Introduction: Extraction is a very common methodused for separating and purifying compounds. This process involves the phenomena of partitioning, which occurs due toa solute’s solubility in two different solvents that are immiscible.
Themixture of two immiscible solvents results in two separate phases. The compounds can be separated due to thedifference in the distribution of the compounds between the phases. A solute will be more soluble in one of thetwo phases than the other, making it possible to separate it from anothercompound in the mixture. Liquid-liquid extractions mostcommonly involve the use of water as a solvent and an organic solvent, such asdiethyl ether or dichloromethane, which are both immiscible with water, butthey are also capable of dissolving other compounds. The two solvents are identified as either theoriginal phase, So, or the extracting phase, Sx.
The partition coefficient, K, is used toexpress the amount of a solute, A, in each phase, in grams per milliliter. The partition coefficient can be approximatedclosely by using the following formula.Thisformula can be used to approximate K by dividing the solubility of A in theextracting solvent by the solubility of A in the original solvent. Since the solute will have a certainsolubility in each solvent, there will be some competition between the two solventsfor solute A, so there will be partitioning of solute A between the twosolvents when it contacts both of them. Using the previous equation, asignificant prediction can be made. Ifthe value of K is greater than 1, then the solute should be found in greateramounts in the extracting solution, but only if the volume of the extractingsolvent is at least equal to the volume of the Original solvent. If the volumes of the two solvents are thesame, then K is just equal to the ratio of the amount in grams of solute in bothof the solvents.
Knowing this, thefollowing equation can be applied.Theratio of grams of A in the extracting solvent, Ax, to grams of A inthe original solution, Ao, is multiplied by the ratio of the volumeof the original solvent, Vo, to the volume of the extractingsolvent, Vx. Although an increase in the volumeof extracting solvent would lead to an increase in the amount of A in theextracting solvent, performing multiple extractions using smaller volumes ofextracting solvent allows for a higher amount of solute extracted compared toone extraction with a large amount of extracting solvent. This can be shown by alteringthe previous equation to accommodate for a series of extractions, in terms of FA, which is the fraction of A remaining in the original solventafter n extractions.
This equation is shown aswhereAo is the amount of solutein So before extraction, A1 is the amount of solute inSo after extraction, Ci=and Cf=. Using this equation, it can be shown thatmultiple extractions with less extracting solvent will extract a higher amountof the solute than one extraction with a larger amount of extracting solvent.For example, if K=5 then FA=1/16when only one extraction is performed with 30 mL of diethyl ether, while FA=1/216 when threeextractions are performed using 10 mL of diethyl ether. This shows that performing more extractionswill result in a lower fraction of the solute remaining in the originalsolvent. When choosing solvents for anextraction, multiple criteria must be met in order to successfully extract thedesired compounds. The first requirement is that the solvent must not reactchemically in an irreversible way with any component in the mixture. Second, the solvent must be immiscible, ornearly immiscible with the original solvent and must selectively extract thedesired compound in the mixture so that the partition coefficient is high forone solvent while the partition coefficient for the other is low.
Finally, thesolvent must be readily separable from the solute by means of distillation orother previously performed procedures which involve separating a solute from asolvent. Percent recovery is reliant upon the selection of a proper extractingsolvent as well as polarity of the compound to be extracted. A proper solvent must have a partitioncoefficient which is significantly higher than the original solvent, so thatthe least amount of solvent possible will remain in the original solvent afterextraction.
On the other hand, polar andnonpolar compounds will partition differently in polar and nonpolarsolvents. If one solvent is nonpolar andthe other is polar, a nonpolar solute will partition into the nonpolar phasewhile a polar solute will partition into the polar phase. A one base extraction can beperformed when trying to extract organic compounds such as phenols andcarboxylic acids.
If a phenol orcarboxylic acid contains more than around six carbons atoms, it will behydrophobic. Such molecules are readilysoluble in organic solvents such as diethyl ether or dichloromethane, but cannotbe extracted with water alone due to the fact that they are hydrophobic and nonpolar. An aqueous base can be used to extract suchcompounds by converting the phenol or carboxylic acid to its conjugatebase.
