O Chem Lab final – Flashcards

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Melting Points
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An impurity _______the melting point of an organic compound.
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decreases
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What is the eutectic point?
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Eutectic point is the percent composition of the mixtures that have a very sharp melting point.
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Does a eutectic mixture have a broad or narrow melting point?
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A eutectic mixture has a narrow melting point.
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At the melting point of a sample, the temperature should not be raised more than _______°C per _______.
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1°C per minute
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Label the percent composition of the eutectic point and record the temperature that 40% silicon melts at. Hint this will be a single temp not a range.
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Eutectic point 20% silicon & 80% 40% silicon melts at 800°C
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The melting points of pure benzoic acid and pure 2-napthol are 122.5°C and 123°C respectively. Given an unknown pure sample that is known to be either pure benzoic acid or pure 2-napthol, describe a procedure you might use to determine the identity of the sample.
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1. First I would mix the unknown with pure benzoic acid. 2. I would take the melting point (using the slow heating method). 3. I would compare the mixture melting point with the m.p. of pure benzoic acid. If the m.p. is decreased than I know it is not benzoic acid and I would need to mix the unknown with the second known pure sample. 4. After mixing the unknown w/ 2-napthol, I would take the melting point of the sample. 5. Last, I would compare the m.p. mixture w/ that of the pure sample of 2-napthol. If the m.p's are the same it is 2-napthol.
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Define melting point
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The melting point of a substance is defined as the temperature at which the liquid and solid phases exist in equilibrium with one another w/o change in temperature.
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The melting point is expressed as...
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the temperature range over which the solid starts to melt and then is completely converted to liquid. Used for: -Characterization (identification) of the compound technique called mixture melting point determination. -Determination of purity (if pure it should melt over a normal range - no more than 1°)
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Melting phase diagram
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What are the errors that would cause a higher than correct m.p?
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This error could be caused by the calibration of the thermometer being slightly off.
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What are the errors that would cause a broad range.
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A broad range could also be a result of an impurity in the sample.
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Vapor pressure as applied to melting
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When a solid has the same vapor pressure as the liquid that it is in equilibrium with the solid, the solid will melt.
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Melting point or melting-point range
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This is a temperature range that begins when the solid begins to melt until the solid has melted completely.
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Mixture or mixed melting point
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A mixed melting point occurs when an impurity is mixed into a pure compound and as a result will broaden and lower the range of the pure compounds melting point.
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Eutectic point
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The eutectic point occurs at which both compounds melt at exactly the same rate at equilibrium composition
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Eutectic mixture
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A eutectic mixture is an impure solid that consists of two compounds that will melt at exactly the same rate causing a sharp melting point range. Eutectic mixtures can sometimes be mistaken for pure compounds because of the sharp melting point.
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Compound A and compound B have approximately the same melting point. State two ways in which a mixed melting point of these two compounds would be different from the melting point of either pure A or pure B.
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The first way in which the mixed melting points of both pure compounds would be different is by the depression of the melting point, causing it to be lower than both of the pure compounds melting points (A and B). The second way in which the melting points of both pure compounds would be different is by a broader range in the melting point. Instead of a sharp melting point within .5 to 1°C the impurities would cause the melting point to be about 5 - 10°C range.
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The melting points of pure benzoic acid and pure 2-napthol are 122.5°C and 123°C, respectively. Given a pure sample that is known to be either pure benzoic acid or 2-napthol, describe a procedure you might use to determine the identity of the sample.
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The first step would be to mix the unknown pure sample with benzoic acid. If the melting point is lower than 122.5°C and has a broader range then the unknown sample is not benzoic acid. If the melting point is the same as benzoic acid, it could be assumed that the unknown pure sample is benzoic acid. If the first procedure does not identify that the sample is benzoic acid then it would be mixed with 2-napthol and the melting point would be obtained. If the melting point is lower than 123°C and has a broader range then the unknown sample is not 2-napthol. If the melting point is the same as 2-napthol, it can be assumed that the unknown pure sample is 2-napthol.
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13) The melting-point-composition diagram for two substances, Q and R, is provided in Figure 3.2, which should be used to answer the following questions. a. What are the melting points of pure Q and R? b. What are the melting point and the composition of the eutectic mixture? c. Would a mixture of 20 mol % Q and 80 mol % R melt if heated to 120°C? d. A mixture of Q and R was observed to melt at 105-110 °C. What can be said about the composition of this mixture? Explain briefly.
