CHEM 219 Final Exam (Portage Learning) 2026/2027 | Principles of Organic Chemistry
Q&A
Organic Chemistry II Comprehensive Review | Key Domains: Aromatic Compounds & Reactions,
Carbonyl Chemistry (Aldehydes, Ketones, Carboxylic Acids & Derivatives), Amines & Nitrogen-Containing
Compounds, Spectroscopy (IR, NMR, Mass Spec) for Structural Analysis, Multistep Synthesis, and
Reaction Mechanisms | Expert-Aligned Structure | Comprehensive Final Exam Format
Introduction
This structured CHEM 219 Final Exam for Portage Learning (2026/2027) provides a comprehensive set
of multiple-choice questions with correct answers and rationales covering the entire second-semester
organic chemistry curriculum. It emphasizes the synthesis of knowledge, requiring students to apply
principles from all modules to predict products, deduce structures, propose reaction pathways, and
analyze spectroscopic data.
Exam Structure:
● Comprehensive Final Exam: (75 MULTIPLE-CHOICE QUESTIONS)
Answer Format
The correct answer for each question is indicated in bold and lime green, followed by a concise
rationale explaining the reaction mechanism, spectroscopic interpretation, synthetic logic, or why other
choices are incorrect.
Multiple-Choice Questions (1–75)
1. What is the major product of the nitration of toluene?
A) Nitrobenzene
B) meta-Nitrotoluene
C) ortho-Nitrotoluene and para-nitrotoluene
, D) 2,4,6-Trinitrotoluene
C) ortho-Nitrotoluene and para-nitrotoluene
The methyl group in toluene is an activating ortho/para-director. Electrophilic aromatic substitution
favors these positions due to resonance and hyperconjugation. Meta substitution is disfavored, and
trinitration requires harsher conditions.
2. Which reagent converts a ketone to an alkane via the Wolff–Kishner reduction?
A) LiAlH₄
B) Zn(Hg), HCl
C) NH₂NH₂, KOH, heat
D) H₂, Pd/C
C) NH₂NH₂, KOH, heat
Wolff–Kishner reduction uses hydrazine (NH₂NH₂) and strong base (KOH) at high temperature to
deoxygenate carbonyls to alkanes. Clemmensen reduction (Zn/Hg, HCl) achieves the same result under
acidic conditions.
3. In IR spectroscopy, a strong, sharp peak at ~1710 cm⁻¹ indicates the presence of:
A) O–H stretch
B) N–H bend
C) C=O stretch
D) C≡C stretch
,C) C=O stretch
The 1710 cm⁻¹ region is characteristic of carbonyl (C=O) stretching vibrations. The exact position varies
slightly with conjugation and substituents, but the strong, sharp peak is diagnostic for aldehydes,
ketones, carboxylic acids, or derivatives.
4. Which carboxylic acid derivative undergoes nucleophilic acyl substitution most readily?
A) Amide
B) Ester
C) Acid anhydride
D) Acid chloride
D) Acid chloride
Acid chlorides are the most reactive due to the strong electron-withdrawing effect of chlorine and the
stability of Cl⁻ as a leaving group. Reactivity order: acid chloride > anhydride > ester > amide.
5. What is the product when benzaldehyde reacts with CH₃MgBr followed by H₃O⁺?
A) Benzyl alcohol
B) 1-Phenylethanol
C) Acetophenone
D) Benzoic acid
B) 1-Phenylethanol
, Grignard reagents add to aldehydes to form secondary alcohols after workup. Benzaldehyde (PhCHO) +
CH₃MgBr → PhCH(OMgBr)CH₃ → (H₃O⁺) → PhCH(OH)CH₃ (1-phenylethanol).
6. Which amine is the strongest base?
A) Aniline
B) p-Nitroaniline
C) Ethylamine
D) Diphenylamine
C) Ethylamine
Alkylamines like ethylamine (pKb ~3.3) are stronger bases than arylamines because the lone pair on
nitrogen in aniline is delocalized into the aromatic ring, reducing basicity. Electron-withdrawing
groups (e.g., NO₂) further reduce basicity.
