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Organic Chemistry - Janice Gorzynski Smith |
300-Question Comprehensive Practice Exam
Full-course mixed practice with answers and detailed explanations
QUESTIONS FORMAT COVERAGE
300 A-D + Explanations Full Course
Original practice questions for independent study and exam preparation.
,300-Question Comprehensive Practice Exam
Practice Questions
Choose the best answer for each item. The answer and a detailed explanation follow each question.
Hybridization and bonding
1. Which option is most consistent with the organic-chemistry principle governing the carbon atom directly
bearing a hydroxyl group in methanol?
A. sp hybridized
B. sp2 hybridized
C. a carbonyl carbon
D. sp3 hybridized
Answer: D
Explanation: The carbon of methanol forms four sigma bonds: three C-H bonds and one C-O bond. Four electron domains correspond
to sp3 hybridization. There is no pi bond at that carbon.
Aldehydes and ketones
2. Which option is most consistent with the organic-chemistry principle governing an aldehyde treated with
NaBH4?
A. a carboxylic acid is formed
B. an ester is formed
C. an alkane forms directly
D. a primary alcohol is formed
Answer: D
Explanation: Hydride attacks the electrophilic carbonyl carbon and creates an alkoxide intermediate. Protonation converts the alkoxide
to an alcohol. Because an aldehyde carbonyl carbon bears one hydrogen, the product is a primary alcohol.
Acid-base chemistry
3. A student is reviewing phenol compared with cyclohexanol, propane, and a typical ether. Which conclusion is
correct?
A. the ether is strongest
B. cyclohexanol is strongest
C. propane is strongest
D. phenol is the strongest acid
Answer: D
Explanation: Deprotonation of phenol gives phenoxide, whose negative charge is delocalized by resonance into the aromatic ring. A
cyclohexoxide ion lacks that resonance stabilization, and propane would give a very unstable carbanion. A typical ether has no O-H
bond to donate in the same way.
Acid-base chemistry
4. Which option is most consistent with the organic-chemistry principle governing an alcohol next to strong
electron-withdrawing fluorine substituents?
A. fluorine has no inductive effect
B. electron-withdrawing groups increase alcohol acidity
C. electron-withdrawing groups always decrease acidity
D. the O-H bond becomes nonpolar
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,300-Question Comprehensive Practice Exam
Answer: B
Explanation: Electron-withdrawing substituents stabilize the negatively charged alkoxide conjugate base by induction. Greater
stabilization of the conjugate base makes proton loss more favorable. The effect is strongest when the withdrawing groups are close to
the O-H group.
Substitution and elimination
5. A student is reviewing tert-butyl bromide with sodium ethoxide and heat. Which conclusion is correct?
A. SN2 substitution dominates
B. no beta hydrogens exist
C. SN1 is the only possible mechanism
D. E2 elimination is favored
Answer: D
Explanation: A tertiary substrate is too hindered for SN2 attack. A strong base such as ethoxide can remove a beta hydrogen in a
concerted E2 step. Heat further favors elimination over substitution.
Hybridization and bonding
6. A student is reviewing a carbon atom of an isolated alkene. Which conclusion is correct?
A. sp3 hybridized
B. sp2 hybridized with one unhybridized p orbital
C. sp3d hybridized
D. sp hybridized
Answer: B
Explanation: Each alkene carbon makes three sigma bonds or electron-domain interactions and retains one p orbital. The p orbitals
overlap to form the pi bond of C=C. This is the characteristic bonding picture for sp2 carbon.
1H NMR spectroscopy
7. A student is reviewing a quartet in a simple first-order proton NMR spectrum. Which conclusion is correct?
A. the observed proton set commonly has three equivalent neighboring protons
B. it must have one neighboring proton
C. it cannot arise from an ethyl group
D. it must have six neighboring protons
Answer: A
Explanation: The n+1 rule predicts four lines when a proton set is coupled to three equivalent neighboring protons. This is commonly
seen for the CH2 portion of an ethyl group. Coupling details can become more complex when neighbors are nonequivalent.
Organometallic chemistry
8. Which option is most consistent with the organic-chemistry principle governing a Grignard reagent exposed to
water before it reaches the intended carbonyl?
