Lab+V+Extraction+and+Separation+of+a+Mixture


 * __ Extraction and Separation of a Mixture __**

__** ﻿Introduction **__ Often chemists will have to extract an organic compound derived from other reactions or syntheses. Extraction is a process in which one selectively dissolves an organic compound in an acceptable solvent. Organic acids and bases can be separated from each other by using a variety of aqueous solutions. The solutions have varying levels of pH which in turn effects the solubility of the organic compound. Therein lies the basis for extraction; the factor of solubility.

Extraction has many real world applications. Numerous organic compounds are extracted from plants for medicinal purposes such as morphine from the opium poppy, aspirin from willow bark, and digitalis (a heart drug) from foxglove to name a few. Also, it is not uncommon to find items at the store that are products of extractions such as spices, flavors, and countless perfumes. When coffee is brewed, one is extracting a soluble compound from a matrix of crystals. Solvent-Solvent/Acid-Base extraction and separation is one of the simplest and most common ways to isolate a compound from another product.

__** Procedure **__ **__ ﻿ __** The procedure for this lab can be found in Williamson’s __Macroscale and Microscale Organic Experiments__, 4th edition, Ch. 8, expt. 1.


 * Compounds used in this lab**

Each group (carboxylic acid, phenol, and a neutral substance) had two possibilities, with only one of the compounds in the actual mixture used in lab.
 * Solid**:



__Solid not included in the initial mix__:


 * Name: Calcium chloride
 * IUPAC Name: Calcium dichloride
 * CAS Number: 10043-52-4
 * Compound Formula: CaCl 2
 * Structure: oooh I like the little ions dancing on your page.


 * Solvents**:


 * Name: Methyl //tert//-butyl ether
 * IUPAC Name: 2-methoxy-2-methylpropane
 * CAS Number: 1634-04-4
 * Molecular Formula: (CH 3 ) 3 COCH 3
 * Compound Formula: C 5 H 12 O
 * Skeletal Structure:


 * Name: Water
 * CAS Number: 7732-18-5
 * Molecular Formula: H2O
 * Skeletal Structure:


 * Aqueous**:
 * Name: Sodium hydroxide
 * CAS Number: 1310-73-2
 * Molecular Formula: NaOH
 * Compound Formula: HNaO


 * Name: Sodium chloride
 * CAS Number: 7647-14-5
 * Molecular Formula:NaCl
 * Compound Formula: ClNa


 * Name: Sodium bicarbonate
 * IUPAC Name: Sodium hydrogen carbonate
 * CAS Number: 144-55-8
 * Molecular Formula: NaHCO 3
 * Compound Formula: CHNa O 3

__** Data **__
 * Overview**


 * 0.180g of the mixture was weighed out and placed in reaction tube designated #1. The mixture consisted of a carboxylic acid, a phenol, and a neutral substance. Then 2 mL of Methyl //tert//-butyl ether (MTBE) was added to #1 and mixed thoroughly for three minutes. Tube #1 was then placed in the rack and observed for separation. After five minutes, no separation was observed. It was then discovered that a step requiring the addition of sodium bicarbonate had been skipped. It was added at this point, and the solution separated immediately.
 * The lower level of liquid in tube #1 was pipetted off and deposited into reaction tube #2.
 * 1.5mL of sodium bicarbonate was added to tube #1 and mixed as before. The solution again separated and the lower level of liquid in tube #1 was extracted and placed into tube #2.
 * 0.2mL MTBE was added to tube #2 and mixed thoroughly. Solution separated into two distinct levels; the superior of the two was pipetted off and placed into a waste container.
 * 1.0 mL 3M NaOH added to tube #1 and mixed. Solution separated into layers, with a large mass of crystals separating the two aqueous layers. The formation of crystals was a result of the common ion effect; the solution was pushed beyond the saturation point. 0.75mL of de-ionized H 2 O was added to dissolve the crystals. The effect was almost immediate; the major concentration of crystals dissolved leaving three aqueous layers. The lowest layer still contained fine crystals in suspension so an additional 0.25 mL of de-ionized H 2 O was added: this fully dissolved the crystals leaving three distinct aqueous layers.
 * Both superior and inferior layers were clear and colorless while the medial layer was slightly milky and semi-transluscent. As there were only two substances in the reaction tube, it was determined that the two clear layers were simply being separated by force (is this the word we should use?); it was gently rotated end over end and then allowed to re-separate The solution then divided into two distinct layers. The inferior layer was siphoned off and placed into reaction tube #3.
 * 0.015mL of de-ionized H 2 O was added to #1 and agitated. The liquid separated and the H 2 O was siphoned off and added to tube #3. The H 2 O was on the bottom of tube #1. This process was repeated one additional time. During the second addition of H 2 O into tube #1, a small quantity of the other liquid made it into tube #3. This showed up in tube #3 as a foggy/milky layer in the top 1/2 cm of tube #3.
 * 0.015ml of MTBE added to tube #3 and agitated. The liquid assumed a milky white color and appearance without any evidence of separation.
 * 0.5mL NaCl added to reaction tube #1 and mixed. Two separate layers formed. The NaCl, being more dense, assumed the inferior position in the tube. One additional drop of NaCl added to identify the aqueous layer, which was the superior layer.
 * Tube #3 had not separated at this point so two drops of MTBE were added to encourage separation. Please see discussion section for possible errors . The solution separated, but a milky suspension remained.
 * The aqueous layer of reaction tube #1 was removed and placed into a test tube labeled tube 1 aq.
 * The aqueous layer of reaction tube #3 was removed and placed in the waste container.
 * One CaCl 2 stone was placed in tube #1. Bubbles escaped from the surface of the stone, which eventually formed a globule. Two more stones were added with similar results. The solution began to assume a milky white appearance. Three more stones were added with the same result. Because the contents of this tube did not form as they were supposed to, the liquid was extracted and placed into another tube, also labeled #1.
 * 0.010mL MTBE was added to the solution in tube #1. The two substances did not mix.
 * 0.015mL 12M HCl was added to reaction tube #2. White foam formed in the tube.
 * 0.25mL 12M HCl was added to reaction tube #3. When the HCl hit the solution in the tube, a white solution formed that was of a different density than the original clear solution. An additional 0.4mL of 12M HCl was added to the solution. This volume of HCL was sufficient to change the entire solution to the milky white solution. A pH test was performed using blue litmus paper, which showed an acidic solution. The tube was then placed in the sand bath with the heat set at 40% of the maximum output voltage.