This conjugate base will be polarand hydrophilic, which will cause it to partition into the aqueous phase andout of the organic phase. Once extracted, it can be re-protonated in order toseparate it from the aqueous phase because it will once again be hydrophobicwhen protonated. A two base extraction is the methodused involving a neutral compound and two acidic compounds. A base is selectedbased upon the pKa value of its respective conjugate acid.
In order to selectively extract an acid intothe aqueous phase, the first base must have a conjugate acid whose pKa value ishigher than that of the acid with the lower pKa value, but must be less thanthat of the acid with the higher pKa value. The second base used in extraction must have a conjugate base whose pKavalue is higher than that of the acid with the higher pKa value. The reasoning behind this is that a basewhose conjugate acid’s pKa value is greater than the pKa of the acid to beextracted will greatly deprotonate the acid and cause it to partition in theaqueous phase. Usually this extraction involves a compound containing anaromatic ring and the carboxylic acid functional group. Acid-Base extractions involve theseparation of a mixture of a neutral compound, an acidic compound, and a basiccompound.
In this extraction a base willfirst be extracted using an aqueous acid, bringing the base into the aqueousphase. Second the acid will be extractedfrom the organic solvent using an aqueous base. This will allow the acid to partition into the aqueous phase and theneutral compound will be left in the organic phase. Materials and Methods:Refer to attached blue sheets Data and Observations:See blue sheets attached containing datarecorded Discussion: The reaction equations are shown onthe blue sheets showing the mechanisms of the reactions which occurred duringthe extractions. In the two base extraction, twodifferent acids can be extracted separately because the first extraction isdone with sodium bicarbonate. Performingthis extraction deprotonates the benzoic acid to a high degree, while2-naphthol will not be protonated highly. This will leave the 2-naphthol in the organic phase while the benzoicacid will partition into the aqueous phase. This allows the benzoic acid to be extracted first and then a secondextraction can be performed using a hydroxide, which will highly deprotonate2-naphthol and allow it to partition into the aqueous phase.
By performing these extractions sequentially,two different acids can be selectively extracted from the mixture. Once an acidic or basic compound hasbeen extracted into aqueous solution it must be taken back out of aqueoussolution. For example, an acid that hasbeen extracted into aqueous solution has been deprotonated in order to make itwater soluble, so in order to take it out of solution it must be re-protonatedby adding an acid. In order to ensurethat all of the compound is re-protonated, enough acid must be added in orderto make the solution distinctly acid so that there is excess acid in solution,which will be at a pH of around 2-4. Similarly, bases that have been extracted, have been protonated. In order to separate them from aqueoussolution they must be deprotonated by adding enough of a base to make thesolution distinctly basic, which will be around pH 10. In the one base extraction the yieldof naphthalene was 57.4%, which in reality, should be higher due to the factthat it was not extracted or chemically reacted in any way and is waterinsoluble so it will remain in the organic phase.
The benzoic acid was relatively higher inyield, with a percent yield of 66.44%. This also shows that there was some experimental error when performingthe extraction because the two compounds were mixed together in 1:1proportions, so the masses yielded, when added together do not add up to theoriginal mass of the crude sample. Inthe two base extraction, only 2-Naphthol was analyzed, with a percent yield of73.21% which is a relatively higher percent yield compared to the one baseextraction. In the acid base extraction,only 4-nitroaniline was analyzed. Thepercent yield was 64.31% for this extraction.
In these extractions, a 100% yield cannot be expected. This is due to the fact that it is hard for100% of the acidic compounds to be deprotonated and 100% of the basic compoundsto be protonated. Using an acid or abase will not cause all of the compound to be extracted to partition into theaqueous phase, but it will cause a certain amount to partition depending uponthe pKa of the extracting solution and the pKa of the compound being extracted. Conclusion Extractions can be performed inorder to separate compounds in a mixture. A one base extraction should be used when a mixture contains a neutralcompound and an acidic compound, when both of them are water insoluble. A two base extraction should be performedwhen a mixture contains a neutral compound and two different acidic compoundswith different pKa values.
Finally, anacid base extraction is useful when separating a neutral compound, and acidiccompound, and a basic compound from a mixture. All of these extractions are performed by selectively reacting thecompounds in order to make them water soluble while leaving the neutralcompound in organic solution so that the compounds will be found in separatesolutions. From there the compounds canbe individually separated from their respective solutions.