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a. Q = 180°C, R = 155°C b. 35% mol R/ 65% mol Q at a temperature of 80°C c. No, to 160°C? Yes, to 75°C? No d. The composition of the mixture would have been either 50% mol R and 50% mol Q or it could have been 30% mol R and 70 % mol Q. Both Q and R in different compositions are acting as impurities to the other and lowering and broadening the melting point of the pure compounds.
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Recrystallization
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WHat are the five criteria for selecting a solvent for re-crystallization?
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1. The compound (being purified) needs to be soluble in a hot solvent and insoluble in a cold solvent. 2. The impurities need to be insoluble at all temperatures or mostly soluble in cold temperatures. 3. In order for the solute to crystallize, the b.p. of the solvent needs to be low. 4. The b.p. of the solvent needs to be lower than the m.p of the compound being purified. 5. The solvent should not react chemically with the compound being purified.
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What are the 7 steps of re-crystallization?
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1. Selection 2. Dissolution 3. Decoloration 4. Hot filtration 5. Formation 6. Isolation 7. Drying
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Why was hot filtration used?
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Hot filtration was used because impurities were present. In order to separate the insoluble impurities from the pure compound the solution needed to be filtered through filter paper.
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List the steps in the systematic procedure for miniscale recrystallization briefly explaining the purpose of each step
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1. Selection - is used to select the appropriate solvent. In selecting a solvent, it is important that the compound (that is being recrystallized) is soluble in the hot solvent and insoluble in the cold solvent. The impurities should be opposite of this and be insoluble in all temperatures of a solvent or slightly soluble in a cold solvent. Another factor that affects selection of appropriate solvent is the boiling point of the solvent, which needs to be low so that the crystals can be removed. The boiling point of the solvent also needs to be lower than the melting point of the solid. It is also important that the solvent does not react chemically with the solid. 2. Dissolution - to dissolve the solid, the mixture is heated to the boiling point of the solvent. This method completely dissolves the solid (if the correct solvent is used), while the impurities remain insoluble in the hot solvent. 3. Decoloration - if the solution is colored, there could be impurities present, so it would need decolorizing carbon to remove any colored impurities. 4. Hot filtration - is used when there are impurities present. This process separates the impurities from the pure solid by pouring the solution through a filter. 5. Formation - allows the hot solution to cool and begin to form crystals. 6. Isolating - using a vacuum filtration removes the excess solvent trapped within the crystals. 7. Drying the crystals - air-drying the crystals helps remove any last traces of solvent from the crystal.
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In performing a hot filtration at the miniscale level, what might happen if the filter funnel is not preheated before the solution is poured through it?
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If the filter funnel is not preheated, the solution would begin to cool and some of the dissolved pure compound would cool down as it was poured through the filter. As you filter out the impurities, having a cooled filter would also filter out some of the pure compound.
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Briefly explain how a colored solution may be decolorized.
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In order to decolorize a solution (as a result of impurities present,) adding a decolorizing carbon to the solution is the first step. The next steps would be to heat the solution to a boil, allow it to cool before adding a filter-aid (celite) to absorb the carbon, and then reheat the solution to a boil. Then, using hot filtration, filter out the impurities and the decolorizer from the pure solid.
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Briefly explain how insoluble particles can be removed from a hot solution.
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The insoluble impurities (particles) are insoluble in the solvent and the pure compound is soluble in the solvent. This results in easy separation of the two using the hot filtration process. After heating the solvent to its boiling point and allowing the pure soluble compound to dissolve, it is cooled. Using a funnel filter and filter paper, the solution is poured through the filter paper. The filter paper catches the impurities and separates them from the solution that now contains the pure compound and the solvent.
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List five criteria that should be used in selecting a solvent for a recrystallization.
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1. The compound needs to be soluble in a hot solvent and insoluble in a cold solvent. 2. The impurities should be insoluble in all temperatures of solvent or slightly soluble in a cold solvent. 3. In order for the pure solid to be easily removed from the crystals, the boiling point of the solvent needs to be low. 4. The solvent also needs to have a boiling point that is lower than the melting point of the solid. 5. The solvent should not chemically react with the solid.
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A 10g sample containing 9g of compound A and 1g of compound B is dissolved in 100 mL of solvent at 60°C. When the solution is cooled to 21°C it becomes supersaturated in compound A but is still unsaturated in compound B. Crystallization of compound A takes place. After crystallization and filtration obtains: (PIC)
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6g of compound A (crystals) 3g of compound A and 1g of compound B in solution % recover of A = 6/3 x 100 = 67%
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A 10g sample containing 9g of compound A and 1g of compound B is dissolved in 100 mL of solvent at 60°C. When the solution is cooled to 21°C it becomes supersaturated in compound A but is still unsaturated in compound B. Crystallization of compound A takes place. After crystallization and filtration obtains: (PIC)
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6g of compound A (crystals) 3g of compound A and 1g of compound B in solution % recover of A = 6/3 x 100 = 67%
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Distillation and Boiling Points
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Distillation and Boiling Points
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Define Boiling Point
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The temperature at which the vapor pressure of a liquid just equals 760m of Hg called "standard boiling point of liquid" -The temp at which the vapor pressure of a liquid just equals the applied pressure is the boiling point of the the liquid.
