Patent Publication Number: US-2011060048-A1

Title: Removing alcohol in vivo through esterification

Description:
BACKGROUND 
     Ethanol (C 2 H 5 OH) is a depressant that is present in all alcoholic beverages in varying percentages. On an empty stomach, 20% of the ethanol consumed almost immediately diffuses into the bloodstream through the walls of the stomach. Within 30 minutes, all of the ethanol will have entered the blood, and clear symptoms can be observed if the BAC (blood alcohol content) of the drinker is over 0.08. Years of chronic drinking leads to lasting damage on vital organs. 
     Effects of alcohol on various parts of the body are summarized below:
         1. The liver: The majority of alcohol consumed is transported to the liver to be converted into harmless products. As a result, 15% of all alcohol-related deaths are attributable to liver failure. Liver damage begins with the hepatotoxic effect (aka fatty liver), and inflicts moderate, reversible damages on the liver. If drinking continues, scarring and inflammation will deteriorate, and eventually culminate in liver cirrhosis, which is irreversible and usually fatal.   2. The brain: Short term effects include drowsiness, loss of coordination, dizziness, and mood swings. Long-term effects include memory loss, strokes, and concussions.   3. Reproductive system: In males, alcohol abuse results in prostate inflammation and erectile disorder. In women, it results in irregular menstrual cycles, hormonal imbalance, infertility, and—if pregnant—physical deformities and mental retardation in the fetus.
 
Alcoholism has far-reaching ramifications on both individuals and society. Although it is impractical to forbid people from drinking altogether, there is a need to limit its effects.
       

     Prior Arts: 
     In January 2008, a patent was filed under the name “Method for the Accelerated in Vivo Removal of Ethanol” (U.S. Pat. No. 7,323,452). The patent stated that the liver naturally converts ethanol into acetaldehyde using an enzyme called alcohol dehydrogenase, but that the equilibrium constant of 1 indicated inefficiency. It suggested that a pill, capsule, or nasal spray containing chemicals accelerators in the direction of acetaldehyde (such as zinc ions) could be added in order to increase the proportion of ethanol that was being eliminated. However, this proposal is based on the assumption that ethanol has already entered the bloodstream, and must be eliminated from there. By the time ethanol has entered the bloodstream, it is extremely difficult to eliminate it in large quantities. Ethanol is a very small molecule (molar mass 46 g/mol) compared to the organic macromolecules, and even molecules such as glucose and amino acids, that normally circulate throughout the bloodstream. Its solubility in water additionally allows it to circulate rapidly throughout the body. Clearly, the removal process must take before ethanol enters the bloodstream. 
     In 1984, a patent was filed under the title of “Alcohol removal from blood with alcohol oxidase (U.S. Pat. No. 4,450,153). Alcohol oxidase is an enzyme that converts ethanol into acetaldehyde in the following method: primary alcohol+O 2 ⇄an aldehyde+H 2 O 2 . After the aldehyde is produced, it naturally decays into water and carbon dioxide, which are eliminated through urination and respiration. However, the alcohol oxidase does not naturally exist in the human body in sufficient amounts, and is very costly. A solution containing approximately 17,200 EU of alcohol oxidase is needed to neutralize the ethanol in a can of beer within 15 minutes. This amount of enzyme will cost nearly $370. A method that requires nearly four hundred dollars to reverse the effects of a single serving of a mild alcoholic beverage cannot be applied commercially, or developed into an easily-accessible sobriety drug. 
     Another method was published in a 1975 volume of the American Journal of Clinical Nutrition, for removing ethanol from the body through ingesting sugars. Eight male volunteers 25-50 years old, of approximately the same height and weight, were each given a few servings of alcoholic beverage, followed by 30 grams of fructose or sucrose. Measurements were taken each hour, and the percent of ethanol decrease flattened out at 29% on average. This experiment is dubious, because a). It only eliminated 29% of the ethanol from the bodies of the drinkers, and thus does not appear to be a very efficient antidote, b). Fructose is very expensive, and c). It promotes a high-sugar diet. 
     As will be discussed below, the use of the acetate ion to neutralize ethanol is a much more efficient method. 
     SUMMARY OF INVENTION 
     The key principle of this invention for ethanol removal is ethanol esterification with acetic acid, as described in the following reaction: 
       C 2 H 5 OH+CH 3 COOH→←CH 3 CH 2 OCOCH 3 K eq =45 M −1  
 
