2 exam questions: identify the item shown

What: 11 pictures in the midterm PDF. Two appear on the exam, and you name them.

Covers: 3 structures, 6 apparatus drawings, 1 graph, 1 device.

Why care: These are the cheapest 2 points on the exam. You only need to recognize each picture.

Connects to: Every lab unit. Each picture is a setup you meet again in a later topic.

Sets up all lab units

What: How chemists clean up a compound, and how they prove it is clean.

Covers: Physical properties. Four purification methods. Criteria of purity.

Why care: Every lab unit is one of these methods, followed by one of these checks.

Connects to: Distilleries and oil refineries use distillation. Melting point returns in Units 1, 2 and 5.

Notes sections I and II; lab technique questions

What: How to run something hot or cold safely.

Covers: Why heat or cool. Bumping and boiling stones. Heat sources. Cooling baths. Reflux.

Why care: You will be asked to recognize these setups, and they explain many lab steps.

Connects to: Boiling water for pasta. Reflux is used in Units 1, 2 and 5.

General Q1; notes section III

What: Dissolve one compound out of a solid mixture.

Covers: Isolating trimyristin from nutmeg. Steps. What makes a good solvent.

Why care: General Q1 asks for the steps in order.

Connects to: Brewing coffee or tea is a solid–liquid extraction. Practice Q4 extracts caffeine from coffee.

General Q2; Unit 1 Q5; Unit 2 Q4, Q6, Q7, Q8

What: Why some compounds dissolve and others do not.

Covers: The four weak attractions, ranked. Like dissolves like. Ranking solubility.

Why care: This idea decides up to 5 practice themes across Units 1 and 2.

Connects to: Solvent choice in extraction and recrystallization. Why soap needs an ionic end.

General Q3; Unit 1 Q1, Q2; Table 1

What: Using the temperature range where a solid melts to judge identity and purity.

Covers: What impurities do. Running a Mel-Temp. Identifying a compound by mixing.

Why care: Two of the five Unit 1 themes are melting-point logic. It is also the purity check for Units 1, 2 and 5.

Connects to: Every recrystallization ends with a melting point.

Unit 1 Q3, Q4; Unit 2 Q1

What: How much product you should get, and how much you did get.

Covers: Limiting reagent. Theoretical, experimental and percent yield. Neutralization.

Why care: These are direct calculations. Bring a calculator.

Connects to: General chemistry. Lab reports for each unit.

Table 1 structures; ester chemistry

What: Convert an ester into a carboxylic acid with base, then acid.

Covers: The Unit 2 reaction. What you see in the vial. Ester and carboxylic acid links.

Why care: Two Table 1 structures come from this reaction.

Connects to: Making soap. Salicylic acid is the starting material for aspirin. Unit 11 Claisen must avoid this reaction.

General Q4–8, Q11; Unit 2 Q5

What: Purify a solid by dissolving it hot and letting it crystallize as it cools.

Covers: Three steps. Cooling rate. Impurities. Choosing a solvent. Troubleshooting.

Why care: Six General questions come from this topic.

Connects to: Solubility rises with temperature, like sugar in hot tea. It uses the rules from intermolecular forces.

Unit 2 Q2, Q3

What: Calculate how much solvent to use and how much solid you can recover.

Covers: Minimum volume. Amount left in solution. Percent recovery. Mixtures of two solids.

Why care: Two practice themes are direct numbers. Other solved versions extend them.

Connects to: The Unit 2 lab uses exactly this calculation for your own crude mass.

General Q9, Q10, Q12–15

What: Separate liquids by boiling and condensing them.

Covers: Vapor pressure. Boiling point. Running a distillation. What changes in the flask.

Why care: Six General questions come from here, and they appear as written.

Connects to: Distilleries and oil refineries. The Unit 3 and 4 lab separates methanol from water.

Units 3/4 Q1, Q2

What: Calculate the composition of the vapor above a liquid mixture.

Covers: Raoult's law. Dalton's law. Mole fraction in liquid and in vapor.

Why care: Two practice themes use exactly this table.

Connects to: It proves in numbers why distillation works.

Units 3/4 Q3–7

What: A graph of temperature against composition for a two-liquid mixture.

Covers: The two curves. Finding boiling temperature. Finding vapor composition. Working backward. Theoretical plates.

Why care: Five of the seven Units 3/4 themes use this graph. Bring your ruler.

Connects to: It is the visual version of Raoult's law. Each staircase step is one cycle in a distillation column.

General Q16; Table 1

What: Two versions of distillation and why one separates better.

Covers: Theoretical plates. Column efficiency. Heat control. When simple is enough.

Why care: General Q16 and two Table 1 pictures come from here.

Connects to: Each theoretical plate is one step on the phase diagram.

Study 1; notes pp. 2–3; lecture 11.1, 11.2

What: Why the carbon next to a carbonyl group gives up a proton easily.

Covers: α-carbons and α-hydrogens. Counting them. pKa values. Resonance.

Why care: Almost every Unit 11 question starts by finding α-hydrogens.

Connects to: Earlier courses focused on the δ+ carbonyl carbon. Unit 11 moves one atom over. The same resonance logic explains why carboxylic acids are acidic.

Study 2; notes pp. 4–7; Q1, Q2

What: One compound exists in two rapidly interconverting forms.

Covers: Definitions. Drawing enols. Drawing keto forms. Base-catalyzed mechanism.

Why care: Two practice themes ask you to draw these. The mechanism is a learning objective.

Connects to: E/Z isomers and R/S stereocenters from earlier courses. The enolate is the shared intermediate for everything later.

Study 3; notes p. 8; Q3, Q4

What: The conjugate base of a carbonyl compound, which is the star of Unit 11.

Covers: Drawing enolates. Where they react. How much of it forms with each base.

Why care: All remaining Unit 11 reactions start with an enolate. Base choice decides which reaction you get.

