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Organic Chemistry II Midterm Study Guide

Exam: Wednesday, October 7, 2026 · 90 min · closed-book · mixed format Window: Wed Sep 23 to Tue Oct 6 (14 study days, about 48 hours total)

Your details were complete, so I haven't asked clarifying questions. I've made these assumptions:

• Format: "Mixed" means multiple choice, predict-the-product, mechanism arrows, a short synthesis and spectroscopy structure problems.
• Topic weights: The weights below are typical for OChem II midterms. If your professor has posted a past exam or a topic breakdown, swap them in.

1. Diagnostic & Priorities

Highest-leverage move: Master the SN/E decision tree first (Days 1–2). It directly powers about 30–35% of the exam, because alcohol and ether reactions use the same logic.

2. Day-by-Day Study Schedule

Methods used in the schedule:

• BPR (blank-page recall): write everything you know about a topic from memory, then check your notes.
• Drill: timed practice problems.
• Interleave: mixed-topic sets in shuffled order.
• FC: flashcards (Anki or paper), practiced in both directions.

Error log: Keep one throughout. Each entry records the question, your wrong answer, the correct answer and the type of error (e.g., "forgot bulky base → Hofmann").

3. Key Concepts Cheat Sheet

A. SN1 / SN2 / E1 / E2 (Priority #1)

Core idea: First classify the substrate (methyl, 1°, 2°, 3°). Then classify the reagent: strong nucleophile, strong base, bulky base, or weak nucleophile/base.

Decision tree:

• Methyl or 1°: SN2. The exception is a bulky base (t-BuOK, LDA), which gives E2.
• 2° substrate:
  • Strong base (NaOH, NaOEt, NaOMe) → E2 major.
  • Good nucleophile that is a weak base (I⁻, Br⁻, CN⁻, N₃⁻, RS⁻) in polar aprotic solvent → SN2.
  • Weak Nu/base in protic solvent → SN1/E1 mixture.
• 3° substrate:
  • Strong base → E2.
  • Weak Nu/solvent → SN1 (E1 grows with heat).
  • Never SN2.

Common mistake: Calling 2° halide + NaOEt an SN2 reaction. Strong base on a 2° or 3° carbon means E2.

Memory hooks:

• "SN2 hates crowds."
• "The number tells you who's in the rate law. '1' = only the substrate. Weak reagents give 1."
• "Bulky base goes for the easy H (Hofmann)."

B. Alcohols & Ethers

Making OH into a leaving group:

• TsCl/pyridine: retention (the C–O bond isn't broken).
• PBr₃ or SOCl₂: 1°/2° alcohols, SN2, inversion.
• HX: 3° alcohols react by SN1, and rearrangements are possible.

Oxidation:

• PCC: 1° → aldehyde.
• H₂CrO₄/Jones: 1° → carboxylic acid; 2° → ketone.
• 3° alcohols: no reaction.

Other key reactions:

• Dehydration: H₂SO₄ with heat gives E1 (Zaitsev).
• Williamson ether synthesis: RO⁻ + methyl or 1° R–X (SN2).
• Ether cleavage: excess HI or HBr with heat.
• Epoxide opening:
  • Basic/strong Nu: attack at the less substituted carbon.
  • Acidic: attack at the more substituted carbon.
  • Both give anti opening.

Common mistake: Running a Williamson synthesis with a 3° halide. You get E2 (an alkene), not an ether. The bulky piece must be the alkoxide.

Memory hook: "Base goes Less, Acid goes More."

C. Aldehydes & Ketones (Nucleophilic Addition)

Core idea: Nucleophiles attack the δ⁺ carbonyl carbon to give a tetrahedral alkoxide, which is then protonated. Aldehydes react faster than ketones because they are less hindered and have more δ⁺ character.

Must-know reactions:

• NaBH₄ or LiAlH₄: aldehyde → 1° alcohol; ketone → 2° alcohol.
• RMgBr then H₃O⁺: formaldehyde → 1°, aldehyde → 2°, ketone → 3° alcohol.
• 2 ROH with H⁺ (remove water) → acetal, which works as a protecting group because it is stable to base and hydride.
• R–NH₂ at pH 4–5 → imine (C=N–R).
• R₂NH → enamine.
• Ph₃P=CR₂ (Wittig) → alkene.
• HCN or NaCN → cyanohydrin.

Common mistake: Adding a Grignard reagent to a molecule that has an O–H, N–H or COOH. The Grignard is destroyed by the acid–base reaction, so protect the group first.

