CHM 2210 Exam 4 V1 | CHM 2210 Organic
Chemistry I | Actual Q&A with Rationale
(CHM2210 Exam 4) | University of Central
Florida
1. Which of the following describes the hybridization and geometry of the carbon atoms in
ethyne (acetylene)?
A. sp2, trigonal planar
B. sp2, linear
C. sp3, tetrahedral
D. sp, linear
Answer: D
Rationale: The carbon atoms in a triple bond are sp hybridized because they are bonded to
only two other atoms. This hybridization results in a bond angle of 180 degrees, which
corresponds to a linear geometry. The remaining two p orbitals on each carbon overlap to
form the two pi bonds of the alkyne.
2. What is the approximate pKa of a terminal alkyne, such as propyne?
A. 25
B. 45
C. 35
,D. 15
Answer: A
Rationale: Terminal alkynes are significantly more acidic than alkanes or alkenes due to
the high s-character of the sp hybridized carbon. An sp orbital is closer to the nucleus,
which stabilizes the resulting conjugate base, the acetylide ion, more effectively. The pKa of
approximately 25 allows terminal alkynes to be deprotonated by very strong bases like
NaNH2.
3. Which reagent is most suitable for the deprotonation of 1-hexyne to form a hexynyl
sodium salt?
A. NaOH
B. NaOCH3
C. NH3
D. NaNH2
Answer: D
Rationale: Sodium amide (NaNH2) is a sufficiently strong base to deprotonate a terminal
alkyne because its conjugate acid, NH3, has a pKa of 38. Hydroxide or alkoxide bases are
not strong enough as their conjugate acids have pKa values around 15.7 to 18, which is
lower than the pKa of the alkyne.
4. What is the major product formed when 1-butyne reacts with excess HBr?
A. 1,1-dibromobutane
, B. 2,2-dibromobutane
C. 1,2-dibromobutane
D. 2,3-dibromobutane
Answer: B
Rationale: The addition of HBr to an alkyne follows Markovnikov’s rule where the
hydrogen adds to the carbon with more hydrogens. After the first addition, the second
equivalent of HBr adds to the same carbon to stabilize the resulting carbocation via
resonance with the bromine lone pairs. This results in a geminal dihalide located at the C2
position.
5. The reaction of an internal alkyne with sodium in liquid ammonia (Na, NH3) produces
which type of alkene?
A. trans-alkene
B. cis-alkene
C. Terminal alkene
D. Alkane
Answer: A
Rationale: The dissolving metal reduction of alkynes proceeds via a radical anion
mechanism. This mechanism favors the formation of the more stable trans-alkene because
Chemistry I | Actual Q&A with Rationale
(CHM2210 Exam 4) | University of Central
Florida
1. Which of the following describes the hybridization and geometry of the carbon atoms in
ethyne (acetylene)?
A. sp2, trigonal planar
B. sp2, linear
C. sp3, tetrahedral
D. sp, linear
Answer: D
Rationale: The carbon atoms in a triple bond are sp hybridized because they are bonded to
only two other atoms. This hybridization results in a bond angle of 180 degrees, which
corresponds to a linear geometry. The remaining two p orbitals on each carbon overlap to
form the two pi bonds of the alkyne.
2. What is the approximate pKa of a terminal alkyne, such as propyne?
A. 25
B. 45
C. 35
,D. 15
Answer: A
Rationale: Terminal alkynes are significantly more acidic than alkanes or alkenes due to
the high s-character of the sp hybridized carbon. An sp orbital is closer to the nucleus,
which stabilizes the resulting conjugate base, the acetylide ion, more effectively. The pKa of
approximately 25 allows terminal alkynes to be deprotonated by very strong bases like
NaNH2.
3. Which reagent is most suitable for the deprotonation of 1-hexyne to form a hexynyl
sodium salt?
A. NaOH
B. NaOCH3
C. NH3
D. NaNH2
Answer: D
Rationale: Sodium amide (NaNH2) is a sufficiently strong base to deprotonate a terminal
alkyne because its conjugate acid, NH3, has a pKa of 38. Hydroxide or alkoxide bases are
not strong enough as their conjugate acids have pKa values around 15.7 to 18, which is
lower than the pKa of the alkyne.
4. What is the major product formed when 1-butyne reacts with excess HBr?
A. 1,1-dibromobutane
, B. 2,2-dibromobutane
C. 1,2-dibromobutane
D. 2,3-dibromobutane
Answer: B
Rationale: The addition of HBr to an alkyne follows Markovnikov’s rule where the
hydrogen adds to the carbon with more hydrogens. After the first addition, the second
equivalent of HBr adds to the same carbon to stabilize the resulting carbocation via
resonance with the bromine lone pairs. This results in a geminal dihalide located at the C2
position.
5. The reaction of an internal alkyne with sodium in liquid ammonia (Na, NH3) produces
which type of alkene?
A. trans-alkene
B. cis-alkene
C. Terminal alkene
D. Alkane
Answer: A
Rationale: The dissolving metal reduction of alkynes proceeds via a radical anion
mechanism. This mechanism favors the formation of the more stable trans-alkene because