Connects to: Predicting acid–base reactions from pKa. Alkene stability from earlier courses.

Study 4; notes pp. 9–10; Q5

What: Replace α-hydrogens with halogens. Methyl ketones go further and split.

Covers: Two-step cycle. Why it runs to completion. Haloform cleavage. Iodoform test.

Why care: One practice theme, and you must spot whether cleavage happens.

Connects to: Before spectroscopy, the yellow iodoform precipitate was a test for methyl ketones. Its cleavage step resembles ester hydrolysis.

Study 5; notes pp. 11–15; Q6, Q7, Q8

What: Replace an α-hydrogen with an alkyl group. This forms a new C–C bond.

Covers: Why LDA. Three scenarios. Speed against stability. Ring-forming version.

Why care: Three practice themes. A small change in conditions flips the answer.

Connects to: The SN2 reaction from earlier courses. A real synthesis of gymnomitrol used this.

Study 6; notes pp. 16–23; Q9, Q11, Q12, Q14, Q16

What: An enolate attacks the C=O of another aldehyde or ketone. This forms a new C–C bond.

Covers: Mechanism. Counting products. Intramolecular rings. Directed aldol.

Why care: With dehydration, this is 8 of the 18 Unit 11 themes. It is the largest block.

Connects to: Grignard addition has the same attack pattern. The Unit 5 product is made by this reaction. Rhizoxin D, an anti-tumor compound from OSU's Professor White, used it.

Notes pp. 24–27; Q10, Q13, Q15

What: A β-hydroxy product can lose water and form a C=C.

Covers: The two-question rule. Mechanism. E,E-dibenzalacetone from Unit 5.

Why care: Every aldol answer needs this check. Three practice themes turn on it.

Connects to: Conjugation. Your Unit 5 lab product. Its melting point is 110–111 °C.

Study 7; notes pp. 28–29; Q17, Q18

What: An ester enolate attacks another ester and forms a β-ketoester.

Covers: Conditions. Why NaOH fails. Five-part mechanism. Aldol comparison.

Why care: Two practice themes, and the first check gives you NR questions quickly.

Connects to: Saponification from Unit 2 is the reaction it must avoid. Its opening matches the aldol reaction.

2 exam questions: identify the item shown

• Three structures: trimyristin (Unit 1 product), methyl salicylate (Unit 2 starting material), salicylic acid (Unit 2 product).
• Methyl salicylate and salicylic acid look alike. Read the group at the top right of the ring: OCH₃ is the ester, OH is the acid.
• Reflux setups have a vertical condenser above the flask. Distillation setups have a condenser that slopes away to a collecting cylinder.
• Fractional distillation is simple distillation with a tall column added below the three-way adapter.

Sets up all lab units

• Nature and reactions both give mixtures. A compound is rarely pure when you first get it.
• There are four main purification methods: distillation, recrystallization, extraction, chromatography.
• Each method exploits a physical difference between compounds.
• A method is often used in sequence with another. For example, extract first, then recrystallize.
• To prove success, measure a number that changes a lot when impurities are present.
• Solids: melting point. Liquids: boiling point.

Notes sections I and II; lab technique questions

• Heat slow reactions. Cool reactions that are too exothermic.
• Liquids in clean glassware tend to bump. Boiling stones or magnetic stirring prevent this.
• Round-bottomed flasks go in a heating mantle. Flat-bottomed flasks go on a hot plate.
• Do not heat in an open container, because volatile material is lost. Do not heat in a closed container, because pressure builds and it can explode.
• Reflux solves both problems. Vapor condenses and runs back into the flask.

General Q1; notes section III

• The solid is the nutmeg. The liquid is the extraction solvent.
• The solvent dissolves the compound you want and leaves the rest behind.
• Unit 1: about 4 g of nutmeg refluxes in acetone for about an hour.
• Trimyristin dissolves. Other components stay solid and are filtered off.
• Cooling the solution crystallizes the trimyristin.
• Trimyristin is a triglyceride. Because it is solid at room temperature, it is a fat. Liquid triglycerides are oils.

General Q2; Unit 1 Q5; Unit 2 Q4, Q6, Q7, Q8

• There are four categories of weak attraction. Know their order.
• Melting and dissolving both work against these attractions.
• Like dissolves like. A compound dissolves best in a solvent that resembles it.
• A compound has two personalities: its polar groups and its hydrocarbon part.
• In water: ionic compounds beat neutral polar compounds, and these beat hydrocarbons.
• In benzene, the order flips.

General Q3; Unit 1 Q1, Q2; Table 1

• Melting point tells you identity and purity.
• Two numbers are reported: where melting starts, and where it ends.
• A range of 2 °C or less means pure. More than 2 °C means impure.
• Impurities always do two things together: lower the values and broaden the range.
• This happens whether the impurity melts higher or lower than your compound.

Unit 1 Q3, Q4; Unit 2 Q1

• Theoretical yield: the amount formed under ideal conditions, if the reaction goes to completion.
• Experimental yield: the amount you actually isolate.
• Percent yield: (experimental ÷ theoretical) × 100%.
• The limiting reagent runs out first and sets the theoretical yield.
• Yields rarely reach 100%.

Table 1 structures; ester chemistry

• Starting material: methyl salicylate, a liquid with the smell of oil of wintergreen.
• Step 1: 5 M aqueous NaOH at 100 °C. This saponifies the ester.
• Step 2: 1 M aqueous H₂SO₄ at room temperature. This acidifies the mixture and salicylic acid crystallizes.
• Products in a 1:1 ratio: salicylic acid and methanol. Methanol is not isolated.
• Both steps happen in one vial.

General Q4–8, Q11; Unit 2 Q5

• Recrystallization is the main way to purify solids (General Q4).
• Dissolve the sample in the minimum hot solvent. This makes a saturated solution.
• Cool it slowly. The compound crystallizes because cold solvent holds less.
• Impurities stay in solution, so they do not join the crystal.
• Collect the crystals by vacuum filtration.
• Slow cooling gives larger, purer crystals. Fast cooling gives smaller, less pure crystals.