The final solids were then assessed by weight and melting point.


 * Tube 1: Did not melt. Not any of the suspected compounds perhaps a salt.
 * Tube 2: Melted at 121.8°C to 144.9°C. Encompasses melting points for both Benzoic acid (123°C) and 2-chlorobenzoic acid (139ºC).
 * Tube 3: Melted at 52.7ºC to 56.4ºC. Encompasses no melting points, but closest to 1,4-dimethyloxybenzene.

__** Analysis **__
 * Percent Recovery:




 * Melting Points:



these results are actually quite good for this experiment. well done. and I like the tables. **__ ﻿ __****__ Discussion __** Due to the notoriously cryptic instruction of the Williamson text, be careful! while you and I can snicker in private about how we don't like the way Williamson writes procedures, you don't want that judgement to creep into your report. in conjunction with the sheer precision required in extracting layers, this experiment had a heightened level of difficulty in comparison to others. Sometimes it is difficult to distinguish between layers, or to know which layer was in fact aqueous; when this situation arose, a drop of water was added to see which layer it would deposit itself in.

Tube 1 was unsuccessful in extracting one of the mixture compounds. During the melting process to assess the compound's identity, the compound did not melt at all. The temperature was increased to 300°C and still no sign of melting was seen. At that point, the experiment was ceased. The most likely explanation was that only Calcium Chloride (CaCl 2 ) was recovered. CaCl 2  has a melting point of 772°C, so the amount of heat added was not even remotely close to the temperature required. This significant error would have occurred during the drying process of tube 1, after the CaCl 2  brick was added. The drying reaction did not go as required, and any of the original mixture was contaminated and/or removed with the other waste.

Tube 2 had more success, but is still largely unreliable, because the melting point range is so wide and thus overlaps the melting point of more than one of the potential substances. This most likely was caused the presence of a significant level of impurity, depressing the melting point and widening the range. Tube 3 has the tightest melting point range, but is still below that of the lowest-melting potential compound. Again, impurities present would depress the melting point. This is good interpretation--which is far more important to me than getting the product in high yield and high purity. I like your thinking.

﻿ **__Conclusion __** While the results were extremely objectionable the crux of the experiment was understood. Once again, the implication for "like dissolves like" can be seen in the preceding lab. The properties of solubility were utilized in order to extract and separate various organic compounds. Essentially there is a solubility switch. Ionic and polar compounds are more soluble in polar solvents, such as water. These solubility differences can be used to separate non-polar compounds from ionic or polar compounds.

The acid-base extraction is done with the knowledge that both solvents are insoluble in one another. The organic acids that are used in the following experiment are insoluble in water but can be separated and extracted with the use of strong base (such as NaOH). Organic bases can be combined with a strong acid (such as HCl) for separation and extraction; these are usually insoluble in water as well. The ionization of these compounds will produce a corresponding salt. These simple acid-base reactions help isolate and identify the compounds during the lab experiment.

This report earned the following scores for: format (2/2) style (2/2) data (3/3) quality of result (1/1) quality of reported data (1/1) conclusion (2/2) error (1/1) post-lab Q (2 freebie points) for a total of 14/14.

**__ ﻿Notes __** The chemical structures, formulas, and vital statistics of each compound was researched using [|Wolfram Alpha computational knowledge engine] and confirmed using the [|CRC Handbook of Chemistry and Physics [90th Edition.]]