1
CHEM 210 Exam 8 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 SN1
reaction at a chiral center?
A. Complete inversion of configuration
B. Complete retention of configuration
C. Racemization with some inversion
D. Complete racemization with no stereoselectivity
☑ Correct Answer: C
☑ Explanation: SN1 reactions proceed through a planar carbocation intermediate,
allowing attack from either face. However, because the leaving group shields one face,
complete racemization does not occur; instead, partial racemization with slight inversion is
observed. Option A describes SN2, B describes neighboring group participation, D is the
theoretical prediction but not experimentally observed.
,2
2. A compound with molecular formula C₇H₁₄O shows a strong IR absorption at 1715 cm⁻¹ and
no absorption above 3000 cm⁻¹. The ¹H NMR shows a triplet at δ 0.9 (3H), triplet at δ 2.4 (2H),
and multiplet at δ 1.5 (2H). What is the most likely structure?
A. 2-heptanone
B. 3-heptanone
C. 4-heptanone
D. Heptanal
☑ Correct Answer: C
☑ Explanation: The IR shows carbonyl (1715 cm⁻¹) and no OH or C=C absorptions. The
NMR pattern with a single triplet at δ 2.4 indicates a symmetrical ketone where both α-carbons
are equivalent. 4-heptanone (CH₃CH₂CH₂COCH₂CH₂CH₃) has equivalent CH₂ groups adjacent to
carbonyl and terminal methyl groups, giving the observed pattern. Options A and B would show
different splitting patterns, D would show aldehyde proton at δ 9-10.
3. In an aromatic electrophilic substitution reaction, which of the following substituents
would be meta-directing and deactivating?
A. -OH
,3
B. -NH₂
C. -NO₂
D. -CH₃
☑ Correct Answer: C
☑ Explanation: -NO₂ is a strong electron-withdrawing group that is meta-directing and
deactivating due to resonance and inductive effects. -OH and -NH₂ are ortho/para directing and
activating, -CH₃ is ortho/para directing and weakly activating. The nitro group withdraws
electron density from the ring, making electrophilic substitution more difficult and directing
incoming electrophiles to the meta position.
4. Calculate the wavelength (in nm) of maximum absorption for a compound with λmax = 215
nm in ethanol but shifts to 230 nm when the solvent is changed to water. What is this
phenomenon called?
A. Bathochromic shift
B. Hypsochromic shift
C. Hyperchromic effect
D. Hypochromic effect
, 4
☑ Correct Answer: A
☑ Explanation: A shift to longer wavelength (lower energy) from 215 nm to 230 nm is a
bathochromic or red shift. The solvent change from ethanol to water increases polarity,
stabilizing the excited state more than the ground state for many chromophores. Option B is
opposite (blue shift to shorter wavelength), C and D describe changes in absorption intensity,
not wavelength.
5. Which NMR technique would be most useful for determining connectivity between protons
separated by three bonds?
A. COSY
B. HSQC
C. HMBC
D. DEPT
☑ Correct Answer: A
☑ Explanation: COSY (Correlation Spectroscopy) shows proton-proton couplings, typically
through three bonds (³JHH). HSQC shows one-bond C-H correlations, HMBC shows long-range C-
CHEM 210 Exam 8 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 SN1
reaction at a chiral center?
A. Complete inversion of configuration
B. Complete retention of configuration
C. Racemization with some inversion
D. Complete racemization with no stereoselectivity
☑ Correct Answer: C
☑ Explanation: SN1 reactions proceed through a planar carbocation intermediate,
allowing attack from either face. However, because the leaving group shields one face,
complete racemization does not occur; instead, partial racemization with slight inversion is
observed. Option A describes SN2, B describes neighboring group participation, D is the
theoretical prediction but not experimentally observed.
,2
2. A compound with molecular formula C₇H₁₄O shows a strong IR absorption at 1715 cm⁻¹ and
no absorption above 3000 cm⁻¹. The ¹H NMR shows a triplet at δ 0.9 (3H), triplet at δ 2.4 (2H),
and multiplet at δ 1.5 (2H). What is the most likely structure?
A. 2-heptanone
B. 3-heptanone
C. 4-heptanone
D. Heptanal
☑ Correct Answer: C
☑ Explanation: The IR shows carbonyl (1715 cm⁻¹) and no OH or C=C absorptions. The
NMR pattern with a single triplet at δ 2.4 indicates a symmetrical ketone where both α-carbons
are equivalent. 4-heptanone (CH₃CH₂CH₂COCH₂CH₂CH₃) has equivalent CH₂ groups adjacent to
carbonyl and terminal methyl groups, giving the observed pattern. Options A and B would show
different splitting patterns, D would show aldehyde proton at δ 9-10.
3. In an aromatic electrophilic substitution reaction, which of the following substituents
would be meta-directing and deactivating?
A. -OH
,3
B. -NH₂
C. -NO₂
D. -CH₃
☑ Correct Answer: C
☑ Explanation: -NO₂ is a strong electron-withdrawing group that is meta-directing and
deactivating due to resonance and inductive effects. -OH and -NH₂ are ortho/para directing and
activating, -CH₃ is ortho/para directing and weakly activating. The nitro group withdraws
electron density from the ring, making electrophilic substitution more difficult and directing
incoming electrophiles to the meta position.
4. Calculate the wavelength (in nm) of maximum absorption for a compound with λmax = 215
nm in ethanol but shifts to 230 nm when the solvent is changed to water. What is this
phenomenon called?
A. Bathochromic shift
B. Hypsochromic shift
C. Hyperchromic effect
D. Hypochromic effect
, 4
☑ Correct Answer: A
☑ Explanation: A shift to longer wavelength (lower energy) from 215 nm to 230 nm is a
bathochromic or red shift. The solvent change from ethanol to water increases polarity,
stabilizing the excited state more than the ground state for many chromophores. Option B is
opposite (blue shift to shorter wavelength), C and D describe changes in absorption intensity,
not wavelength.
5. Which NMR technique would be most useful for determining connectivity between protons
separated by three bonds?
A. COSY
B. HSQC
C. HMBC
D. DEPT
☑ Correct Answer: A
☑ Explanation: COSY (Correlation Spectroscopy) shows proton-proton couplings, typically
through three bonds (³JHH). HSQC shows one-bond C-H correlations, HMBC shows long-range C-