Unit 2 Q2, Q3

• Two solubility numbers matter: one hot, one cold.
• Use the hot value to find the volume of solvent.
• Use the cold value to find what stays dissolved.
• Crystals = starting mass − mass that stays dissolved.
• Extra solvent always lowers the recovery.

General Q9, Q10, Q12–15

• Separation depends on boiling point differences (General Q9).
• Vapor pressure is the pressure a liquid exerts on its environment.
• A pure liquid boils when its vapor pressure equals atmospheric pressure.
• A lower boiling point means more volatile, and a higher vapor pressure at a given temperature.
• The vapor above a mixture is richer in the more volatile component. This makes distillation possible.
• So the liquid left behind is richer in the less volatile component.

Units 3/4 Q1, Q2

• Raoult's law: partial pressure = mole fraction in the liquid × vapor pressure of the pure component.
• Dalton's law: total pressure = sum of the partial pressures.
• Mole fraction in the vapor = partial pressure ÷ total pressure.
• The vapor always has more of the volatile component than the liquid does.

Units 3/4 Q3–7

• The graph has a lens shape made of two curves.
• Lower curve: liquid composition curve (LCC).
• Upper curve: vapor composition curve (VCC).
• The two curves meet at the boiling points of the two pure liquids.
• Vertical moves change temperature. Horizontal moves go between liquid and vapor at one temperature.
• One step across and down is one theoretical plate.

General Q16; Table 1

• The difference is the number of vaporization–condensation cycles, also called theoretical plates.
• Simple distillation: one or two plates. Fractional distillation: three or more.
• A fractionating column adds these extra cycles. It is a tall tube packed with glass beads.
• More plates give better separation.
• Unit 3 is simple distillation. Unit 4 is fractional distillation.

Study 1; notes pp. 2–3; lecture 11.1, 11.2

• An α-carbon is bonded directly to a carbonyl carbon. An α-hydrogen is bonded to an α-carbon.
• α-Hydrogens are far more acidic than alkane hydrogens: pKa about +20 against +50.
• Both lose a proton from a C–H bond, so the bond type is not the reason.
• The reason is the conjugate base. In an alkane anion, the charge is localized on one carbon. In an enolate, it is shared between carbon and oxygen.
• Always look for α-hydrogens first. Their number and position decide whether a reaction happens.

Study 2; notes pp. 4–7; Q1, Q2

• Tautomers are structural isomers that rapidly interconvert at equilibrium.
• The keto form has C=O. The enol form has OH directly on an alkene carbon: "ene" plus "ol".
• One H moves between the α-carbon and oxygen, and the π bond moves.
• The keto form dominates at equilibrium.
• Tautomers are real, different compounds. Resonance structures are drawings of one compound.

Study 3; notes p. 8; Q3, Q4

• An enolate anion is the conjugate base of a keto or enol tautomer. Both give the same anion.
• It has two resonance structures. Either drawing is correct.
• It can react at carbon or oxygen. In this course, the α-carbon lone pair always attacks.
• Weak base (NaOH, NaOR): a small amount of enolate, surrounded by neutral carbonyl compound.
• Strong base (LDA): about 100% enolate, with no neutral carbonyl compound left.
• This difference separates aldol conditions from alkylation conditions.

Study 4; notes pp. 9–10; Q5

• Conditions: H₂O, excess NaOH, excess X₂. X is Cl, Br or I, not F.
• Each H replaced takes two steps: OH⁻ removes the proton, then the enolate attacks X₂.
• With excess reagents, every α-hydrogen is replaced.
• Methyl ketones and acetaldehyde go further. This is the haloform reaction.
• Haloform products: a carboxylate and CHX₃.

Study 5; notes pp. 11–15; Q6, Q7, Q8

• Two steps in one pot: LDA forms the enolate, then the enolate does SN2 on an alkyl halide.
• LDA is strong enough to give about 100% conversion.
• LDA is too bulky to attack the carbonyl carbon, so it can only act as a base.
• Use ketones. Aldehydes give poor yields.
• Alkyl halide: methyl, primary or secondary. Never tertiary.
• 1 equivalent LDA at −78 °C: R goes on the less substituted α-carbon.
• Less than 1 equivalent, then warm: R goes on the more substituted α-carbon.

Study 6; notes pp. 16–23; Q9, Q11, Q12, Q14, Q16

• Conditions: NaOH, H₂O. Assume room temperature unless told otherwise.
• Requirement: at least one α-hydrogen.
• Product: a β-hydroxy aldehyde or β-hydroxy ketone.
• Two molecules join, even if the equation shows no "2".
• NaOH gives a low enolate concentration, so the enolate meets a neutral carbonyl molecule.
• With several enolates and partners, you get a mixture.
• With two C=O groups in one molecule, consider only the intramolecular reaction.
• To get one chosen cross product, use a directed aldol with LDA.

Notes pp. 24–27; Q10, Q13, Q15

• Some aldol products dehydrate at room temperature. Some need heat. Some never do.
• Ask two questions, in order.
• Q1: Is there an H on the carbon between C=O and C–OH? If no, it never dehydrates.
• Q2: Will the new C=C be conjugated with a π system other than the existing C=O? If yes, it dehydrates at room temperature. If no, it needs heat.

Study 7; notes pp. 28–29; Q17, Q18

• Substrate: an ester with at least two α-hydrogens. With one or none, write NR.
• Step 1: NaOR or KOR in ROH. The OR must match the ester's OR.
• Step 2: H₃O⁺.
• Product: a β-ketoester.
• The first two mechanism steps match the aldol. After that, it follows its own path.

2 exam questions: identify the item shown

Sets up all lab units

Physical properties listed in the overview: melting point (mp), boiling point (bp), density (d), refractive index, specific rotation.

