for. B. used in rotary hydrocarbon, with a vapor damper, falling film evaporator and thin film volume ratio of hydrocarbon phase to the damper free from solvent,
alcoholic-aqueous phase extracted from about 0.1 to 1. In the subsequent extraction stage D extra- 3s driving by means of countercurrent extraction Kolonhiert one consisting of the extraction stage C hervorgehen- NEN for the combination of the extraction stages A de hydroalcoholic phase with an extraction and B is the flow diagram in Fig. 1 erläumittel4 , soft from a mixture of aliphatic iert: The primary extract solution with aromatic hydrocarbons, which is about 10 to the extraction column A , is passed through line 1. Extraction stage A takes place in column A with 50% by volume of aromatic hydrocarbons, with holding in countercurrent. The volume of the extractant 1 is preferably fed to column A through line 2 (carbon ratio of the hydrocarbon phase to the alcoholic-hydrogen phase) and the aqueous phase compared to the extraction stage C to an alcoholic-aqueous solvent mixture that is at least about five to at least that Approximately Zwand the solvent composition increased by the primary xyfold and a volume ratio of approx. Column A is fed.

In this case, one can at same Tannins remain in the alcoholic-wäßchen Lösungsmiuelsystemen in the extraction stages membered phase and are removed from the bottom of column C and D, only the volume ratio of carbon A through line 4 in the extraction stage D. The bitter and new hydrogen phase to the alcoholic-aqueous phase tral substances get into the extract and are elevated at the top, preferably an increase to about the column A taken via line 5 and five to ten times the corresponding value of the column B supplied.
Extraction stage C takes place In column B the extraction stage β takes place.

It has also proven to be particularly advantageous where the extractant 2 (mixture of alcohol /

Wiesen, in the extraction stage D 55 water) is fed through line 7. Via line

a) with the same water content of the alcoholic-aqueous 6, a hydrocarbon or a hydrocarbon phase of the extraction stages C and D is added to the mixture of substances, which corresponds in the solvent content to the composition of the hydrocarbon phase to be extracted from the hydrocarbon phase to be extracted

matic hydrocarbons by approx. 10 to 60. The bitter substances are increased in alcoholic-aqueous 20% by volume or removed from column B via line 9.

b) with the same solvent composition of the The neutral substances remain in the hydrocarbon phase of the extraction stage C substance phase and the water content of the alcoholic-aqueous is removed from column B via lines 8 and D.

Phase compared to the water content of the aqueous 65 Furthermore, the flow diagram in FIG. 2 will increase an embodiment of the veruni according to the invention by approx. 5 to 20% by volume from the extraction stage C. driving for the combination of the extraction stages C

Of course, the aforementioned dimensions and D can be explained:

The alcoholic-aqueous primary extract solution is passed through line 11 into the extraction stage C. In the column C ei.Olgt the extraction stage C. whereby the extractant 3 is fed in countercurrent through line 12 (hydrocarbon phase) into the bottom of the column C and a solvent mixture which corresponds to the solvent composition of the primary extract solution, is fed through line 13 at the top of column C.

The bitter and tannins remain in the alcoholic-aqueous phase and are removed from column C through line 14 and fed to column D. The neutral substances enter the extract and are removed at the top of column C via line 15.

In the column D the extraction stage D is carried out, wherein the extraction agent is 4 (hydrocarbon mixture) is fed through line sixteenth An alcohol-water mixture which corresponds to the solvent composition of the alcoholic-aqueous phase to be extracted is fed in via line 17.

The bitter substances are extracted into the hydrocarbon phase and are removed from column D via line 18. The tannins are removed from column D via line 19 in an aqueous alcoholic solution.

The extracts obtained by the process according to the invention can be used for the preparation of beer can be used. You can use the extracts z. B. in the form of those obtained in the individual extraction stages Use alcoholic solutions or preferably after evaporation of the solvent. According to the invention The extracts obtained have the advantage that the individual extracts, which contain in their entirety all essential ingredients for the brewing process, in a precisely defined manner Quantity and can be used in different phases of the brewing process in the preparation of beer. It is thus possible to prevent the ingredients from having a detrimental effect during the brewing process to be influenced. For example, you can add bitter and neutral extracts at different times the hop flavoring with full utilization of the bitter substances.

The inventive method also offers the advantage that this separation method in contrast to other separation processes, such as B. Steam distillation or precipitation, is extremely gentle on the product, because neither worked with higher temperatures nojh in the presence of salts, acids or alkalis so that the sometimes very sensitive ingredients of the hops are not changed. Farther get by the process according to the invention no foreign substances in the extracts, so that the Use of the extracts for the preparation of beer is harmless to health.

It is to be regarded as extremely surprising that the hops ingredients, which are a complex mixture represent from the most varied of compounds, according to the process according to the invention through a series of extractions into the three characteristic groups of neutral, bitter and tannins let open.

The process establishes a significant technical advance by enabling it entirely new techniques in beer preparation. Thanks to the dosing that is reproducible in terms of time and quantity of the three extracts during the brewing process, a large number of wedges for possible variations are opened up.

Examples 1 to 8e describe the preparation of the primary extract solutions used in Examples 9 to 30f and 58 to 83m.

example 1

To produce a primary extract solution, 35 kg of hops of the Hallertau Mittelfrüher 1970, which was available in the form of commercially available pressed hops and at the time of processing as a whole Had been stored for 3 months, extracted with 80 kg of methanol For this purpose, the hops were crushed, filled in 4 portions of 8.75 kg each into 4 cylindrical extraction columns and

'5 solvent percolated

This percolation was carried out so that the entire Added solvent to the first column and the draining solution successively in 10 equal Volume parts was collected. At the same time, after all of the solvent had been added, there was still enough Solvent also added to the column, as the hops hold back as a bound solution, see above that the volume of the solution drained off below corresponds to the volume of the total solution added above

solvent was the same. The 10 thus obtained are the same Volume parts, starting with the first part of the solution that ran off, were moved to the next one one after the other Pillar abandoned. After adding all 10 aliquots from the first column to the second column

} o again on the second pillar so much more Solvent given as the hops retain as a bound solution, so that the volume of the below expired solution is equal to the volume of the solution added above. With the rest of the pillars was proceeded in the same way.

The 4th column became 112.21 primary extract solution with a weight of 93.7 kg, the 9.65 kg of hop extract corresponding to 10.2% by weight and 4.05 kg Water, corresponding to 4.32% by weight. 9.65 kg of hop extract corresponded to an extraction yield of 27.6% (based on the hops used). 5.64 kg (58.5%) of this extract was soluble in η-hexane. This The n-hexane-soluble part contains the bitter and neutral substances. 4.01 kg (41.5%) was insoluble in η-hexane. This n-hexane-insoluble part contains the tannins.

Example 2

To produce a primary extract solution, 50 kg of hops of the Tettnanger variety 1971, which were in the form of unpressed bale hops was present and at the time of processing for a total of 5 months had been stored, extracted with 61.28 kg of methanol. For this purpose, the hops were crushed in 6 portions of 8.33 kg each in 6 cylindrical extraction columns and filled with the solvent such as percolated in Example 1

From the 6th column Primärextraktlösunf 92.681 kg were obtained with a weight of 78.75 to 12.05 kf hop extraction substances, according to 15.3 wt ⁰ / and 3.75 kg of water, corresponding to 4.76 wt .-% contained The 12.05 kg hop extract corresponded to an extraction yield of 24.1% and contained 6.09 kg (50.5% neutral and bitter substances and 5.96 kg (49.5%) tannins.

Example 3

To produce a primary extract solution, 65 kg of Hallertauer Nordbrauer hops were used in 1970

which was available in the form of commercially available pressed hops and at the time of processing a total of 11 months was stored for a long time, extracted with 63.0 kg of methanol. For this purpose, the hops were crushed, in 8 portions of 8.13 kg each filled into 8 cylindrical extraction columns and filled with the solvent percolated as in Example 1

The 8th column became 98.021 primary extract solution with a weight of 85.8 kg, the 16.08 kg of hop extract, corresponding to 18.75% by weight and 7.22 kg Water contained 8.42% by weight. The 16.08 kg of hop extract corresponded to an extraction yield of 24.75% and contained 93 kg (57.5%) neutral and Bitter substances and 6.834 kg (423%) tannins.

Example 4

To produce a primary extract solution 45 kg of hop varieties splitter 1968 of the commercially available in the form of van Preßhopfen were was present and had been stored for 18 months at the time of processing as a whole, extracted 69.5 kg of methanol. For this purpose, the hops, broken up in 6 portions of 7.5 kg each, were placed in 6 cylindrical extrusion columns and percolated with the solvent as in Example 1.

From the 6th column, 99.951 primary extract solution with a weight of 84.92 kg was obtained, the 10.62 kg The hop extract contained 12.51% by weight and 4.80 kg of water, corresponding to 6.46% by weight 10.62 kg of hop extract corresponded to an extraction yield of 23.6% and contained 3.95 kg (37.2%) neutral and bitter substances and 6.67 kg (62.8%) tannins.

Example 5

To produce a primary extract solution, 50 kg of the Hallertauer Mittelfrüher 1970, in the form of commercially available pressed hops and a total of 8 months at the time of processing had been stored for a long time, extracted with 68.0 kg of methanol. For this purpose the hops were crushed Filled in 6 portions of 8.33 kg each into 6 cylindrical extraction columns and filled with the solvent percolated as in Example 1.

From the 6th column 99.801 primary extract solution with a weight of 84.94 kg was obtained, the 12.05 kg Hop extract corresponding to 14.18% by weight and 4.89 kg of water corresponding to 5.76% by weight. The 12.05 kg Hop extract corresponded to an extraction yield of 24.2% and contained 6.14 kg (51.0%) neutral and Bitter substances and 5.91 kg (49.0%) tannins.

Example 6

To produce a primary extract solution, 50 kg of hops of the Tettnanger 1968 variety, which in the form of unpressed bale hops was available and a total of 2 months at the time of processing had been stored, extracted with 75.0 kg of ethanol. For this purpose the hops were crushed, in 6 portions of 8.33 kg each filled into 6 cylindrical extraction columns and filled with the solvent such as percolated in Example 1.

From the 6th column there were 110.3 liters of primary extract solution with a weight of 92.28 kg which contained 11.25 kg of hop extract, corresponding to 5.45% by weight. the 11.25 kg of hop extract corresponded to an extraction yield of 22.5% and contain 6.83 kg (60.7%) neutral and bitter substances and 4.42 kg (31.3%) tannins.

Example 7

To produce a primary extract solution, 50 kg of hops of the Nordbrauer 1970 variety, which in Form of commercially available pressed hops and at the time of processing for a total of 1.5 months had been stored, extracted with 82.5 kg of n-propanol. For this purpose, the hops were crushed, in 7 portions of 7.15 kg each filled into 7 cylindrical extraction columns and filled with the solvent such as percolated in Example 1

From the 7th column, 117.11 primary extract solutions were obtained with a weight of 98.63 kg obtained the 10.90 kg of hop extract, corresponding to 11.05% by weight and 5.73 kg Water contained 5.81% by weight. The 10.90 kg of hop extract corresponded to an extraction yield of 21.8% and contained 7.44 kg (68.2%) neutral and bitter substances and 3.46 kg (31.8%) tannins.

