Solvent Assignment

Solvent Assignment Words: 1720

Solubility of Organic Compounds Objectives: understanding the relative solubility of organic compounds in various solvents. Exploration of the effect of polar groups on a monopoly hydrocarbon skeleton. Introduction: The solubility of a solute (a dissolved substance) In a solvent (the dissolving medium) is the most Important chemical principle underlying three major techniques you will study in the organic chemistry laboratory: crystallization, extraction, and chromatography.

In this experiment on solubility you will gain an understanding of the structural features of a substance that determine its solubility in various elevens. This understanding will help you to predict solubility behavior and to understand the techniques that are based on this property. In one part of this experiment, you will determine whether a solid organic compound is soluble or insoluble in a given solvent. You should keep in mind that this is actually an over simplification since some solids may be partially soluble in a given solvent.

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If the organic compound being dissolved in a solvent is a liquid, then it is sometimes more appropriate to say that the compound and the solvent are miscible (mix homogeneously In all proportions). Likewise, If the liquid organic impound Is Insoluble In the solvent, then they are Immiscible (do not mix, and form two liquid phases). A major goal of this experiment is to learn how to make predictions about whether or not a substance will be soluble in a given solvent. This is not always easy to do, even for an experienced chemist.

However there are some guidelines which will often make It possible for you to make a good guess about the collectibles of compounds In specific solvents. In discussing these guidelines It Is helpful to separate the types of solutions we will be looking at into two categories: (1 ) Solutions n which both the solvent and the solute are covalent (molecular). (2) Ionic solutions in which the solute unionizes and dissociates. L. Solutions In which the Solvent and Solute are Molecular A generalization which is very useful in predicting solubility behavior is the widely used rule, “Like dissolves like. This rule is most commonly applied to polar and monopoly compounds. According to this rule, a polar solvent will dissolve polar (or ionic) compounds and a monopoly solvent will dissolve monopoly compounds. The reason why this rule works has to do with the nature of Intermolecular forces of attraction. Although we will not be focusing on the attraction between polar molecules is called dipole-dipole interaction (or hydrogen bonding under certain circumstances); between monopoly molecules it is called London dispersion forces.

In both cases these attractive forces can occur between molecules of the same compound or different compounds In this experiment, you will be testing the solubility of several compounds in several solvents. You will then be asked to explain these results primarily in terms of the polarity of the solvent and the solute. This can be done only if you are able to determine whether a substance is polar or monopoly. The polarity of a compound is dependent on both the polarities of the individual bonds and the shape of the molecule.

For most organic compounds, evaluating these factors can become quite complicated because of the complexities of the molecules. However, it is possible to make some reasonable predictions by Just looking at the types of atoms which a compound possesses. As you read the following guidelines, it is important to understand that, although we often describe compounds as being polar or monopoly, polarity is a matter of degree, ranging from monopoly to highly alarm. Guidelines. These guidelines will be helpful in completing this laboratory experiment.

They will help you to determine if a substance is polar or monopoly and to make predictions about solubility. 1. All hydrocarbons are monopoly. Examples: Hexane Benzene 2. Compounds possessing the electronegative elements oxygen or nitrogen are generally polar. Acetone Ethyl alcohol Ethyl acetate Ethylene Ethyl ether Water The polarity of these compounds depends on the presence of polar C-O, C=O, OH, NH and CNN bonds. The compounds which are most polar are capable of forming hydrogen bonds (see #4) and have NH or OH bonds. Although all of these polar.

This is due to the effect of the size of the carbon chain on polarity and whether or not the compound can form hydrogen bonds. 3. The presence of halogen atoms, even though their electromagnetisms are relatively high, does not alter the polarity of an organic compound in a significant way. Therefore, these compounds are only slightly polar. The polarities of these compounds are more similar to hydrocarbons, which are monopoly, than to water which is highly polar. Examples Methyl chloride (disintermediation) Schoolchildren 4.

When comparing organic compounds within the same family, adding carbon atoms to the chain decreases the polarity. For example, methyl alcohol (COACH) is more polar than propel alcohol (CHICHI’S). This is because hydrocarbons are monopoly, and increasing the length of a carbon chain makes the compound more hydrocarbon-like. The general rule of thumb is that each polar group (groups containing nitrogen or oxygen) will allow up to 4 carbons to be soluble in water. 5. As mentioned earlier, the force of attraction between polar molecules is dipole- dipole interaction. A special case of dipole-dipole interaction is hydrogen bonding.

