ochem lab final post lab ques
Refractive Index Questions
A compound has a refractive index of 1.3191 at 20.1°C. Calculate its refractive index at 25.0°C.
Calculation of Refractive Index at 25.0°C
Given:
Refractive index at 20.1°C: n1 = 1.3191
Initial temperature: T1 = 20.1°C
Final temperature: T₂ = 25.0°C
Formula:
n2 = n1 - 0.00045 (T2 - T1)
Step 1: Temperature difference
T2 - T1 = 25.0 - 20.1 = 4.9
Step 2: Apply correction factor
0.00045 × 4.9 = 0.002205
Step 3: Calculate n2.
n2 = 1.3191 - 0.002205 = 1.3169
Answer:
n2 = 1.3169
2. To clean the glass surfaces of a refractometer, ethanol or isopropyl alcohol, but not acetone or water, is usually recommended. Why?
Answer: Ethanol and isopropanol dissolve residues and evaporate quickly without damaging the prism. Acetone can degrade adhesives or coatings on the prism, while water can leave streaks, residues, or mineral deposits that distort measurements.
1. Describe the characteristics of a good recrystallization solvent.
A good solvent for recrystallization should dissolve a lot of the solute when it's hot, but only a little when it's cold. Impurities should be able to stay either fully soluble or completely insoluble for simple removal without undergoing a chemical reaction with the solute.
2. The solubility of a compound is 59 g/100 mL in boiling methanol and 30 g/100 mL in cold methanol, whereas in water it is 7.2 g/100 mL at 95 °C and 0.22 g/100 mL at 2 °C. Which solvent would be better for recrystallization? Explain.
Water would be a better solvent because its solubility varies significantly between hot and cold conditions (7.2 g vs 0.22 g). This allows for more efficient crystallization than methanol, which has a lower contrast in solubility.
3. Explain how the rate of crystal growth can affect the purity of a recrystallized compound.
Impurities may become trapped within the crystal lattice, and lower purity may result if crystals form too quickly. Pure solute molecules form ordered crystals, while impurities stay in solution due to slow, controlled crystal growth.
4. In what circumstances is it necessary to filter a hot recrystallization solution?
Hot filtration is necessary when insoluble impurities are present or when a compound may crystallize prematurely during cooling, which could trap impurities.
5. Why should a hot recrystallization solution be filtered by gravity rather than vacuum filtration?
Gravity filtration avoids premature cooling of the solution. Vacuum filtration can cause rapid cooling, leading to unwanted crystallization in the filter apparatus.
6. Low-melting solids often 'oil out' of a recrystallization solution rather than crystallizing. If this were to happen, how would you change the recrystallization procedure to ensure good crystals?
To prevent oiling out, the solution can be cooled more slowly, a different solvent can be used, or the mixture can be seeded with pure crystals of the compound. Using a solvent in which the compound has a higher melting point than the solvent’s boiling point can also help.
1. Explain why the observed boiling point for the first drops of distillate collected in the simple distillation of a 1:1 molar solution of pentane and hexane, illustrated in Figure 12.5, will be above the boiling point of pentane.
Pentane has a tougher time boiling because hexane dilutes the initial drops of pentane, raising the temperature. This indicates that its vapor pressure is lower.
2. The molar composition of a mixture is 80% hexane and 20% pentane. Use the phase diagram in Figure 12.4 to estimate the composition of the vapor over this liquid.
This vapor is condensed, and the resulting liquid is heated. What is the composition of the vapor above the second liquid?
The vapor above the second liquid would consist of 80% pentane and 20% hexane.
3 . A student carried out a simple distillation on a compound known to boil at 124°C and
reported an observed boiling point of 116°-117°C. Gas chromatographic analysis of the product showed that the compound was pure, and a calibration of the thermometer indicated that it was accurate. What procedural error might the student have made in setting up the distillation apparatus?
The student may have committed a procedure error during setting up the distillation apparatus, which is a thermometer issue. He or she might have positioned the thermometer too high in the device, preventing it from being completely immersed in the hot vapor, which would have resulted in a lower reading.
4. The directions in an experiment specify that the solvent, diethyl ether, be removed from the product by using a simple distillation. Why should the heat source for this
distillation be a steam bath, not an electrical heating mantle?
We use a steam bath as the heat source for this distillation because diethyl ether is highly flammable.