Example 8

To produce a primary extract solution, 50 kg of the Hallertauer Mittelfrüher 1971 variety, which was available in the form of commercially available pressed hops and had been stored for a total of 4 months at the time of processing, were extracted with 71.5 kg of i-propanol. For this purpose, the hops were crushed, filled in 7 portions of 7.15 kg each into 7 cylindrical extraction columns and percolated with the solvent as in Example 1.
From the 7th column, 104.741 primary extract solution with a weight of 87.02 kg was obtained, which contained 9.92 kg of hop extract, corresponding to 11.4% by weight and 5.6 kg of water, corresponding to 6.94% by weight. The 9.92 kg of hop extracts corresponded to an extraction yield of 19.83% and contained 6.289 kg (63.4%) of neutral and bitter substances and 3.630 kg (36.6%) of tannins.

Example 8 a

To produce a primary extract solution, 20 kg of Hallertauer Mittelfrüher hops were used in 1972, which was available in the form of commercially available pressed hops and at the time of processing a total of 6 months had been stored, extracted with 33.755 kg of methanol. For this purpose the hops were crushed, filled in 6 portions of 3.333 kg each into 6 cylindrical extraction columns and filled with the solvent percolated as in Example 1.

From the 6th column, 48.4081 primary extract solution with a weight of 40.905 kg were obtained, the 5.064 kg Containing hops extraction substances, corresponding to 12.38% by weight and 2.086 kg of water, corresponding to 5.10% by weight. The 5.064 kg of hop extract corresponded to an extraction yield of 25.32% and contained 2.719 kg (53.7%) neutral and bitter substances and 2.345 kg (46.3%) tannins.

Example 8 b

To produce a primary extract solution, 20 kg of hops of the Hallertauer Nordbrauer variety 1972, which was available in the form of commercially available pressed hops and at the time of processing a total of 6 months had been stored for a long time with 35.827 kg of methanol

extracted. For this purpose the hops were crushed, filled in 6 portions of 3.333 kg each into 6 cylindrical extraction columns and filled with the solvent percolated as in Example 1

The 6th column became 51.4101 primary extract solution Weighing 43.441 kg obtained the 5.226 kg Containing hops extraction substances, corresponding to 12.03% by weight and 2.388 kg of water, corresponding to 5.50% by weight. The 5.226 kg of hop extract corresponded to an extraction yield of 26.13% and contained 3.465 kg (66.3%) neutral and bitter substances and 1.761 kg (33.7%) tannins.

Example 8 c

To produce a primary extract solution, 23 kg of Hallertauer Mittelfrüher hops were used in 1971, which was available in the form of commercially available pressed hops and at the time of processing a total of 3 months had been stored for a long time, extracted with 36.293 kg of ethanol. For this purpose, the hops were crushed, Filled in 8 partial amounts of 2.875 kg each into 8 cylindrical extraction columns and filled with the solvent percolated as in Example 1.

The 8th column became 52.7941 primary extraction solution with a weight of 44.295 kg, the 5.187 kg of hop extractants, accordingly Contained 11.71% by weight and 2.815 kg of water, corresponding to 6.36% by weight, which corresponded to 5.187 kg of hop extract an extraction yield of 22.55% and contained 3.055 kg (58.9%) of neutral and bitter substances and 2.132 kg (41.1%) tannins.

Example 8 d

To produce a primary extract solution, 35 kg of hops of the Hallertau Mittelfrüher 1972, which was available in the form of commercially available pressed hops and a total of 6 months at the time of processing had been stored for a long time, extracted with 71.776 kg of methylene chloride. For this purpose, the Chopped hops, filled in 7 portions of 5 kg each into 7 cylindrical extraction columns and with the Solvent percolated as in Example 1

From the 7th column, 58.5531 primary extract solutions were obtained with a weight of 76.997 kg obtained that 5.128 kg Containing hops extraction substances, corresponding to 6.66% by weight and 0.093 kg of water, corresponding to 0.12% by weight. The 5.123 kg of hop extract corresponded to an extraction yield of 14.65% and contained 4.972 kg (96.96%) neutral and bitter substances and 0.156 kg (3.04%) tannins.

Example 8 e

To produce a primary extract solution, 40 kg of hops of the Hallertauer Nordbrauer variety 1972, which was available in the form of commercially available pressed hops and at the time of processing a total of 6 months had been stored for a long time, extracted with 51.737 kg of benzene. For this purpose the hops were crushed, Filled in 5 portions of 8 kg each into 5 cylindrical extraction columns and filled with the solvent percolated as in Example 1.

From the 5th column 63.8321 primary extract solution with a weight of 57.449 kg were obtained, the 5.676 kg Containing hops extraction substances, corresponding to 9.88% by weight and 0.036 kg of water, corresponding to 0.06% by weight The 5.676 kg of hop extract corresponded to an extraction yield of 14.19% and contained 5.103 kg (71.24%) neutral and bitter substances and 0.573 kg (10.10%) tannins.

Example 9 to 110

In the following examples nos. 9 to 110, the separation of the hop constituents with the separation stages A, B, C and D is described in tabular form, ίο The details of the individual examples relate to a sequence of extractions in the separating funnel. The results of these examples can also be transferred to other discontinuously operating distribution apparatus or can be used as a basis for the engineering design of continuously operating extraction columns.

In the sequence of extractions on which the individual examples are based, according to the method shown in FIG follow the given scheme.

In Fig. 3 stands for “5” for the extraction agent (solvent), “//” for the solution to be extracted and “Λ” for a solvent mixture which corresponds to the solvent composition of the solution to be extracted extracting phase H is extracted with the extractant S in the extraction stage £ 1. The phases S _M (solvent phase obtained in the extraction) and R _n (raffinate phase obtained in the extraction) which form in the process are each fed to a further extraction in the extraction stage £ 2.

In the extraction stage El , the phase R _{u is} extracted with fresh extractant S and the phase Sn with fresh solvent "Ä", which corresponds to the solution to be extracted. The resulting solvent phases S ₁₂ and S ₂₂ and the raffinate phases R _n and R ₂₂ are according to the arrows further extractions in the extraction stage £ 3 supplied. R _n is extracted with solvent “S”, S ₂ 2 with fresh “Λ” and R ₂₂ with S _n. The phases S, ₃ , S ₂₃ , S ₃₃ as well as Λ ₁₃ , Λ ₂₃ and R _{} i formed in the process} can be fed to a further extraction stage EA and subsequently in an analogous manner to the extraction stages £ 5 and £ 6 etc. .

Tables 1 to 8 contain the following information for Examples 9 to 110:

The origin of the extract solution is indicated in the column »solution used«. The solutions used have the same solvent composition as the corresponding phase of the solvent system used for the example in question. When using primary extract solutions (Examples 1 to 8), the primary extract solution is brought to the water content indicated in the column "alcoholic phase" and the indicated amount is used. When using a solution from one of Examples 9 to 110, the ingredients of the raffinate or fio solvent phase, if necessary, by evaporation z. B. at 40 ° C / 10 to 100 mm Hg in a rotary evaporator) freed from the solvent and then dissolved in the specified amount of solvent. The “Solvent system” column shows the solvents and solvent quantities used for the <> 5 extraction. The solvent quantities given in the column "Quantity in liters" relate to one extraction stage (El, El etc.).

19th

20th

The result is the content of neutral substances ("N"), bitter substances ("B") and tannins ("G") in the solvent and raffinate phase, whereby the figures for ζ, χ and y are the values on which the values are based Solvent phases S ^ ₂ or raffinate phases R _n , which were selected to determine the content of ingredients, result. (The remaining fractions were discarded.)

Table 1

Separation level C Separation of the neutral substances from the bitter and tannic substances

example
No.

Implemented solution: ¹ )

Solvent system

Hydrocarbon phase

Primary extract solution from composition example (parts by volume)

alcoholic phase lot
in
Ltr. lot
in
Ltr. composition
(Mixture with water) 1 0.44 90% by volume of methanol 5 3.50 90% by volume of methanol 5 2.75 90% by volume ethanol 1 1.16 89.5% by volume of ethanol 1 0.46 80% by volume of methanol 1 0.91 90% by volume ethanol 1 0.60 90% by volume of methanol 1 1.16 90% by volume ethanol 1 1.35 80% by volume n-propanol 1 1.04 85% by volume of methanol

9
10
11th
12th
13th
14th
15th
16
17th
18th

1 2 6 6 1 6 2 6 7 2

Pentane fraction

Hexane fraction

Hexane fraction

Heptane fraction

Octane fraction

Octane fraction

Decane fraction

Decane fraction

Decane fraction

Cyclohexane

Table 1 (continued)

With- Number of Solvent Phase Won ³ )

game extracts r. ionic χ hydrocarbon solution levels ² )

ζ Content of neutral substances

Bitter substances tannins recovered raffinate phase ³ )

γ alcoholic-aqueous solution

Content of neutral substances bitter substances
Tanning agents

11th

11th

26th

11th

14th

26th

5-9

6-11

6-11

14 -

6-11

8-14

14 -

N = 5.23 G = 96.2% B. = 0.21 G = 3.8% G = 0 G = 0% N = 34.21 G - 94.6% B. = 1.93 G = 5.3% G = 0.02 G = 0.1% N = 34.75 G = 95.8% B. = 1.51 G = 4.2% G = 0 G = 0% N = 5.95 G = 95.4% B. = 0.29 G = 4.6% G = 0.01 G = 0.1% N = 4.71 G = 95.6% B. = 0.21 G = 4.4% G = 0 G = 0% N = 6.64 G = 97.4% B. = 0.18 G = 2.6% G = 0 G = 0% N = 6.22 G = 99.1% B. = 0.05 G = 0.9% G = 0 G = 0%

- 5th

- 6

N = 0.04 g
B = 33.43 g
G = 33.33 g

N = 0.31 g
B = 211.88 g
G = 280.27 g

N = 0.49 g

- 6 B = 203.99 g

G = 187.36 g

- 13

N = 0.04 g
B = 34.6 g
G = 36.91 g

- 7th

- 13

0.03 g
29.84 g
30.41 g

0.02 g
37.59 g
35.85

N = 0.02 g
B = 43.78 g
G = 58.82 g

N
B.
G

N
B.
G

0.1% 50.0% 49.9%

0.1% 43.0% 56.9%

0.1% 52.1% 47.9%

0.1% 42.0% 52.0%

0.1% 49.5% 50.4%

0.1% 51.2% 48.8%

0.1% 42.7% 57.3%

continuation number of
Extrak
functional Won
.Y Solvent phase ³ )
Hydrocarbon! solution Recovered raffinate phase ¹ )
r alcoholic-aqueous solution Content of neutral substances
BitterstolTen
Tanning agents at
game
No. Z Content of natural substances
Bitter substances
Tanning agents N =
B =
G = 26th 14- N =
B =
G = 1 - N =
B =
G = 16 16 9 - N =
B =
G = 1 - N =
B =
G = 17th 31 15 - N =
B =
G = 1 - = 0.01 g = 0.1%
= 35.57 g = 50.0%
= 35.50 g = 50.0% 18th = 6.23 g = 97.9%
= 0.13 g = 2.1%
= 0 g = 0% = 0.02 g = 0.1%
= 34.38 g = 59.8%
= 23.1Og = 40.2% -26 = 5.76 g = 97.7%
= 0.17 g = 2.8%
= 0 g = 0% - 12 = 0.40 g = 0.4%
= 31.7Og = 36.0%
= 56.9 g = 63.6% - 16 = 5.15 g = 70.2%
= 2.18 g = 29.8%
= 0 g = 0% - 8th - 31 - 17th

') The primary extract solution used is used in the same amount and the same solvent composition as ( indicated in the "alcoholic phase" column.

