CHM 2210 Exam 4 V2 | CHM 2210 Organic Chemistry I | Actual Q&A
with Rationale (CHM2210 Exam 4) | University of Central Florida
1. Which reagent is most suitable for the conversion of 3-hexyne into cis-3-hexene?
A. H2, Lindlar’s catalyst
B. Na, NH3 (liq)
C. H2, Pd/C
D. BH3, THF followed by H2O2, NaOH
Answer: A
Explanation: Lindlar’s catalyst is a poisoned palladium catalyst that facilitates the partial
hydrogenation of alkynes to alkenes. It specifically promotes syn-addition of hydrogen,
resulting in the formation of the cis-isomer. Using sodium in liquid ammonia would instead
yield the trans-alkene via a radical mechanism.
2. What is the major organic product of the reaction between 1-butyne and two equivalents
of HBr?
A. 1,1-dibromobutane
B. 2,2-dibromobutane
C. 1,2-dibromobutane
D. 2,3-dibromobutane
Answer: B
Explanation: The addition of HBr to an alkyne follows Markovnikov’s rule where the
hydrogen adds to the less substituted carbon. The first addition creates a vinyl bromide,
and the second addition proceeds to place the second bromine on the same carbon to
stabilize the intermediate carbocation. This results in a geminal dibromide, specifically 2,2-
dibromobutane.
3. In the radical bromination of 2-methylpropane, which product is formed in the greatest
yield?
A. 1-bromo-2-methylpropane
B. 2-bromobutane
C. 2-bromo-2-methylpropane
D. 1-bromobutane
Answer: C
,Explanation: Bromination is highly selective for the more stable radical intermediate. A
tertiary radical formed at the C2 position of 2-methylpropane is significantly more stable
than a primary radical. Consequently, 2-bromo-2-methylpropane is the dominant product
despite the statistical advantage of the primary hydrogens.
4. Which of the following compounds will show a sharp absorption peak at approximately
2250 cm-1 in an IR spectrum?
A. 1-hexyne
B. 2-hexanone
C. 1-hexene
D. Hexanoic acid
Answer: A
Explanation: The region around 2100-2260 cm-1 is characteristic of triple bond stretches.
Specifically, a terminal alkyne like 1-hexyne shows a C-C triple bond stretch in this range.
Alkenes and ketones appear at lower frequencies, typically around 1650 and 1715 cm-1
respectively.
5. Predict the product of the reaction of an epoxide with CH3MgBr followed by an aqueous
workup.
A. A primary alcohol
B. A substituted alcohol
C. A diol
D. An ether
Answer: B
Explanation: Grignard reagents act as strong nucleophiles that attack the less sterically
hindered carbon of an epoxide. The ring opens to form an alkoxide intermediate, which is
then protonated during the aqueous workup. The resulting product is an alcohol where the
carbon chain has been extended by the alkyl group of the Grignard reagent.
6. Which reagent set converts 1-methylcyclopentanol into 1-chloro-1-methylcyclopentane?
A. Cl2, light
B. HCl
C. NaCl, H2O
D. SOCl2, Pyridine
Answer: B
, Explanation: Tertiary alcohols react readily with concentrated hydrogen halides like HCl
via an SN1 mechanism. The hydroxyl group is protonated to form a good leaving group
(water), which then departs to form a stable tertiary carbocation. Chloride then attacks the
carbocation to yield the tertiary alkyl chloride.
7. What is the major product of the reaction of 1-pentyne with HgSO4, H2SO4, and H2O?
A. Pentanal
B. 3-pentanone
C. 2-pentanone
D. 1-pentanol
Answer: C
Explanation: Acid-catalyzed hydration of terminal alkynes using mercuric sulfate follows
Markovnikov addition. The water molecule adds to the more substituted carbon of the
triple bond, forming an enol. The enol rapidly tautomerizes to the more stable ketone form,
yielding 2-pentanone.
8. In mass spectrometry, the presence of a M+2 peak that is approximately the same height
as the M+ peak indicates the presence of which element?
A. Bromine
B. Chlorine
C. Iodine
D. Nitrogen
Answer: A
Explanation: Bromine has two naturally occurring isotopes, Br-79 and Br-81, which exist
in an approximately 1:1 ratio. This results in two molecular ion peaks separated by two
mass units of nearly equal intensity. Chlorine also shows an M+2 peak, but in a 3:1 ratio
due to the abundance of Cl-35 and Cl-37.
