1
CHEM 210 Exam 6 Advanced Organic Chemistry
Examination Questions Comprehensive Assessment |
100% Pass Guaranteed | Graded A+
1. Which of the following statements best describes the stereochemical outcome of an SN2
reaction at a chiral center?
A. Retention of configuration with complete inversion
B. Racemization with partial retention
C. Complete inversion of configuration with Walden inversion
D. Scrambling of stereochemistry with racemization
☑ Correct Answer: C
☑ Explanation: SN2 reactions proceed via a backside attack mechanism, resulting in
complete inversion of configuration at the stereocenter. This is known as Walden inversion.
Option A describes retention (SN1 with neighboring group participation), B describes partial
racemization (SN1), and D describes complete racemization (SN1 without stereochemical
control).
,2
2. A compound with molecular formula C₆H₁₀O shows IR absorption at 1715 cm⁻¹ and 1640
cm⁻¹. The ¹H NMR shows a triplet at δ 2.4 (2H), quartet at δ 2.5 (2H), and a multiplet at δ 5.3
(1H). What is the most likely structure?
A. 3-hexen-2-one
B. 4-hexen-3-one
C. 2-hexen-4-one
D. 5-hexen-2-one
☑ Correct Answer: A
☑ Explanation: IR absorptions at 1715 cm⁻¹ (conjugated C=O) and 1640 cm⁻¹ (C=C) indicate
an α,β-unsaturated ketone. The NMR pattern shows CH₃CH₂COCH=CH- system, consistent with
3-hexen-2-one. The triplet at δ 2.4 corresponds to CH₃COCH₂, quartet at δ 2.5 to CH₂ adjacent to
C=O, and multiplet at δ 5.3 to vinylic protons. Options B-D would show different splitting
patterns.
3. In the Diels-Alder reaction, which of the following dienophiles would exhibit the highest
reactivity?
A. Ethylene
,3
B. Maleic anhydride
C. 1,3-butadiene
D. Acrylonitrile
☑ Correct Answer: B
☑ Explanation: Maleic anhydride is an excellent dienophile due to its electron-withdrawing
carbonyl groups that lower the LUMO energy, enhancing reactivity. Ethylene is unactivated and
shows poor reactivity. 1,3-butadiene is a diene, not a dienophile. Acrylonitrile has one electron-
withdrawing group, making it less reactive than maleic anhydride which has two.
4. Calculate the degree of unsaturation for a compound with molecular formula C₈H₇NO₂.
A. 5
B. 6
C. 7
D. 4
☑ Correct Answer: B
☑ Explanation: Degree of unsaturation = (2C + 2 + N - H - X)/2 = (16 + 2 + 1 - 7)/2 = 12/2 =
, 4
6. The compound has 6 degrees of unsaturation, which could include aromatic rings, carbonyl
groups, and other unsaturations. Option A (5) would be incorrect calculation, C (7) would ignore
nitrogen contribution, D (4) would be missing two units.
5. Which spectroscopic technique would be most useful for distinguishing between cis- and
trans-2-butene?
A. IR spectroscopy
B. UV-Vis spectroscopy
C. Mass spectrometry
D. NMR spectroscopy
☑ Correct Answer: A
☑ Explanation: IR spectroscopy is most useful because cis-alkenes show C-H out-of-plane
bending absorptions in the 700-750 cm⁻¹ region, while trans-alkenes absorb in the 960-980 cm⁻¹
region. NMR would show different coupling constants but IR provides definitive distinction. UV-
Vis is not useful for simple alkenes, mass spectrometry would give identical molecular ions, and
standard NMR without specialized techniques might not be as definitive.
CHEM 210 Exam 6 Advanced Organic Chemistry
Examination Questions Comprehensive Assessment |
100% Pass Guaranteed | Graded A+
1. Which of the following statements best describes the stereochemical outcome of an SN2
reaction at a chiral center?
A. Retention of configuration with complete inversion
B. Racemization with partial retention
C. Complete inversion of configuration with Walden inversion
D. Scrambling of stereochemistry with racemization
☑ Correct Answer: C
☑ Explanation: SN2 reactions proceed via a backside attack mechanism, resulting in
complete inversion of configuration at the stereocenter. This is known as Walden inversion.
Option A describes retention (SN1 with neighboring group participation), B describes partial
racemization (SN1), and D describes complete racemization (SN1 without stereochemical
control).
,2
2. A compound with molecular formula C₆H₁₀O shows IR absorption at 1715 cm⁻¹ and 1640
cm⁻¹. The ¹H NMR shows a triplet at δ 2.4 (2H), quartet at δ 2.5 (2H), and a multiplet at δ 5.3
(1H). What is the most likely structure?
A. 3-hexen-2-one
B. 4-hexen-3-one
C. 2-hexen-4-one
D. 5-hexen-2-one
☑ Correct Answer: A
☑ Explanation: IR absorptions at 1715 cm⁻¹ (conjugated C=O) and 1640 cm⁻¹ (C=C) indicate
an α,β-unsaturated ketone. The NMR pattern shows CH₃CH₂COCH=CH- system, consistent with
3-hexen-2-one. The triplet at δ 2.4 corresponds to CH₃COCH₂, quartet at δ 2.5 to CH₂ adjacent to
C=O, and multiplet at δ 5.3 to vinylic protons. Options B-D would show different splitting
patterns.
3. In the Diels-Alder reaction, which of the following dienophiles would exhibit the highest
reactivity?
A. Ethylene
,3
B. Maleic anhydride
C. 1,3-butadiene
D. Acrylonitrile
☑ Correct Answer: B
☑ Explanation: Maleic anhydride is an excellent dienophile due to its electron-withdrawing
carbonyl groups that lower the LUMO energy, enhancing reactivity. Ethylene is unactivated and
shows poor reactivity. 1,3-butadiene is a diene, not a dienophile. Acrylonitrile has one electron-
withdrawing group, making it less reactive than maleic anhydride which has two.
4. Calculate the degree of unsaturation for a compound with molecular formula C₈H₇NO₂.
A. 5
B. 6
C. 7
D. 4
☑ Correct Answer: B
☑ Explanation: Degree of unsaturation = (2C + 2 + N - H - X)/2 = (16 + 2 + 1 - 7)/2 = 12/2 =
, 4
6. The compound has 6 degrees of unsaturation, which could include aromatic rings, carbonyl
groups, and other unsaturations. Option A (5) would be incorrect calculation, C (7) would ignore
nitrogen contribution, D (4) would be missing two units.
5. Which spectroscopic technique would be most useful for distinguishing between cis- and
trans-2-butene?
A. IR spectroscopy
B. UV-Vis spectroscopy
C. Mass spectrometry
D. NMR spectroscopy
☑ Correct Answer: A
☑ Explanation: IR spectroscopy is most useful because cis-alkenes show C-H out-of-plane
bending absorptions in the 700-750 cm⁻¹ region, while trans-alkenes absorb in the 960-980 cm⁻¹
region. NMR would show different coupling constants but IR provides definitive distinction. UV-
Vis is not useful for simple alkenes, mass spectrometry would give identical molecular ions, and
standard NMR without specialized techniques might not be as definitive.