• A purity test must be easy to measure precisely, and impurities must change it significantly.
• Research labs now also use infrared and NMR spectroscopy to judge purity.

Notes sections I and II; lab technique questions

General Q1; notes section III

General Q1: the seven steps, in order.

1. Add the solid mixture to a suitable flask.
2. Add the extraction solvent.
3. Outfit the flask for heating under reflux.
4. Reflux the suspension.
5. Cool the suspension.
6. Isolate the filtrate by gravity filtration.
7. Remove the solvent by evaporation.

An effective extraction solvent must:

• Dissolve the desired component well and not dissolve the others.
• Not react with the components of the mixture in an undesirable way.
• Be easy to separate from the solute afterward.

The lecture says you must be able to tick all three.

General Q2; Unit 1 Q5; Unit 2 Q4, Q6, Q7, Q8

General Q2, strongest to weakest:

1. Hydrogen bonding
2. Dipole–dipole interactions
3. Dipole–induced dipole interactions
4. Induced dipole–induced dipole interactions (van der Waals)

Ranking in water, in this order:

1. Ionic species are most soluble.
2. Among neutral compounds, more hydrogen-bonding groups means more soluble.
3. With equal groups, the less hydrocarbon-like compound is more soluble. Count the carbons.
4. Pure hydrocarbons are least soluble.

Ranking in benzene: hydrocarbon first, then neutral compounds with the larger hydrocarbon part, ionic species last.

Course data (Unit 2 notes, item 7):

Other generalizations in the notes:

• Low-melting solids are generally more soluble than high-melting compounds of similar structure.
• In a homologous series, higher members become more hydrocarbon-like.

General Q3; Unit 1 Q1, Q2; Table 1

Pure melting points to know: trimyristin 56–57 °C, salicylic acid 157–159 °C, E,E-dibenzalacetone 110–111 °C.

Reading the device. First sign of liquid gives the lower number. Fully melted gives the upper number. The dial runs 0 to 10, so 39 volts is a dial setting of 3.9.

Unit 1 Q3, Q4; Unit 2 Q1

Method.

1. Convert each reactant to moles.
2. Use the balanced equation to find the limiting reagent.
3. Use the mole ratio to get the moles of product.
4. Convert to grams.
5. If a volume is needed, divide by density.
6. For percent yield, divide the real mass by this value.

Why yields fall short (notes section V):

• A side reaction used up some reactant.
• Some reactant was lost during handling.
• A reactant or reagent was impure.
• Product changed chemically before it was isolated.
• Mechanical losses occurred during separation and purification.

Table 1 structures; ester chemistry

Procedure.

1. Combine the reactants. A white solid forms at once. This is not the product. It comes from an acid–base reaction.
2. Reflux at 100 °C. Watch for the white solid to disappear.
3. After the solid dissolves, boil for 10 more minutes.
4. Cool to room temperature.
5. Add aqueous sulfuric acid. Crude salicylic acid precipitates.
6. Collect by vacuum filtration and record the crude mass.
7. Keep a little crude for a melting point.
8. Recrystallize the rest from water, in an Erlenmeyer flask. Volume = crude mass × 15 mL/g.
9. Collect the purified crystals by vacuum filtration.

Setup. Hot plate stirrer, aluminum block, small reaction vial, magnetic spinner, microscale condenser.

General Q4–8, Q11; Unit 2 Q5

Three basic steps (notes item 3).

1. Prepare a hot, saturated solution at the solvent's boiling point. Remove any insoluble impurities by hot gravity filtration.
2. Cool the hot, saturated solution gradually.
3. Collect the crystalline material by vacuum filtration.

Impurities (General Q7). Best case: they dissolve in the hot solvent and remain in solution as it cools. Alternative: they do not dissolve in the hot solvent, and a hot gravity filtration removes them.

An effective solvent (General Q8) must:

• Be chemically inert.
• Have a very high capacity for the solute at the solvent's boiling point, and a very low capacity just above its freezing point.
• Dissolve impurities at all temperatures, or not dissolve them at any temperature.

Low recovery, apart from mechanical losses:

• Too much solvent was used.
• Crystallization occurred during hot gravity filtration.
• Vacuum filtration was done before crystallization was complete.

Unit 2 Q2, Q3

Two ways to raise recovery: cool to a lower temperature, or use less solvent. You cannot use less than the minimum volume, or the sample will not all dissolve.

General Q9, Q10, Q12–15

Units 3/4 Q1, Q2

Equations.

Steps.

1. If given weight %, assume 100 g total.
2. Convert to moles with molar masses.
3. Find the mole fraction of each in the liquid.
4. Multiply by the pure vapor pressure.
5. Add for the total.
6. Divide each partial pressure by the total.

Units 3/4 Q3–7

Conversions. Mole fraction × 100 = mole %. For two components, the percentages add to 100.

General Q16; Table 1

Study 1; notes pp. 2–3; lecture 11.1, 11.2

An α-carbon can have zero α-hydrogens. Check each one separately.

Course pKa values. Use these on the exam.

A lower pKa means a stronger acid. Each unit is a factor of 10, so +50 to +20 is 30 orders of magnitude.

Study 2; notes pp. 4–7; Q1, Q2

Keto to enol.

1. Mark the carbonyl carbon. It becomes the alkene carbon holding OH.
2. Find each α-carbon with at least one hydrogen.
3. For each one, draw C=C between it and the old carbonyl carbon.
4. Move one α-hydrogen to oxygen.
5. Check for E and Z forms.
6. Cross out duplicates. It is fine to draw both first.

Enol to keto.

1. The alkene carbon with OH becomes the carbonyl carbon.
2. The other alkene carbon is the α-carbon. It gains one H.
3. Check that α-carbon for a new stereocenter.

Mechanism: three lines (notes p. 5).

Study 3; notes p. 8; Q3, Q4

Drawing enolates.