² ) Extraction stages are understood to mean extraction stages £ 1, E1 , etc., in accordance with the scheme in FIG.

³ ) The values given relate to the values according to the scheme in FIG. 3 obtained solvent phase Ä, _ (for x, y and ζ see the values given for the individual examples).

Table 2

Separation stage C Separation of the neutral substances from the bitter and tannins

at
game Deployed
Solution: ') Solvent system lot alcoholic phase lot No. Hydrocarbon phase in in Primary Ltr. composition LIr. extract composition (Mixture with water) solution off (Parts by volume) example

19th
20th
21
22nd
23
24
25th
26th
27
28
29
30th

Cyclohexane 0.34

Hexane fraction benzene (10: 1) 1.44

Hexane fraction benzene (10: 1) 0.28

Hexane fraction benzene (7: 3) 0.22

Hexane fraction benzene (10: 1) Ό.5

Hexane fraction toluene (6: 4) 0.24 hexane fraction cydohexane (7: 3) 0.26

Hexane fraction cyclohexane (10: 1) 0.55

Heptane fraction benzene (15: 1) 3.44

Heptane fraction benzene (10: 3) 1.82

Octane fraction xylene (6: 4) 0.5

Decane fraction benzene (7: 3) 0.51

70% by volume of methanol 90% by volume of methanol 80% by volume of methanol 80% by volume of methanol 80% by volume of ethanol 90% by volume of methanol 80% by volume of methanol 80% by volume of ethanol 93.2% by volume of methanol 90% by volume ethanol 80% by volume ethanol 90% by volume methanol

Table 2 (continued)

at number of Solvent phase recovered ³ ) game Extra No. functional χ hydrocarbon solution level ² ) Content of neutral substances Z Bitter substances Tanning agents N = 7.4Og = 994% 19th 26th 15-26 B = 0.04 g = 04% G = O g = 0% N = 9.47 g = 96.9% 20th 16 9-16 B = 030 g = 3.1% G = O g = 0%

Recovered raffinate phase ³ )

ν alcoholic-aqueous solution

Content of neutral substances, bitter substances, tannins

1-8

N
B.
G

N
B.
G

0.01 g = 0.1% 46.93 g = 47.0% 52.98 g = 53.0%

0.05 g = 0.1% 59.95 g = 474% 66.28 g = 524%

continuation number of
Extra
functional
level ² ) 23 23 50 Deployed
Solution: )
Primary 5.56 g = 99.6%
0.02 g = 0.4%
0 g = 0% 633 24 0.01 g = 0.1%
34.00 g = 38.7%
53.77 g = 61.3% Methanol
Methanol
Methanol
Methanol
Methanol
Methanol
Ethanol
Methanol Lot !
in
LU. = at
game
No. 26th Solvency phase won ^{1 year}
λ hydrocarbon solution
Content of neutral substances
Bitter substances η
Tanning agents ΓΙ llllol ~
extract-
solution off
example 8.61 g = 97.4%
O _; 23g = 2.6%
0 g = 0% Recovered raffinate phase - ¹ )
y alcoholic-aqueous solution
Content of neutral substances
Bitter substances
Tanning agents 0.04 g = 0.1%
49.64 g = 45.4%
59.62 g = 54.6% 0.500
0.500
1,000
0.500
0.750
1,000
1,000
0.500 21 26th N =
15-26 B =
G = 3
2
8b
1
8b
2
6th
8e 5.37 g = 95.3%
0.26 g = 4.7%
0 g = 0 ¹ Co N =
1-12 B =
G = 0.04 g = 0.1%
30.7 g = 47.8%
33.54 g = 52.1% 230 222/179 22nd 26th N =
14-26 B =
G = 5.62 g = 93.8%
0.37 g = 6.2%
0 g = 0% N =
1-13 B =
G = 0.05 g = 0.1%
40.74 g = 42.4%
57.5 g = 57.5% 23 11th N =
14-26 B =
G = 9.57 g = 96.1%
0.39 g = 3.9%
0 g = 0% N =
1-13 B =
G = 0.07 g = 0.1%
54.67 g = 47.7%
59.87 g = 52.2% 24 13th N =
6-11 B =
G = 5.62 g = 99.6%
0.02 g = 0.4%
0 g = 0% N =
1-6 B =
G = 0.01 g = 0.1%
32.29 g = 48.9%
33.71 g = 51.1% 25th 26th N =
7-13 B =
G = 26.43 g = 97.6%
0.66 g = 2.4%
0 g = 0% N =
1-7 B =
G = 0.12 g = 0.1%
164.01 g = 40.5%
241.12 g = 59.5% 26th 16 N =
15-26 B =
G = 11.83 g = 85.8%
1.96 g = 14.2%
0 g = 0% N =
1-12 B =
G = 0.30 g = 0.2%
67.89 g = 47.9%
73.67 g = 51.9% 27 26th N =,
9-16 B =
G = 5.95 g = 96.0%
0.25 g = 4.0%
0 g = 0% N =
1-8 B =
G = 0.04 g = 0.1%
34.20 g = 51.9%
31.60 g = 48.0% 28 15th N =
13-26 B =
G = 9.5Ig = 36.9%
0.30 g = 3.1%
0 g = 0% N =
1-14 B =
G = 0.05 g = 0.1%
54.69 g = 45.2%
66.36 g = 54.5% 29 26th
See footnote N =
8-15 B =
G = of the neutral substances N =
1-8 B =
G = the bitter and tannins 30th
V) ³ ) N =
14-20 B =
G =
to Table 1. Solvent system
Hydrocarbon phase N =
1-13 B =
G = alcoholic phase Table 2 a separation stage Γ separation composition
(Parts by volume) Amount of composition
in (mixture with water)
Ltr. at
game
No. Cyclooctane
Octane fraction-Cydohexane (7: 3)
Hexane-xylene (6: 4)
benzene
benzene
benzene
benzene
benzene 0.425 90% by volume
0.170 80% by volume
0.196 90% by volume
0.045 70% by volume
0.068 70% by volume
0.070 40% by volume
0.090 70% by volume
0.043 70% by volume 30a
30b
30c
3Od
3Oe
30f
30g
3Of

25th

26th

Table 2a (continued)

example

number of
Extraction
stubborn ² )

Solvent phase recovered ^} )

x hydrocarbon solution

Content of neutral substances bitter substances Tanning agents

Won ralfinate phase ■ ')

r alcoholic-aqueous solution

salary at Neulralsloffen g = 0.0% Bitter substances g = 45.0% Tanning agents g = 55.0% N = 0.00 g = 0.1% B = 26.92 g = 40.3% G = 32.88 g = 59.7% N = 0.01 g - 0.1% B = 18.25 g = 60.7% G = 07/27 g = 39.3% KT _ Λ ni g - 0.0% B = 49.46 g = 37.6% G = 32.09 g = 62.4% N = 0.00 g = 0.0% B = 21.34 g = 37.6% G = 35.48 g = 62.4% N = 0.00 g = <0.1% B = 21.34 g = 35.8% G = 35.48 g = 64.1% N = 0.02 g = 0.1% B = 26.45 g = 45.3% G = 47.34 g = 54.7% N = 0.03 g = 0.1% B = 24.33 g = 82.9% G = 29.39 g = 17.1% N = 0.01 B = 21.33 G = 4.41

20 - 35

N = 4.67 g
B = 0.01 g
G = 0.00 g

30b 16

9-16

5-8

30d 50 28-50

N =

B =

G =

N =

B =

G =

50

3Of 60

-50

32-60

2.94 g
0.06 g
0.00 g
5.20 g
0.06 g
0.00 g
= 3.43 g
B = <0.01 g
G = 0.00 g

N = 3.43 g

B = 0.01 g

G = 0.00 g

N = 4.26 g

B = 0.10 g

G = 0.00 g

30g 65 34-65

50

28-50

') ² ) ³ ) See footnote to Table 1.

N =

B =

G =

N =

B =

G =

4.24 g
0.17 g
0.00 g
5.59 g
0.01g
0.00 g

99.7% 0.3% 0.0%

98.1% 1.9% 0.0%

98.8% 1.2% 0.0%

99.9% 0.1% 0.0%

99.7% 0.3% 0.0%

97.8% 2.2% 0.0%

96.1% 3.9% 0.0%

99.9% 0.1% 0.0% 1 - 16

1 - 4 1 - 23 1 - 23 1 - 29 1 - 32 1 - 23

Table 3

Separation stage D Separation of the bitter substances from the tannins

at
game
No Deployed
Solution: ⁷ ) Solvent system lot
in
Ltr. alcoholic phase lot
in
Ltr. M 1. ingredients
Hf * r R afTinaf Hydrocarbon phase 4.97 composition
(Mixture with water) 0.7 UCl l ^ dllinaL-
phise out
example composition
(Parts by volume) 2.85 90% by volume of methanol 1 31 9 Pentane fraction 3.5 70% by volume of methanol 0.7 32 10 Hexane fraction 5.6 80% by volume of ethanol 1 33 11th Hexane fraction 3.85 76.9% by volume of ethanol 0.7 34 12th Heptane fraction 7.7 80% by volume of methanol 1 35 13th Octane firing 3.4 80% by volume of ethanol 1 36 14th Octane fraction 5.18 80% by volume of methanol 0.7 37 15th Decane fraction 6.09 90% by volume ethanol 0.7 38 16 Decane fraction 3.73 80% by volume n-propanol 1 39 17th Decane fraction 3.15 70% by volume of methanol 1 40 18th Cyclohexane 4.1 60% by volume of methanol 1 41 19th Cyclohexane 3.6 90% by volume of methanol 1 42 20th Hexane fraction benzene (5: S) 75% by volume of methanol 43 20th Hexane fraction benzene (8: 2)