Hydrogen bonding is a possibility when a compound possesses a hydrogen atom bonded too nitrogen, oxygen, or fluorine atom. It is the attraction between this hydrogen atom and a nitrogen, oxygen, or fluorine atom in another molecule. Hydrogen bonding may occur between two molecules of the same compound or between molecules of different compounds. Hydrogen bonding is the strongest type of dipole-dipole interaction. When hydrogen bonding between solute and solvent is possible, solubility is greater than one would expect for compounds of similar polarity that cannot form hydrogen bonds.

Hydrogen bonding is very important in organic chemistry and biochemistry, and you should be alert for situations in which hydrogen bonding may occur. II. Solutions in which the Solute Unionizes and Dissociates Many ionic compounds are highly soluble in water because of the strong attraction between ions and the highly polar water molecules. This also applies to organic compounds that can exist as ions. For example, sodium acetate consists of An+ and CHOC- ions, which are highly soluble in water. Although there are some exceptions, you may assume in this experiment that all organic compounds that are n the ionic form will be water-soluble. Sections. For example, carboxylic acids can be converted to water-soluble salts when they react with dilute aqueous Noah: water-insoluble carboxylic acid water-soluble salt The water-soluble salt can then be converted back to the original carboxylic acid by adding another acid (usually aqueous HCI) to the solution of the salt: Amines, which are organic bases, can also be converted to water-soluble salts when they react with dilute aqueous HCI: amine This salt can be converted back to the original amine by adding a base (usually aqueous Noah) to the solution of the salt. Pre-Lab: .

Several portions of this lab request that you make predictions regarding outcomes of solubility experiments. Make these predictions before you arrive in lab and record them on this worksheet. 2. Write a paragraph describing hydrogen bonding and draw an illustration of hydrogen bonding between water molecules. Do not forget to discuss electronegative, lone pairs and bond polarity. Complete this assignment on a separate sheet of paper and submit it at the beginning of lab. Lab Report Guide: 1. Results (5 pets) Tables neatly filled out Questions legibly answered Results: (Please fill out and submit with the lab report.

Recopy if messy) Part A. Solubility of Solid Compounds Place about 40 MGM (0. 040 g) of phenyl into each of two dry test tubes. (Don’t try to be exact: you can be 1-2 MGM off and the experiment will still work. ) Label the test tubes and then add 1 ml of water to the 1st tube and 1 ml of hexane to the 2nd tube. Determine the solubility of each sample in the following way. Using the rounded end of a spatula, stir each sample continuously for 60 seconds by twirling the spatula rapidly. After 60 seconds (do not stir longer), note whether the compound is soluble (dissolves completely) or insoluble (none of it dissolves).

If all but a couple granules have dissolved, then you should state that the sample is soluble. Record these results in the following table. Now repeat the directions given above for with Miltonic acid. Record these results. Compound Water Prediction Water Result Hexane Prediction Hexane Result Phenyl Miltonic acid Problems 1. For each of the four solubility tests, explain your results in terms of the polarities. A. Phenyl in water b. Phenyl in hexane d. Miltonic acid in hexane Is hydrogen bonding possible for any of the 4 pairs of solids and solvent? If so, 2. Draw a picture showing a hydrogen bond between the solvent and solute.

Part B. Solubility of Different Alcohols For each solubility test (see table), add 1 ml of solvent (water or hexane) to a test tube. Then add one of the alcohols drowses. Shake the tube after adding each drop. Continue adding the alcohol until you have added a total of 10 drops. If you see one layer, the liquids are miscible (soluble). If you see two layers, they are immiscible. Record your results (miscible or immiscible) in your notebook in table form. Methyl alcohol 1 -Octagon . Explain each of these results. Methyl alcohol in water a. B. Methyl alcohol in hexane . I-octagon in water d. I-octagon in hexane

Part C. Miscible or Immiscible Pairs For each of the following pairs of liquid compounds, predict if the pairs will be miscible or immiscible. (It is best to make your prediction for each pair after you have tested the previous pairs. ) Then add 1 ml of both liquids to the same test tube. Use a different test tube for each pair. Shake the test tube for 10-20 seconds to determine if the two liquids are miscible (form one layer) or immiscible (form two layers). Record your results in the table. Mixture Prediction Result Water and Ethyl Alcohol Water and Ethyl Ether Water and Methyl Chloride Water and Hexane.

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