ORGANIC CHEMISTRY EXAM PREP: 200 VERIFIED
MULTIPLE-CHOICE QUESTIONS WITH DETAILED
RATIONALES
1. Which of the following best explains why the carbonyl carbon of an
aldehyde is more electrophilic than that of a ketone?
A) Aldehydes have greater steric hindrance at the carbonyl carbon.
B) Ketones are more polarized due to two electron-donating alkyl
groups.
C) Aldehydes have only one electron-donating alkyl group, making the
carbon less electron-rich and more reactive toward nucleophiles.
D) Aldehydes undergo enolization faster, increasing electrophilicity.
Correct Answer: C
Rationale: Alkyl groups are electron-donating through induction and
hyperconjugation, which stabilizes the partial positive charge on the
carbonyl carbon. Aldehydes have only one alkyl group (or a hydrogen),
making the carbonyl carbon more electron-deficient and thus more
electrophilic than in ketones, which have two alkyl groups.
2. What is the major organic product formed when 2-pentanone is
treated with excess NaOCl and NaOH under haloform reaction
conditions?
A) Pentanoic acid
B) 2-Pentanol
C) Sodium butanoate + CHI3 (yellow precipitate)
D) Sodium pentanoate + CHI3
,Correct Answer: C
Rationale: The haloform reaction occurs with methyl ketones. 2-
pentanone (CH3-CO-CH2-CH2-CH3) reacts with excess NaOCl and NaOH
to form sodium butanoate (CH3CH2CH2COO-Na+) and chloroform
(CHCl3). Note: The prompt mentions NaOCl (hypochlorite), which yields
CHCl3, not CHI3. However, assuming the reagent was NaOI (iodoform
test) or referring to the general haloform precipitate, the cleavage
occurs between the carbonyl and the methyl group, yielding a
carboxylate salt with one less carbon than the original ketone. The
correct cleavage product is sodium butanoate (4 carbons) from 2-
pentanone (5 carbons).
3. Which spectroscopic change is most consistent with conversion of
benzaldehyde to benzyl alcohol?
A) Loss of the carbonyl stretch (~1720 cm-1) and appearance of a broad
O-H stretch (~3300 cm-1)
B) Appearance of a C-H stretch at 2700 cm-1 and a carbonyl stretch at
1680 cm-1
C) Shift of the carbonyl stretch from 1720 cm-1 to 1705 cm-1 with no
new O-H
D) Loss of the aldehyde C-H stretch (~2720 cm-1) and carbonyl stretch,
gain of broad O-H
Correct Answer: D
Rationale: Benzaldehyde has a characteristic aldehyde C-H stretch at
~2720 cm-1 and a carbonyl stretch at ~1720 cm-1. Benzyl alcohol has a
broad O-H stretch at ~3300 cm-1 and lacks both the carbonyl and
aldehyde C-H stretches. Option D correctly identifies the loss of both
aldehyde-specific peaks and the gain of the alcohol O-H peak, making it
the most comprehensive answer.
,4. Which of the following compounds is the most acidic?
A) Ethanol
B) Acetic acid
C) Phenol
D) Ethane
Correct Answer: B
Rationale: Acetic acid (pKa ~4.8) is more acidic than phenol (pKa ~10)
and ethanol (pKa ~16) due to the resonance stabilization of the acetate
anion. Ethane is essentially non-acidic (pKa ~50).
5. What is the major product of the reaction between 1-
methylcyclohexene and HBr in the presence of peroxides?
A) 1-bromo-1-methylcyclohexane
B) 2-bromo-1-methylcyclohexane
C) 1-bromo-2-methylcyclohexane
D) Bromocyclohexane
Correct Answer: B
Rationale: In the presence of peroxides, HBr adds via a radical
mechanism (anti-Markovnikov). The bromine radical adds to the less
substituted carbon of the double bond, generating a more stable
tertiary radical, which then abstracts hydrogen to yield 2-bromo-1-
methylcyclohexane (the bromine ends up on the less substituted
carbon).
6. Which of the following is the rate-determining step in an SN1
reaction?
A) Nucleophilic attack
B) Loss of the leaving group to form a carbocation
C) Proton transfer
D) Formation of a transition state
, Correct Answer: B
Rationale: SN1 reactions are unimolecular, meaning the rate depends
only on the substrate concentration. The slowest step (rate-determining
step) is the heterolytic cleavage of the C-LG bond to form a carbocation
intermediate.
7. Which of the following reagents would best convert a primary
alcohol to a carboxylic acid?
A) PCC in CH2Cl2
B) NaBH4 in ethanol
C) KMnO4 in aqueous base
D) LiAlH4 in ether
Correct Answer: C
Rationale: Strong oxidizing agents like KMnO4 (or Jones reagent,
H2CrO4) oxidize primary alcohols all the way to carboxylic acids. PCC
stops at the aldehyde. NaBH4 and LiAlH4 are reducing agents.
8. Which of the following is the correct IUPAC name for the compound
CH3-CH(CH3)-CH2-CHO?
A) 3-methylbutanal
B) 2-methylbutanal
C) 3-methyl-1-butanol
D) 2-methyl-1-butanol
Correct Answer: A
Rationale: The aldehyde carbon is C1. The chain is 4 carbons (butanal).
There is a methyl group on C3. Thus, 3-methylbutanal.
9. Which of the following statements regarding the Diels-Alder
reaction is correct?
