Learn & Review: Organic Chemistry 2 Final Exam

Organic Chemistry II Practice Test Summary

This document summarizes a practice test covering various topics in Organic Chemistry II, including reaction mechanisms, functional group transformations, spectroscopy, and aromaticity.

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Question 1: Aldehyde Oxidation

• Main Idea: Tollen's reagent (Ag+ in ammonia or Ag2O) oxidizes aldehydes to carboxylic acids.
• Key Points:
  • Aldehydes are oxidized to carboxylic acids.
  • Ketones are generally not oxidized by Tollen's reagent unless they are alpha-hydroxy ketones.
  • The reaction conditions (acidic in step 2) determine the final form of the carboxylic acid (protonated vs. deprotonated).
• Outcome: Under acidic conditions (H3O+), the carboxylic acid is protonated.

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Question 2: Radical Bromination

• Main Idea: Radical bromination using Br2 and UV light selectively replaces benzylic hydrogens.
• Key Points:
  • Br2 with UV light is a radical bromination.
  • This reaction does not occur on the benzene ring itself; it targets benzylic hydrogens.
  • To brominate the benzene ring, Br2 with a Lewis acid (e.g., FeBr3) is required.
  • Radical bromination favors the formation of the most stable radical.
• Stability of Radicals: Tertiary radicals are more stable than secondary radicals.
• Outcome: Bromine selectively replaces the tertiary benzylic hydrogen, forming a tertiary radical, which is more stable.

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Question 3: Acidity and pKa

• Main Idea: The acidity of a proton is determined by the stability of its conjugate base, influenced by electronegativity, resonance, and inductive effects.
• Key Concepts:
  • pKa: A measure of acidity; lower pKa means a stronger acid.
  • Alpha Hydrogens: Hydrogens on the carbon adjacent to a carbonyl group.
  • Electron-Withdrawing Groups (EWGs): Increase acidity (lower pKa).
  • Electron-Donating Groups (EDGs): Decrease acidity (higher pKa).
• Trends:
  • Acidity increases across a period (e.g., C-H < N-H < O-H < F-H).
  • Acidity increases down a group (e.g., C-H < Si-H).
  • Hydrogens on positively charged atoms are highly acidic (e.g., H3O+ vs. H2O).
• Comparison:
  • Hydrogen between two carbonyls: pKa ~ 9.
  • Hydrogen next to one ketone: pKa ~ 20.
  • Hydrogen between a ketone and an ester (with EDG): pKa ~ 11.
  • Carboxylic acid O-H: pKa ~ 4-5.
  • Protonated carboxylic acid: pKa < 0 (highly acidic).
• Outcome: The proton on a protonated carboxylic acid has the lowest pKa (is the most acidic).

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Question 4: IR Spectroscopy

• Main Idea: Infrared (IR) spectroscopy identifies functional groups based on their characteristic absorption frequencies.
• Key Signals:
  • Carboxylic Acid O-H: Very broad, strong peak around 2500-3300 cm⁻¹.
  • Carbonyl (C=O): Strong peak around 1700 cm⁻¹.
  • Alcohol O-H: Broad peak around 3300-3500 cm⁻¹, less broad than carboxylic acid O-H.
  • C≡C (Alkyne): Peak around 2200 cm⁻¹.
  • C=C (Alkene): Peak around 1600 cm⁻¹.
  • C-O stretch (Alcohol/Ether): Around 1000-1150 cm⁻¹.
  • C-O stretch (sp² hybridized carbon, e.g., carboxylic acid): Around 1200 cm⁻¹.
• Analysis:
  • A broad peak at 2500-3300 cm⁻¹ and a carbonyl peak at 1700 cm⁻¹ strongly indicate a carboxylic acid.
  • A C≡C peak at 2200 cm⁻¹ distinguishes between an alkyne and an alkene.
• Outcome: The spectrum is consistent with a carboxylic acid containing an alkyne functional group.

