CHM 2210 Exam 4 V3 | CHM 2210 Organic
Chemistry I | Actual Q&A with Rationale
(CHM2210 Exam 4) | University of Central
Florida
1. Identify the major organic product formed when 1-pentanol is treated with pyridinium
chlorochromate (PCC) in dichloromethane.
A. Pentanal
B. Pentanoic acid
C. 2-pentanone
D. 1-pentene
Answer: A
Rationale: Pyridinium chlorochromate is a mild oxidizing agent that converts primary
alcohols to aldehydes without further oxidation to carboxylic acids. The reaction occurs in
anhydrous conditions, typically using dichloromethane as a solvent to prevent the
formation of hydrates. This specific selectivity is a critical component of synthetic organic
chemistry strategies taught in the CHM 2210 curriculum.
2. Which of the following reagents would be most suitable for the synthesis of 2-methyl-2-
butanol from acetone?
A. NaBH4 in methanol
,B. CH3Li followed by H2O
C. CH3CH2MgBr followed by H3O+
D. LiAlH4 in ether
Answer: C
Rationale: The synthesis of 2-methyl-2-butanol from acetone requires the addition of a
two-carbon alkyl group to a ketone. Ethylmagnesium bromide acts as a nucleophile,
attacking the carbonyl carbon to form an alkoxide intermediate. Subsequent protonation
with hydronium ion yields the tertiary alcohol, which is the standard protocol for Grignard
additions to ketones.
3. What is the predicted product when (R)-2-butanol reacts with PBr3?
A. (R)-2-bromobutane
B. * (S)-2-bromobutane*
C. Racemic 2-bromobutane
D. 1-bromobutane
Answer: B
Rationale: The reaction of a secondary alcohol with phosphorus tribromide proceeds via
an SN2 mechanism. This results in a complete inversion of configuration at the
stereocenter containing the hydroxyl group. Therefore, the (R) enantiomer of the starting
material will yield the (S) enantiomer of the resulting alkyl bromide.
, 4. In the Williamson ether synthesis, which combination of reagents would efficiently
produce methyl tert-butyl ether?
A. Methanol and tert-butyl bromide
B. Tert-butanol and methylmagnesium bromide
C. Sodium methoxide and tert-butyl bromide
D. Sodium tert-butoxide and methyl iodide
Answer: D
Rationale: The Williamson ether synthesis is an SN2 reaction that requires an unhindered
alkyl halide to be successful. Using a tertiary alkyl halide like tert-butyl bromide would lead
to E2 elimination as the major pathway rather than substitution. Consequently, the
alkoxide should be the hindered species and the halide should be methyl or primary.
5. Predict the major product of the acid-catalyzed ring-opening of 1,2-epoxypropane with
methanol.
A. 1-methoxy-2-propanol
B. 2-methoxy-2-propanol
C. 1,2-dimethoxypropane
D. 2-methoxy-1-propanol
Answer: D
Chemistry I | Actual Q&A with Rationale
(CHM2210 Exam 4) | University of Central
Florida
1. Identify the major organic product formed when 1-pentanol is treated with pyridinium
chlorochromate (PCC) in dichloromethane.
A. Pentanal
B. Pentanoic acid
C. 2-pentanone
D. 1-pentene
Answer: A
Rationale: Pyridinium chlorochromate is a mild oxidizing agent that converts primary
alcohols to aldehydes without further oxidation to carboxylic acids. The reaction occurs in
anhydrous conditions, typically using dichloromethane as a solvent to prevent the
formation of hydrates. This specific selectivity is a critical component of synthetic organic
chemistry strategies taught in the CHM 2210 curriculum.
2. Which of the following reagents would be most suitable for the synthesis of 2-methyl-2-
butanol from acetone?
A. NaBH4 in methanol
,B. CH3Li followed by H2O
C. CH3CH2MgBr followed by H3O+
D. LiAlH4 in ether
Answer: C
Rationale: The synthesis of 2-methyl-2-butanol from acetone requires the addition of a
two-carbon alkyl group to a ketone. Ethylmagnesium bromide acts as a nucleophile,
attacking the carbonyl carbon to form an alkoxide intermediate. Subsequent protonation
with hydronium ion yields the tertiary alcohol, which is the standard protocol for Grignard
additions to ketones.
3. What is the predicted product when (R)-2-butanol reacts with PBr3?
A. (R)-2-bromobutane
B. * (S)-2-bromobutane*
C. Racemic 2-bromobutane
D. 1-bromobutane
Answer: B
Rationale: The reaction of a secondary alcohol with phosphorus tribromide proceeds via
an SN2 mechanism. This results in a complete inversion of configuration at the
stereocenter containing the hydroxyl group. Therefore, the (R) enantiomer of the starting
material will yield the (S) enantiomer of the resulting alkyl bromide.
, 4. In the Williamson ether synthesis, which combination of reagents would efficiently
produce methyl tert-butyl ether?
A. Methanol and tert-butyl bromide
B. Tert-butanol and methylmagnesium bromide
C. Sodium methoxide and tert-butyl bromide
D. Sodium tert-butoxide and methyl iodide
Answer: D
Rationale: The Williamson ether synthesis is an SN2 reaction that requires an unhindered
alkyl halide to be successful. Using a tertiary alkyl halide like tert-butyl bromide would lead
to E2 elimination as the major pathway rather than substitution. Consequently, the
alkoxide should be the hindered species and the halide should be methyl or primary.
5. Predict the major product of the acid-catalyzed ring-opening of 1,2-epoxypropane with
methanol.
A. 1-methoxy-2-propanol
B. 2-methoxy-2-propanol
C. 1,2-dimethoxypropane
D. 2-methoxy-1-propanol
Answer: D