27

28

Table 3 (continued)

at
game
No. number of
PV t Π l Solvent phase recovered) 8th - Hydrocarbon solution
Content of neutral substances
Bitter substances
Tanning agents 0.04 g = 0.1% ΠΛΙΓοΚ-
functional _Λ
stages') V N = 25.67 g = 99.4% 31 B = 0.12 g = 0.5% 14th 6 - G = 0.06 g = 0.2% N = 33 g = 98.6% 32 B = 0.42 g = 1.2% 11th 5 - G = 33.20 g = 0.3% N = 33.20 g = 98.8% 33 ο 0.26 g = 0.8% 9 4 - G - 0.04 g = 0.1% N = 29.80 g = 99.5% 34 B = 0.13 g = 0.4% 7th 8th - G = 0.03 g = ^: 0.2% N = 22.08 g = 97.9% 35 B = 0.44 g = 1.9% 16 5 - G = 0.02 g = 0.1% N = 31.67 g = 99.7% 36 B = 0.07 g = ^: 0.2% 8th G = - 14 - 11 - 9 - 7th - 16 - 8th

Recovered raffinate phase ³ )

γ alcoholic-aqueous solution

salary of neutral substances 0 G = 0% Bitter substances 0.36 G = 1.4% Tanning agents 25.68 G = 98.6% N = 0 G = 0% B = 0.31 G = 0.7% G = 44.27 G = ■ - 99.3% N = 0 G = 0% B = 0.28 G = 0.9% G = 30.89 G = 99.1% N = 0 G = 0% B = 0.11 G = 0.3% 32.33 G = 99.7% N = 0 G = 0% D. 0.17 G = 0.8% G = 22.54 G = 99.2% N = 0 G = 0% B = 0.06 G = 0.2% G = 30.09 G = 99.8% N = 0 G = 0% B = 0.22 G = 0.5% G = 47.69 G = 99.5% N = 0 G = 0% B = 0.14 G = 0.6% G = 25.02 G - 99.4% N = 0 G = 0% B = 1.05 G = 6.5% C ^ - 15.00 G = 93.5% N = 0 G = 0% 0.08 G = 0.2% G = 49.23 G = 99.8% N = 0 G = 0% B = 0.16 G = 0.3% G = 46.26 G = 99.7% N = 0 G = 0% B = 0.17 G = 0.6% G = 26.37 G = 99.4% N = B = G =

11th

26th

26th

7th

7th

N = 0.02 g = 0.1%

6-11 B = 35.29 g = 99.2%

G = 0.27 g = 0.7%

N = 0.01 g = <0.1%

14-26 B = 24.89 g = 99.4%

G = 0.14 g = 0.6%

N = 0.02 g = 0.1%

13-26 B = 22.54 g = 96.8%

G = 0.71 g = 3.1%

N = 0.39 g = 1.4%

4-7 B = 27.97 g = 98%

G = 0.16 g = 0.5%

N = 0.01 g = <0.1%

4-7 B = 41.03 g = 99.5%

G = 0.19 g = 0.5%

N = 0.02 g = 0.1%

5-9 B = 24.78 g = 99.2%

0.17 g = 0.7%

N = 0.02 g = 0.1%

4-7 B = 25 ^ 9 g = 99.2%

G = 0.19 g = 0.7%

1 - 6 1 - 5 - 4

1 - 8 1 -4 1 -6 1 -

1 1 -4 1 -4 1 -5 1 -4

N = 0 g = 0% B = 0.18 g = 0.6% G = 28.26 g = 99.4%

² ) ³ ) See footnotes to Table 1.

) The solution used is a solution of the contents of the raffinate phase mentioned in the same amount and type in the column "alcoholic phase" specified solvent.

29

Table 4

Separation stage D Separation of the bitter substances from the tannins

Example

Embedded solvent system solution: ⁴ )

Ingredients Hydrocarbon phase of the raffinate phase from composition
Example (parts by volume)

alcoholic phase lot
in
Ltr. lot
in
Ltr. composition
(Mixture with water) 1 2.8 80% by volume of methanol 1 2.2 80% by volume of methanol 1 7.4 80% by volume of ethanol 1 4.0 80% by volume of methanol 1 3.7 80% by volume of methanol 1 2.9 80% by volume ethanol 1 8.81 90% by volume of methanol 1 4.71 8ü VoI.- ⁰ /. Methanol 1 2.17 70% by volume of methanol 1 2.02 60% by volume of methanol 1 5.61 80% by volume of ethanol 1 3.0 70% by volume ethanol 0.7 2.6 90% by volume ethanol 2 3.68 70% by volume of methanol

44 21 45 22nd 46 23 47 24 48 25th 49 26th 50 27 51 27 52 27 53 27 54 28 55 28 56 29 57 30th

Hexane fraction benzene (5: 5) hexane fraction benzene (2: 8) hexane fraction benzene (10: 1) Hexane fraction toluene (6: 4) hexane fraction cyclohexane (7: 3) hexane fraction cyclohexane (3: 7) Heptane fraction benzene (8: 2) heptane fraction benzene (10: 1) Heptane fraction benzene (15: 1) heptane fraction benzene (15: 1) heptane fraction benzene (10: 1) Heptane fraction benzene (10: 1) octane fraction cyclohexane (5: 5) decane fraction benzene (7: 3)

Table 4 (continued)

at
game
No. number of
Extrak Won 3 - 6th Solvent phase ³ ) 0.01 g = <0.1% functional
level ² ) X 30.05 g = 97.7% 44 Hydrocarbon solution
Content of neutral substances
Bitter substances
Tanning agents 0.69 g = 2.3% 6th 5 - 9 N = 0.04 g = C, l% B = 42.06 g = 99.4% 45 G = 0.22 g - 0.5% 9 4 - 7th N = 0.04 g = 0.1% B = 26.37 g = 99.2% 46 G = 0.19 g - 0.7% 7th - 5 N = 0.05 g = = 0.1% B = 37.89 g = = 99.6% 47 G = 0.12 g = = 0.3% 5 6 - - 11 N = 0.07 g = ■ - 0.2% B = 43.10 g = = 98.9% 48 G = 0.41 g = = 0.9% 11th 14 - - 26th N = 0.01 g = = 0.1% B = 22.84 g = = 99.4% 49 G = 0.13 g = = 0.6% 26th N = B = G =

Recovered raffinate phase ³ )

y alcoholic-aqueous solution

Content of neutral substances, bitter substances, tannins

- 4th

- 5th

- 4th

- 6

- 13

N
B.
G

N
B.
G

N
B.
G

N B. G

N
B.
G

N
B.
G

0 g ^;

0.42 g =

47.74 g ■■

0 g

0.18 g

50.54 g

0 g

0.17 g

28.71 g

0 g

0.07 g

51.09 g

0 g

0.41 g

47.50 g

0 g

0.10 g

23.67 g

0%

0.9%

99.1 ° /,

0%

0.3%

99.7%

0%

0.6%

99.4%

- 0.1% = 99.9%

= 0% = 0.9% = 99.1%

- 0% = 0.4% = 99.6%

continuation

31

example

Number of extraction stages ² )

Shall gain ie Solvent Phase ³⁾ Hydrocarbon solution

11th Content of neutral substances Bitter substances g = 0.1% Tanning agents g = 98.5% = 0.03 g = 1.4% 5 N = 35.83 g = 0.1% 6- B. = 0.53 g = 99.4% G = 0.03 g = 0.5% 5 N = 37.56 g = 0.1% 3 - B. = 0.18 g = 99.5% G = 0.03 g = 0.4% 5 N = 37.97 g = 0.1% 3 - B. = 0.14 g = 99.8% G = 0.03 g = 0.1% 8th N = 39.17 g = 0.5% 3 - B. = 0.04 g = 99.3% G = 0.15 g = 0.2% 6th N = 28.69 g = 0.5% 5 - B. = 0.05 g = 99.4% G = 0.15 g = 0.1% 26th N = 30.45 g = 0.2% 4 - B. = 0.03 g = 96.3% G = 0.04 g = 3.5% 5 N = 22.73 g = 0.1% 13 - B. = 0.84 g = 99.7% G = 0.05 g = 0.2% N = 50.82 3 - B. = 0.13 G

Recovered Raffinate Pliase ¹ )

y alcoholic-aqueous solution

Content of neutral substances, bitter substances, tannins

11th

26 - 6

- 3

- 3

- 4th

- 14

- 3

N
B.
G

N
B.
G

N B. G

N
B.
G

N
B.
G

N
B.
G

N
B.
G

N
B.
G

O g = O%

0.36 g = 0.7%

47.36 g = 99.3%

0 g

0.36 g

55.19 g

0 g

0.09 g

55.84 g

0 g

0.02 g

57.39 g

0 g

0.05 g

31.35 g

0 g

0.03 g

32.98 g

0 g

0.91g

21.04 g

0 g

0.10 g

61.71 g

= 0.7%

* 99.8%

* 0.2% = 99.8%

= 0.1% = 100%

= 0.2%

= 99.8%

= 0.1%

= 99.9%

= 4.1%

= 95.9%

= 0.2%

= 99.8%

Table 4 a Separation stage I) Separation of the bitter substances from the tannins

Example no.

Solution used:)

Solvent system

Hydrocarbon phase

Ingredients of the raffinate phase from composition example (parts by volume) alcoholic phase

Amount composition amount

in (mixture with water) in

L tr. L tr.

57a 30 a Cyclooctane 6.395 57 b 30 c Heptane fraction benzene (15: 1) 4,619 57 c 30 a Octane fraction cyclohexane (7: 3) 28.780 57 d 30d benzene 1.054 57 e 30e benzene 2,293 57 f 30f benzene 0.784 57 g 30 g benzene 1.790 57 h 30 h benzene 7,840

Vol .-% methanol wt .-% methanol vol .-% methanol vol .-% methanol Vol .-% methanol Vol .-% methanol Vol .-% ethanol Vol .-% methanol

0.800 4.480 0.400 0.500 5.850 2.000 2.000 20.507

33

Table 4 a (continued)

game No.

57a 12 7-12 B = 21.16 g = 99.6% 1-6

57b 7 4- 7 B = 43.39g = 99.8% 1-4

57c 8 5-8B = 15.62 g = 99.7% 1 - 4

number of
Pvtral Solvent phase recovered ³ ) 12th Hydrocarbon solution = 0.0Og =
= 21.16 g -
= 0.09 g = = 0.0%
= 99.6%
= 0.4% functional
stages ) X 7th Content of neutral substances
Bitter substances
Tanning agents = <0.01g =
= 43.39 g =
= 0.10 g = = 99.8%
= 0.2% ζ 8th N
B.
G = 0.0Ig =
= 15.62 g =
= 0.03 g - = 0.1%
= 99.7%
= 0.2% 12th T- 10 N
B.
G = 0.0Og =
= 8.71 g -
= 0.02 g = 0.0%
99.7% 7th 4- 5 N
B.
G = <D, 01g =
= 9.00g =
= <D, 01 g = 100.0% 8th 5- 5 N
B.
G = 0.02 g =
= 24.54 g =
= 0.09g = 0.1%
99.5%
0.4% 10 6- 9 N
B.
G = 0.03 g =
= 20.61 g =
= 0.10 g - 0.1%
99.4%
0 ^% 5 3- 5 N
B.
G = 0.0Og =
= 31.99g =
= 0.6Og = 0.0%
99.8%
0.2% 5 3- N
B.
G 9 5- N
B.
G 5 3-

Recovered raffinate phase ³ )

y alcoholic-aqueous solution

57d 10 6-10 B = 8.7Ig = 99.7% 1-5

57e 5 3-5B = 9.00 g = 100.0% 1-3

57Γ 5 3-5B = 2444 g = 994% 1 - 3

57g 9 5-9B = 20.61 g = 99.4% 1 - 5

57h 5 3- 5 B = 31.99g = 99.8% 1-3

² ) *) See footnote to Table 1. ⁴ ) See footnote to Table 3.