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Define Boiling Point
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The temperature at which the vapor pressure of a liquid just equals 760m of Hg called "standard boiling point of liquid" -The temp at which the vapor pressure of a liquid just equals the applied pressure is the boiling point of the the liquid.
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Example: The b.p. of water in Denver, Colorado
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Denver is 1609 meters above sea level. The average atmospheric pressure is 630 mm of Hg. The boiling point of water at 630 mm of Hg is 95°C
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Example: The b.p. of water in Denver, Colorado
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Denver is 1609 meters above sea level. The average atmospheric pressure is 630 mm of Hg. The boiling point of water at 630 mm of Hg is 95°C
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Raoult's Law
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PA = P°AXA (for liquid A) PA = partial vapor pressure of pure A P°A = vapor pressure of pure A XA = mole fraction of A PB = P°BXB (liquid B) PTotal = PA + PB (Only valid for Raoult's law if it is homogeneous mixture)
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Raoult's Law
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PA = P°AXA (for liquid A) PA = partial vapor pressure of pure A P°A = vapor pressure of pure A XA = mole fraction of A PB = P°BXB (liquid B) PTotal = PA + PB (Only valid for Raoult's law if it is homogeneous mixture)
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A mixture of 75 mole % toluene and 25 mole% cycloexane is distilled through a simple distillation apparatus. The boiling temperature is found to be 100°C as the first amount of distillate is collected. The standard vapor pressurs of toluene and cyclohexane are known to be 436 mm Hg and 1732 mm Hg, respectively at 100°C. Calculate the percentages of each of the two components in the first few drops of distillate.
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PT = P°TXT PT = 436 mm Hg x 0.75 PT = 327 mmhg Pc = P°cXc Pc = 1732 mmHg x 0.25 Pc = 433 mmHg Ptotal = 327 + 433 = 760 mmHg Mole Fraction is found by: % Cyclohexane = 433/760 x 100 = 57% % toluene = 327/760 x 100 = 43%
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A mixture of 75 mole % toluene and 25 mole% cycloexane is distilled through a simple distillation apparatus. The boiling temperature is found to be 100°C as the first amount of distillate is collected. The standard vapor pressurs of toluene and cyclohexane are known to be 436 mm Hg and 1732 mm Hg, respectively at 100°C. Calculate the percentages of each of the two components in the first few drops of distillate.
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PT = P°TXT PT = 436 mm Hg x 0.75 PT = 327 mmHg Pc = P°cXc Pc = 1732 mmHg x 0.25 Pc = 433 mmHg Ptotal = 327 + 433 = 760 mmHg Mole Fraction is found by: % Cyclohexane = 433/760 x 100 = 57% % toluene = 327/760 x 100 = 43%
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Boiling-point-composition curves for cyclohexane-toluene mixture
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Boiling-point-composition curves for cyclohexane-toluene mixture
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Vapor pressure vs. temperature for cyclohexane and toluene
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Vapor pressure vs. temperature for cyclohexane and toluene
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Refer to figure 4.1 and answer the following: a) What total pressure would be required in a system in order for the liquid to boil at 45°C? b) At about what temperature would the liquid boil when the total pressure in the system is 300 torr?
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a. 400 torr b. 40°C
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Refer to figure 4.1 and answer the following: a) What total pressure would be required in a system in order for the liquid to boil at 45°C? b) At about what temperature would the liquid boil when the total pressure in the system is 300 torr?
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a. 400 torr b. 40°C
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Using Dalton's law, explain why a fresh cup of tea made with boiling water is not as hot at higher altitudes as it is at sea level.
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Dalton's law is defined by Ptotal = Psample + Pair The molecules of the vapor are mixed with air, at higher altitudes the air pressure is lowered. This lowers the total pressure in the system. The boiling point is dependent upon the total pressures. (reduction in the total pressure of the system reduces the boiling point of the sample to a temperature at which it no longer decomposes)
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Using Dalton's law, explain why a fresh cup of tea made with boiling water is not as hot at higher altitudes as it is at sea level.