Memory hook: Acid-catalyzed acetal and imine formation both follow PADPED: Protonate, Add, Deprotonate, Protonate, Eliminate, Deprotonate.

D. Carboxylic Acids & Derivatives

Core idea: Nucleophilic acyl substitution goes by addition–elimination through a tetrahedral intermediate. A derivative can only be converted down the reactivity ladder.

Reactivity ladder (most to least reactive):

1. Acid chloride
2. Anhydride
3. Ester ≈ acid
4. Amide
5. Carboxylate

Must-know reactions:

• Fischer esterification: RCOOH + R'OH with H⁺ catalyst ⇌ ester + H₂O. It is an equilibrium, so use excess alcohol or remove water.
• Amide formation: acid chloride + 2 equiv amine. The 2nd equivalent traps the HCl.
  • Acid + amine directly gives an ammonium salt. It needs strong heat or DCC to form the amide.
• Saponification: ester + NaOH → carboxylate + alcohol. It is irreversible because the carboxylate is deprotonated.
• LiAlH₄:
  • Ester → 1° alcohol.
  • Amide → amine (the C=O becomes CH₂).
• Excess RMgBr + ester: adds twice to give a 3° alcohol.

Common mistake: Trying to make an acid chloride from an amide. That goes uphill on the ladder, so it won't happen.

Memory hook: "Clumsy Ants Eat Apples" (Chloride → Anhydride → Ester → Amide). Water flows downhill.

E. Aromatic Substitution (Priority #2)

EAS mechanism:

1. An electrophile is attacked by the ring, giving the arenium ion (sigma complex). This step is rate-determining.
2. The ring loses H⁺, which restores aromaticity.

The 5 core EAS reactions:

Useful follow-ups:

• Reduce the acyl C=O to CH₂: Clemmensen (Zn(Hg)/HCl) or Wolff–Kishner (H₂NNH₂/KOH).
• Reduce NO₂ to NH₂: Fe/HCl or Sn/HCl.

Directing groups:

Rules when two substituents are present:

1. The strongest activator wins.
2. Don't substitute between two groups that are meta to each other (steric hindrance).
3. Para usually beats ortho.

Friedel–Crafts fails on:

• Rings with a meta director (they are too deactivated).
• Rings bearing –NH₂ or –NHR (the nitrogen complexes with AlCl₃).

NAS (nucleophilic aromatic substitution):

• Needs a strong EWG (NO₂) ortho or para to the leaving group.
• Goes through a Meisenheimer anion.
• F is the fastest leaving group, because the addition step is rate-limiting.
• Benzyne (NaNH₂) needs no EWG but can give product mixtures.

Common mistake: Choosing the wrong step order in a synthesis. The current substituent directs the next one, so install the directing group you need first.

Memory hooks:

• "Lone pair on the attached atom → o/p."
• "δ⁺ or a π-bond to O/N on the attached atom → meta."
• "Halogens are the oddballs: they deactivate but still direct o/p."

F. Spectroscopy (Priority #3)

IR peaks:

¹H NMR chemical shifts (ppm):

Three kinds of NMR information:

1. Number of signals = number of sets of equivalent H.
2. Integration = how many H the signal itself represents.
3. Multiplicity (the n+1 rule) = n non-equivalent H on adjacent carbons, plus 1.

Splitting rules:

• Equivalent H don't split each other.
• O–H usually appears as a broad singlet and doesn't split its neighbors.

Must-know splitting fingerprints:

Degrees of unsaturation (DoU): DoU = (2C + 2 + N − H − X)/2. DoU ≥ 4 suggests a benzene ring.

Common mistake: Counting the H on the same carbon, or the O–H, as neighbors when applying n+1.

• Multiplicity tells you about the neighbors.
• Integration tells you about the signal itself.

Memory hooks:

• "Neighbors plus one."
• "Ethyl is 3–2 (t–q), isopropyl is 6–1 (d–sept)."

4. Practice Set (answers at the end of each set)

Set A: SN1/SN2/E1/E2

1. (Easy) 1-bromobutane + NaCN in DMSO. Mechanism and product?
2. (Easy) 2-bromo-2-methylpropane in ethanol, 25 °C. Major mechanism and product?
3. (Med) 2-bromobutane + NaOEt/EtOH, heat. Major product?
4. (Med) 2-bromo-2-methylbutane + t-BuOK. Major product?
5. (Med) (R)-2-bromobutane + NaN₃ in DMF. Product with stereochemistry?
6. (Med) For an SN2 reaction, [RX] is tripled and [Nu⁻] is doubled. How does the rate change? What if the reaction were SN1?
7. (Hard) Rank by SN2 rate with NaI/acetone: bromomethane, 1-bromobutane, 2-bromobutane, 1-bromo-2,2-dimethylpropane.
8. (Hard) cis- and trans-1-bromo-2-methylcyclohexane are each treated with NaOEt. Give the major E2 product of each.