1. Find each α-carbon with at least one hydrogen. Each distinct one gives a distinct enolate.
2. Remove H⁺. Put a lone pair and a negative charge on that carbon.
3. Draw the second resonance structure, with C=C and O⁻.
4. Check for E and Z forms.

Which enolate is more stable (practice Q4). Look at the resonance structure with O⁻. It contains an alkene. More alkyl groups on that alkene means more stable. Answer to Q4: A is more stable than B.

Study 4; notes pp. 9–10; Q5

Base-promoted halogenation.

• Each H replaced uses 1 NaOH and 1 X₂. It forms 1 H₂O and 1 NaX.
• An α-carbon with no hydrogens does not change.
• Enolate attack on X₂: the nearby anion distorts the X–X electrons, so the closer X becomes δ+. This is an induced dipole.

Haloform reaction.

• Works for all methyl ketones and acetaldehyde. No other aldehyde works.
• Uses 4 NaOH and 3 X₂ per molecule.
• Products: chloroform (CHCl₃), bromoform (CHBr₃) or iodoform (CHI₃), plus a carboxylate.

The four stages.

1. Three halogenation cycles replace the CH₃ hydrogens. This is six steps. It uses 3 NaOH and 3 X₂.
2. With no α-hydrogens left, the fourth OH⁻ acts as a nucleophile and attacks the δ+ carbonyl carbon.
3. C=O reforms. The C–C bond breaks and ⁻CX₃ leaves.
4. ⁻CX₃ takes the proton from the carboxylic acid, giving CHX₃ and the carboxylate.

Study 5; notes pp. 11–15; Q6, Q7, Q8

Less substituted: speed decides. Call the less substituted carbon a and the other b. Their enolates are A and B. Bulky LDA reaches the hydrogens on a more easily. A forms faster, so A > B. At −78 °C the ratio is locked in. RX converts A to the major product.

More substituted: stability decides. With less than 1 equivalent, some ketone stays unreacted. A > B at first. On warming, that leftover ketone acts as a catalyst and lets A and B interconvert. The mixture reaches equilibrium, which favors the more stable B. RX converts B to the major product.

Intramolecular version. Add no RX. LDA forms the enolate at an α-carbon that has a hydrogen. The enolate attacks the carbon bearing the halogen in the same molecule. Count ring atoms by the shortest path. Rings of 5 and 6 form easily.

Study 6; notes pp. 16–23; Q9, Q11, Q12, Q14, Q16

Mechanism: three steps.

1. OH⁻ removes an α-hydrogen. The enolate concentration is low.
2. The enolate α-carbon attacks a δ+ carbonyl carbon of a neutral molecule. This forms the new C–C bond and an O⁻.
3. Water protonates O⁻ to give OH. This returns the OH⁻ catalyst.

Reading a product. The new bond joins the α-carbon and the β-carbon. The β-carbon was the attacked carbonyl carbon. Its OH was the carbonyl oxygen.

Counting products: the "fates" method.

1. List every possible enolate. Each distinct α-carbon with a hydrogen gives one.
2. List every carbonyl compound that can be attacked.
3. Pair each enolate with each carbonyl compound. Each pair gives one product, ignoring stereoisomers.
4. Draw each. Copy the groups on the attacking carbon and the attacked carbon. Add OH to the attacked carbon.

Intramolecular aldol.

1. Label every α-carbon. Cross out those with no hydrogens.
2. Draw each remaining enolate.
3. Sketch a dashed arrow to the other carbonyl carbon.
4. Number from the attacking carbon (1) to the attacked carbon by the shortest path.
5. If the count ends at 5 or 6, the path works. Draw the product.
6. Otherwise, that enolate only returns to starting material.
7. If no path works, write NR. There is no intermolecular product.

Directed aldol.

1. LDA (1 equivalent), dry THF, −78 °C. Convert one carbonyl compound fully to its enolate.
2. Add the other carbonyl compound.
3. Add H₃O⁺.

Notes pp. 24–27; Q10, Q13, Q15

How to answer question 2.

1. Draw the dehydrated product on paper.
2. List the three groups on the new alkene, other than the carbonyl group. Call them X, Y and Z.
3. If at least one connects through an sp or sp² atom, the answer is yes. Typical groups are a phenyl ring, an alkene or an alkyne.

Mechanism: two steps.

1. OH⁻ removes the α-hydrogen, forming an enolate.
2. The lone pair forms the C=C, and OH⁻ leaves from the β-carbon. This returns the catalyst.

Geometry. Open-chain products can be cis or trans (Z or E). Trans has less steric strain, so it is the major product.

Study 7; notes pp. 28–29; Q17, Q18

Recognize it. Sodium or potassium alkoxide matching the ester, in the matching alcohol, then H₃O⁺.

Why not NaOH? Hydroxide would saponify the ester. That is what NaOH did to methyl salicylate in Unit 2.

Mechanism.

1. RO⁻ removes an α-hydrogen. Enolate concentration is low. The ester's pKa is about +25.
2. The enolate attacks the carbonyl carbon of a second ester. This forms the new C–C bond and an O⁻.
3. A lone pair on O⁻ reforms C=O, and RO⁻ leaves. This is unique to the Claisen.
4. RO⁻ removes the hydrogen between the two carbonyl groups. Its pKa is about +10. The equilibrium lies far to the product side, and the reaction rests here.
5. Step 2: H₃O⁺ protonates the carbon, giving the neutral β-ketoester.

A new stereocenter forms as a racemic mixture.

2 exam questions: identify the item shown

Fast sorting rules:

The letters come from the Unit 1 lecture for the Unit 1 reflux drawing (lecture 1, 26:56). The lectures do not name every letter on the other drawings. The exam asks you to identify the item, not each letter.

Sets up all lab units

A pure compound in this course means no detectable impurities. It does not mean zero impurities. A mole contains about 10²³ molecules, so a few stray molecules are always present. They are too few to change the melting behavior (lecture 2, 02:59). An analytical chemist would use a far stricter standard.