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Dalton's law is defined by Ptotal = Psample + Pair The molecules of the vapor are mixed with air, at higher altitudes the air pressure is lowered. This lowers the total pressure in the system. The boiling point is dependent upon the total pressures. (reduction in the total pressure of the system reduces the boiling point of the sample to a temperature at which it no longer decomposes)
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At a given temperature, liquid A has a higher vapor pressure than liquid B. Which liquid has the higher boiling point?
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Liquid B will have a higher boiling point.
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At a given temperature, liquid A has a higher vapor pressure than liquid B. Which liquid has the higher boiling point?
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Liquid B will have a higher boiling point.
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Explain why a packed fractional distillation column is more efficient than an unpacked column for separating two closely boiling liquids.
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A packed fractional distillation column is more efficient than an unpacked column because it has more theoretical plates. Each of the theoretical plates is where distillations can occur as vapors move up the column. These mini distillations continue enriching the vapors so it becomes more pure as it goes higher and higher in the column, which allows for better separation.
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Explain why a packed fractional distillation column is more efficient than an unpacked column for separating two closely boiling liquids.
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A packed fractional distillation column is more efficient than an unpacked column because it has more theoretical plates. Each of the theoretical plates is where distillations can occur as vapors move up the column. These mini distillations continue enriching the vapors so it becomes more pure as it goes higher and higher in the column, which allows for better separation.
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A mixture of 60 mol % n-propylcyclohexane and 40 mol % n-propylbenzene is distilled through a simple distillation apparatus; assume that no fractionation occurs during distillation. The boiling temperature is found to be 157°C (760 torrs) as the first small amount of distillate is collected. The standard vapor pressure of n-propylcyclohexane and n-propylbenzene are known to be 769 torr and 725 torr, respectively at 157.3°C. Calculate the percentage of each of the 2 compounds in the first few drops of distillate.
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PC = P°CXC PC = 769 torr x .60 PC = 461.4 torr PB = P°BXB PB = 725 torr x .40 PB = 290 torr Ptotal = 461.4 + 290 torr = 751.4 torr XC = 461.4/751.4 = 61% XB = 290/751.4 = 39%
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A mixture of 60 mol % n-propylcyclohexane and 40 mol % n-propylbenzene is distilled through a simple distillation apparatus; assume that no fractionation occurs during distillation. The boiling temperature is found to be 157°C (760 torrs) as the first small amount of distillate is collected. The standard vapor pressure of n-propylcyclohexane and n-propylbenzene are known to be 769 torr and 725 torr, respectively at 157.3°C. Calculate the percentage of each of the 2 compounds in the first few drops of distillate.
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PC = P°CXC PC = 769 torr x .60 PC = 461.4 torr PB = P°BXB PB = 725 torr x .40 PB = 290 torr Ptotal = 461.4 + 290 torr = 751.4 torr XC = 461.4/751.4 = 61% XB = 290/751.4 = 39%
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Steam Distillation
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Steam Distillation
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Define Steam distillation
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Steam distillation is a technique that is very useful for the separation of slightly volatile, water-insoluble substances from nonvolatile substances. It is especially useful in cases where a relatively small amount of volatile material is to be separated from a large amount of nonvolatile material. This technique is often used in the isolation of natural products.
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Define Steam distillation
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Steam distillation is a technique that is very useful for the separation of slightly volatile, water-insoluble substances from nonvolatile substances. It is especially useful in cases where a relatively small amount of volatile material is to be separated from a large amount of nonvolatile material. This technique is often used in the isolation of natural products.
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What are the principles of steam distillation? (PIC)
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- Heterogeneous distillation -This method requires that the two components are immisible (two phases). If the two compound are completely insoluble in each other, each component vaporizes independently of the other. In steam distillation the vapor pressure of the mixture is the sum of the vapor pressure of water and the insoluble organic compound
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What are the principles of steam distillation? (PIC)
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- Heterogeneous distillation -This method requires that the two components are immisible (two phases). If the two compound are completely insoluble in each other, each component vaporizes independently of the other. In steam distillation the vapor pressure of the mixture is the sum of the vapor pressure of water and the insoluble organic compound
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Important points concerning steam distillation
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1. The b.p. of the mixture is below the b.p. of either component. 2. Since the vapor pressure of each substance and of the mixture is independent of concentration, it follows that the b.p. of the mixture will remain constant as long as there is some of each component present. If either component becomes exhausted the b.p. will suddenly change to that of the remaining component. 3. Since the molar concentration of each substance in the vapor is proportional to its vapor pressure, the composition of the distillate (which remains constant throughout the distillation)
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Important points concerning steam distillation
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1. The b.p. of the mixture is below the b.p. of either component. 2. Since the vapor pressure of each substance and of the mixture is independent of concentration, it follows that the b.p. of the mixture will remain constant as long as there is some of each component present. If either component becomes exhausted the b.p. will suddenly change to that of the remaining component. 3. Since the molar concentration of each substance in the vapor is proportional to its vapor pressure, the composition of the distillate (which remains constant throughout the distillation)
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Determined by the following ratio
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(PIC)
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Determined by the following ratio
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(PIC)
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Why is it important that a drop of condensate be suspended from the thermometer during a distillation?