9. Which of the following describes the first step in the mechanism of radical halogenation?
A. Propagation
B. Initiation
C. Termination
D. Elimination
Answer: B
Explanation: Radical halogenation begins with the initiation step, where heat or light
causes the homolytic cleavage of a halogen-halogen bond. This step generates two halogen
with Rationale (CHM2210 Exam 4) | University of Central Florida
1. Which reagent is most suitable for the conversion of 3-hexyne into cis-3-hexene?
A. H2, Lindlar’s catalyst
B. Na, NH3 (liq)
C. H2, Pd/C
D. BH3, THF followed by H2O2, NaOH
Answer: A
Explanation: Lindlar’s catalyst is a poisoned palladium catalyst that facilitates the partial
hydrogenation of alkynes to alkenes. It specifically promotes syn-addition of hydrogen,
resulting in the formation of the cis-isomer. Using sodium in liquid ammonia would instead
yield the trans-alkene via a radical mechanism.
2. What is the major organic product of the reaction between 1-butyne and two equivalents
of HBr?
A. 1,1-dibromobutane
B. 2,2-dibromobutane
C. 1,2-dibromobutane
D. 2,3-dibromobutane
Answer: B
Explanation: The addition of HBr to an alkyne follows Markovnikov’s rule where the
hydrogen adds to the less substituted carbon. The first addition creates a vinyl bromide,
and the second addition proceeds to place the second bromine on the same carbon to
stabilize the intermediate carbocation. This results in a geminal dibromide, specifically 2,2-
dibromobutane.
3. In the radical bromination of 2-methylpropane, which product is formed in the greatest
yield?
A. 1-bromo-2-methylpropane
B. 2-bromobutane
C. 2-bromo-2-methylpropane
D. 1-bromobutane
Answer: C
,Explanation: Bromination is highly selective for the more stable radical intermediate. A
tertiary radical formed at the C2 position of 2-methylpropane is significantly more stable
than a primary radical. Consequently, 2-bromo-2-methylpropane is the dominant product
despite the statistical advantage of the primary hydrogens.
4. Which of the following compounds will show a sharp absorption peak at approximately
2250 cm-1 in an IR spectrum?
A. 1-hexyne
B. 2-hexanone
C. 1-hexene
D. Hexanoic acid
Answer: A
Explanation: The region around 2100-2260 cm-1 is characteristic of triple bond stretches.
Specifically, a terminal alkyne like 1-hexyne shows a C-C triple bond stretch in this range.
Alkenes and ketones appear at lower frequencies, typically around 1650 and 1715 cm-1
respectively.
5. Predict the product of the reaction of an epoxide with CH3MgBr followed by an aqueous
workup.
A. A primary alcohol
B. A substituted alcohol
C. A diol
D. An ether
Answer: B
Explanation: Grignard reagents act as strong nucleophiles that attack the less sterically
hindered carbon of an epoxide. The ring opens to form an alkoxide intermediate, which is
then protonated during the aqueous workup. The resulting product is an alcohol where the
carbon chain has been extended by the alkyl group of the Grignard reagent.
6. Which reagent set converts 1-methylcyclopentanol into 1-chloro-1-methylcyclopentane?
A. Cl2, light
B. HCl
C. NaCl, H2O
D. SOCl2, Pyridine
Answer: B
, Explanation: Tertiary alcohols react readily with concentrated hydrogen halides like HCl
via an SN1 mechanism. The hydroxyl group is protonated to form a good leaving group
(water), which then departs to form a stable tertiary carbocation. Chloride then attacks the
carbocation to yield the tertiary alkyl chloride.
7. What is the major product of the reaction of 1-pentyne with HgSO4, H2SO4, and H2O?
A. Pentanal
B. 3-pentanone
C. 2-pentanone
D. 1-pentanol
Answer: C
Explanation: Acid-catalyzed hydration of terminal alkynes using mercuric sulfate follows
Markovnikov addition. The water molecule adds to the more substituted carbon of the
triple bond, forming an enol. The enol rapidly tautomerizes to the more stable ketone form,
yielding 2-pentanone.
8. In mass spectrometry, the presence of a M+2 peak that is approximately the same height
as the M+ peak indicates the presence of which element?
A. Bromine
B. Chlorine
C. Iodine
D. Nitrogen
Answer: A
Explanation: Bromine has two naturally occurring isotopes, Br-79 and Br-81, which exist
in an approximately 1:1 ratio. This results in two molecular ion peaks separated by two
mass units of nearly equal intensity. Chlorine also shows an M+2 peak, but in a 3:1 ratio
due to the abundance of Cl-35 and Cl-37.
9. Which of the following describes the first step in the mechanism of radical halogenation?
A. Propagation
B. Initiation
C. Termination
D. Elimination
Answer: B
Explanation: Radical halogenation begins with the initiation step, where heat or light
causes the homolytic cleavage of a halogen-halogen bond. This step generates two halogen