Notes sections I and II; lab technique questions

Why bumping happens. New glassware is very smooth inside. Bubbles need rough spots to form. Without them, a liquid can superheat and then boil violently without warning. Boiling stones are inert and very rough, so bubbles form on them gently.

Rule of thumb from the lecture. Reaction speed roughly doubles for every 10 °C rise (lecture 1, 14:29).

Ice bath method. Put the ice in first, then add just enough water to cover it (lecture 1, 24:07).

Water direction. Water enters the lower port of the condenser and leaves by the upper port, so the jacket stays full (lecture 1, 26:56 and 55:30).

General Q1; notes section III

Unit 1 as you run it.

1. Reflux nutmeg, boiling stones and acetone for about an hour.
2. First filtration, on a Büchner funnel. The undissolved solid stays in the funnel. The solution with trimyristin passes through.
3. Cool the solution. Trimyristin crystallizes.
4. Second filtration, on a smaller Hirsch funnel. Collect the off-white solid.
5. Dry until the next lab session. Measure the mass and the melting point.

Quantify the result. Percent of nutmeg = (mass of trimyristin ÷ mass of nutmeg) × 100. Practice Q4 uses the same form: (0.14 g ÷ 10.06 g) × 100 = 1.4%.

General Q2; Unit 1 Q5; Unit 2 Q4, Q6, Q7, Q8

Worked rankings from the practice questions.

The solutions box the answers to Unit 1 Q5 and Unit 2 Q6 and give no reason. The reasons above are applied from the course rules.

Solved questions, Unit 1. Q1: the less hydrocarbon-like alcohol is more soluble. Q2: the compound with two hydrogen-bond donors and two hydrogen-bond acceptors beats the one with one of each. Count both when carbons are similar.

General Q3; Unit 1 Q1, Q2; Table 1

The exam logic. A contaminated sample must melt below the pure compound and with a range over 2 °C.

A purified sample can jump a lot. A sample melting at 110–115 °C is impure. After recrystallizing, 178–180 °C is possible. The values rise and the range tightens as impurities leave (lecture 1, 40:06).

Identifying a compound by mixing. Compound X is either benzoic acid or ortho-fluorobenzoic acid. Both melt at 122 °C, so the melting point alone cannot decide.

1. Mix X with authentic benzoic acid. Call this sample A.
2. Mix X with authentic ortho-fluorobenzoic acid. Call this sample B.
3. Take the melting points of A and B.
4. The sample that still melts sharply at 122 °C contains two portions of the same compound. That tells you X.
5. The other sample is a mixture. It melts below 122 °C with a range over 2 °C.

Solved Q5 applies the same method to two vials, using authentic pipacycline only.

Voltage example. For 120 °C: (120 + 52.5) ÷ 4.45 = 38.8 volts. Use a dial setting of about 3.9.

Unit 1 Q3, Q4; Unit 2 Q1

Notes example. 27.0 g BF₃ and 9.00 g NaBH₄ react: 4 BF₃ + 3 NaBH₄ → 3 NaBF₄ + 2 B₂H₆.

1. Moles BF₃ = 27.0 ÷ 67.8062 = 0.398. Moles NaBH₄ = 9.00 ÷ 37.83268 = 0.238.
2. 0.398 mol BF₃ needs 0.299 mol NaBH₄. Only 0.238 mol is present, so NaBH₄ is limiting.
3. Moles B₂H₆ = 0.238 × 2 ÷ 3 = 0.159.
4. Mass = 0.159 × 27.66982 = 4.40 g.

Unit 1 Q3. 350 g of A (269.18 g/mol) gives 1.300 mol. One mole of A gives one mole of B (192.30 g/mol). Theoretical yield = 250 g. The excess LiAlH₄ means A is limiting.

Unit 2 Q1. H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O.

1. Moles NaOH = 8.40 M × 0.265 L = 2.226 mol.
2. Moles H₂SO₄ = 2.226 ÷ 2 = 1.113 mol, because sulfuric acid is diprotic.
3. Volume = 1.113 ÷ 6.75 M = 0.1649 L = 165 mL.

Solved questions, Unit 1.

Table 1 structures; ester chemistry

Why two steps. Base breaks the ester and leaves a carboxylate salt, which is ionic and stays dissolved. Acid adds a proton and gives the neutral carboxylic acid. The neutral acid is much less soluble in water, so it precipitates. This is the solubility ranking in action.

General Q4–8, Q11; Unit 2 Q5

Testing a solvent by trial and error (notes item 12). Unit 2 Q5 tests this.

Unit 2 Q5. Compound E does not dissolve in n-hexanol cold, or at its boiling point. It fails at step 2, so n-hexanol is not suitable.

Why seed crystals work. New glassware is smooth, so crystals have no ledge to grow on. A seed crystal gives them a place to start.

Why set aside crude. The notes give seed crystals. The lecture adds a second use: you can compare the crude melting point with the purified one (lecture 2, 24:37).

Why cool in ice. Solubility drops as temperature drops, but it never reaches zero. The colder the solution, the less stays dissolved. The usable range runs from the solvent's boiling point down to just above its freezing point. For water, that is 100 °C to about 0 °C.

How to cool slowly. Set the flask aside undisturbed. Benchtop cooling is slow enough. Putting a hot flask straight into ice is too fast.

Why water is the first choice. It is cheap, plentiful and non-toxic. It can be too polar for some compounds. Unit 2 uses water. The large Unit 5 product needs ethanol.

Unit 2 Q2, Q3

Worked example from the notes. 1.20 g of salicylic acid. Solubility: 1 g/460 mL at 25 °C, 1 g/15 mL at 100 °C.

1. Volume = 1.20 g × 15 mL/g = 18.0 mL.
2. Stays dissolved at 25 °C = 18.0 mL × (1 g ÷ 460 mL) = 0.04 g.
3. Crystallizes = 1.20 − 0.04 = 1.16 g.
4. Maximum recovery = (1.16 ÷ 1.20) × 100 = 96.7%.