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It ensures that it is not heating up too quickly -A drop of liquid needs to be seen suspended from the end of the thermometer, this ensures steady distillation and a constant temperature.
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Why is it important that a drop of condensate be suspended from the thermometer during a distillation?
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It ensures that it is not heating up too quickly -A drop of liquid needs to be seen suspended from the end of the thermometer, this ensures steady distillation and a constant temperature.
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The top of the mercury bulb of the thermometer placed at the head of a distillaion apparatus should be adjacent to the exit opening in the condenser. Explain the effect on the observed temperature reading if the bulb is placed: a. below the opening to the condensor.
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This would cause a head temperature reading that was too high.
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The top of the mercury bulb of the thermometer placed at the head of a distillaion apparatus should be adjacent to the exit opening in the condenser. Explain the effect on the observed temperature reading if the bulb is placed: a. below the opening to the condensor.
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This would cause a head temperature reading that was too high.
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b. above the opening to the condenser.
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This would cause a head temperature reading that was too low.
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b. above the opening to the condenser.
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This would cause a head temperature reading that was too low.
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Using the temperature-composition diagram below, answer the following questions for a mixture that contains 25 mole % cyclohexane and 75 mole % Toluene. a. At what temp will it boil? b. What is the composition, in mole % of the first small amount of vapor that forms? c. If the vapor in b) is condensed and the resulting liquid
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a. 100°C b. 40% Toluene & 60% cyclohexane (boils first line, next line vaporizes - stair step) c. 11% Toluene & 89% cyclohexane
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Using the temperature-composition diagram below, answer the following questions for a mixture that contains 25 mole % cyclohexane and 75 mole % Toluene. a. At what temp will it boil? b. What is the composition, in mole % of the first small amount of vapor that forms? c. If the vapor in b) is condensed and the resulting liquid
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a. 100°C b. 40% Toluene & 60% cyclohexane (boils first line, next line vaporizes - stair step) c. 11% Toluene & 89% cyclohexane
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Why was a steam distillation rather than a simple distillation performed in the isolation of citral from lemon grass oil?
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Simple distillation is used to isolate a pure liquid that is non-volatile from another substance that is also non-volatile. Steam distillation is used to separate liquid and solid components that are moderately volatile and are immiscible in water. Citral is moderately volatile, so steam distillation was chosen.
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Why was a steam distillation rather than a simple distillation performed in the isolation of citral from lemon grass oil?
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Simple distillation is used to isolate a pure liquid that is non-volatile from another substance that is also non-volatile. Steam distillation is used to separate liquid and solid components that are moderately volatile and are immiscible in water. Citral is moderately volatile, so steam distillation was chosen.
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What type of product is expected from the reaction of citral with Br2/CH2Cl2 -With chromic acid?
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What type of product is expected from the reaction of citral with Br2/CH2Cl2 -With chromic acid?
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Both geranial and vitamin A are members of the class of natural products called terpenes. This group of compounds has a common characteristic of being biosynthesized by linkage of the appropriate number of five-carbon units having the skeletal structure shown below. Determine the number of such units present in each of these terpenes and indicate the bonds linking the various individual units. C-C-C-C-C
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Geranial = 2 five-carbon units Vitamin A = 4 five-carbon units
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Both geranial and vitamin A are members of the class of natural products called terpenes. This group of compounds has a common characteristic of being biosynthesized by linkage of the appropriate number of five-carbon units having the skeletal structure shown below. Determine the number of such units present in each of these terpenes and indicate the bonds linking the various individual units. C-C-C-C-C
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Geranial = 2 five-carbon units Vitamin A = 4 five-carbon units
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Extraction
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List the order of extractions with a base, neutral, weak acid and strong acid.
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Anhydrous (drying agent) always goes in solid organic. (True/False)
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False
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After shaking and venting during extractions, the stopper is always removed once ready to separate layers (True/False)
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True
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Using equation 5.9, show that three extractions with 5-mL portions of a solvent give a better recovery than a single extraction with 15 mL of solvent when K = 0.5.