     In an in-vitro test, a known volume (17.15 mL) and concentration (0.4 M, 0.6 M, 0.8 M, 1.0 M) of ethanol reacted with the same volume and concentration of acetic acid. Titrations with potassium permanganate (KMnO 4 ) were later performed after the contents of each beaker received 5 minutes each of vigorous stirring and 25 minutes of inactivity to compare the final amount of ethanol left with the original amount. In agreement with the calculations done using the known equilibrium constant of K=45 M −1 , a 79-86% decrease in the amount of ethanol was demonstrated when the system reaches equilibrium. Average results calculated after 5 trials were within a 4% margin of error, as shown in Table 1. 
                     TABLE 1                  In vitro experiment of esterification with different values of initial       ethanol and acetic acid concentrations. The last column indicates       the amount of decrease in the ethanol concentration after the       system reaches equilibrium.                         Initial               C 2 H 5 OH =       Initial   Equilibrium Concentrations   % Ethanol                             CH 3 COOH   C 2 H 5 OH═CH 3 COOH   CH 3 CH 2 OCOCH 3     Decrease               0.4 M   0.084 M   0.316 M   79.0%       0.5 M   0.095 M   0.405 M   81.0%       0.6 M   0.105 M   0.495 M   82.5%       0.7 M   0.114 M   0.586 M   83.7%       0.8 M   0.123 M   0.677 M   84.6%       0.9 M   0.131 M   0.769 M   85.4%       1.0 M   0.138 M   0.862 M   86.2%                    
The amount of alcohol intake before reaching the DUI level could vary widely among individuals and depends on many factors such as weight, age, gender, and race. It is generally believed that, for most people, four cans of beer or an equivalent amount of ethanol taken over a time period of 1-2 hours can influence people&#39;s health, behaviors, and judgment. Four cans of beer contain approximately 1 mole of ethanol. According to the above in vitro experiment, it will take about an equal amount (i.e. 1 mole) of acetic acid to substantially reduce the amount of ethanol through esterification. A 100 ml of off-the-counter vinegar (5% acidity), on the other hand, contains only 0.086 mole of acetic acid. It is unconceivable that anyone is able to take 1160 ml of off-the-shelf vinegar in order to effectively remove the ethanol. Both the sheer volume and the acidity of the vinegar make the approach impractical.
 
     An important part of the invention is about practical methods to orally take the required amount of acetic acid to react with ethanol. A practical approach has to meet two requirements: (a) substantially reduce the intake volume of acetic acid containing liquid, and (b) introducing the required amount (around 1.0 mole) of acetic acid in a harmless and least intrusive manner, meaning to create no acid burn or discomfort. In accordance with the invention, these practical approaches are developed based on one key concept: reducing the acidity of concentrated acetic acid with a proper amount of strong base to yield a final product having a proper level of acidity, i.e. for a range of PH value between 3-4. For references, the PH value is 2.9-3.3 for apple juice, 3-4 for orange juice, and 4.2 for tomato juice. Possible choices of strong base include NaOH and KOH. Added to concentrated acetic acid, the solution contains CH 3 COONa or CH 3 COOK and CH 3 COOH with lower acidity. The mixed solution contains sodium and/or potassium, and the amount should be kept well below the FDA recommended daily allowance. Since it is less desirable to create high sodium-containing drinks, KOH becomes a more attractive choice of base to partially neutralize the acetic acid. 
     The detailed analysis, to be discussed in the following section, suggests the following practical recipe: 
     Adding 0.02 mole of KOH (20% FDA recommended daily dosage for potassium) to 1 mole of 50% concentrated acetic acid, one can take 1 mole of acetic acid in a 75 mil drink at a PH-value of around 3.3. 
     Numerous recipes following the similar principle can be created to produce essentially the same effect of turning ethanol into ester in stomach. Additional flavors and components can also be introduced to the drink without altering its function while improving the taste. Furthermore, the same ingredients can also be incorporated into solid food such as cakes, snack bars, cookies, and other bakeries in manners that the heating process of making such goods will not degrade the performance. It is also conceivable that the mixed KOH/CH 3 COOH solution may be turned into powder or gel with extra additives so that it can be packaged into pills that are easier to carry. 
     The invention produces a solution for removing the harmful effects of alcohol and is inexpensive, safe and effective. Compared to approaches that use enzymes, acetic acid and bases such as KOH and NaOH are easy to obtain in large quantities at very low cost. In addition, all ingredients (CH 3 COOK and CH 3 COOH) are safe to use and their daily dosages (e.g. K and Na) are well established. Therefore, the product can be off the counter without prescription. It does not require FDA approval either. 
     Although we have used acetic acid as an example to turn the harmful ethanol into a product that is more benign, one can choose other organic acid such as citric acid to achieve the similar effect. Each citric acid molecule contains three functional sites to react with three ethanol molecules to generate a product of triethylcitrate. Since citric acid is a weak acid too, the above mentioned technique of partial neutralization with strong base is applicable too. It should be understood that all approaches based on the similar principle discussed in this disclosure are covered by this invention. 
    