An extraction procedure specifies that an aqueous solution containing dissolved organic material be extracted twice with 10-mL portions of diethyl ether. A student removes the lower layer after the first extraction and adds the second 10-mL portion of ether to the upper layer remaining in the separatory funnel. After shaking the funnel, the student observes only one liquid phase with no interface. Explain.
After the initial extraction, the student only had the ether layer left after removing the lower aqueous layer. An interface did not form when an additional 10 mL of ether was added because it produced a single organic phase without an aqueous layer. For extraction to work correctly, two immiscible layers—aqueous and organic—must separate.
A crude nonacidic product mixture dissolved in diethyl ether contains acetic acid. Describe an extraction procedure that could be used to remove the acetic acid from the ether.
An acid–base extraction can remove the acetic acid:
Place the ether solution in a separatory funnel.
Add a portion of saturated aqueous sodium bicarbonate (NaHCO₃) solution and gently shake while venting frequently (CO₂ gas forms).
Allow the layers to separate and drain the aqueous layer, which now contains the water-soluble sodium acetate.
Repeat the extraction once more to ensure all acetic acid is removed.
Wash the combined ether layers with water, then dry them with anhydrous sodium sulfate (Na₂SO₄).
Filter and evaporate the ether to obtain the purified, nonacidic organic product.
3. What precautions should be observed when an aqueous sodium carbonate solution is used to extract an organic solution containing traces of acid?
When using aqueous sodium carbonate, several precautions must be followed:
Vent the separatory funnel frequently because neutralization produces CO₂ gas, which builds pressure quickly.
Shake gently to prevent excessive foaming and emulsions.
Cool the funnel or keep it at room temperature since gas formation and neutralization can be exothermic.
Avoid using too strong a base if the organic compound is sensitive; bicarbonate is milder than carbonate.
Never point the funnel toward yourself or others when venting.
4. When two layers form during a petroleum ether / water extraction, what would be an easy, convenient way to tell which layer is which if the densities were not available?
Place a small sample of the unknown layer in a test tube and add a few drops of water:
If the drop mixes or dissolves, that layer is aqueous.
If it does not mix and forms a separate droplet, that layer is organic (petroleum ether).
This simple test quickly identifies which layer is water without needing density data.
5. You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be 0.30 g. The distribution coefficient of benzoic acid in diethyl ether and water is approximately 10. Calculate the amount of benzoic acid that would be left in the water solution after four 20-mL extractions with ether. Do the same calculation using one 80-mL extraction with ether to determine which method is more efficient.
Given:
- K = 10
- Vw = 100 mL (water)
- Initial benzoic acid = 0.30 g
Formula: Fraction remaining in water after one extraction:
Vw / (Vw + KVo)
A. Four 20-mL extractions (each Vo = 20):
Fraction remaining after one extraction = 100 / (100 + 10×20) = 100 / 300 = 1/3
After four extractions: (1/3)⁴ = 1/81 = 0.0123
Mass remaining = 0.30 × 0.0123 = 0.0037 g = 3.7 mg
B. One 80-mL extraction:
Fraction remaining = 100 / (100 + 10×80) = 100 / 900 = 1/9 = 0.111
Mass remaining = 0.30 × 0.111 = 0.0333 g = 33.3 mg
Conclusion:
- Four 20-mL extractions → 3.7 mg benzoic acid left
- One 80-mL extraction → 33.3 mg benzoic acid left
6. What component of the mixture was extracted from the ether solution following extraction with NaOH solution?
Benzoic acid was extracted as sodium benzoate, as shown below:
C₆H₅COOH (s)+NaOH (aq)→C₆H₅COONa (aq)+H₂O (l){C₆H₅COOH (s)} + {NaOH (aq)} → {C₆H₅COONa (aq)} + {H₂O (l)}C₆H₅COOH (s)+NaOH (aq)→C₆H₅COONa (aq)+H₂O (l)
7. What happened when the above extracts were treated with HCl?
The sodium benzoate converted back to benzoic acid, which precipitated:
C₆H₅COONa (aq)+HCl (aq)→C₆H₅COOH (s)+NaCl (aq){C₆H₅COONa (aq)} + {HCl (aq)} → {C₆H₅COOH (s)} + {NaCl (aq)}C₆H₅COONa (aq)+HCl (aq)→C₆H₅COOH (s)+NaCl (aq)
8. Provide a flowchart for both extraction exercises.
Mixture (Benzoic acid + Dibenzalacetone)
│
Dissolve in Dichloromethane
│
Extract twice with 1 M NaOH
┌───────────────────────┐
│ │
Aqueous Layer (sodium benzoate) Organic Layer (dibenzalacetone)
│ │
Acidify with 6 M HCl Wash with sat. NaCl → Dry with Na₂SO₄
│ │
Benzoic acid precipitates → Filter Evaporate Dichloromethane → Solid product
1. When 2-propanol was used as the developing solvent, two substances moved with the solvent front (Rf = 1) during TLC analysis on a silica gel plate. Can you conclude that they are identical?
