CHM 2210 Exam 3 V1 | CHM 2210 Organic Chemistry I | Actual Q&A
with Rationale (CHM2210 Exam 3) | University of Central Florida
1. Which reagent set is required to convert an internal alkyne into a trans-alkene?
A. H2, Lindlar’s catalyst
B. Na, NH3 (liq)
C. H2, Pd/C
D. 1) BH3, 2) H2O2, NaOH
Correct Answer: B
Explanation: The dissolving metal reduction using sodium in liquid ammonia proceeds
through a radical anion intermediate. This mechanism favors the formation of the more
stable trans-alkene via anti-addition of hydrogen atoms. In contrast, Lindlar’s catalyst
would produce a cis-alkene, while Pd/C would reduce the alkyne completely to an alkane.
2. What is the major product formed when 1-methylcyclohexene reacts with NBS and heat or
light?
A. 1-bromo-1-methylcyclohexane
B. 3-bromo-1-methylcyclohexene
C. 1,2-dibromo-1-methylcyclohexane
D. 2-bromo-1-methylcyclohexane
Correct Answer: B
Explanation: NBS (N-bromosuccinimide) is a reagent used specifically for allylic
bromination via a radical mechanism. The reaction proceeds through the most stable allylic
radical, which is the one formed at the secondary allylic position in this case. This allows
the double bond to remain in the molecule while substituting a hydrogen at the allylic
position.
3. Which of the following radicals is the most stable?
A. Primary radical
B. Tertiary radical
C. Secondary radical
D. Methyl radical
Correct Answer: B
,Explanation: Radical stability follows the same trend as carbocation stability because
radicals are electron-deficient species. Tertiary radicals are the most stable due to the
greatest degree of hyperconjugation from adjacent alkyl groups. Methyl radicals are the
least stable because they lack any stabilizing alkyl groups.
4. What reagents are used for the oxymercuration-demercuration of an alkyne?
A. 1) BH3, 2) H2O2, NaOH
B. HgSO4, H2O, H2SO4
C. H2, Pt
D. Na, NH3
Correct Answer: B
Explanation: Acid-catalyzed hydration of an alkyne using mercury(II) sulfate results in the
Markovnikov addition of water. The initial product is an enol, which is unstable and rapidly
tautomerizes to a ketone. This reaction is a standard method for converting terminal
alkynes into methyl ketones.
5. In the reaction of 1-pentyne with 1 equivalent of HBr, what is the primary product?
A. 1-bromopentene
B. 1,1-dibromopentane
C. 2,2-dibromopentane
D. 2-bromopentene
Correct Answer: D
Explanation: The addition of hydrogen halides to alkynes follows Markovnikov’s rule
where the halogen adds to the more substituted carbon. In 1-pentyne, the C2 position is the
more substituted carbon compared to C1. Therefore, the formation of the vinyl bromide
occurs preferentially at the second carbon of the chain.
6. Which reagent will oxidize a primary alcohol to a carboxylic acid?
A. Na2Cr2O7, H2SO4, H2O
B. PCC, CH2Cl2
C. DMP, CH2Cl2
D. 1) DMSO, (COCl)2, 2) Et3N
Correct Answer: A
Explanation: Sodium dichromate in aqueous acid, also known as Jones reagent, is a strong
oxidizing agent. It oxidizes primary alcohols past the aldehyde stage all the way to
, carboxylic acids. Other reagents like PCC or the Swern oxidation are milder and stop at the
aldehyde.
7. What is the product of the reaction between phenylmagnesium bromide and ethylene
oxide, followed by aqueous workup?
A. Phenol
B. Benzyl alcohol
C. 2-phenylethanol
D. 3-phenylpropanol
Correct Answer: C
Explanation: A Grignard reagent acts as a nucleophile and attacks the less hindered carbon
of the epoxide ring. Ethylene oxide is a two-carbon epoxide, so the phenyl group adds to
one carbon and the ring opens to leave a hydroxyl group on the second carbon. Upon acidic
workup, the resulting alkoxide is protonated to form 2-phenylethanol.
8. Which of the following is the best leaving group for a nucleophilic substitution reaction?
A. F-
B. Cl-
C. Br-
D. I-
Correct Answer: D
Explanation: Leaving group ability generally increases with the size of the atom and the
stability of the conjugate base. Iodide is the largest of the halides listed and has the weakest
bond to carbon, making it the most easily displaced. Consequently, alkyl iodides are
typically the most reactive in SN1 and SN2 reactions.