A. water acts as an inert solvent
B. the carbonyl addition accelerates
C. the Grignard reagent becomes more nucleophilic
D. the Grignard reagent is quenched by acid-base reaction
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, 300-Question Comprehensive Practice Exam
Answer: D
Explanation: Grignard reagents are extremely strong bases as well as nucleophiles. Even weak proton donors such as water transfer a
proton to the carbon attached to magnesium, destroying the organometallic reagent. Grignard reactions therefore require dry solvents
and apparatus.
Enolates and carbon-carbon bond formation
9. A student is reviewing malonic ester synthesis. Which conclusion is correct?
A. it is useful for preparing substituted acetic-acid derivatives after alkylation, hydrolysis, and decarboxylation
B. it cannot form C-C bonds
C. it always yields tertiary alcohols
D. it is a method for making only aromatic nitro compounds
Answer: A
Explanation: The malonate anion is stabilized by two carbonyl groups and can be alkylated by suitable electrophiles. Hydrolysis
converts the esters to acids, and heating promotes decarboxylation. The sequence provides substituted acetic-acid products.
Substitution and elimination
10. Which option is most consistent with the organic-chemistry principle governing an E2 elimination?
A. the nucleophile adds to a carbonyl
B. two discrete protonation steps are required
C. a free carbocation must form first
D. the C-H and C-leaving-group bonds are broken in one concerted step
Answer: D
Explanation: E2 is a bimolecular concerted elimination. Base removes a beta hydrogen as the leaving group departs and the pi bond
forms. Because no carbocation intermediate exists, rearrangements are not part of the normal mechanism.
Aldehydes and ketones
11. A student is reviewing hydration of a carbonyl compound. Which conclusion is correct?
A. a vicinal dibromide forms
B. a nitrile forms
C. an alkyne forms
D. water adds reversibly to give a geminal diol
Answer: D
Explanation: Water can add to a carbonyl carbon while the oxygen is protonated or activated. The product has two OH groups on the
same carbon, a geminal diol. Equilibrium position depends strongly on carbonyl substituents and electron-withdrawing groups.
Functional groups
12. A student is reviewing the functional group R-CO2H. Which conclusion is correct?
A. a carboxylic acid
B. an aldehyde
C. an ester
D. an alcohol only
Answer: A
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Organic Chemistry - Janice Gorzynski Smith |
300-Question Comprehensive Practice Exam
Full-course mixed practice with answers and detailed explanations
QUESTIONS FORMAT COVERAGE
300 A-D + Explanations Full Course
Original practice questions for independent study and exam preparation.
,300-Question Comprehensive Practice Exam
Practice Questions
Choose the best answer for each item. The answer and a detailed explanation follow each question.
Hybridization and bonding
1. Which option is most consistent with the organic-chemistry principle governing the carbon atom directly
bearing a hydroxyl group in methanol?
A. sp hybridized
B. sp2 hybridized
C. a carbonyl carbon
D. sp3 hybridized
Answer: D
Explanation: The carbon of methanol forms four sigma bonds: three C-H bonds and one C-O bond. Four electron domains correspond
to sp3 hybridization. There is no pi bond at that carbon.
Aldehydes and ketones
2. Which option is most consistent with the organic-chemistry principle governing an aldehyde treated with
NaBH4?
A. a carboxylic acid is formed
B. an ester is formed
C. an alkane forms directly
D. a primary alcohol is formed
Answer: D
Explanation: Hydride attacks the electrophilic carbonyl carbon and creates an alkoxide intermediate. Protonation converts the alkoxide
to an alcohol. Because an aldehyde carbonyl carbon bears one hydrogen, the product is a primary alcohol.
Acid-base chemistry
3. A student is reviewing phenol compared with cyclohexanol, propane, and a typical ether. Which conclusion is
correct?
A. the ether is strongest
B. cyclohexanol is strongest
C. propane is strongest
D. phenol is the strongest acid
Answer: D
Explanation: Deprotonation of phenol gives phenoxide, whose negative charge is delocalized by resonance into the aromatic ring. A
cyclohexoxide ion lacks that resonance stabilization, and propane would give a very unstable carbanion. A typical ether has no O-H
bond to donate in the same way.
Acid-base chemistry
4. Which option is most consistent with the organic-chemistry principle governing an alcohol next to strong
electron-withdrawing fluorine substituents?