Answers A:

1. SN2 (1° substrate, strong Nu, aprotic solvent) → pentanenitrile.
2. SN1 (3° substrate, weak Nu, protic solvent, no heat) → tert-butyl ethyl ether, with minor E1 alkene.
3. E2 (2° substrate + strong base) → trans-2-butene (Zaitsev; trans is more stable).
4. E2 Hofmann (bulky base) → 2-methyl-1-butene.
5. SN2 with inversion → (S)-2-azidobutane. N₃ takes Br's top priority, and the geometry flips.
6. SN2: rate goes up ×6. SN1: rate goes up ×3, because only [RX] is in the rate law.
7. Fastest to slowest: bromomethane > 1-bromobutane > 2-bromobutane > neopentyl bromide. Neopentyl is 1° but its β-branching blocks backside attack.
8. E2 needs axial Br and an axial anti H.
   • cis: The chair with Br axial puts CH₃ equatorial, so the C2–H is axial. Elimination toward C2 gives 1-methylcyclohexene (Zaitsev).
   • trans: Br is axial only when CH₃ is also axial, so the C2–H is equatorial. The only anti H is on C6, giving 3-methylcyclohexene (non-Zaitsev, and slower).

Set B: Alcohols & Ethers

1. (Easy) 1-butanol + PCC? 1-butanol + Na₂Cr₂O₇/H₂SO₄?
2. (Easy) Best Williamson route to tert-butyl methyl ether?
3. (Med) 3-methyl-2-butanol + HBr. Major product?
4. (Med) 2,2-dimethyloxirane with (a) NaOCH₃/CH₃OH and (b) CH₃OH/H₂SO₄. Products?
5. (Hard) (R)-2-butanol + TsCl/pyridine, then NaBr in DMF. Final product with stereochemistry?

Answers B:

1. PCC gives butanal. Chromic acid gives butanoic acid.
2. (CH₃)₃CO⁻Na⁺ + CH₃I. The reverse pairing (CH₃O⁻ + t-BuBr) gives E2 and isobutylene.
3. 2-bromo-2-methylbutane. The 2° cation undergoes a 1,2-hydride shift to a 3° cation (SN1).
4. (a) Base attacks the less hindered carbon → 1-methoxy-2-methyl-2-propanol. (b) Acid attacks the more substituted carbon → 2-methoxy-2-methyl-1-propanol.
5. Tosylation keeps the configuration: (R)-tosylate. SN2 with Br⁻ inverts it: (S)-2-bromobutane.

Set C: Aldehydes & Ketones

1. (Easy) Cyclohexanone + CH₃MgBr, then H₃O⁺?
2. (Easy) Rank by reactivity toward nucleophilic addition: acetone, formaldehyde, acetaldehyde.
3. (Med) Acetone + CH₃NH₂ with catalytic acid. Product, and why pH ~4–5?
4. (Med) Cyclohexanone + Ph₃P=CH₂?
5. (Hard) Convert methyl 4-oxopentanoate to 5-hydroxypentan-2-one.

Answers C:

1. 1-methylcyclohexanol (a 3° alcohol).
2. Formaldehyde > acetaldehyde > acetone. Alkyl groups add steric bulk and donate electrons, which lowers reactivity.
3. The imine (CH₃)₂C=NCH₃, formed by PADPED.
   • Too acidic: the amine is protonated and stops being a nucleophile.
   • Too basic: the O–H can't be protonated to leave as water.
4. Methylenecyclohexane (Wittig).
5. The goal is to reduce the ester but keep the ketone:
   1. Protect the ketone as an acetal with ethylene glycol and TsOH.
   2. Reduce the ester to CH₂OH with LiAlH₄.
   3. Remove the acetal with H₃O⁺.
   • LiAlH₄ alone would reduce both carbonyls.