Table 5 Separation stage Λ Separation of the neutral and bitter substances (= soft resins) from the tannins

Content of neutral substances
Bitter substances
Tanning agents 0.00 g
0.07 g
25.85 g = 0.0%
= 0.3%
= 99.7% OWZ
Il Il Il 0.00 g
0.16 g
28.15 g = 0.0%
= 0.6%
= 99.4% Il Il Il
Z OQ O 0.00 g
0.02 g
23.17 g = 0.0%
= 0.1%
= 99.9% OWZ
Il Il Il 0.00 g
0.02 g
10.72 g = 0.0%
= 0.2%
= 99.8% Il Il Il
Z OQ O 0.00 g
0.02 g
1.86 g = 0.0%
= 1.0%
= 99.0% Il Il Il
Z ca C 0.00 g
0.05 g
43.92 g = 0.0%
= 0.1%
= 99.9% OWZ
Il Il Il 0.00 g
0.08 g
24.89 g = 0.0%
= 0.0%
= 99.7% Il Il Il
Z SQ U 0.00 g
0.07 g
26.82 g = 0.0%
= 0.2%
= 99.8% OWZ
Il Il Il

at F in v & cct7tp Carbon dioxide phase lot
in
LUr. alcoholic phase lot
in
Ltr. game
No. Solution: ¹ )
Primary composition
(Parts by volume) 2.44 composition
(Mixture with water) 1 CAU9KI-
solution from
example Heptane fraction 5.0 70% by volume n-propanol 04 58 7th Octane fraction 3.0 85% by volume n-propanol 04 59 7th Octane fraction 3.6 80% by volume i-propanol 3.6 60 8th Nonane fraction 2 ^ 0 80% by volume of methanol 1 61 8th Cyclopentane 1.67 80% by volume of methanol 1 62 2 Cyclooctane 4.05 70% by volume of methanol 04 63 1 Hexane fraction benzene (10: 1) 34 90% by volume of methanol 1 64 3 Hexane fraction benzene (8: 2) 24 80% by volume of methanol 1 65 4th Hexane fraction benzene (6: 4) 75% by volume of methanol 66 4th

continuation

35

example
No.

Deployed
Solution: ¹ )

Solvent sys) em

Coals were itofl phase

Primary extract solution from composition Example (parts by volume)

67 4th 68 6th 69 6th 70 6th

Hexane fraction benzene (6: 4) Hexane fraction benzene (10: 1) Hexane fraction benzene (7: 3) Hexane fraction benzene (S: 5)

alcoholic phase lot
in
Ltr. lot
in
Ltr. composition
(Mixture with water) 1 2.8 70% by volume of methanol 1 6.0 90% by volume ethanol 1 4.3 90% by volume ethanol 1 3.3 80% by volume of ethanol

Table 5 (continued)

example
No.

Number of Solvent Phase Recovered ³ ) Extraction χ Hydrocarbon ^{Solution stages 2} )

ζ Soft resin content

Tanning agents Recovered raffinate phase ³ )

y alcoholic-aqueous solution

Soft resin content of tanning agents

58 7th 4th - 7th 59 9 5 - 9 60 11th 6th - 11 61 6th 4th - 6 62 5 3 - 5th 63 5 3 - 5th 64 13th 7th - 13 65 8th 5 - 8th 66 7th 4th - 7th 67 5 3 - 5th 68 12th 7th - 12 69 26th 15th -26 70 8th 5 - 8th

W.
G

W.
G

W.
G

W.
G

W.
G

W.
G

W.
G

W.
G

W.
G

W.
G

W.
G =

W =
G =

W =
G =

= 42.59 g = 0.10 g

24.85 g 0.05 g

21.66 g 0.55 g

73.55 g 0.88 g

51.05 g 0.16 g

34.16 g 0.04 g

37.11 g 0.21 g

29.10 g 0.07 g

27.97 g 0.19 g

26.98 g 0.16 g

47.08 g 0.12 g

44.12 g 0.02 g

45.73 g 0.27 g

99.8% 0.2%

= 99.8% - 0.2%

= 974%

99.9% 0.1%

= 99.7%

99.9% 0.1%

99.4% 0.6%

99.8% 0.2%

99.3% 0.7%

99.4% 0.6%

99.8% 0.2%

100% 0.1%

99.4% 0.6% - 4th

- 5th

- 6

- 3

- 3

- 3

- 7th

- 4th

- 4th

- 3

- 6

- 12

- 3

W =

G =

W =

G =

W =

G =

W =

G =

W =

G =

W =

G -

W =

G =

W =

G =

W =

G =

W =

G =

W =

G =

W =

G =

W =

G =

0.18 g 19.74 g

0.09 g lU7g

0.11 g 10.33 g

0.09 g 53.54 g

0.18 g 50.01 g

0.08 g 24.95 g

0.24 g 26.39 g

0.04g 4831g

0.09 g 46.66 g

0.07 g 4549 g

0.16 g 29.50 g

0.02 g 26.83 g

035 g 28% g

= 0.9% = 99.1%

= 0.7% = 99.3%

= 1.0% = 99.0%

= 0.2% = 99.8%

= 0.4% = 99.6%

= 0.3% = 99.7%

= 0.9% = 99.1%

= 0.1% = 99.9%

= 0.2% = 99.8%

= 0.1% = 99.9%

= 04% = 99.5%

= 0.1% = 99.9%

= 1.2% = 98.8%

') See footnote to Table 1.
hh See footnote to Table 1.

6th 23 50
37 633 Deployed
Solution ') Solvent system , W = 46.80 g = 99.9%
G = 0.03 g = 0.1% lot
in
Ltr. 38 , W = 0.05 g = 0.1%
G = 29.59 g = 99.9% lot
in
LU. Tabel Primary
extract
solution off
example Hydrocarbon phase
composition
(Parts by volume) , W = 50.23 g = 99.9%
G = 0.05 g = 0.1% 4.1 , W = 0.04 g = 0.1%
G = 48.14 g = 99.9% 1 Separation level A Separation of neutral and bitter substances (= soft resins ¹ 6th Hexane fraction toluene (6: 4) W = 70.43 g = 99.9%
G = 0.31 g = 0.4% 3.2 of the tannins W = 0.33 g = 0.6%
G = 50.55 g = 99.4% 1 at
game
No 2 Hexane fraction xylene (6: 4) , W = 46.51 g = 99.9%
° G = 0.04 g = 0.1% 3.0 , W = 0.06 g = 0.2%
G = 29.81 g = 99.8% 1 in r. 4th Hexane fraction xylene (5: 5) -W = 30.92 g = 99.5%
G = 0.17 g = 0.5% 4.8 alcoholic phase
composition
(Mixture with water) W = 0.10 g = 0.2%
G = 51.94 g = 99.8% 1 I 72 6th Hexane fraction xylene (5: 5) W = 43.43 g = 99.8%
G = 0.11 g = 0.2% 2.73 80% by volume of methanol ₃ W = 0.08 g = 0.2%
G = 41.33 g = 99.8% 1 73 4th Heptane fraction benzene (8: 2) W = 33.83 g = 99.8%
G = 0.08 g = 0.2% 2.13 80% by volume of methanol W = 0.09 g = 0.4%
G = 24.42 g = 99.6% 1 74 5 Heptane fraction benzene (10: 1) _q W = 47.52 g = 99.6%
G = 0.17 g = 0.4% 1.79 80% by volume of methanol W = 0.19 g = 0.6%
G = 29.87 g = 99.4% 1 75 1 Heptane fraction benzene (8: 2) W = 37.24 g = 99.8%
G = 0.06 g = 0.2% 7.51 80% by volume of ethanol W = 0.09 g = 0.4%
G = 23.92 g = 99.6% 1 76 6th Heptane fraction benzene (10: 1) , W = 33.95 g = 99.8%
G = 0.07 g = 0.2% 2.01 80% by volume of methanol , W = 0.09 g = 0.4%
G = 2447 g = 99.6% 1 77 6th Heptane fraction benzene (10: 1) 1.40 70% by volume of methanol 1 78 1 Octane fraction benzene (7: 3) 4.5 70% by volume of methanol 1 79 5 Octane fraction cyclohexane (5: 5) 5.69 85% by volume ethanol 1 80 4th Cyclopentane benzene (7: 3) 1.36 60% by volume ethanol 1 81 5 Cyclohexane benzene (10: 1) 70% by volume of methanol 82 80% by volume of methanol 83 Solvent phase recovered ³ )
χ hydrocarbon solution
Soft resin content
Raw materials 90% by volume of methanol Tabel 4th 70% by volume of methanol at
game
No. 4th 71 5 Recovered raffinate phase ³ )
ν alcoholic-aqueous solution
Soft resin content
Tanning agents 72 4th
H 1 - 73 3 1 - 74 3 1 - 75 3 1 - 76 5 1 - 77 3 1 - 78 3 1 - 79 1 - 80 1 - 1 - 6 (continued) number of
Extra
functional
level ² ) 6th 6th 9 6th 5 5 5 9 5 5

