- Aldol: Which workup step most directly removes ionic base catalyst before recrystallization?
Answer: Wash solid with water
This question tests intermediate organic chemistry skills, focusing on reaction workup and yield calculations. Yield calculations benefit from removing catalysts to avoid impure products. In this aldol scenario, washing removes soluble base before purification. The correct answer reflects water wash for ionic removal. A common distractor involves confusing with drying or reaction adjustments. To help students, instructors should emphasize catalyst removal in workups. Practice aldol workup procedures.
- Grignard: 0.0600 mol benzaldehyde limiting; theoretical mass of 1-phenyl-1-propanol (MW 136.19)?
Answer: 8.17 g
This question tests intermediate organic chemistry skills, focusing on reaction workup and yield calculations. Yield calculations involve determining the theoretical maximum product based on the limiting reagent's moles and the product's molecular weight. In this Grignard scenario, the workup steps ensure the product is isolated properly, but here the focus is on theoretical mass computation. The correct answer reflects an understanding of straightforward multiplication of moles by MW for the given limiting reagent. A common distractor involves misapplying the MW or forgetting the limiting aspect. To help students, instructors should emphasize the importance of precise calculations and units in theoretical yields. Practice similar problems with varying limiting reagents to build proficiency.
- A student performs a Fischer esterification and purifies the product by distillation. An IR spectrum of the final product shows a strong C=O stretch at 1735 cm⁻¹, but also a significant, broad absorption centered at 3400 cm⁻¹. This indicates the product is contaminated with an O-H containing compound. How does this impurity affect the calculated percent yield?
Answer: The calculated yield is artificially high because the measured mass includes the mass of the impurity.
When analyzing experimental results in organic chemistry, you need to distinguish between what spectroscopic data tells you about product identity versus how impurities affect quantitative measurements like percent yield.
The IR spectrum confirms you've made the desired ester (C=O stretch at 1735 cm⁻¹), but the broad absorption at 3400 cm⁻¹ indicates contamination with an O-H containing compound—likely unreacted carboxylic acid or alcohol from the Fischer esterification. Since percent yield is calculated as (actual mass of product/theoretical mass of product) × 100%, any extra mass from impurities will inflate your measured "product" mass, making your calculated yield artificially high.
Looking at the wrong answers: Answer B incorrectly assumes IR data doesn't relate to quantitative analysis—but impurities detected by IR directly affect the mass you're weighing. Answer C suggests the yield appears low because IR signals interfere, but spectroscopic interference doesn't change the actual mass measurements used in yield calculations. Answer D claims you need the impurity's molar mass to determine the effect, but you don't need specific molecular weights to know that extra mass from any impurity will increase your measured product mass.
The key insight is that percent yield calculations assume your isolated material is pure product. When impurities are present, the "product" mass you measure includes both desired product and contaminants, leading to an overestimated yield. Always consider how impurities affect quantitative measurements, not just qualitative identification.
- A student prepares a Grignard reagent from bromobenzene and then reacts it with excess acetone. After an acidic workup, the isolated crude product weighs 125% of the theoretical yield. Which of the following is the most plausible explanation for this observation?
Answer: The crude product is contaminated with residual solvent (e.g., diethyl ether or water) from the workup.
When you encounter a question about yields exceeding 100%, you're dealing with experimental error rather than miraculous chemistry. This scenario tests your understanding of practical laboratory considerations in Grignard reactions.
The correct answer is D because crude products commonly retain solvents from the reaction and workup process. Grignard reactions typically use anhydrous diethyl ether as solvent, and the acidic workup involves aqueous solutions. Even after standard isolation procedures, the crude product often contains residual ether, water, or other solvents that haven't been completely removed. This extra mass makes the apparent yield exceed the theoretical value based on pure product alone.
Let's examine why the other options don't explain yields over 100%. Option A involves calculation errors that would affect your computed theoretical yield, but wouldn't actually increase the physical mass of isolated product beyond what's theoretically possible. Option B suggests CO₂ contamination forming byproducts - while Grignard reagents do react with CO₂, this would typically be minimized under proper inert atmosphere conditions, and wouldn't commonly produce 25% excess mass. Option C incorrectly implies that exothermic reactions somehow create extra product, which violates stoichiometry and conservation of mass.