    
     DETAILED DESCRIPTION OF INVENTION 
     The first part of the invention is the process of turning ethanol into ester from an oral intake of acetic acid. The reaction of esterification occurs inside stomach. The second part of the invention is the methodology of making oral intake products that contain a sufficient amount (e.g. 1 mole) of acetic acid in a relatively small volume (e.g. 75 mil) of proper acidity (e.g. PH: 3-4). 
     In the following we describe the detailed procedure of in vitro experiment to verify the validity of reducing the amount of ethanol by esterification:
         1. Label four 100-mL beakers, 0.4 M, 0.6 M, 0.8 M, and 1.0 M.   2. Put 17.15 mL of distilled water into each.   3. Add 0.4 mL of concentrated ethanol to the first beaker, 0.6 mL to the second, 0.8 mL to the third, and 1.0 mL to the fourth. Swivel each beaker gently for 20-30 seconds to ensure mixing. See calculations below for explanation.   4. Add another 17.15 mL of water to each beaker   5. Add 0.4 mL of glacial acetic acid to the first beaker, 0.6 mL to the second, 0.8 mL to the third, and 1.0 mL to the fourth. See calculations below for explanation.   6. Stir each sample vigorously for five minutes in succession. Each beaker will have had a total of twenty minutes for the reaction to proceed.   7. Put on latex gloves before handling potassium permanganate.   8. Set up two burettes, and fill each with approximately 100 mL of 0.2 M potassium permanganate. This will enable two titrations to be simultaneously done.   9. Record the initial volume of potassium permanganate present in both burettes.   10. Place 1.0 M and 0.8 M beakers on hot plate, and turn to power level 3. Heating quickens the titration process.   11. Slowly drip KMnO 4  into beakers while stirring. Make sure that each beaker is consistently titrated with the same burette to ensure that the exact volumes of KMnO 4  that each beaker received can be easily calculated.   12. Potassium permanganate is purple in color. When reacted with ethanol, a brown color (manganese oxide, MnO 2 ) appears. The reaction is as follows:       

       3C 2 H 5 OH+4KMnO 4 →3CH 3 COOH+4MnO 2 +4KOH+H 2 O
 
     When potassium permanganate ceases to be converted into manganese oxide, and the purple color stays, the titration is finished. There is no ethanol left in the beaker to react with potassium permanganate.
         13. Remove both beakers from hot plate. Measure the volume of potassium permanganate remaining in each burette. Subtract from the original volumes to find the volume of KMnO 4  consumed in each reaction.   14. Repeat steps 9-13 for the 0.6 M and 0.4 M beakers.   15. Use stoichiometry to calculate the molar amount of 0.2 M potassium permanganate used in each titration.   16. Ethanol and potassium permanganate react in a 3:4 molar ratio, according to the equation 3C 2 H 5 OH+4KMnO 4 →3CH 3 COOH+4MnO 2 +4KOH+H 2 O.       