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Question 5: Electrophilic Aromatic Substitution (EAS) and Directing Groups

• Main Idea: The order of reactions in EAS is crucial for achieving the desired regiochemistry (para substitution) due to the directing effects of substituents.
• Key Concepts:
  • Nitration: HNO3/H2SO4 introduces -NO2.
  • Friedel-Crafts Alkylation: R-Cl/AlCl3 introduces an alkyl group (-R).
  • Oxidation: KMnO4 oxidizes alkyl side chains to -COOH.
  • Reduction of Nitro Group: Sn/HCl or similar reduces -NO2 to -NH2.
  • Directing Groups:
    • Ortho/Para Directors: Activating groups (e.g., -CH3, -NH2, -OH) or halogens.
    • Meta Directors: Deactivating groups (e.g., -NO2, -COOH, -CN).
• Analysis of Options:
  • Option B: Nitration first (-NO2 is meta-directing), then alkylation leads to meta-nitrobenzoic acid.
  • Option C: Alkylation first (-CH3 is o/p directing), then oxidation to -COOH (meta-directing), then nitration leads to meta-nitrobenzoic acid.
  • Option D: Nitration, reduction to -NH2 (o/p directing but deactivates ring significantly when protonated by Lewis acid), then Friedel-Crafts alkylation is problematic.
  • Option A: Friedel-Crafts alkylation first (-CH3 is o/p directing), then nitration (methyl directs o/p), then oxidation of methyl to -COOH. This sequence yields the para product.
• Outcome: Option A provides the greatest yield of para-nitrobenzoic acid because the activating methyl group directs the nitration to the ortho/para positions, and the para product is desired. Subsequent oxidation converts the methyl to a carboxylic acid.

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Question 6: Acid Chloride Reactions

• Main Idea: Acid chlorides react with different nucleophiles to form various products.
• Reactions:
  • Grignard Reagent (RMgX): Reacts twice with acid chlorides to form tertiary alcohols (after aqueous workup).
  • Gilman Reagent (R2CuLi): Reacts once with acid chlorides to form ketones.
  • Alcohol (ROH): Reacts with acid chlorides to form esters.
  • Ester: Does not typically react with acid chlorides in this manner.
• Outcome: An alcohol reacting with an acid chloride produces an ester.

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Question 7: PCC Oxidation and Wittig Reaction

• Main Idea: PCC oxidizes alcohols, and the Wittig reaction converts ketones to alkenes.
• Key Reactions:
  • PCC (Pyridinium Chlorochromate): Oxidizes primary alcohols to aldehydes and secondary alcohols to ketones. Cyclohexanol (secondary alcohol) will form cyclohexanone.
  • Wittig Reaction: Reagents (triphenylphosphine, alkyl halide, strong base like butyllithium) convert a ketone into an alkene.
    • Triphenylphosphine reacts with an alkyl halide to form a phosphonium salt.
    • A strong base deprotonates the phosphonium salt to form a phosphorus ylide.
    • The ylide reacts with the ketone, forming a four-membered ring intermediate (betaine or oxaphosphetane), which then collapses to form the alkene and triphenylphosphine oxide.
• Regiochemistry: The carbon atom that was part of the carbonyl group in the ketone becomes one of the carbons in the double bond. The other carbon in the double bond comes from the alkyl halide used to form the ylide.
• Outcome: Cyclohexanone reacts with the ylide derived from 2-iodobutane (specifically, the carbon bearing the iodine) to form an alkene where the carbonyl oxygen is replaced by the carbon chain from the ylide. The product will have a double bond formed between the carbonyl carbon and the carbon that bore the iodine in 2-iodobutane.

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Question 8: Diels-Alder Reaction

• Main Idea: The Diels-Alder reaction is a [4+2] cycloaddition between a conjugated diene and a dienophile to form a six-membered ring.
• Key Concepts:
  • Diene: Contains a conjugated system of four pi electrons (two double bonds).
  • Dienophile: Contains a double or triple bond (two pi electrons).
  • Stereochemistry: The relative stereochemistry of substituents on the diene and dienophile is preserved in the product.
  • Cyclic Diene: Using a cyclic diene leads to a bicyclic product.
• Analysis:
  • The product is bicyclic, indicating that the diene must have been cyclic.
  • The dienophile has two nitro groups that are trans to each other.
  • The diene has a methyl group.
• Outcome: The diene must be cyclic (like 1,3-cyclohexadiene derivative) and the dienophile must be trans-2-butene with nitro groups. Option B fits these criteria.