number of
lixtrak-
lions
level ² ) 23 50 633
39 40 Solvent system
Hydrocarbon phase ⁶ ) Solvent phase recovered i
λ hydrocarbon solution lot
in
Ltr. . W = 0.16 g = 0.4%
G = 44.06 g = 99.6% Soft resin content
Tanning agents Amount ■
in '·
Ltr. ;; ■; 7th composition
(Parts by volume) Soft resin content
Tanning agents 1.613 , W = 0.10 g = 0.2%
G = 53.37 g = 99.8% N = <0.01 g = <0.1%
3 B = 0.03 g = 0.2%
G = <18.60 g = 99.8% 0.750 continuation 9 Solvent phase recovered)
ν hydrocarbon solution
Soft resin content
Tanning agents Recovered raffinate phase ³ )
· alcoholic-aqueous solution
Soft resin content
Tanning agents Octane fraction benzene (7: 3) N = 5.05 g = 15.5%
4-6B = 27.50 g = 84.4%
G = 0.02 g = 0.1% 3,450 ₃ W = 0.07 g = 0.2%
G = 42.16 g = 99.8% N = 0.01 g = <0.1%
4 B = 0.03 g = 0.2%
G = 21.41 g = 99.8% 0.500 at
game
No. 5 . W = 46.54 g = 99.6%
G = 0.17 g = 0.4% 1 - Octane fraction cyclohexane (7: 3) N = 3.1Ig = 13.7%
5-8 B = 19.58 g = 86.2%
G = 0.04 g = 0.2% 2.225 N = <0.01 g = <% y%
3 B = 0.18 g = 0.1%
G = 135.21 g = 99.9% 2,500 81 W = 32.36 g = 99.4%
G- 0.20 g = 0.6% 1 - benzene N = 28.21 g = 15.3%
4-6 B = 155.85 g = 84.6%
G = 0.15 g = <0.1% 0.82 1,000 82 ₃ W = 43.82 g = 99.9%
G = 0.05 g = 0.1% 1 - benzene « 1.045 0.500 83 6a Separation level A Separation of neutral and bitter substances (= benzene 0.894 1,000 Tabel Deployed
Solution: )
Primary- benzene 0.405 0.500 at
game
No. extrakl-
solution off
example benzene 0.235 0.500 8th toluene 1.180 Soft resins) from tanning agents; 0.500 ρ 83 a 5 Xylene 0.200 alcoholic phase 0.500 jf 83 b 3 Methylene chloride 1,000 composition
(Mixture with water) 2,500 S 83c 2 Methylene chloride 0.500 75% by volume isopropanol 1,320 83 d 6th Methylene chloride 80% by volume of ethanol 83 e 8th 6a (continued) 80% by volume of methanol 83 f 8th number of
Extra
functional
stages ) 60% by volume ethanol 83 g 1 80% by volume of ethanol 83 h 6th 6th 80% by volume of methanol 83 i 2 8th 60% by volume ethanol 83 k 8d 6th 70% by volume of methanol 230222/179 831 8d 80% by volume of ethanol 83 m 70 volume percent methanol Tabel 70% by volume of methanol at
game
No. 60% by volume of methanol 83a Won raf tinate phase)
i-alcoholic-aqueous solution 83b 83c 1 - 1 - 1 -

■ Continuation number of
K ν Ι γι L 23 50
41 633 42 f ' ⁵ Table 7
Ll separation stage ß Separation of the neutral substances from the bitter substances Employees
Solution ⁵ ) Solvent system lot
in
Ltr. alcoholic-warmth. solution Methanol 0.0%
0.4%
99.6% lot
in
Ltr. Bci-
. game
No. C All tlK ~
functional
stages-) Solvent phase recovered ¹ ) RaITinat phase won) L at-
Ef game
B IMr ingredients Hydrocarbon phase 1 Salary of soft resins
Tanning agents n-propanol 0.0%
0.1%
99.9% 2.70 5 ν hydrocarbon solution V B. phase of
example composition
(Vohnnentefle) 0.70 N =
3 B =
G = <0.01 g =
0.13 g =
34.99 g = n-propanol 0.1%
0.8%
99.2% 1.19 83d 6th Soft resin content
Tannins I. 61 Nonanfraknbn 0.50 N =
3 B =
G = 0.00 g =
0.02 g =
15.09 g = i-propanol 2.9%
97.1% 0.85 83e 6th N = 5.25 g = 14.5%
3-5 B = 30.75 g = 85.1%
G = 0.15 g = 0.4% 1 - I ** 58 Heptane fraction 1 N =
3 B =
G = <0.01 g =
0.06 g =
7.97 g = Methanol <0.1%
0.1%
99.9% 1.61 83f 5 N = 3.66 g = 15.6%
4-6 B = 19.86 g = 84.4%
G = 0.02 g = 0.1% 1 - I 85 59 Octane fraction 1 N =
3 B =
G = <0.01 g =
0.12 g =
4.08 g = 0.1%
1.6%
98.3% 3.01 83g 6th N = 15.36 g = 21.5%
4-6 B = 56.21 g = 78.5%
G = 0.01 g = <0.1% 1 - I 86 60 Octane fraction N =
3 B =
G - <0.01 g =
0.01 g =
12.02 g = 0.3%
99.7% 83h 4th N = 7.99 g = 21.7%
3-5 B = 28.75 g = 78.2%
G = 0.02 g = 0.1% 1 - 1 ⁸⁷ 62 Cyclopentane N =
3 B =
G - 0.01 g =
0.26 g =
15.85 g = 1.7%
98.2% 83i 7th N = 2.17 g = 13.2%
4-6 B = 14.27 g = 86.8%
G = 0.01 g = <0.1% 1 - 1 88 N =
4 B =
G = <0.01 g =
0.06 g =
21.20 g = 0.0%
0.9%
99.1% 83k 8th N = 3.77 g = 15.2%
2-4 B = 20.86 g = 84.0%
G = 0.20 g = 0.8% 1 - N =
4 B =
G = <0.01 g =
0.04 g =
2.34 g = 831 4th N = 3.09 g = 14.2%
4-7 B = 18.63 g = 85.5%
G = 0.07 g = 0.3% 1 - N =
2 B =
G = 0.00 g =
0.01 g =
1.26 g = |; 83 m N = 10.20 g = 13.6%
5-8 B = 64.77 g = 86.4%
G = <0.01 g = 0.0% 1 - J- '■ » ^{:) 3} )
ti ^b ) or N = 5.52 g = 13.6%
3-4 B = 32.38 g = 86.4%
G = 0.00 g = 0.0% 1 - See footnote to Table 1.
Halogenated hydrocarbon phase. alcoholic phase composition
(Mixture with water) 80% 80% 80% 80% 90 full%

43

continuation

example

Solution used: ^> )

Solution milking system

Hydrocarbon phase

Ingredients of the solvent phase of the composition Example (parts by volume)

alcoholic phase lot
in
Llr. lot
in
Ltr. composition
(Mixture with water) 4.55 0.50 80% by volume of methanol 1.39 1 90% by volume of methanol 1.67 1 90% by volume of methanol 2.56 1 85 vol .- "/ l methanol 5.00 1 70% by volume of methanol 0.95 1 90% by volume ethanol 1.27 1 90 vol. /., Ethanol 2.56 1 8Ü vol .-% ethanol 4.00 1 80% by volume of ethanol

39 90 91 92 93 94 95 96 97

63 64 66 6S 67 68 69 70 64

Cyclooctane fraction toluene (7: 3) hexane fraction benzene (10: 1) Hexane fraction benzene (8: 2) Hexane fraction benzene (8: 2) Hexane fraction benzene (8: 2) Hexane fraction benzene (15: 1) Hexane fraction benzene (8: 2) Hexane fraction benzene (7: 3) Hexane fraction benzene (6: 4)

Table 7 (continued)

at
game
No. number of Raffinate phase recovered 6-11 Hydrocarbon solution of neutral substances
Bitter substances
Tanning agents Won - 6 Solvent phase ³ ) of neutral substances
Bitter substances
Tanning agents = 0.2%
= 99.7%
= 0.1% functional
stages ) V 15-26 salary 9.34 g = 96.5%
0.34 g = 3.5%
0 g = 0% Λ - 12 0.09 g
49.61 g
0.08 g = 0.1%
= 99.6%
= 0.4% 84 15-26 O co Z
Ii Il Il 4.54 g = 99.5%
0.02 g = 0.5%
0 g = 0% - 12 alcoholic-aqueous solution 0.01 g
26.36 g
0.09 g = 0.1%
= 99.7%
= 0.3% 85 11th 13-26 N =
B =
G = 2.85 g = 99.6%
0.01 g = 0.4%
0 g = 0% 1 - 14 salary 0.01 g
15.17 g
0.05 g = 0.5%
= 95.5%
= 04.0% 86 26th 8-16 Il Il Il
Z PQ O 2.49 g = 87.5%
0.36 g = 12.5%
0 g = 0% 1 - 8 N =
B =
G = 0.07 g
12.32 g
0.52 g = 0%
= 99.5%
= 0.5% 87 26th 13-26 Il Il Il
ZeO 5.5Ie = 88.5%
0.71 g = 11.5%
0 g = 0 % 1 - 14 Il Il Il
Z M O 0 g
32.04 g
0.16 g = 0.8%
= 99.0%
= 0.2% 88 26th 9-16 Il Il Il
Z PQ Ü 3.09 g = 81.6%
0.70 g = 18.4%
0 g = 0 % 1 - 8 O 03 Z
Il Il Il 0.15 g
18.74 g
0.04 g = 0.1%
= 99.0%
= 0.9% 89 16 15-26 Il Il Il
Z PQ O 4.88 g = 97.7%
0.11 g = 2.3%
0 g = 0 % 1 -12 CWZ
Il Il Il 0.02 g
22.72 g
0.20 g = 0.1%
= 99 _: 0%
= 1.0% 90 26th Il Il Il
Z CQ O 2.92 g - 99.5%
0.01 g = 04%
0 g = 0% 1 Il Il Il
Z PQ O 0.01g
17.45 g
0.18 g 91 16 N =
B =
G = 1 Il Il Il
Z PQ O 26th 1 owz
Il Il Il Il Il Il

45

continuation number of
Pvtral Recovered raffinate phase 8-16 Hydrocarbon solution 3.25 g = 87.5%
0.47 g = 12.5%
0 g = 0% Solvent phase recovered ³ ) alcoholic-aqueous solution = 0.07 g = 0.4%
= 17.90 g = 99.2%
= 0.07 g = 0.4% at
game
No. HXUeK-
ions
level ² ) y 14-26 Content of neutral substances
Bitter substances
Tanning agents 2.8Ig = 97.6%
0.07 g = 2.4%
0 g = 0% X Content of neutral substances
Bitter substances
Tanning agents = 0.01 g = 0.1%
= 16.42 g = 99.0%
= 0.16 g = 0.9% 16 14-26 OWZ
Il Il Il 5.52 g = 96.1%
0.23g = 3.4Vo
0 g = 0% N
B.
G = 0.05 g = 0.2%
= 26.92 g = 99.4%
= 0.1Ig = 0.4% 92 26th 15-26 Il Il Il 4.94 g = 96.5%
0.18 g = 3.5%
0 g = 0% .1-8 N
B.
G = 0.22 g = 0.9%
= 23.84 g = 99.1%
= 0.01 g = 0.1% 93 26th 14-26 Il Il Il
Z OQ O 5.16 g = 96.4%
0.19 g = 3.6%
0 g = 0% 1 - 13 N
B.
G = 0.04 g = 0.1%
= 25.94 g = 98.9%
= 0.26 g = 1.0% 94 26th 15-26 Il Il Il
Z CQ O 5.38 g = 99.6%
0.02 g = 0.4%
0 g = 0% 1 - 13 N
B.
G = 0.0Ig = 0.1%
= 28.29 g = 99.8%
= 0.04 g = 0.2% 95 26th Il Il Il
Z CQ O 1 - 14 N
B.
G 96 26th Il Il Il
Z oa O I - 13 N
B.
G 97 1 - 13

² ) ³ ) See footnote to Table 1.
^s ) See footnote to Table 4.