9. Which solvent is most suitable for a Williamson ether synthesis?
A. CH3OH
B. H2O
C. THF
D. CH3COOH
Correct Answer: C
Explanation: Williamson ether synthesis involves the reaction of an alkoxide nucleophile
with an alkyl halide. Polar aprotic solvents like THF or DMF are preferred because they do
not solvate the nucleophile strongly, keeping it reactive. Protic solvents like water or
methanol would protonate the alkoxide, rendering it ineffective as a nucleophile.
with Rationale (CHM2210 Exam 3) | University of Central Florida
1. Which reagent set is required to convert an internal alkyne into a trans-alkene?
A. H2, Lindlar’s catalyst
B. Na, NH3 (liq)
C. H2, Pd/C
D. 1) BH3, 2) H2O2, NaOH
Correct Answer: B
Explanation: The dissolving metal reduction using sodium in liquid ammonia proceeds
through a radical anion intermediate. This mechanism favors the formation of the more
stable trans-alkene via anti-addition of hydrogen atoms. In contrast, Lindlar’s catalyst
would produce a cis-alkene, while Pd/C would reduce the alkyne completely to an alkane.
2. What is the major product formed when 1-methylcyclohexene reacts with NBS and heat or
light?
A. 1-bromo-1-methylcyclohexane
B. 3-bromo-1-methylcyclohexene
C. 1,2-dibromo-1-methylcyclohexane
D. 2-bromo-1-methylcyclohexane
Correct Answer: B
Explanation: NBS (N-bromosuccinimide) is a reagent used specifically for allylic
bromination via a radical mechanism. The reaction proceeds through the most stable allylic
radical, which is the one formed at the secondary allylic position in this case. This allows
the double bond to remain in the molecule while substituting a hydrogen at the allylic
position.
3. Which of the following radicals is the most stable?
A. Primary radical
B. Tertiary radical
C. Secondary radical
D. Methyl radical
Correct Answer: B
,Explanation: Radical stability follows the same trend as carbocation stability because
radicals are electron-deficient species. Tertiary radicals are the most stable due to the
greatest degree of hyperconjugation from adjacent alkyl groups. Methyl radicals are the
least stable because they lack any stabilizing alkyl groups.
4. What reagents are used for the oxymercuration-demercuration of an alkyne?
A. 1) BH3, 2) H2O2, NaOH
B. HgSO4, H2O, H2SO4
C. H2, Pt
D. Na, NH3
Correct Answer: B
Explanation: Acid-catalyzed hydration of an alkyne using mercury(II) sulfate results in the
Markovnikov addition of water. The initial product is an enol, which is unstable and rapidly
tautomerizes to a ketone. This reaction is a standard method for converting terminal
alkynes into methyl ketones.
5. In the reaction of 1-pentyne with 1 equivalent of HBr, what is the primary product?
A. 1-bromopentene
B. 1,1-dibromopentane
C. 2,2-dibromopentane
D. 2-bromopentene
Correct Answer: D
Explanation: The addition of hydrogen halides to alkynes follows Markovnikov’s rule
where the halogen adds to the more substituted carbon. In 1-pentyne, the C2 position is the
more substituted carbon compared to C1. Therefore, the formation of the vinyl bromide
occurs preferentially at the second carbon of the chain.
6. Which reagent will oxidize a primary alcohol to a carboxylic acid?
A. Na2Cr2O7, H2SO4, H2O
B. PCC, CH2Cl2
C. DMP, CH2Cl2
D. 1) DMSO, (COCl)2, 2) Et3N
Correct Answer: A
Explanation: Sodium dichromate in aqueous acid, also known as Jones reagent, is a strong
oxidizing agent. It oxidizes primary alcohols past the aldehyde stage all the way to
, carboxylic acids. Other reagents like PCC or the Swern oxidation are milder and stop at the
aldehyde.
7. What is the product of the reaction between phenylmagnesium bromide and ethylene
oxide, followed by aqueous workup?
A. Phenol
B. Benzyl alcohol
C. 2-phenylethanol
D. 3-phenylpropanol
Correct Answer: C
Explanation: A Grignard reagent acts as a nucleophile and attacks the less hindered carbon
of the epoxide ring. Ethylene oxide is a two-carbon epoxide, so the phenyl group adds to
one carbon and the ring opens to leave a hydroxyl group on the second carbon. Upon acidic
workup, the resulting alkoxide is protonated to form 2-phenylethanol.
8. Which of the following is the best leaving group for a nucleophilic substitution reaction?
A. F-
B. Cl-
C. Br-
D. I-
Correct Answer: D
Explanation: Leaving group ability generally increases with the size of the atom and the
stability of the conjugate base. Iodide is the largest of the halides listed and has the weakest
bond to carbon, making it the most easily displaced. Consequently, alkyl iodides are
typically the most reactive in SN1 and SN2 reactions.
9. Which solvent is most suitable for a Williamson ether synthesis?
A. CH3OH
B. H2O
C. THF
D. CH3COOH
Correct Answer: C
Explanation: Williamson ether synthesis involves the reaction of an alkoxide nucleophile
with an alkyl halide. Polar aprotic solvents like THF or DMF are preferred because they do
not solvate the nucleophile strongly, keeping it reactive. Protic solvents like water or
methanol would protonate the alkoxide, rendering it ineffective as a nucleophile.