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Question 9: Intramolecular Aldol Reaction

• Main Idea: An intramolecular aldol reaction occurs when a molecule with two carbonyl groups reacts with itself under basic or acidic conditions to form a ring. Dehydration often follows, especially with heat.
• Key Steps:
  1. Deprotonation: A base removes an alpha-hydrogen to form an enolate.
  2. Nucleophilic Attack: The enolate attacks another carbonyl group within the same molecule. This forms a beta-hydroxy aldehyde or ketone (Aldol addition product).
  3. Dehydration (with heat): Elimination of water occurs, forming an alpha, beta-unsaturated aldehyde or ketone (Aldol condensation product). The double bond forms between the alpha and beta carbons, conjugated with the carbonyl.
• Analysis:
  • The starting material is a diketone.
  • Heat is applied, indicating dehydration will occur.
  • The product is a six-membered ring with a double bond conjugated to a carbonyl group.
• Outcome: The intramolecular aldol reaction followed by dehydration yields an alpha, beta-unsaturated cyclic ketone. Option C represents this product.

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Question 10: Enolate Formation (Kinetic vs. Thermodynamic)

• Main Idea: The choice of base and temperature influences whether a kinetic or thermodynamic enolate is formed.
• Key Concepts:
  • Kinetic Enolate: Formed under conditions that favor the fastest reaction (low temperature, strong, bulky base). It is typically the less substituted enolate.
  • Thermodynamic Enolate: Formed under conditions that favor the most stable product (higher temperature, smaller base). It is typically the more substituted enolate.
  • Bases:
    • LDA (Lithium Diisopropylamide): Strong, bulky base. Favors kinetic enolate formation by abstracting the most accessible (less hindered) proton.
    • NaH (Sodium Hydride): Strong, small base. Tends to favor thermodynamic enolate formation.
  • Temperature: Low temperature favors kinetic product; high temperature favors thermodynamic product.
• Analysis:
  • The product shown has the methyl group added to the less substituted side of the original ketone (kinetic product).
  • This requires a bulky base (LDA) and a low temperature.
• Outcome: LDA at low temperature (-80°C) will produce the kinetic enolate, which upon reaction with methyl bromide, yields the desired kinetic product.

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Question 11: EAS - Directing Groups and Reaction Order

• Main Idea: The order of reactions and the relative activating/deactivating strengths of substituents determine the regiochemistry of EAS.
• Key Concepts:
  • Friedel-Crafts Acylation: Introduces an acyl group (-COR). The product is a ketone, which is a meta-director and deactivator.
  • Wolff-Kishner Reduction: Reduces a ketone to an alkane (-CH2R). Alkyl groups are weak activators and ortho/para directors.
  • Bromination: Introduces -Br. Halogens are deactivators but ortho/para directors.
  • Activating Strength: -OR > -NR2 > -R > Halogens > -NO2 > -COR.
• Analysis:
  • The left ring is deactivated by the carbonyl group (meta-director).
  • The right ring is activated by the oxygen atom (ortho/para director).
  • Therefore, reactions will occur on the right ring.
  • Step 1 (Acylation): Occurs on the right ring, likely para to the oxygen due to steric hindrance.
  • Step 2 (Wolff-Kishner): Reduces the ketone to a propyl group.
  • Step 3 (Bromination): Both the oxygen and the propyl group are ortho/para directors. Oxygen is a much stronger activator than the propyl group.
• Outcome: The stronger activating group (oxygen) dictates the position of bromination. Bromine will be directed ortho to the oxygen atom. Option B is the correct product.

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Question 12: ¹H NMR Spectroscopy

• Main Idea: The number of signals in a ¹H NMR spectrum corresponds to the number of chemically distinct types of protons in a molecule. Symmetry is key to determining equivalence.
• Key Concepts:
  • Chemical Equivalence: Protons are chemically equivalent if they are interchangeable by symmetry operations (rotation, reflection) or rapid processes.
  • Symmetry: Look for planes of symmetry or rotational axes that make protons identical.
• Analysis:
  • Compound 1: Four distinct types of protons (methyl, two different CH2 groups, CH). Expected: 4 signals.
  • Compound 2: Symmetry exists. The two methyl groups are equivalent. The two CH2 groups are equivalent. Expected: 2 signals.
  • Compound 3: Symmetry exists. The three methyl groups are equivalent. The remaining CH2 group is unique. Expected: 2 signals.
  • Compound 4: Five distinct types of protons (tert-butyl methyls, the CH, the two CH2 groups, and the isopropyl methyls). Expected: 5 signals.
• Outcome: Compounds 2 and 3 will show two signals in their ¹H NMR spectra.