Table 8

Separation level B Separation of the neutral substances from the bitter substances

example
No.

Implemented solution: ⁵ )

Solvent system

Ingredients Hydrocarbon phase of the solvent phase of composition example (parts by volume)

alcoholic phase lot
in
Ltr. lot
in
Ltr. composition
(Mixture with water) 3.70 1 90% by volume of methanol 5.00 1 90% by volume of methanol 4.76 1 80% by volume of methanol 1.18 1 93.2 volume percent methanol 1.18 1 93.2% by volume methanes! 0.65 0.5 93.2 volume percent methanol 1.32 1 90% by volume ethanol 1.10 1 90% by volume ethanol 2.5 0.5 80% by volume of methanol 1 0.66 90% by volume of methanol 1.79 0.50 80% by volume of methanol 2.50 0.70 80% by volume of methanol 1.25 0.5 90% by volume of methanol

98 71 99 72 100 73 101 76 102 77 103 75 104 78 105 79 106 82 107 80 108 81 109 83 110 82

Hexane fraction toluene (6: 4) hexane fraction xylene (6: 4) hexane fraction xylene (5: 5) Heptane fraction benzene (10: 1) heptane fraction benzene (10: 1) heptane fraction benzene (8: 2) Heptane fraction benzene (15: 1) heptane fraction benzene (10: 1) cyclopentane fraction benzene (5: 5) Octane fraction benzene (10: 1) Octane fraction cyclohexane (110: 1) Cyclohexane fraction benzene ("10: 1) Cyclopentane benzene (7: 3)

47

Table 8 (continued)

example

Number of extraction stages ² )

31

26th

16

16

16

31

26th

26th

26th

26th

26th

26th

Raffinate phase recovered

y Hydrocarbon solution

Content of neutral substances, bitter substances, tannins

Solvent phase recovered ³ )

λ- alcoholic-aqueous solution

Content of neutral substances bitter substances Tanning agents

-31

-8th

-26

9-16

9-16

8-16

N
B.
G

N
B.
G

N
B.
G

N
B.
G

N
B.
G

4.15 g
0.21g
0 g

6.45 g
0.05 g
0 g

8.57 g
0.14 g
0 g

4.06 g
0.17 g
0 g

3.56 g
0.13 g
0 g

95.1% 4.9% 0%

99.2% 0.8% 0%

98.5% 1.5% 0%

95.4% 4.1% 0%

96.4% 3.6% 0%

N = 3.28 g = 89.2% B = 0.40 g = 10.8% G = O g = 0%

N = 5.02 g = 77.5% 15 - 31 B = 1.44 g = 22.3% G = 0 g = 0% N = 3.97 g = 86.5% 13-26 B = 0.62 g = 13.5% G = 0 g = 0% N = 1.69 g = 98.0% 14-26 B = 0.04 g = 2.0% G = 0 g = 0% N = 3.48 g = 97.8% 14-26 B = 0.09 g = 2.4% G = 0 g = 0% N = 4.49 g = 97.6% 14-26 B = 0.11 g = 2.4% G = 0 g = 0% N = 3.95 g = 97.4% 14-26 B = 0.11 g = 2.6% G = 0 g = 0% N = 1.74 g = 97.5% 14-26 B = 0.05 g = 2.5% G = 0 g = 0%

- 17th

- 4th

- 13

- 9

- 17th

- 14

- 13

- 13

- 13

- 13

- 13

N
B.
G

N
B.
G

N
B.
G

N
B.
G

N
B.
G

N =

B =

G =

N =

B =

G =

N =

B =

G =

N =

B =

G =

N =

B =

G =

N =

B =

G =

N =

B =

G =

0.40 g

25.22 g

0.01g

0.01 g

36.56 g

0.05 g

0.03 g

42.68 g

0.31 g

0.02 g

27.92 g

0.11 g

0.02 g

21.35 g

0.08 g

1.5%

98.5%

0.1%

0.1% 99.9% 0.01%

0.1%

99.2%

0.7%

0.1%

99.5%

0.4%

0.1%

99.5%

0.4%

N = 0.07 g = 0.4% B = 19.64 g = 98.8% G = 0.16 g = 0.8%

0.52 g = 2.1%

24.73 g = 97.3%

0.17 g = 0.6%

0.10 g = 0.4%

21.48 g = 99.3%

0.06 g = 0.3%

0.01 g = 0.1%

10.18 g = 98.9%

0.10 g = 1.0%

0.01 g = 0.1%

20.72 g = 99.6%

0.07 g = 0.3%

0.02 g = 0.1%

28.82 g = 99.3%

0.17 g = 0.6%

0.01 g = 0.1%

27.42 g = 99.8%

0.05 g = 0.2%

0.01 g = 0.1%

9.70 g = 98.9%

0.10 g = 1.0%

² ) ■ ') See footnote to table

The solution used is a solution of the ingredients of the named solvent phase in the same amount and type as in in the column "Hydrocarbon phase" specified solvent.

49

Table 8 a

Separation level B Separation of the neutral substances from the bitter substances

at
game
KIr Deployed
Solution:') Solvent system lot alcoholic phase lot 1ΝΓ. Ingredients Hydrocarbon phase ⁶ ) in
Ltr. in
Ltr. the solvent 1.534 composition 2.290 phase of composition 0.636 (Mixture with water) 0.800 example (Parts by volume) 0.808 90% by volume ethanol 1,000 110a 83 f benzene 0.045 90% by volume ethanol 0.500 110b 83 a Octane fraction benzene (10: 1) 0.042 90% by volume ethanol 0.450 110c 83 b Octane fraction cyclohexane (5: 5) 0.017 70% by volume of methanol 0.180 HOd 83 c benzene 0.017 70% by volume ethanol 0.186 HOe 83 d benzene 0.031 70% by volume ethanol 0.300 Court 83 e benzene 0.047 70% by volume ethanol 0.500 110g 83 g benzene 0.075 80% by volume of ethanol 1,500 110h 83 h Toiuoi 0.075 80% by volume of methanol 1,500 HOi 83 i Xylene 0.100 60% by volume of methanol 2,000 110k 83 k Methylene chloride 60% by volume of methanol HOI 831 Methylene chloride 60% by volume of methanol HOm 83 m Methylene chloride

Table 8 a (continued)

example

No.

number of
Extraction stages ² )

Solvency phase won ³ )

Hydrocarbon to ITI solution

Content of neutral bitter tannins

I. HOa 35 H 110b 23 HOc 35 HOd 45 HOe 70 Court 70 110g 70 HOh 65

-35

- 23

- 35

-45

- 70

- 70

- 70

-65

N
B.
G

N
B.
G

N
B.
G

N
B.
G

N
B.
G

N
B.
G

N
B.
G

N
B.
G

2.69 g
0.02 g
0.00 g

3.59 g
0.05 g
0.00 g

2.13 g
0.03 g
0.00 g

3.16 g
0.01g
0.00 g

1.45 g
0.04 g
0.00 g

2.25 g
0.05 g
0.00 g

0.75 g
0.01g
0.00 g

1.33 g
0.02 g
0.00 g

Recovered raffinate phase- ¹ )

y alcoholic-aqueous solution

Content of neutral substances Bitters to (Ten tannins

- 17th

- 11

- 17th

- 21

- 34

- 34

- 34

- 31

N = 0.01 g B = 9.82 g G = <0.01 g

N
B.
G

N
B.
G

N
B.
G

N
B.
G

N
B.
G

N
B.
G

N
B.
G

0.01 g

19.56 g

0.02 g

<0.01 g

13.41 g

0.04 g

<0.01 g

17.48 g

0.03 g

<0.01 g 8.52 g 0.07 g

0.01 g

12.23 g

0.02 g

<0.01 g 2.71 g 0.03 g

<0.01 g 8.77 g 0.01 g

0.1% 99.9%

99.9% 0.1%

<0.1%

99.7%

0.3%

0.1%

99.8%

0.2%

<0.1%

99.1%

0.8%

<0.1%

99.8%

0.2%

0.1%

98.8%

1.1%

<0.1%

99.8%

0.1%

continuation

With- Number of Solvent Phase Won ³ )

game extra

No. ion χ hydrocarbon solution

level ² )

Content of neutral substances ζ bitter substances

Tanning agents Recovered raffinate phase ³ )

y alcoholic-aqueous solution

Content of neutral substances, bitter substances, tannins

HOi 55 29-55

HOk 70 37-70

HOi 70 37-70

110m 70

37-70

NB
GN
B.
G

N
B.
G

N
B.
G

2.41 g 0.06 g 0.00 g 1.86 g 0.08 g 0.00 g

6.12 g 0.26 g 0.00 g

0.33 g 0.01 g 0.00 g

: 97.5% 2.5% 0.0% 95.9% 4.1% 0.0%

95.9% 4.1% 0.0%

97.1% 2.9% 0.0%

² ) ³ ) See footnote to Table 1.

⁵ ) See footnote to Table 4.

^b ) or halogen hydrocarbon phase.

Example 11 1

870 g of Hallertauer Mittelfrüh 1970 pressed hops were divided into 4 stages, each with 217.7 g of hops and hops extracted with a mixture consisting of 4354 ml of methanol and 86 ml of η-hexane. The extraction was made carried out according to the principle of a carousel extractor in such a way that the 4.44 1 solvent means a pump were sent successively from the 1st stage to the 4th stage, the solids content rose steadily in the solvent front. Ultimately, 2,641 primary extract solutions were obtained Containing 4.15% water and 10.8% by weight solids. The solid was separated from the primary extract solution. 79.06 g of 100% strength could be obtained from the primary extract solution by evaporating the solvent Gain primary extract.

Example 112

75.0 g of the primary extract obtained according to Example 111 were dissolved in methanol with the addition of 15 ml hexane fraction (bp. 67-69 ⁰ C) and then related by the addition of distilled water, a water content of 10 vol .-% adjusted to methanol. This resulted in 625 g (= 721 ml) of solution with a solids content of 12.0% by weight.

1 -27

1 -34

1 -34

1 -34

N
B.
G
N
B.
G

N
B.
G

N
B.
G

0.01 g 13.31 g

0.20g

0.01 g 11.18 g

0.07 g

0.04 g

38.88 g

0.01 g

<0.01 g 1.97 g 0.00 g

0.1% 98.4%

1.5%

0.1% 99.3%

0.6%

0.1%

99.9%

0.1%

<ö, l%

99.9%

0.0%

In a rotating split tube column of 134 cm length and in the area of the 100 cm long rotor of 4 mm Inside width, 216 to 232 ml of hexane fraction (boiling point 67-69X) per hour were at the level of the lower end of the rotor dosed. The hexane fraction had previously been saturated with 90% by volume of methanol. At the height of the upper end of the rotor 310 to 340 ml / h of 90% by volume of hexane-saturated methanol were fed in. The column rotor was set to a speed of 430 rpm. The column itself was thermostatted to 15C.