Study tip: When you see yields significantly above 100% in organic chemistry problems, immediately think "contamination" rather than reaction mechanisms. The most common culprits are residual solvents, water, or unreacted starting materials. This pattern appears frequently on exams testing practical laboratory knowledge.
- After carrying out a reaction in an organic solvent, a student needs to remove a byproduct, p-toluic acid. The student washes the organic layer with 5% aq. NaHCO₃. In which layer will the p-toluic acid primarily reside after this wash, and why?
Answer: In the aqueous layer, because it is converted to its conjugate base, sodium p-toluate, which is an ionic salt.
p-Toluic acid is a carboxylic acid. Sodium bicarbonate (NaHCO₃) is a base that is strong enough to deprotonate the carboxylic acid, forming the corresponding carboxylate salt (sodium p-toluate). This salt is ionic and therefore much more soluble in the polar aqueous layer than in the nonpolar organic layer. This acid-base reaction is the basis for extractive separation.
- Grignard setup: Which precaution most prevents reaction failure before workup?
Answer: Exclude moisture with dry ether and glassware
This question tests intermediate organic chemistry skills, focusing on reaction workup and yield calculations. Yield calculations can fail if the reaction doesn't initiate due to moisture. In this Grignard setup scenario, anhydrous conditions are critical to prevent quenching. The correct answer reflects excluding water with dry materials. A common distractor involves allowing moisture or air exposure. To help students, instructors should emphasize anhydrous techniques. Practice Grignard setups with emphasis on dryness.
- Grignard: 0.0800 mol RMgBr and 0.0500 mol cyclohexanone; theoretical moles alcohol produced?
Answer: 0.0500 mol
This question tests intermediate organic chemistry skills, focusing on reaction workup and yield calculations. Yield calculations involve determining theoretical product moles based on the limiting reagent in organometallic additions. In this Grignard scenario, workup converts the adduct to alcohol, but the theoretical is set by stoichiometry. The correct answer reflects the carbonyl as limiting, producing equal moles of alcohol. A common distractor involves adding the reagent moles incorrectly. To help students, instructors should emphasize 1:1 stoichiometry in Grignard reactions. Practice limiting reagent problems for various carbonyl additions.
- Grignard workup: After ether reaction, which reagent quenches excess RMgX to alcohol-compatible products?
Answer: Dilute HCl(aq)
This question tests intermediate organic chemistry skills, focusing on reaction workup and yield calculations. Yield calculations involve accurate product isolation, which depends on proper workup to avoid losses or impurities. In this Grignard workup scenario, the steps such as quenching excess reagent are crucial for converting intermediates to the desired alcohol without side reactions. The correct answer reflects an understanding of using dilute acid to safely quench excess Grignard to compatible byproducts. A common distractor involves confusing quenching with drying or washing steps. To help students, instructors should emphasize the role of each workup reagent in Grignard reactions. Practice sequencing workup steps and explaining their purposes to enhance comprehension.
- In the workup for the reduction of a ketone with LiAlH₄, the reaction is cautiously quenched by the sequential addition of H₂O, then 15% aq. NaOH, then more H₂O (Fieser workup). What is the purpose of adding aqueous NaOH in this procedure?
Answer: To convert the aluminum salts into a granular, easily filterable solid (Al(OH)₃).
When you encounter questions about LiAlH₄ workup procedures, focus on understanding what happens to the aluminum-containing species formed during the reaction. The Fieser workup is specifically designed to handle the complex mixture of aluminum salts that result from hydride reductions.
During LiAlH₄ reduction, aluminum forms various alkoxide complexes with the alcohol product. When water is first added, these complexes are partially hydrolyzed, creating a mixture of aluminum salts, hydroxides, and alkoxides that form a gelatinous, difficult-to-filter mess. The key insight is that adding 15% aqueous NaOH converts all these aluminum species into aluminum hydroxide (Al(OH)₃), which precipitates as a granular, crystalline solid that filters easily and completely separates from your organic product.
Looking at the wrong answers: (A) incorrectly suggests the goal is to deprotonate the alcohol product—but alcohols are actually easier to extract in their neutral form, and alkoxides would complicate workup. (B) misidentifies the timing—unreacted LiAlH₄ is destroyed by the initial water addition, not the NaOH step. (C) wrongly characterizes alcohols as acidic products needing neutralization—alcohols are actually neutral compounds that don't require acid-base treatment for isolation.
