CHM 2210 Exam 4 V2 | CHM 2210 Organic
Chemistry I | Actual Q&A with Rationale
(CHM2210 Exam 4) | University of Central
Florida
1. Which of the following reagent sets will convert 2-hexyne into cis-2-hexene?
A. H2, Lindlar’s catalyst
B. H2, Pd/C
C. Na, NH3 (liq)
D. O3 followed by Zn/H2O
Answer: A
Rationale: Lindlar’s catalyst is a deactivated palladium catalyst that facilitates the partial
hydrogenation of alkynes to alkenes. Because the hydrogen atoms are added to the same
face of the triple bond, the reaction is stereoselective for the cis-alkene. Using sodium in
liquid ammonia would instead yield the trans-alkene through a radical-anion mechanism.
2. Identify the major product of the reaction between 1-pentyne and H2SO4, H2O, and
HgSO4.
A. Pentanal
B. 1-pentanol
C. 2-pentanone
,D. 2-pentanol
Answer: C
Rationale: The acid-catalyzed hydration of a terminal alkyne in the presence of mercuric
sulfate follows Markovnikov’s rule. The water molecule adds to the more substituted
carbon of the triple bond to form an enol intermediate. This enol rapidly undergoes keto-
enol tautomerism to produce a methyl ketone, specifically 2-pentanone.
3. Which radical species is the most stable?
A. Primary alkyl radical
B. Tertiary alkyl radical
C. Secondary alkyl radical
D. Methyl radical
Answer: B
Rationale: Radical stability follows the same trend as carbocation stability because
radicals are electron-deficient species. Tertiary radicals are stabilized by the inductive
effect and hyperconjugation from three surrounding alkyl groups. Consequently, they have
lower bond dissociation energies compared to secondary, primary, or methyl radicals.
4. What is the expected product when 1-butyne reacts with excess HBr?
A. 2,2-dibromobutane
B. 1,2-dibromobutane
, C. 1,1-dibromobutane
D. 2,3-dibromobutane
Answer: A
Rationale: The addition of HBr to an alkyne follows Markovnikov’s rule twice when the
reagent is in excess. The first addition forms a vinyl bromide with the bromine on the
second carbon. The second addition places the second bromine on the same carbon,
resulting in a geminal dibromide.
5. Which of the following steps is classified as a propagation step in the radical chlorination of
methane?
A. Cl2 + hv -> 2 Cl.
B. CH4 + Cl. -> .CH3 + HCl
C. .CH3 + .CH3 -> CH3CH3
D. .CH3 + Cl. -> CH3Cl
Answer: B
Rationale: Propagation steps are characterized by the reaction of a radical with a neutral
molecule to produce a new radical and a new neutral molecule. In this specific step, a
chlorine radical abstracts a hydrogen atom from methane to form a methyl radical. This
maintains the chain reaction by providing the species needed for the subsequent step.
Chemistry I | Actual Q&A with Rationale
(CHM2210 Exam 4) | University of Central
Florida
1. Which of the following reagent sets will convert 2-hexyne into cis-2-hexene?
A. H2, Lindlar’s catalyst
B. H2, Pd/C
C. Na, NH3 (liq)
D. O3 followed by Zn/H2O
Answer: A
Rationale: Lindlar’s catalyst is a deactivated palladium catalyst that facilitates the partial
hydrogenation of alkynes to alkenes. Because the hydrogen atoms are added to the same
face of the triple bond, the reaction is stereoselective for the cis-alkene. Using sodium in
liquid ammonia would instead yield the trans-alkene through a radical-anion mechanism.
2. Identify the major product of the reaction between 1-pentyne and H2SO4, H2O, and
HgSO4.
A. Pentanal
B. 1-pentanol
C. 2-pentanone
,D. 2-pentanol
Answer: C
Rationale: The acid-catalyzed hydration of a terminal alkyne in the presence of mercuric
sulfate follows Markovnikov’s rule. The water molecule adds to the more substituted
carbon of the triple bond to form an enol intermediate. This enol rapidly undergoes keto-
enol tautomerism to produce a methyl ketone, specifically 2-pentanone.
3. Which radical species is the most stable?
A. Primary alkyl radical
B. Tertiary alkyl radical
C. Secondary alkyl radical
D. Methyl radical
Answer: B
Rationale: Radical stability follows the same trend as carbocation stability because
radicals are electron-deficient species. Tertiary radicals are stabilized by the inductive
effect and hyperconjugation from three surrounding alkyl groups. Consequently, they have
lower bond dissociation energies compared to secondary, primary, or methyl radicals.
4. What is the expected product when 1-butyne reacts with excess HBr?
A. 2,2-dibromobutane
B. 1,2-dibromobutane
, C. 1,1-dibromobutane
D. 2,3-dibromobutane
Answer: A
Rationale: The addition of HBr to an alkyne follows Markovnikov’s rule twice when the
reagent is in excess. The first addition forms a vinyl bromide with the bromine on the
second carbon. The second addition places the second bromine on the same carbon,
resulting in a geminal dibromide.
5. Which of the following steps is classified as a propagation step in the radical chlorination of
methane?
A. Cl2 + hv -> 2 Cl.
B. CH4 + Cl. -> .CH3 + HCl
C. .CH3 + .CH3 -> CH3CH3
D. .CH3 + Cl. -> CH3Cl
Answer: B
Rationale: Propagation steps are characterized by the reaction of a radical with a neutral
molecule to produce a new radical and a new neutral molecule. In this specific step, a
chlorine radical abstracts a hydrogen atom from methane to form a methyl radical. This
maintains the chain reaction by providing the species needed for the subsequent step.