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FA = Cf/Ci = (Vo/KVx+Vo)n Extraction 1: 15 mL 5mL/(0.5*15mL + 5mL) = 40% remains in solution and 60% is extracted from solution Extraction 2: 3 extractions of 5mL each (5mL/(0.5*5mL + 5mL))³ = 29.6% remains in solution and 70.4% is total extracted from solution.
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Assume that the partition coefficient K, for partitioning of compound A between diethyl ether and water is 3; that A preferentially partitions into Ether. a. Given 400mL of an aqueous solution containing 12g of compound A, how many grams of A could be removed from the solution by a single extraction with 200mL of diethyl ether.
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CE/CW = 3 = WE/200mL / Ww/400mL = 3 WE + Ww = 12g Ww = 12 - WE =WE/200mL / 12-WE/400mL = WE/200mL *400mL/12-WE 2WE/12-WE = 12 2WE = 3(12-WE) 2WE = 36 - 3WE WE = 7.2 g Ww = 12 - 7.2 g = 4.8g = 7.2 grams of compound was removed from water
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How many total grams of A can be removed from the aqueous solution with three successive extractions of 67 mL each?
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1st 67mL ether extract CE/CW = WE/67mL / Ww/400mL = 3 WE + Ww = 12g = Ww = 12-WE WE/67mL / 12-WE/400mL = 3 WE/67mL * 400mL/12-WE = 3 5.97WE/12-WE = 3(12-WE) Ww = 12 - 4.01 = 7.99g WE = 4.01 grams 2nd 67/mL ether extract CE/CW = WE/67mL / Ww/400mL = 3 WE + Ww = 7.99g = Ww = 7.99-WE WE/67mL * 400mL/7.99-WE = 3 5.97WE/7.99-WE = 3(7.99 - WE) 5.97WE = 23.97 - 3WE Ww = 7.99 - 2.67 g = 5.23 g WE = 2.67 grams 3rd 67mL ether extract CE/CW = WE/67mL / Ww/400mL = 3 WE + Ww = 5.23 g We = 5.32 - WE WE/67mL * 400mL/5.32 - WE 5.97WE = 3(5.32 - WE) 5.97WE = 15.96 - 3WE Ww = 5.32 - 1.78 grams = 3.54 grams WE = 1.78 grams Total Extracted = 8.46 grams of compound A removed
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Column Chromatography
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List the four types of chromatography
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1. Thin-layer chromatography 2. Column chromatography 3. High-performance liquid chromatography 4.Gas-liquid chromatography
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In the separation of fluorene and fluorenone by column chromatography, what is the: a. Stationary phase b. Mobile phase
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a. Alumina b. Petroleum Ether and dichloromethane
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To run the column chromatography experiment the fluorene and fluorenone mixture was diluted in dichloromethane with a few drops of ether (true/false)
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False
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What were you supposed to do with the columns after the lab? (what were the waste removal procedures)
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The funnel was to be removed and the columns were to be placed in the beaker that was inside of the waste hood.
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What are the structures for Fluorene and Fluorenone
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What is chromatography?
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Chromatography is a physical method of separation in which the components of a mixture to be separated are distributed between two phases, one of which is stationary (the stationary phase) while the other moves (the mobile phase) in a definite direction. The partitioning of the components of a mixture into the stationary phase is an equilibrium process. As the mobile phase moves past the stationary phase, repeated adsorption and desorption of the solute occurs at a rate determined by the distribution between the two phases (mobile and stationary). If the distribution between the two phases is different for the different components of a mixture, the components will undergo separation as they pass through the column containing the stationary phase.
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Balancing redox equations using the half-reaction method.
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1. Remove spectator ions and separate the equation into two half-reactions. 2. Balance all atoms other than oxygen and hydrogen 3. Balance oxygen atoms with water molecules 4. Balance hydrogen atoms with hydrogen ions (H+) 5. Balance the charge with electrons (e-) 6. Multiply each half-reaction by a whole number so that the electrons will cancel out when the two half-reactions are added together. 7. Add the two half-reactions; cancel water and H+ 8. Add hydroxide ion (OH-) to neutralize H+ when necessary 9. Add spectator ions to the equation
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Redox reaction
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Explain why it is unwise to use column chromatography on compounds having boiling points below about 150°C (760 torr).