     Again, use a chain calculation to calculate the molar amount of ethanol remaining in each beaker, then divide by 0.01715 mL H 2 O to convert to molarity. Compare to the molarity of ethanol originally in each beaker. Calculate % error.
         17. Results (see Table 2)       

                     TABLE 2                  Summary of the results of 5 repeats of in vitro experiment.       The last column indicates the amount (in percentage) of reduction       in ethanol with the addition to acetic acid.                     Initial   C 2 H 5 OH % Decrease at Equilibrium                                         C 2 H 5 OH   Trial 1   Trial 2   Trial 3   Trial 4   Trial 5   Average               0.4 M   —   78.1%   81.5%   81.2%   80.0%   80.2%       0.6 M   —   82.7%   83.8%   83.7%   82.5%   83.2%       0.8 M   89.1%   86.2%   84.7%   84.6%   85.5%   86.0%       1.0 M   91.3%   88.8%   86.5%   91.1%   90.4%   89.6%                    
Table 2 summarizes the results of 5 repeats of experiment. The results indicate that over a wide range of ethanol concentration relevant to our application, adding an equal molar number of acetic acid to the ethanol solution can reduce the ethanol amount by as much as 80-90%.
 
     The second part of the invention is regarding to the methodology of creating products that contain a sufficient amount (e.g. 1 mole) of acetic acid in a relatively small volume (e.g. 75 mil) with proper acidity (e.g. PH: 3-4). The design principle and analysis are discussed in the following: 
     We use KOH as an example to reduce the acidity of a large amount of CH 3 COOH so that one can comfortably drink enough amount of acetic acid by mouth to remove alcohol in the stomach. 
     The chemical reaction for dissociation of CH 3 COOH: 
     
       
         
           
             
               
                 
                   
                     
                       
                         CH 
                         3 
                       
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                           - 
                         
                       
                       + 
                       
                         H 
                         + 
                       
                     
                   
                    
                   
                     
 
                   
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                         K 
                         α 
                       
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                             [ 
                             
                               
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                                 - 
                               
                             
                             ] 
                           
                            
                           
                             [ 
                             
                               H 
                               + 
                             
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                           [ 
                           
                             
                               CH 
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                           ] 
                         
                       
                     
                     ; 
                     
                       
                         K 
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                         1 
                         × 
                         
                           10 
                           
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                          
                         M 
                       
                     
                   
                 
               
               
                 
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                   ) 
                 
               
             
           
         
       
     
     If we do not add any KOH, then the PH value of CH 3 COOH becomes 
     
       
         
           
             
               
                 
                   
                     
                       PH 
                       
                         
                           CH 
                           3 
                         
                          
                         COOH 
                       
                     
                     ≅ 
                     
                       
                         - 
                         
                           log 
                           10 
                         
                       
                        
                       
                         
                           
                             K 
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                            
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                     2.5 
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                         2 
                       
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                         log 
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                        
                       
                           
                       
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                   ( 
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     Where X is the molar concentration of CH 3 COOH. For 50% acetic acid, X=13.5 M. From Eq. (2), the acetic solution has a PH value of 2.0. Note that this is 10 times stronger acetic acid than the off-the-counter white vinegar and the solution is corrosive and could cause acid burn, not suitable for oral intake. 
     For 0.1% highly diluted vinegar, X≈0.03 M, the PH value, according to Eq. (2), becomes 3.3. This is close to the PH value of apple juice and orange juice, a PH value we are used to in food and beverage. Our approach is to create an acetic solution containing a large enough amount of acetic acid while keeping the PH value at the level of fruit juice. 
     Next we calculate how the PH value of the acetic solution changes with the addition of a small amount of KOH. Before reaction, we assume the initial molar concentration of CH 3 COOH and KOH is X and Y, respectively. We also assume that we have a large amount of CH 3 COOH so the final PH value of the solution is slightly acidic. This assumption works for our purpose because we try to maximize the amount of CH 3 COOH but control the amount of KOH so the potassium intake can be kept well within the FDA recommended daily dosage (i.e. 4 g or 0.1 mole of potassium). 
     After equilibrium is reached, the final concentration for each component becomes: 
     
       
         
           
             
               
                 
                   
                     