A. fluorine has no inductive effect
B. electron-withdrawing groups increase alcohol acidity
C. electron-withdrawing groups always decrease acidity
D. the O-H bond becomes nonpolar
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,300-Question Comprehensive Practice Exam
Answer: B
Explanation: Electron-withdrawing substituents stabilize the negatively charged alkoxide conjugate base by induction. Greater
stabilization of the conjugate base makes proton loss more favorable. The effect is strongest when the withdrawing groups are close to
the O-H group.
Substitution and elimination
5. A student is reviewing tert-butyl bromide with sodium ethoxide and heat. Which conclusion is correct?
A. SN2 substitution dominates
B. no beta hydrogens exist
C. SN1 is the only possible mechanism
D. E2 elimination is favored
Answer: D
Explanation: A tertiary substrate is too hindered for SN2 attack. A strong base such as ethoxide can remove a beta hydrogen in a
concerted E2 step. Heat further favors elimination over substitution.
Hybridization and bonding
6. A student is reviewing a carbon atom of an isolated alkene. Which conclusion is correct?
A. sp3 hybridized
B. sp2 hybridized with one unhybridized p orbital
C. sp3d hybridized
D. sp hybridized
Answer: B
Explanation: Each alkene carbon makes three sigma bonds or electron-domain interactions and retains one p orbital. The p orbitals
overlap to form the pi bond of C=C. This is the characteristic bonding picture for sp2 carbon.
1H NMR spectroscopy
7. A student is reviewing a quartet in a simple first-order proton NMR spectrum. Which conclusion is correct?
A. the observed proton set commonly has three equivalent neighboring protons
B. it must have one neighboring proton
C. it cannot arise from an ethyl group
D. it must have six neighboring protons
Answer: A
Explanation: The n+1 rule predicts four lines when a proton set is coupled to three equivalent neighboring protons. This is commonly
seen for the CH2 portion of an ethyl group. Coupling details can become more complex when neighbors are nonequivalent.
Organometallic chemistry
8. Which option is most consistent with the organic-chemistry principle governing a Grignard reagent exposed to
water before it reaches the intended carbonyl?
A. water acts as an inert solvent
B. the carbonyl addition accelerates
C. the Grignard reagent becomes more nucleophilic
D. the Grignard reagent is quenched by acid-base reaction
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, 300-Question Comprehensive Practice Exam
Answer: D
Explanation: Grignard reagents are extremely strong bases as well as nucleophiles. Even weak proton donors such as water transfer a
proton to the carbon attached to magnesium, destroying the organometallic reagent. Grignard reactions therefore require dry solvents
and apparatus.
Enolates and carbon-carbon bond formation
9. A student is reviewing malonic ester synthesis. Which conclusion is correct?
A. it is useful for preparing substituted acetic-acid derivatives after alkylation, hydrolysis, and decarboxylation
B. it cannot form C-C bonds
C. it always yields tertiary alcohols
D. it is a method for making only aromatic nitro compounds
Answer: A
Explanation: The malonate anion is stabilized by two carbonyl groups and can be alkylated by suitable electrophiles. Hydrolysis
converts the esters to acids, and heating promotes decarboxylation. The sequence provides substituted acetic-acid products.
Substitution and elimination
10. Which option is most consistent with the organic-chemistry principle governing an E2 elimination?
A. the nucleophile adds to a carbonyl
B. two discrete protonation steps are required
C. a free carbocation must form first
D. the C-H and C-leaving-group bonds are broken in one concerted step
Answer: D
Explanation: E2 is a bimolecular concerted elimination. Base removes a beta hydrogen as the leaving group departs and the pi bond
forms. Because no carbocation intermediate exists, rearrangements are not part of the normal mechanism.
Aldehydes and ketones
11. A student is reviewing hydration of a carbonyl compound. Which conclusion is correct?
A. a vicinal dibromide forms
B. a nitrile forms
C. an alkyne forms
D. water adds reversibly to give a geminal diol
Answer: D
Explanation: Water can add to a carbonyl carbon while the oxygen is protonated or activated. The product has two OH groups on the
same carbon, a geminal diol. Equilibrium position depends strongly on carbonyl substituents and electron-withdrawing groups.
Functional groups
12. A student is reviewing the functional group R-CO2H. Which conclusion is correct?
A. a carboxylic acid
B. an aldehyde
C. an ester
D. an alcohol only
Answer: A
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