Set D: Carboxylic Acids & Derivatives

1. (Easy) Rank by reactivity: ethyl acetate, acetamide, acetyl chloride, acetic anhydride.
2. (Easy) Acetic acid + ethanol with H₂SO₄. Product? Give two ways to raise the yield.
3. (Med) Benzoyl chloride + 2 equiv diethylamine. Product, and why 2 equivalents?
4. (Med) Ethyl benzoate + excess CH₃MgBr, then H₃O⁺?
5. (Med) N-methylacetamide + LiAlH₄, then H₂O?
6. (Hard) Fischer esterification uses ¹⁸O-labeled ethanol. Does the label end up in the ester or in the water? Why?

Answers D:

1. Acetyl chloride > acetic anhydride > ethyl acetate > acetamide.
2. The product is ethyl acetate. Raise the yield with excess ethanol, or by removing water (Dean–Stark trap or drying agent).
3. N,N-diethylbenzamide. The 2nd equivalent of amine neutralizes the HCl. Otherwise the HCl protonates the amine and it can no longer act as a nucleophile.
4. 2-phenyl-2-propanol (+ ethanol). Two CH₃ groups add: the first displaces OEt to give a ketone, and the second adds to that ketone.
5. N-methylethylamine (CH₃CH₂NHCH₃). The C=O becomes CH₂.
6. The label ends up in the ester. The alcohol oxygen is the nucleophile, and the acid's O–H is what leaves as water.

Set E: Aromatic Substitution

1. (Easy) Nitration of anisole. Major product?
2. (Easy) Br₂/FeBr₃ on nitrobenzene?
3. (Med) Rank by EAS rate: nitrobenzene, phenol, chlorobenzene, toluene, benzene, benzaldehyde.
4. (Med) Nitration of 4-methylanisole. Major product?
5. (Med) Br₂/FeBr₃ on 4-nitrotoluene?
6. (Med) Benzene + 1-chloropropane/AlCl₃ gives mainly isopropylbenzene. Why? How do you make propylbenzene instead?
7. (Hard) Synthesize m-bromoaniline from benzene.
8. (Hard) Which reacts fastest with NaOCH₃: chlorobenzene, 3-nitrochlorobenzene, 4-nitrochlorobenzene or 2,4-dinitrochlorobenzene? What is the product?

Answers E:

1. 4-nitroanisole (para major, ortho minor). OMe is a strong o/p activator.
2. 3-bromonitrobenzene. NO₂ is a meta director.
3. Fastest to slowest: phenol > toluene > benzene > chlorobenzene > benzaldehyde > nitrobenzene.
4. 4-methyl-2-nitroanisole. Substitution goes ortho to OMe, because the stronger activator wins.
5. 2-bromo-4-nitrotoluene. The directing effects cooperate: this position is ortho to CH₃ and meta to NO₂.
6. The 1° carbocation rearranges to a 2° cation by hydride shift. For propylbenzene:
   1. FC acylation with propanoyl chloride/AlCl₃ → propiophenone.
   2. Clemmensen or Wolff–Kishner reduction → propylbenzene.
7. Three steps, in this order:
   1. HNO₃/H₂SO₄ → nitrobenzene.
   2. Br₂/FeBr₃ → m-bromonitrobenzene (NO₂ directs meta).
   3. Fe/HCl, then NaOH workup → m-bromoaniline.
   • If you reduce first, NH₂ directs o/p, which is wrong.
8. 2,4-dinitrochlorobenzene reacts fastest. Its ortho and para NO₂ groups stabilize the Meisenheimer anion. The product is 2,4-dinitroanisole.

Set F: Spectroscopy

1. (Easy) Cyclohexanol vs. cyclohexanone. What IR difference tells them apart?
2. (Easy) How many ¹H NMR signals does p-xylene give? What are their integrations and multiplicities?
3. (Easy) Predict the ¹H NMR of 2-chloropropane.
4. (Med) Predict the ¹H NMR of ethyl acetate.
5. (Med) What is the splitting of the middle CH₂ in 1-bromopropane?
6. (Med) C₄H₈O; IR 1715 cm⁻¹; NMR δ 2.4 (q, 2H), 2.1 (s, 3H), 1.05 (t, 3H). Structure?
7. (Med) C₃H₆O₂; IR very broad 2500–3300 cm⁻¹ and 1710 cm⁻¹; NMR δ 11.5 (br s, 1H), 2.35 (q, 2H), 1.15 (t, 3H). Structure?
8. (Hard) C₈H₁₀; NMR δ 7.2 (m, 5H), 2.6 (q, 2H), 1.2 (t, 3H). Structure?