In the first hour of the experiment, 224 ml of primary extract solution were placed in the middle of the column (= 23.30 g of solid) are metered in. Subsequently, the dosage of the 12.0% primary extract solution with a rate of 124-128 ml / h (13.10 g / h solids) continued. After 1.5 hours the extraction process reached A steady equilibrium in the column, i.e. the supply of solids by means of the primary extract solution and the amounts of solids flowing out of the separation zones at the column ends were total practically the same.

Table 9 shows the result of this extraction, with the effluent collected in fractions and its content was analyzed for neutral or bitter and tannins.

Tabel 9 Neutral substances Fixed Wt% Bitter- and tannins Wt% Vs'Vr total Time material η -acid α -acid fixed (H) V _s (G) Vr Fixed material 1.06 4.33 material 26.80 (ml) 1.49 4.18 (ml) (G) 30.14 247 1.35 1.18 430 13.22 28.68 0.58 14.28 1.5 273 1.47 0 450 12.29 30.18 0.61 13.78 2.5 ' 292 5.37 2.31 155 11,534 29.00 0.64 12.88 3.5 248 = 10% 405 11.303 0.61 12.78 4.5 1060 1740 48.34 0.61 53.72 = 90% = 100%

From Table 9 it can be seen that if the extraction is continued continuously according to cn. 4 hours the content of α-acid and bitter substances in the neutral fraction becomes negligibly small and can no longer be determined analytically. The extraction was terminated after a test duration of 4.5 hours. _. . ,, ..,
Example 113

1740 ml of the alcoholic-aqueous phase, which remains after the extraction in Example 112 and which still contains the bitter and tannins by adding water to a mixing ratio of Brought 80% by volume of methanol and 20% by volume of water.

This solution was in a further rotary split tube column of the same type as in Example 112 fed at 15 ° C and at a rotor speed of 650 Lipm in the middle of the column, with the feed was 190 ml / h in the first hour and about 131.3 ml / h thereafter. At the bottom of the column were an average of 838.8 ml / h of a mixture of 7 parts by volume of hexane and 3 parts by volume of benzene, which Was saturated with 80 vol .-% aqueous methanol, fed and on average at the top of the column 71 ml / h of pure methanol are metered in.

After about 1.5 hours the extraction process reached steady equilibrium in the column.

On average, a volume ratio of the absorbing phase V _s (hydrocarbon phase) to the releasing phase V _R (alcoholic-aqueous phase) of = 4.37 = V ₅ IVn was used .

After a running time of 3.17 hours, all of the solution used had passed through.

From the absorbing phase V _s , which contained the bitter substances, and from the releasing phase V _R , which contained the tannins, the solvent was drawn off ^{at 40 ° C. under reduced pressure.} The amount and composition of the tannin and bitter substances obtained are shown in the following list:

a) For the absorbing phase ΙΛ, (bitter substance extract solution)

Pure extract-100% 19.17 g (= 35.7%

related to

lawful solution)

Bitter substances 19.17 g (= 91.5% rel.

on bitter extract), of it
was

Proportion of «-acids 9.43 g ■ {= 49.2% rel.

on bitter extract)

Neutral substances traces

No tannins detectable

b) For the outgoing phase V _R (tannin extract solution)

Pure extract-100%

Tannins

Bitter substances

29.17 g (= 60.3% rel.
on deployed
Solution)

27.55 g (= 944% rel.
on tannin extract)

1.62 g (= 5.5% rel.
on Gerstoffextrakt)

60

65 For the implementation of the procedure with the series connection of examples 112 and 113 the following overall result is obtained:

For the content of neutral substances in the neutral substance extract:
practically completely contained in the extract.

For the content of bitter substances in the bitter substance extract:

Content in the pure extract
Content in the neutral substance extract
Content in the tannin extract

Total salary

19.17 g 0.23 g 1.62 g

21.02 g

This means that approx. 91.2% of the bitter substances contained in the primary extract solution are left in the bitter substance extract.

For the content of tannins in the tannin extract: contained practically completely in the extract.

Example 114

40 kg pressed hops Hallertauer Nordbrauer 1970 were in four stages of 10 kg pressed hops with one Solvent mixture consisting of 411.61 methanol and 8.4 1 η-hexane treated, the solvent was passed through the four vessels within 7 hours. 3401 solution were obtained, which were then passed again in 4 stages through 10 kg of fresh pressed hops each time in the manner described. Then 80 liters of fresh methanol were pumped in. 1301 solution were obtained, additionally from the last vessel 601 solution and from the penultimate one Vessel 101 solution. The primary extract solution therefore consisted of 2001 solution with a solids content of 10.83%, which corresponds to 18.049 kg. The water content was found to be 3.6%.

The primary extract solution was at a rate of 7.50 l / h corresponding to 6.25 kg / h together with 0.55 kg / h of water in the middle of a total of 5 m long sieve bottom pulsation column with a nominal width of 50 mm and a tray clearance of 50 mm. 5.20 l / h = 3.51) kg / h η-hexane were pumped into the lower inlet connection and 4.42 l / ha 3.70 kg / h 90% by volume of methanol were pumped into the upper inlet connection. This corresponds to a volume ratio before Φ = ^ V'ä = 0.41- ^The mass transfer at 15 ° C was caused by a pulsation of the column ^{contents with an amplitude of 3-4 mm and a frequency of 108 mm "1.} Yeasts from the upper separator in the steady state 5.10 l / ha 3 ^ 5 kg / h neutral substances in n-hexane solution, while the lower separating vessel would contain 12.55 l / ha 10.55 kg / h bitter and tannins.

The bitter substance and tannins were adjusted to a methanol content of 80% by volume with water and at a rate of 13.831 / η = Μΐ.83 kg / h in the upper third of the sieve tray column 2, which is a total of 9 m long, a nominal width of 724 now and has a floor clearance of 100 mm. The light phase, consisting of n-hexane-benzene (7: 3 v / v), saturated with 80% by volume of methanol, was fed into the central feed station at a rate of 67.80 l / h = 49.22 kg / h introduced, while 5 ^ 3 l / h & 4.74 kg / h 80 vol .-% methanol saturated with n-hexane-benzene (7: 3 v / v) were pumped into the upper inlet nozzle. The mass transfer was generated by a pulsation with a pulse of 6 mm and a frequency of 54 mm " ^1. 18.00 l / t = 15.64 kg / h of tannin dissolved with 80% by volume of methanol ran out of the lower part of the sump

56

from, while from the upper sump part 69.17 l / h either in a third column with 88-96 vol .-%

^ 50.15 kg / h of bitter substances dissolved in hexane-benzene (7: 3 v / v) left the column, ie a volume ratio of φ = V ^ V _R = 3.84 was used.
The bitter substances in n-hexane-benzene (7: 3 v / v) are re-extracted with ethanol or, as in Example 113, freed from the solvent by known methods.

After the solvents have been removed from the individual fractions, the following is obtained:

Neutral substance extract

Of which neutral substances

Bitter substances

The bitter substances contain

an "acids

Bitter extract

The bitter substances contain

of α-acids

Neutral substance

Tannins not detectable

Bitter substances from bitter substance extract

Bitter substances from neutral substance extract

Bitter substances from tannin extract

Total bitter substances

α-acids in the bitter extract

α-acids in the neutral substance extract

α-acids in the tannin extract

Total alpha acid

Tannin extract

Tannins

The tannin extract contains

of bitter substances

These bitter substances are on

α-acid present

1.960 kg (= 10.86% based on solids in the primary extract) 1.825 kg (= 93.11% based on neutral substance extract) 0.135 kg (= 6.89% based on neutral substance extract) 0.0065 kg (= 3.32% based on neutral substance extract)

8.293 kg (= 45.94% based on the primary extract) 4.432 kg (= 53.44% based on the bitter extract)

0.01% 8.293 kg (= 96.50% based on total bitter substances)

0.135 kg

0.166 kg 8.594 kg (= 47.61% based on primary extract) 4.432 kg (= 97.79% of the total a-acids)

0.065 kg

0.035 kg 4.532 kg (= 25.11% based on primary extract)

7.800 kg (= 43.22% based on primary extract) 7.634 kg (= 97.87% based on tannin extract) 0.166 kg (= 2.13% based on tannin extract)

0.035 kg (= 0.45% based on tannin extract)

It can be seen from these data that, analogously to Examples 12 and 113, each of the three fractions was obtained with a purity of more than 90%.

Overall balance

Primary extract
Total neutrals
Total bitter substances
Total tannins

(100%) 18.049 kg

(100%) 1.825 kg (= 10.11% based on primary extract)

(100%) 8.594 kg (= 47.61% based on primary extract)

(100%) 7.654 kg (= 42.41% based on primary extract)

18.073 kg (= 100.13% based on primary extract)

The slight excess is due to small residual water contents example

To carry out a series of brewing tests, Vorderwüize from a commercial brewery were used . By diluting with 32 liters of water, wort with an extract content of 12% was obtained. Four portions of 251 each of the wort produced in this way were boiled under the same conditions with hops or with hop extracts made from them. These four hopped worts were then processed in the same way.They were filtered, fermented with bottom-fermented brewer's yeast at 5 - 9 ° C, the beers for four weeks stored at 1 ° C., filtered under pressure and bottled. The respective amount of hops or extract was chosen so that the same amount of total soft resins got into the wort with each hop. In detail, the following amounts were added: Sud I 50 g

Sud Π Sud III SudlV 6.76 g

6.92 g

6.92 g

10.40 g

Hops, variety Hallertau Middle Early 1970 Neutral substance extract Bitter extract Bitter extract Tannin extract

As expected, the bitter substance contents of the finished beers were different. Foam durability and cloud stability were also not uniform. The extract beers had a more stable foam and slightly less cloud stability than the hops

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beer. Other important analysis data of the beers, such as degree of fermentation, extract content, alcohol content and CO _r content, however, largely agreed, so that their organoleptic comparability was ensured. The taste assessment of the beers by 6 selected testers resulted in the following assessments:

I hop beer

weak hop aroma

medium, slightly lingering bitterness

II neutral beer

fine, distinctive hop aroma
pure, very weak bitterness

III bitter beer

no hop aroma

strong, non-trailing bitterness

IV bitter substance-tannin beer

no hop aroma, but strong, beer-typical smell

very strong, lingering bitterness

ίο These results show a strong differentiation the taste of beer in the extract beers. At the same time it becomes clear that the aroma and bitterness of the beers can be influenced independently of each other depending on the extract used.

drawings

Claims (1)

Claim: Process for obtaining the neutral, bitter and tannin substances in hops by extracting s Hop extracts or hop extract solutions, thereby characterized in that a hop extract solution containing these substances (primary extract solution) subjected to at least two successive liquid-liquid extractions, wherein with exclusively organic solvents of different polarity, possibly water contain, and possibly different volume ratios of polar dissimilar solvents is worked so that in the first extraction stage either the tannins or the neutral substances are separated and in the subsequent extraction stage that in the primary extract solution remaining neutral and bitter substances or tannins and bitter substances separated from each other will.