Using 26.5 mL instead. 26.5 ÷ 460 = 0.06 g stays dissolved. 1.14 g crystallizes. Recovery is 95%.

The limit. 1.20 × 460 = 552 mL. At this volume, everything dissolves and nothing crystallizes at 25 °C.

Practice Q2. 431 g × (500 mL ÷ 223 g) = 966 mL.

Practice Q3. The student used 2125 mL, far more than needed.

1. Stays dissolved = 2125 mL × (75.1 g ÷ 500 mL) = 319.2 g.
2. Crystallizes = 431 − 319.2 = 111.8 g.
3. Recovery = (111.8 ÷ 431) × 100 = 26%.

Mixtures: the solved question set. Each solute behaves as if the other is not there.

General Q9, Q10, Q12–15

Why the boiling point rises. Vapor removes more of the volatile component. The flask moves toward the less volatile one, so more heat is needed to keep it boiling. If you hold the heat constant, the distillation slows and stops. Watch the drop rate and raise the setting as needed (lecture 3, 53:05).

Why stop before dryness. Many organic solvents, such as diethyl ether, can form explosive peroxides over time. Distilling to dryness leaves the peroxides behind. Here, "dryness" means no liquid at all, not "no water".

Why the thermometer position matters. The bulb senses the vapor as it turns into the condenser. If it sits too high or too low, the reading is wrong.

Methanol and water. Methanol boils at about 65–66 °C. Water boils at 100 °C. The Unit 3 and 4 unknown is 30 mL of a methanol–water mixture. You plot temperature against total volume of distillate.

Units 3/4 Q1, Q2

Notes example: 18.0% benzene by weight at 40 °C.

Benzene goes from 0.205 in the liquid to 0.444 in the vapor.

Practice Q1: 70 wt % benzene. Moles 0.896 and 0.326. Mole fractions 0.733 and 0.267. Partial pressures 134.6 and 15.8 torr. Vapor: 89.5 mole % benzene, 10.5 mole % toluene.

Practice Q2: ratio only. 6 mol A, 8 mol B, and A has 5 times the vapor pressure of B. Call B's vapor pressure 1.

1. N_A = 6 ÷ 14 = 0.43. N_B = 0.57.
2. P_A = 0.43 × 5 = 2.15. P_B = 0.57. Total = 2.72.
3. Vapor A = 0.79. Vapor B = 0.21.

Answer: 79 mole % A, 21 mole % B. The actual pressure cancels, so you do not need it.

Solved Q4. 1 mol ethanol and 3 mol 1-butanol, with ethanol's vapor pressure 3 times that of butanol. N = 0.25 and 0.75. P = 0.75 and 0.75. Vapor is 50 mole % each.

Solved Q5. 50 wt % benzene. N = 0.541 and 0.459. Vapor is 78.6 mole % benzene, 21.4 mole % toluene.

Units 3/4 Q3–7

Lecture examples.

Lecture check: methanol–water that boils at 79 °C has about 0.26 mole fraction methanol (lecture 3, 36:58).

Midterm practice, with the course's arrows.

Solved questions. Q1: one plate from X gives 0.64 A. Q8: 54 mole % D gives vapor of 75.5 mole % D. Q9: 70 mole % C begins to boil near 200 °C. Q10: two plates end at 59 mole % D. Q11: three plates end at 53 mole % E.

Why the steps move this way. A horizontal line links a liquid to the vapor in equilibrium with it, at one temperature. Moving down from the vapor point means condensing that vapor. Each cycle moves you toward the more volatile component.

General Q16; Table 1

What happens inside the column.

1. Hot vapor rises and meets cooler condensate coming down.
2. Heat transfers from vapor to condensate.
3. The vapor partly condenses. The condensate partly vaporizes.
4. The new vapor is richer in the more volatile component.
5. This repeats up the column. Ideally, only pure methanol reaches the thermometer.

What sets the plate count. The lecture links it to the column's vertical path length and its internal surface area. The beads provide a large surface area.

Why insulate in Unit 4. The fractional setup has a much larger surface area, so it loses more heat. Wrap from the flask up to the thermometer, first with cotton and then with foil.

What you see in the data.

Example of the second use of simple distillation. Boil acetone away from trimyristin. The solvent distills and the nearly non-volatile solid stays.

Study 1; notes pp. 2–3; lecture 11.1, 11.2

Why resonance explains the acidity.

1. Removing H⁺ from an alkane leaves a full negative charge on one carbon. Carbon is reluctant to hold this.
2. Removing an α-hydrogen also seems to put the charge on carbon. That drawing is only half the picture.
3. A second resonance structure puts the charge on oxygen, which holds negative charge better.
4. Neither drawing exists alone. The real anion is a hybrid. If each contributed equally, carbon and oxygen would each carry −½.
5. A delocalized charge is more stable, so the acid is stronger.

Curved arrows: the carbon lone pair moves in to form C=C, and the C=O π electrons move onto oxygen.

Study 2; notes pp. 4–7; Q1, Q2

Why OH⁻ must come back. The reaction is base-catalyzed. The final step has to regenerate the catalyst. Water is the only proton source available, so water does the last step. The catalyst to reform is OH⁻, not H₃O⁺ (lecture 11.2, 29:57).

Step 2 is a redrawing, not a chemical step. Use a double-headed resonance arrow.

At equilibrium. Interconversion never stops. Only the amounts stop changing.

New stereocenter. If the new α-carbon has four different groups, you get R and S in a 1:1 ratio. The enol was achiral. Draw both with wedge and dash if asked (notes p. 7, example 2).

Why a ring limits geometry. A small ring is not large enough to hold the other alkene isomer. Open chains can rotate, so they can give both E and Z (lecture 11.2, 43:29).