Answers F:

1. Cyclohexanol shows a broad O–H at ~3300 and no C=O. Cyclohexanone shows a strong C=O at ~1715 and no O–H.
2. 2 signals: 7.0 ppm (s, 4H) and 2.3 ppm (s, 6H).
3. δ ~1.5 (d, 6H) and ~4.1 (septet, 1H).
4. δ 2.0 (s, 3H, CH₃C=O), 4.1 (q, 2H, OCH₂) and 1.2 (t, 3H).
5. It has 3 + 2 = 5 neighbors, so a sextet (assuming similar coupling constants).
6. Butanone. DoU = 1 (the C=O). The singlet CH₃ sits next to C=O, and the ethyl group gives the q + t pair.
7. Propanoic acid. The very broad O–H and the 11.5 ppm signal mark a COOH.
8. Ethylbenzene. DoU = 4, which fits a benzene ring; the 5H aromatic signal means monosubstituted.

Timed Mock Exam Outline (90 min, run twice: Oct 3 and Oct 5)

Rules: Closed-book, with a timer and no phone. Build Mock 2 from different questions than Mock 1.

Parts A, B and E: Pull questions from Sets A–F. Save 2 questions per set that you haven't seen for Mock 2.

Part C and D items (new):

• C1. SN1 of t-BuBr in methanol.
  • Answer: C–Br ionizes to the t-Bu⁺ cation. CH₃OH attacks the cation. Deprotonation gives t-Bu–OMe.
• C2. Acid-catalyzed Fischer esterification of acetic acid with methanol.
  • Answer: PADPED. Protonate the C=O, MeOH adds, proton transfer onto an OH, water leaves, deprotonate.
• D. Benzene → m-nitroacetophenone.
  • Answer: CH₃COCl/AlCl₃ first, then HNO₃/H₂SO₄. The acetyl group directs meta.
  • Trap: nitrating first makes FC acylation fail on the deactivated ring.
• E-bonus. C₄H₈O₂; IR 1740 cm⁻¹; NMR δ 4.1 (q, 2H), 2.0 (s, 3H), 1.25 (t, 3H).
  • Answer: ethyl acetate.
  • Contrast with methyl propanoate: δ 3.7 (s, 3H), 2.3 (q, 2H), 1.1 (t, 3H).
  • The 4.1 q means the CH₂ is on the oxygen.

Scoring:

• ≥ 80%: on track.
• 65–79%: re-drill the top 2 error types.
• < 65%: apply the re-prioritization rule in §6.

5. Technique Toolkit

1. Blank-page mechanism recall. Pick a mechanism name, draw every arrow and intermediate from memory, then check it and circle what you missed. How to apply: do 3 mechanisms at the start of each session before opening your notes.
2. Two-way flashcards, interleaved. Make one card for "reagent → product" and one for "product → reagent needed". Shuffle across all topics. How to apply: 15 minutes a day, and every card you miss goes back into tomorrow's pile.
3. Condition-sorting drills for SN/E and EAS. Write 20 scenario cards (substrate + reagent + solvent/temp) and sort them into SN1/SN2/E1/E2 piles as fast as you can. How to apply: time yourself and aim for under 10 s per card with ≥ 90% accuracy by Oct 1.
4. Exam triage + annotate-before-answer. On pass 1, answer everything you're sure of. On pass 2, do the multi-step questions. In every SN/E question, circle the substrate, box the reagent and underline the solvent/temperature before deciding. How to apply: practice this on both mocks so it becomes automatic.

6. Progress Tracker

Re-prioritization rule:

• Promote a topic to the first block of the next day's session if either is true:
  • You score < 70% on a topic in two consecutive practice sets.
  • The same error type shows up 3+ times in your log.
• Make room for it by taking time from your highest-scoring topic.
• Demote a topic to flashcards-only spaced review once you score ≥ 90% twice in a row.

Exam-Day Calm & Focus Tips

1. Start with a 2-minute brain dump. When the exam begins, jot the decision tree, the directing-group list and the n+1 fingerprints in a margin (if your professor allows scratch writing). That offloads memory and steadies your nerves.
2. Use the 2-minute rule. If you're stuck for 2 minutes, mark the question, move on and come back. Momentum protects your score better than stubbornness.
3. Box breathing when you tense up. Breathe in 4 s, hold 4 s, out 4 s, hold 4 s, and repeat twice. Then re-read the question and circle the substrate, reagent and conditions. This resets both your panic and your focus.

You have 14 well-structured days, which is enough time to turn all three weak areas into strengths. Good luck on October 7!