7. In ¹H NMR, a triplet at δ 1.1 ppm (3H) and a quartet at δ 2.4 ppm (2H) suggests the
presence of:
A) Isopropyl group
B) Ethyl group attached to an electron-withdrawing group
C) Methyl group next to carbonyl
D) tert-Butyl group
B) Ethyl group attached to an electron-withdrawing group
Q&A
Organic Chemistry II Comprehensive Review | Key Domains: Aromatic Compounds & Reactions,
Carbonyl Chemistry (Aldehydes, Ketones, Carboxylic Acids & Derivatives), Amines & Nitrogen-Containing
Compounds, Spectroscopy (IR, NMR, Mass Spec) for Structural Analysis, Multistep Synthesis, and
Reaction Mechanisms | Expert-Aligned Structure | Comprehensive Final Exam Format
Introduction
This structured CHEM 219 Final Exam for Portage Learning (2026/2027) provides a comprehensive set
of multiple-choice questions with correct answers and rationales covering the entire second-semester
organic chemistry curriculum. It emphasizes the synthesis of knowledge, requiring students to apply
principles from all modules to predict products, deduce structures, propose reaction pathways, and
analyze spectroscopic data.
Exam Structure:
● Comprehensive Final Exam: (75 MULTIPLE-CHOICE QUESTIONS)
Answer Format
The correct answer for each question is indicated in bold and lime green, followed by a concise
rationale explaining the reaction mechanism, spectroscopic interpretation, synthetic logic, or why other
choices are incorrect.
Multiple-Choice Questions (1–75)
1. What is the major product of the nitration of toluene?
A) Nitrobenzene
B) meta-Nitrotoluene
C) ortho-Nitrotoluene and para-nitrotoluene
, D) 2,4,6-Trinitrotoluene
C) ortho-Nitrotoluene and para-nitrotoluene
The methyl group in toluene is an activating ortho/para-director. Electrophilic aromatic substitution
favors these positions due to resonance and hyperconjugation. Meta substitution is disfavored, and
trinitration requires harsher conditions.
2. Which reagent converts a ketone to an alkane via the Wolff–Kishner reduction?
A) LiAlH₄
B) Zn(Hg), HCl
C) NH₂NH₂, KOH, heat
D) H₂, Pd/C
C) NH₂NH₂, KOH, heat
Wolff–Kishner reduction uses hydrazine (NH₂NH₂) and strong base (KOH) at high temperature to
deoxygenate carbonyls to alkanes. Clemmensen reduction (Zn/Hg, HCl) achieves the same result under
acidic conditions.
3. In IR spectroscopy, a strong, sharp peak at ~1710 cm⁻¹ indicates the presence of:
A) O–H stretch
B) N–H bend
C) C=O stretch
D) C≡C stretch
,C) C=O stretch
The 1710 cm⁻¹ region is characteristic of carbonyl (C=O) stretching vibrations. The exact position varies
slightly with conjugation and substituents, but the strong, sharp peak is diagnostic for aldehydes,
ketones, carboxylic acids, or derivatives.
4. Which carboxylic acid derivative undergoes nucleophilic acyl substitution most readily?
A) Amide
B) Ester
C) Acid anhydride
D) Acid chloride
D) Acid chloride
Acid chlorides are the most reactive due to the strong electron-withdrawing effect of chlorine and the
stability of Cl⁻ as a leaving group. Reactivity order: acid chloride > anhydride > ester > amide.
5. What is the product when benzaldehyde reacts with CH₃MgBr followed by H₃O⁺?
A) Benzyl alcohol
B) 1-Phenylethanol
C) Acetophenone
D) Benzoic acid
B) 1-Phenylethanol
, Grignard reagents add to aldehydes to form secondary alcohols after workup. Benzaldehyde (PhCHO) +
CH₃MgBr → PhCH(OMgBr)CH₃ → (H₃O⁺) → PhCH(OH)CH₃ (1-phenylethanol).
6. Which amine is the strongest base?
A) Aniline
B) p-Nitroaniline
C) Ethylamine
D) Diphenylamine
C) Ethylamine
Alkylamines like ethylamine (pKb ~3.3) are stronger bases than arylamines because the lone pair on
nitrogen in aniline is delocalized into the aromatic ring, reducing basicity. Electron-withdrawing
groups (e.g., NO₂) further reduce basicity.
7. In ¹H NMR, a triplet at δ 1.1 ppm (3H) and a quartet at δ 2.4 ppm (2H) suggests the
presence of:
A) Isopropyl group
B) Ethyl group attached to an electron-withdrawing group
C) Methyl group next to carbonyl
D) tert-Butyl group
B) Ethyl group attached to an electron-withdrawing group