Deuterium exchange (solved Q2). NaOD behaves like NaOH, and D₂O behaves like H₂O. OD⁻ removes an α-hydrogen. D₂O delivers D to the α-carbon. H is replaced by D.

Study 3; notes p. 8; Q3, Q4

Worked Q3. The ketone has two distinct α-carbons with hydrogens. The answer shows two enolates, each with two resonance structures.

Why protons move this way. Protons move from the stronger acid to the weaker acid. Compare the pKa of the carbonyl compound with the pKa of the base's conjugate acid. Ketone +20 against diisopropylamine 36: the equilibrium lies fully toward the enolate. Ketone +20 against water 15.74: it lies toward the ketone.

Solution contents in aqueous base. Mostly keto form, a small amount of enol, and a small amount of enolate.

Alkene stability order (lecture 11.3, 25:38). Tetrasubstituted is most stable, then trisubstituted, then disubstituted, then monosubstituted, then ethylene.

Study 4; notes pp. 9–10; Q5

Practice Q5. The bicyclic ketone is not a methyl ketone, so there is no cleavage. It has three α-hydrogens, so the product has three iodines.

Why it does not stop after one halogen. The halogen pulls electron density away, so the remaining α-hydrogens become more acidic. They react faster (OpenStax 22.6).

Why the fourth NaOH changes role. Hydroxide usually acts as a base. Once no α-hydrogens remain on that carbon, it cannot. It acts as a nucleophile instead.

Iodoform. A yellow solid with a medicinal odor. Forming it suggested a methyl ketone or acetaldehyde.

Balance sheet up to the CX₃ stage. 3 H₂O and 3 NaX have formed. All X₂ is used.

Study 5; notes pp. 11–15; Q6, Q7, Q8

Why not use a carbanion? The methyl anion is an even stronger base than LDA. It is also small, so it attacks the carbonyl carbon instead of removing a proton. This is the Grignard-type addition from earlier courses. A strong base is not enough. It must also be a poor nucleophile.

Why B is more stable. Compare the O⁻ resonance structures. In the notes example, B has three groups on its alkene and A has two.

Why not tertiary halides. Too much crowding around the carbon with the leaving group. The nucleophile cannot reach it.

New stereocenters. A new stereocenter forms as a racemic mixture, because both faces of the enolate are attacked equally.

Notes p. 15 examples. Item i is a page 14 scenario, because it says less than 1 equivalent. Item iii is a page 13 scenario. Item ii is intramolecular. One α-carbon has no hydrogens, so LDA has only one option. The enolate attacks the C–Br carbon and closes a new ring, giving a compact bicyclic ketone.

Practice Q6, Q7, Q8. Q6: 1 equivalent, ethyl goes on the less substituted carbon. Q7: less than 1 equivalent and warming, ethyl goes on the more substituted carbon. Q8: one α-carbon has H, and a five-membered ring forms, giving a spiro ketone.

Solved Q6ii. Only one α-carbon has H, so only one site reacts.

Real use. The gymnomitrol synthesis (Welch, 1980) used NaH (less than 1 equivalent), warming, then CH₃I. This is the more substituted scenario with a different base.

Study 6; notes pp. 16–23; Q9, Q11, Q12, Q14, Q16

Worked ring counts.

Planning a directed aldol backward.

1. Find the C=O, the α-carbon, and the β-carbon (the C–OH carbon).
2. Cut the bond between α and β.
3. The β side becomes the carbonyl partner. Change C–OH to C=O.
4. The α side becomes the enolate. Put H back on the α-carbon to find the starting compound.

Worked example (notes p. 23). Target: CH₃C(O)CH₂C(OH)(CH₃)CH₂CH₃. The enolate comes from acetone. The partner is butanone. Conditions: acetone with 1. LDA (1 equivalent), −78 °C, dry THF; 2. butanone; 3. H₃O⁺.

Alkylation against aldol.

Three LDA patterns. Then RX: alkylation. Then a carbonyl compound and H₃O⁺: directed aldol. Nothing else, with a halogen in the molecule: intramolecular SN2.

No α-hydrogens, no enolate. Formaldehyde and benzaldehyde can only be attacked.

Notes pp. 24–27; Q10, Q13, Q15

Paired practice questions.

Why OH⁻ can leave here. Hydroxide is normally a poor leaving group, so ordinary alcohols do not dehydrate in base. In an aldol product, the enolate lone pair pushes it out (OpenStax 23.3).

Worked synthesis: E,E-dibenzalacetone (Unit 5). 2 benzaldehyde + 1 acetone, with NaOH, H₂O and ethanol, at room temperature. Ethanol only dissolves the benzaldehyde.

1. Only acetone has α-hydrogens, so only acetone forms an enolate.
2. It prefers to attack benzaldehyde. An aldehyde carbonyl carbon is more accessible, and it has a larger δ+.
3. The process repeats on acetone's other methyl group, using the second benzaldehyde.
4. Each side has X, Y, Z = H, H, phenyl. The ring connects through an sp² carbon.
5. Both sides dehydrate at room temperature and give the more stable E,E isomer. The reaction took about 15 minutes.

Study 7; notes pp. 28–29; Q17, Q18

Why two α-hydrogens? One is lost to form the enolate. The second is needed for mechanism step 4, the very favorable deprotonation that drives the reaction. With only one, that step cannot happen (practice Q17).

Why must OR match? The lecture gives the rule, not the reason. Alkoxide can attack the ester C=O. The same OR leaves the ester unchanged. A different OR would swap the ester group and give mixed products.

Esters have one α-carbon, so a single ester gives one enolate and one Claisen product. Practice Q18 follows this pattern.

Mixed Claisen, solved Q6i. Methyl butanoate and methyl benzoate with excess NaOCH₃. Methyl benzoate has no α-hydrogens. One enolate and two partners give two β-ketoesters. The method matches the fates method.

Aldol against Claisen.