The correct answer is (D): the NaOH converts aluminum salts into easily filterable Al(OH)₃ solid.
Study tip: Remember that workup procedures are designed to solve practical separation problems. For LiAlH₄ reductions, the main challenge is dealing with messy aluminum salts—the Fieser workup's NaOH step specifically addresses this by creating a clean, filterable precipitate.
- Grignard: Which workup step most directly removes dissolved water from the ether layer before evaporation?
Answer: Add anhydrous Na₂SO₄
This question tests intermediate organic chemistry skills, focusing on reaction workup and yield calculations. Yield calculations require pure product mass, which relies on effective water removal during workup to prevent inflated or inaccurate measurements. In this Grignard scenario, workup steps like drying the organic layer are essential before solvent evaporation. The correct answer reflects knowledge that adding a drying agent directly absorbs dissolved water. A common distractor involves mistaking washing steps for drying. To help students, instructors should emphasize the distinction between washing and drying in extractions. Practice lab simulations of Grignard workups to reinforce these concepts.
- Aldol recrystallization: Which step most ensures purity of crystals before weighing?
Answer: Rinse crystals with cold solvent
This question tests intermediate organic chemistry skills, focusing on reaction workup and yield calculations. Yield calculations require pure crystals, achieved through effective recrystallization techniques. In this aldol scenario, rinsing removes surface impurities. The correct answer reflects using cold solvent for purity. A common distractor involves improper boiling or addition. To help students, instructors should emphasize recrystallization steps. Practice purification techniques in lab settings.
- A student performs an aldol condensation of acetone, which produces mesityl oxide (C₆H₁₀O) upon dehydration. Starting with 11.6 g of acetone (MW = 58.08 g/mol), the student isolates 6.50 g of purified mesityl oxide (MW = 98.14 g/mol). What is the percent yield?
Answer: 66.2%
First, determine the stoichiometry. Two moles of acetone react to form one mole of mesityl oxide. Calculate the starting moles of acetone: 11.6 g / 58.08 g/mol = 0.200 mol. Based on the 2:1 stoichiometry, the maximum moles of mesityl oxide that can be formed is 0.200 mol / 2 = 0.100 mol. This corresponds to a theoretical yield of 0.100 mol × 98.14 g/mol = 9.814 g. The percent yield is (Actual / Theoretical) × 100 = (6.50 g / 9.814 g) × 100 ≈ 66.2%. A common mistake is to assume a 1:1 ratio, which would give half the correct yield.
- An ether solution contains benzoic acid (pKa ≈ 4.2), p-cresol (a phenol, pKa ≈ 10.2), and anisole (neutral). Which sequence of aqueous extractions would most effectively separate these three components?
Answer: First extract with 5% aq. NaHCO₃ to remove benzoic acid, then with 5% aq. NaOH to remove p-cresol.
To separate acids of different strengths, one should use bases of corresponding strengths. Sodium bicarbonate (NaHCO₃) is a weak base that is strong enough to deprotonate the carboxylic acid (pKa 4.2) but not the much weaker acid, phenol (pKa 10.2). After removing the benzoic acid, a strong base like sodium hydroxide (NaOH) can be used to deprotonate and extract the remaining phenol, leaving the neutral anisole in the ether.
- During the workup of an esterification reaction, the organic layer is washed with a saturated aqueous solution of NaCl (brine). What is the primary purpose of this brine wash?
Answer: To increase the polarity of the aqueous phase, thereby reducing the solubility of water in the organic layer.
The primary purpose of a brine wash is to remove the bulk of dissolved water from the organic layer before using a chemical drying agent. By saturating the aqueous phase with salt (NaCl), its polarity is greatly increased. This high polarity decreases the solubility of the less polar organic compounds in the aqueous layer and also pulls dissolved water out of the organic layer due to osmotic-like effects (the 'salting out' effect).
- Esterification workup: Which wash neutralizes residual H₂SO₄ catalyst in the organic layer?