No, you can't assume that the two chemicals are the same. Poor separation results from an overly polar solvent when both compounds migrate with the solvent front. Repeat the experiment with a less polar solvent solution or perform co-spotting (spotting both compounds together on the same baseline) to determine if they form one or two spots in order to confirm identity.
2. The Rf value of compound A is 0.34 in pentane and 0.44 in diethyl ether. Compound B has an Rf of 0.42 in pentane and 0.60 in diethyl ether. Which solvent would be better for separating a mixture of A and B by TLC? Explain.
Because pentane has a somewhat greater separation than diethyl ether due to a bigger difference in Rf values (ΔRf = 0.10) than diethyl ether, pentane would be preferable. Solvents with lower polarity decrease mobility and enhance resolution between compounds with minor polarity differences.
1. Which compound, 2-bromo-2-methylpropane or 2-chloro-2-methylpropane, reacted faster in your SN1 experiment?
2-bromo-2-methylpropane reacted faster. The relative rate was significantly higher, with a reaction time of about 1–2 minutes versus over 15 minutes for the chloro compound.
2. Based on your answer, which is the better leaving group, Br− or Cl−? Briefly explain.
Bromide (Br−) is the better leaving group because it is a larger, weaker base and can better stabilize the negative charge after leaving.
3. Which compound, 2-bromo-2-methylpropane or 2-bromopropane, reacted faster in your SN1 experiment? Briefly explain.
2-bromo-2-methylpropane reacted faster because it forms a more stable tertiary carbocation, whereas 2-bromopropane forms a less stable secondary carbocation.
4. Which of the two solvents, 40% 2-propanol or 60% 2-propanol, is more polar? Briefly explain.
The 40% 2-propanol solution is more polar because it contains a higher proportion of water, a highly polar solvent.
5. In which of the two solvent mixtures did 2-bromo-2-methylpropane react faster? Briefly explain.
It reacted faster in the 40% 2-propanol mixture, consistent with the fact that more polar solvents stabilize the carbocation intermediate, enhancing the SN1 rate.
6. Which variable (leaving group, alkyl structure, or solvent polarity) has the greatest impact on the rate of an SN1 reaction?
The alkyl structure has the greatest impact because carbocation stability is the key determinant of the reaction rate. Tertiary carbocations form and react much faster than secondary or primary ones.
1. What is the purpose of washing the crude product with aqueous NaHCO3? Give the reactions.
The purpose of washing with NaHCO3 is to neutralize any remaining acids (HBr or H2SO4). Reactions:
HBr + NaHCO3 → NaBr + H2O + CO2
H2SO4 + 2NaHCO3 → Na2SO4 + 2H2O + 2CO2
2. Report the mechanism of the reaction for the synthesis of 1-bromobutane.
1) Protonation of 1-butanol forms R–OH2⁺.
2) Br⁻ performs a backside SN2 attack, displacing H2O.
3) Product formed: 1-bromobutane + H2O.
3. Determine the limiting agent? Calculate the theoretical yield.
Limiting reagent: 1-butanol (0.02623 mol)
Theoretical yield: 3.594 g 1-bromobutane.
Q1. What would be the major product obtained from the E1 dehydration of 2-methylcyclohexanol?
Answer: The major product is 1-methylcyclohexene (Zaitsev product). The mechanism proceeds via protonation of the alcohol, loss of water to form a carbocation, and β-hydrogen elimination forming the more substituted alkene.
Q2. Outline a mechanism for the dehydration of 1-methylcyclohexanol. Would it be faster or slower than the reaction you performed?
Protonation of -OH leads to the formation of carbocations, which are then deprotonated to produce the alkene. Because 1-methylcyclohexanol forms a more stable tertiary carbocation, dehydration occurs faster than that of cyclohexanol.
Q3. The product decolorizes Br₂ in CH₂Cl₂. What product is being formed as a result of reaction of bromine with your product?
Answer: The product is trans-1,2-dibromocyclohexane formed by anti-addition of Br₂ across the C=C bond. Bromine's color fades as it reacts with the alkene.