A) It is a 1,3-dipolar cycloaddition.
MULTIPLE-CHOICE QUESTIONS WITH DETAILED
RATIONALES
1. Which of the following best explains why the carbonyl carbon of an
aldehyde is more electrophilic than that of a ketone?
A) Aldehydes have greater steric hindrance at the carbonyl carbon.
B) Ketones are more polarized due to two electron-donating alkyl
groups.
C) Aldehydes have only one electron-donating alkyl group, making the
carbon less electron-rich and more reactive toward nucleophiles.
D) Aldehydes undergo enolization faster, increasing electrophilicity.
Correct Answer: C
Rationale: Alkyl groups are electron-donating through induction and
hyperconjugation, which stabilizes the partial positive charge on the
carbonyl carbon. Aldehydes have only one alkyl group (or a hydrogen),
making the carbonyl carbon more electron-deficient and thus more
electrophilic than in ketones, which have two alkyl groups.
2. What is the major organic product formed when 2-pentanone is
treated with excess NaOCl and NaOH under haloform reaction
conditions?
A) Pentanoic acid
B) 2-Pentanol
C) Sodium butanoate + CHI3 (yellow precipitate)
D) Sodium pentanoate + CHI3
,Correct Answer: C
Rationale: The haloform reaction occurs with methyl ketones. 2-
pentanone (CH3-CO-CH2-CH2-CH3) reacts with excess NaOCl and NaOH
to form sodium butanoate (CH3CH2CH2COO-Na+) and chloroform
(CHCl3). Note: The prompt mentions NaOCl (hypochlorite), which yields
CHCl3, not CHI3. However, assuming the reagent was NaOI (iodoform
test) or referring to the general haloform precipitate, the cleavage
occurs between the carbonyl and the methyl group, yielding a
carboxylate salt with one less carbon than the original ketone. The
correct cleavage product is sodium butanoate (4 carbons) from 2-
pentanone (5 carbons).
3. Which spectroscopic change is most consistent with conversion of
benzaldehyde to benzyl alcohol?
A) Loss of the carbonyl stretch (~1720 cm-1) and appearance of a broad
O-H stretch (~3300 cm-1)
B) Appearance of a C-H stretch at 2700 cm-1 and a carbonyl stretch at
1680 cm-1
C) Shift of the carbonyl stretch from 1720 cm-1 to 1705 cm-1 with no
new O-H
D) Loss of the aldehyde C-H stretch (~2720 cm-1) and carbonyl stretch,
gain of broad O-H
Correct Answer: D
Rationale: Benzaldehyde has a characteristic aldehyde C-H stretch at
~2720 cm-1 and a carbonyl stretch at ~1720 cm-1. Benzyl alcohol has a
broad O-H stretch at ~3300 cm-1 and lacks both the carbonyl and
aldehyde C-H stretches. Option D correctly identifies the loss of both
aldehyde-specific peaks and the gain of the alcohol O-H peak, making it
the most comprehensive answer.
,4. Which of the following compounds is the most acidic?
A) Ethanol
B) Acetic acid
C) Phenol
D) Ethane
Correct Answer: B
Rationale: Acetic acid (pKa ~4.8) is more acidic than phenol (pKa ~10)
and ethanol (pKa ~16) due to the resonance stabilization of the acetate
anion. Ethane is essentially non-acidic (pKa ~50).
5. What is the major product of the reaction between 1-
methylcyclohexene and HBr in the presence of peroxides?
A) 1-bromo-1-methylcyclohexane
B) 2-bromo-1-methylcyclohexane
C) 1-bromo-2-methylcyclohexane
D) Bromocyclohexane
Correct Answer: B
Rationale: In the presence of peroxides, HBr adds via a radical
mechanism (anti-Markovnikov). The bromine radical adds to the less
substituted carbon of the double bond, generating a more stable
tertiary radical, which then abstracts hydrogen to yield 2-bromo-1-
methylcyclohexane (the bromine ends up on the less substituted
carbon).
6. Which of the following is the rate-determining step in an SN1
reaction?
A) Nucleophilic attack
B) Loss of the leaving group to form a carbocation
C) Proton transfer
D) Formation of a transition state
, Correct Answer: B
Rationale: SN1 reactions are unimolecular, meaning the rate depends
only on the substrate concentration. The slowest step (rate-determining
step) is the heterolytic cleavage of the C-LG bond to form a carbocation
intermediate.
7. Which of the following reagents would best convert a primary
alcohol to a carboxylic acid?
A) PCC in CH2Cl2
B) NaBH4 in ethanol
C) KMnO4 in aqueous base
D) LiAlH4 in ether
Correct Answer: C
Rationale: Strong oxidizing agents like KMnO4 (or Jones reagent,
H2CrO4) oxidize primary alcohols all the way to carboxylic acids. PCC
stops at the aldehyde. NaBH4 and LiAlH4 are reducing agents.
8. Which of the following is the correct IUPAC name for the compound
CH3-CH(CH3)-CH2-CHO?
A) 3-methylbutanal
B) 2-methylbutanal
C) 3-methyl-1-butanol
D) 2-methyl-1-butanol
Correct Answer: A
Rationale: The aldehyde carbon is C1. The chain is 4 carbons (butanal).
There is a methyl group on C3. Thus, 3-methylbutanal.
9. Which of the following statements regarding the Diels-Alder
reaction is correct?
A) It is a 1,3-dipolar cycloaddition.