                       [ 
                       
                         
                           CH 
                           3 
                         
                          
                         COOH 
                       
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                     = 
                     
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                   , 
                   
                     
                       [ 
                       
                         
                           CH 
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                     = 
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                   , 
                   
                     
                       [ 
                       
                         H 
                         + 
                       
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                     = 
                     
                       
                         u 
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                         Y 
                       
                       &gt; 
                       0 
                     
                   
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     [ 
                     
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                       + 
                     
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                   = 
                   
                     
                       
                         
                           
                             K 
                             α 
                           
                            
                           
                             [ 
                             
                               
                                 CH 
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                                
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                             ] 
                           
                         
                         
                           [ 
                           
                             
                               CH 
                               3 
                             
                              
                             
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                               - 
                             
                           
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                       ⇒ 
                       
                         u 
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                         Y 
                       
                     
                     = 
                     
                       
                         
                           K 
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                          
                         
                           ( 
                           
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                             - 
                             u 
                           
                           ) 
                         
                       
                       u 
                     
                   
                 
               
               
                 
                   ( 
                   3 
                   ) 
                 
               
             
           
         
       
     
     Since acetic acid is a weak acid with a very low value of K a &lt;&lt;1, we can find an approximate solution for u: 
     
       
         
           
             
               
                 
                   u 
                   ≅ 
                   
                     
                       Y 
                       2 
                     
                     + 
                     
                       
                         
                           
                             Y 
                             2 
                           
                           + 
                           
                             4 
                              
                             
                               K 
                               α 
                             
                              
                             X 
                           
                         
                       
                       2 
                     
                   
                 
               
               
                 
                   ( 
                   4 
                   ) 
                 
               
             
           
         
       
     
     Then the PH value can be represented as 
     
       
         
           
             
               
                 
                   PH 
                   = 
                   
                     
                       
                         - 
                         
                           
                             log 
                             10 
                           
                            
                           
                             [ 
                             
                               
                                 
                                   
                                     
                                       Y 
                                       2 
                                     
                                     + 
                                     
                                       4 
                                        
                                       
                                         K 
                                         α 
                                       
                                        
                                       X 
                                     
                                   
                                 
                                 2 
                               
                               - 
                               
                                 Y 
                                 2 
                               
                             
                             ] 
                           
                         
                       
                       ≅ 
                       
                         - 
                         
                           
                             log 
                             10 
                           
                            
                           
                             [ 
                             
                               
                                 
                                   K 
                                   α 
                                 
                                  
                                 X 
                               
                               Y 
                             
                             ] 
                           
                         
                       
                     
                     = 
                     
                       5 
                       - 
                       
                         
                           log 
                           10 
                         
                          
                         
                           [ 
                           
                             X 
                             Y 
                           
                           ] 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   5 
                   ) 
                 
               
             
           
         
       
     
     Table 3 shows the PH value of the solution for different concentration ratios between CH 3 COOH and KOH before mixture. 
     
       
         
           
               
             
               
                 TABLE 3 
               
             
            
               
                   
               
               
                 Dependence of PH value of the solution on the ratio of 
               
               
                 CH 3 COOH and KOH before reaction. 
               
            
           
           
               
               
            
               
                 X/Y: initial ratio of [CH 3 COOH] and [KOH] 
                 PH value at equilibrium 
               
               
                   
               
            
           
           
               
               
            
               
                 10 
                 4.0 
               
               
                 50 
                 3.3 
               
               
                 100 
                 3.0 
               
               
                   
               
            
           
         
       
     
     This result is highly significant. It shows that we can mix KOH with 50 times as much as CH 3 COOH while maintaining an acidity level comfortable for drinking (i.e. equivalent to 0.1% highly diluted acetic acid). 
     The daily recommended potassium (K) intake is 4 g, equivalent to 0.1 moles. That means the above approach allows us to take as many as 5 moles of acetic acid without feeling acid burn or discomfort! 5 moles of acetic acid is much more than we need for ethanol esterification since 4 cans of beer contain only a total amount of 1.0 mole of ethanol. 
     While the invention has been described in connection with various embodiments, it is not intended to limit the scope of the invention to the particular form set forth. It is intended to cover such alternatives, modifications, and equivalents as may be included within the spirit and scope of the invention as defined by the appended claims