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In order to observe the separations of the mixtures, the best combinations of solvent and the adsorbent compounds should have a boiling point higher than 150°C (760 torr). Compounds with lower boiling points will elute much faster (as a result of smaller molecular weight) so that the separations between the mixtures will not be as easily observed and the equilibrium between the two phases will not be established. This causes the results to be rounded and wider peaks.
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In normal-phase column chromatography, which solvent has more eluting power: petroleum ether or dichloromethane? In what way is the eluting power of a solvent related to its polarity?
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-Dichloromethane The eluting power of a solvent is related to its polarity because of its ability to remove a compound from a stationary phase. In normal-phase column chromatography, the more polar a solvent the more likely it will displace the polar compound that was bound to the stationary phase, which causes the polar compound to elute through the column.
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When separating a mixture by normal-phase column chromatography, why is it better to change from a less-polar solvent to a more polar solvent rather than the opposite?
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If the opposite was performed, both compounds would move too quickly through the column and equilibration between the mobile phase and stationary phases would not occur. In a normal-phase column chromatography, equilibration (separation between stationary and mobile phase) can only occur if a less-polar solvent is used followed by a more polar solvent.
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Why should care be exercised in the preparation of the column to prevent air bubbles from being trapped in the adsorbent?
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If air bubbles are trapped in the column, it will allow channeling. Channeling causes poor separations between the compounds as they are eluted to give bands that are ragged instead of sharp.
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Does fluorene or 9-fluorenone move faster down the column when petroleum ether is used as the eluent? Why?
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Fluorene will move faster because it is more weakly adsorbed to the stationary (polar) phase and less polar than 9-fluorenone, which is more strongly adsorbed to the stationary phase. So it is more easily removed using the less polar solvent petroleum ether.
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Thin-Layer Chromatography
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In a TLC experiment, why should the spot not be immersed in the solvent in the developing chambers?
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If the spot was immersed into the solvent, it would be dissolved into the solvent and removed from the plate, which would not allow for proper separation and create the need to repeat the experiment from the beginning.
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Explain why the solvent must not be allowed to evaporate from the plate during development.
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If the solvent evaporated from the plate, its compounds would not move up the plate to allow for maximum separation. This evaporation would result in the inaccurate relative mobility calculations.
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Explain why the diameter of the spot should be as small as possible.
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If the diameter of the spot is too large (anything of 1-2 mm), it could cause overlapping if there is more than one compound being tested. The consistency in the size of the small dot will also allow for accurate Rf values.
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A student obtained the silica gel TLC plate shown in figure 6.6 by spotting samples of midol, caffeine, and acetaminophen on the plate and eluting with petroleum ether: chloroform (9:1 v:v). a. What are the Rf values of acetaminophen and caffeine, respectively? b. Based on this TLC analysis, what are the ingredients in a tablet of midol? c. What are the mobile and stationary phases, respectively, in this TLC experiment?
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a. Acetaminophen Rf 32/38 = 0.84 mm Caffeine Rf 13/38 = 0.34 mm b. Acetaminophen and caffeine c. -Mobile phase is petroleum ether: chloroform -Stationary phase is silica gel TLC plate
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d. No spots were observed visually when the TLC plate was removed from the developing chamber. How might the student effect the visualization of the spots?
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If the compounds fluoresce, the student could place the TLC plate under an ultraviolet light to observe any spots. Another option would be to spray the plate with multiple reagents that would react with the compounds and cause the spots to turn a certain color. A last option would be to expose the plate to an iodine vapor. This could be done by placing the TLC plate into a closed chamber with iodine crystals. The iodine will react with organic compounds and cause the spots to brown.
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e. Another student accidentally used Midol PM in her experiment and observed only one spot. Speculate as to which spot was absent and offer a possible explanation for the difference in this student's results.
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The spot that was absent in the Midol PM would be the caffeine spot. The compound Midol PM did not contain the compound caffeine, but did contain acetaminophen, which is why only one spot was observed when the TLC was performed.
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The solid adsorbent in thin layer chromatography is usually Alumina or Silica, both which are: Polar or Non-polar
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Polar
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Thin layer chromatography was preferred on a compound and the Rf value taken. If the distance from the origin to the solvent line was 6.0 cm, and the distance traveled from the origin to the substance was 4.1 cm. What is the Rf value for this compound?
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Rf = 4.1 cm/6.0 cm = 0.68 cm
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List and describe three mistakes that can occur when running a TLC plate in a developing chamber.
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1. The diameter of the spot is too large causing streaking and overlapping 2. The origin line and dot are below the solvent line causing the spot to dissolve into the solvent. 3. Over saturation - if too much of the compound being tested is used it can cause streaking and would give inaccurate results.