Answer: NaHCO₃(aq) wash
This question tests intermediate organic chemistry skills, focusing on reaction workup and yield calculations. Yield calculations depend on pure product, requiring effective removal of impurities like catalysts during workup. In this esterification workup scenario, washing steps neutralize acids to prevent contamination. The correct answer reflects using bicarbonate to neutralize sulfuric acid. A common distractor involves confusing neutralization with drying. To help students, instructors should emphasize chemical reactions in workup washes. Practice sequencing esterification workups to solidify knowledge.
- Aldol: 0.250 mol benzaldehyde + 0.200 mol acetone → chalcone + H₂O; limiting reagent?
Answer: Acetone
This question tests intermediate organic chemistry skills, focusing on reaction workup and yield calculations. Yield calculations require identifying the limiting reagent from given moles and stoichiometry. In this aldol scenario, the condensation is 1:1, making the lesser reactant limiting. The correct answer reflects acetone as limiting due to fewer moles. A common distractor involves ignoring stoichiometry. To help students, instructors should emphasize reagent ratios in condensations. Practice limiting reagent problems for aldol reactions.
- Aldol: Acetone limiting at 0.200 mol; theoretical mass of chalcone (MW 208.26)?
Answer: 41.7 g
This question tests intermediate organic chemistry skills, focusing on reaction workup and yield calculations. Yield calculations involve theoretical mass from limiting moles and product MW. In this aldol scenario, workup isolates the condensed product for weighing. The correct answer reflects multiplication for the given limiting reagent. A common distractor involves using the wrong MW. To help students, instructors should emphasize calculation steps. Practice theoretical yields for condensation reactions.
- Nitration workup: Which step primarily removes residual strong acids from crude nitrobenzene?
Answer: Wash organic layer with NaHCO₃(aq)
This question tests intermediate organic chemistry skills, focusing on reaction workup and yield calculations. Yield calculations depend on removing impurities like acids during workup. In this nitration scenario, washing neutralizes residuals. The correct answer reflects bicarbonate wash for acid removal. A common distractor involves confusing with drying or distillation. To help students, instructors should emphasize workup chemistry. Practice nitration workup sequences.
- Esterification: Which step best improves ester purity before distillation?
Answer: Dry organic layer with MgSO₄
This question tests intermediate organic chemistry skills, focusing on reaction workup and yield calculations. Yield calculations benefit from high purity, achieved through proper pre-distillation steps. In this esterification scenario, drying removes water that could impure the ester. The correct answer reflects drying the layer to improve purity. A common distractor involves confusing reaction conditions with workup. To help students, instructors should emphasize purification techniques. Practice esterification labs with emphasis on purity assessment.
- A reaction workup requires the separation of a desired neutral product from an acidic byproduct in an ether solution. A student mistakenly adds 5% aqueous HCl to the separatory funnel instead of the prescribed 5% aqueous NaHCO₃. What is the most immediate consequence of this error?
Answer: Both the neutral product and the acidic byproduct will remain in the ether layer, resulting in a failed separation.
Acid-base extractions rely on converting compounds between neutral and ionic forms to control their solubility in organic versus aqueous layers. When you have a neutral product and an acidic byproduct in ether, you normally add base (like NaHCO₃) to deprotonate the acid, making it ionic and water-soluble so it moves to the aqueous layer.
By adding HCl instead of NaHCO₃, you've created an acidic aqueous layer that cannot deprotonate the acidic byproduct. Since the byproduct is already in its neutral, protonated form, adding more acid won't change its charge or solubility. Both compounds remain neutral and stay in the organic ether layer, making separation impossible.
Answer A correctly identifies this failed separation. Answer B incorrectly assumes the neutral product will become protonated - but neutral organic products typically aren't basic enough to be protonated by dilute HCl under these conditions. Answer C wrongly suggests the acidic byproduct will be "protonated by HCl" - this is backwards since the byproduct is already protonated (that's why it's called an acid). Answer D mentions pressure buildup, but simply mixing dilute HCl with an ether solution containing typical organic compounds won't cause a dangerous exothermic reaction.
The key study tip: In acid-base extractions, acids need bases to become ionic and move to water, while bases need acids to become ionic. Using the wrong reagent (acid instead of base, or vice versa) prevents the charge conversion that drives the separation, leaving everything in the original layer.