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Question 13: Imine and Enamine Formation

• Main Idea: Ketones react with primary amines to form imines, and with secondary amines to form enamines. The reactivity of nitrogen atoms depends on their environment (amide vs. amine).
• Key Concepts:
  • Imine: C=N double bond, formed from reaction with a primary amine (RNH2).
  • Enamine: C=C-N system, formed from reaction with a secondary amine (R2NH).
  • Amide Nitrogen: Lone pair is delocalized via resonance with the carbonyl group, making it non-nucleophilic.
  • Amine Nitrogen: Lone pair is available for nucleophilic attack.
• Analysis:
  • The reactant is cyclohexanone and a molecule containing both an amide and an amine group.
  • The amide nitrogen is not nucleophilic due to resonance.
  • The amine nitrogen (NH2) is nucleophilic and will attack the carbonyl carbon.
  • Since it's a primary amine (NH2), an imine will form.
• Outcome: The primary amine nitrogen attacks the ketone, leading to the formation of an imine (C=N). Option A represents the correct imine product.

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Question 14: Aromaticity (Hückel's Rule)

• Main Idea: For a compound to be aromatic, it must meet three criteria:
  1. Cyclic: The molecule must be a ring.
  2. Planar: All atoms in the ring must be sp² hybridized (or nearly so) to allow for continuous p-orbital overlap.
  3. Hückel's Rule: The ring must contain (4n + 2) pi electrons (where n = 0, 1, 2, ...), meaning 2, 6, 10, 14, etc., pi electrons.
• Key Concept for Lone Pairs: A lone pair on an atom contributes to the pi system only if it is in a p-orbital that can overlap with the other p-orbitals in the ring and if counting it satisfies Hückel's rule. Generally, count the lone pair if the atom is sp² hybridized and the lone pair makes it part of the conjugated system. Do not count lone pairs if the atom is sp³ hybridized or if counting it would lead to more than 6 pi electrons in a 5-membered ring system (or more than required by Hückel's rule).
• Analysis:
  • A (Pyrrole): Cyclic, planar, 6 pi electrons (2 from each double bond + 2 from lone pair on N). Aromatic.
  • B (Pyridine): Cyclic, planar, 6 pi electrons (2 from each double bond). The lone pair on N is in an sp² orbital, not a p-orbital, and does not contribute to the pi system. Aromatic.
  • C (Imidazole): Cyclic, planar, 6 pi electrons (2 from each double bond + 2 from one of the lone pairs on N). Aromatic.
  • D (Cyclohexadiene with a lone pair): Cyclic, has 6 pi electrons (2 from each double bond + 2 from the lone pair). However, one carbon atom is sp³ hybridized (CH2 group), breaking the continuous p-orbital overlap and planarity required for conjugation around the entire ring. Non-aromatic.
• Outcome: Compound D is not aromatic because it contains an sp³ hybridized carbon atom, which disrupts planarity and conjugation.

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Question 15: ¹H NMR Spectroscopy and Functional Groups

• Main Idea: ¹H NMR chemical shifts and splitting patterns help identify functional groups and the molecular structure.
• Key Chemical Shifts:
  • Aldehyde H: ~9-10 ppm.
  • Benzene Ring H: ~6.5-8.5 ppm.
  • CH3 next to C=O or Benzene: ~2.3 ppm.
  • CH3 next to electronegative atom (O, N, X): ~3-4 ppm.
  • CH2 next to electronegative atom (O, N, X): ~3-4 ppm.
  • Alkyl CH3: ~0.9-1.3 ppm.
• Key Splitting Patterns (n+1 rule):
  • Singlet (s): 0 adjacent H.
  • Doublet (d): 1 adjacent H.
  • Triplet (t): 2 adjacent H.
  • Quartet (q): 3 adjacent H.
• Analysis:
  • The spectrum lacks an aldehyde signal (~9-10 ppm), eliminating option A.
  • The spectrum lacks benzene ring signals (~6.5-8.5 ppm), eliminating option B.
  • Option D: CH3 next to C=O (~2.3 ppm, singlet), CH2 next to O (~3.5 ppm, quartet due to adjacent CH3), CH3 next to CH2 (~1 ppm, triplet due to adjacent CH2). This matches the spectrum.
  • Option C: CH3 next to O (~3.5 ppm), CH2 next to O (~3.5 ppm), CH3 next to CH2 (~1 ppm). This would likely show two signals around 3.5 ppm, not one.
• Outcome: Compound D is consistent with the ¹H NMR spectrum.