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SN2 Reaction - 1-Bromobutane
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What is the reaction of 1-butanol to 1-bromobutane?
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What is the mechanism of 1-butanol to 1-bromobutane?
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What is the side reaction?
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What type of reaction is the synthesis of 1-butanol to 1-bromobutane
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SN2 reaction S = Substitution
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How did the SN2 reaction work?
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To separate the 1-bromobutane, it is steam distilled from the other compounds in the reaction flask. H2SO4, water and hydrobomic acid is also co-distilling -Add H2O separates distillate layer from aqueous layer -Addition of 2 5mL NaOH removes acids (unreacted) HBr and H2SO4 -Addition of H2O removes NaOH into aqueous layer -Addition of NaCl removes any H2O into aqueous layer
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1) Some water was added to the initial reaction mixture in the procedure you performed. a. How might the yield of 1-Bromobutane be affected if the water was not added and what product(s) would be favored?
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The yield of 1-Bromobutane would decrease as the reaction would not be able to occur. The Sodium Bromide added will not react with 1-Butanol and Bromination will not occur
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b. How might the yield of product be affected by adding twice as much water as is specified, while keeping the quantities of the other reagents the same?
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The yield of product would decrease as a result of the diluted solution. The addition of twice as much water would result in the decrease in the displacement of 1-Butanol with by the Bromide ion as a result of the other reagents quantities remaining the same.
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Using curved arrows to symbolize the flow of electrons, propose a mechanism for formation of the by-product(s) from elimination that could be produced when 1-Butanol is treated with HBr. Specify whether your mechanism is E1, E2 or both.
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E2 Reaction is the only mechanism that would work as a result of the alcohol on a primary carbon.
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How would doubling the concentration of the nucleophile affect the rate of an SN2 reaction?
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The rate of an SN2 reaction is = k [Nuc][Substrate]. Doubling the concentration of the nucleophile would double the rate of the SN2 reaction.
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1-Butanol does not undergo SN2 reactions in the absence of acids. If the alcohol were converted to the corresponding p-toluenesulfonate ester shown below, would you expect this ester to undergo an SN2 reaction with NaBr in the absence of acid. Give your reasoning.
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1-Butanol does not undergo an SN2 reaction in the absence of acids because in an SN2 reaction the leaving group must be weakly basic. The leaving group (if the reaction were to go forward) would be a strongly basic hydroxide which would be a poor leaving group. P-Toluenesulfonate ester (strong acid) however, would be expected to undergo an SN2 reaction in the absence of an acid because sulfonate ester is weak base (stabilized by resonance structures). Therefore, it is a good leaving group and the SN2 reaction would go forward.
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Consider the structure of 1-Butanol and explain why this alcohol does not undergo SN1 reactions.
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1-Butanol does not undergo SN1 reaction because it is a primary alcohol with the substrate on the primary carbon. SN1 mechanisms occur with tertiary alcohols, whereas SN2 reactions would be sterically hindered by a tertiary alcohol.
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Dehydration of Alcohols - E1
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Why did we use phosphoric acid instead of sulfuric acid?
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Because its greener
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What does E stand for?
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Elimination
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Why is the boiling point of the parent alcohol higher than that of the product alkene
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The boiling point of the parent alcohol (Cyclohexanol) is higher than that of the alkene (cyclohexene) as a result of difference in their intermolecular forces. Cyclohexanol has hydrogen bonding and dipole-dipole attractions which are stronger forces than the Van der Waals dispersion forces (temporary dipole moments) in Cyclohexene.
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Why is it proposed that the alcohol functional group is protonated by acid before dehydration can occur via either an E1 or E2 mechanism?
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An acid-catalyst is used to protonate the alcohol because hydroxide is a strong base and poor leaving group and water is a weak base much better leaving group allowing the reaction to move forward. However, only certain acids are used to protonate the alcohol functional group and not hydrohalic acids. Hydrohalic acids are not used because of their conjugate bases which act as good nucleophiles and result in substitution products instead of elimination products.
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Why is it particularly important that the crude cyclohexene be dry prior to its distillation?
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Water and cyclohexene form a minimum-boiling azeotrope that occurs as a result of the two compounds distilling out at a boiling point that is lower than both pure compounds. Fractional distillation does not result in a pure compound but a azeotropic mixture of the two. Therefore, it is important that the water be removed from the crude cyclohexene by adding a drying agent to the mixture prior to its distillation.
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Provide a detailed mechanism for the acid-catalyzed dehydration of cyclohexanol. Use curved arrows to symbolize the flow of electrons.
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What is the main reaction?
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What is the mechanism?
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