Department of Chemistry
012 • 3 MAXE
CH College of Natural Sciences & Mathematics
VERITAS ET SCIENTIA
EST. 1869
CHEM 210 — Exam 3
A L KY N E S , O R G A N O M E TA L L I C R E A G E N TS & R E A C T I O N M E C H A N I S M S
INSTITUTION Department of Chemistry EXAM CODE CHEM-210-EX3-2026
PROGRAM CHEM 210 — Organic Chemistry ACADEMIC YEAR
EXAM TITLE Exam 3 — Alkynes, Organometallics & TOTAL QUESTIONS 30 Questions
Oxidation/Reduction
COURSE TITLE Organic Chemistry I FORMAT Multiple Choice — Select the Single Best
Answer
EXAMINATION INSTRUCTIONS
▸ Select the single best answer for each question.
▸ Questions cover terminal alkyne chemistry (acidity, alkylation), hydration and hydroboration-oxidation of alkynes, reduction of
alkynes (Lindlar, dissolving metal, catalytic hydrogenation), Grignard and Gilman reagents, SN2/E2 mechanisms,
stereochemistry, and radical reactions.
▸ Distinguish carefully between related reagents: HgSO₄/H₂SO₄/H₂O vs. bulky borane/H₂O₂/OH⁻; Lindlar catalyst vs. Li/Na in liquid
NH₃; Grignard vs. Gilman reagents.
▸ Correct answers and detailed rationales appear below each question for comprehensive review.
▸ All content is derived from CHEM 210 Exam 3 curriculum.
SECTION I — ALKYNES, ORGANOMETALLICS & REACTION MECHANISMS Questions 1 – 30
1. The pKa of a terminal alkyne is approximately:
A. 10
B. 25
C. 44
D. 50
CORRECT ANSWER B — 25
RATIONALE Terminal alkynes have a pKa of ~25, making them significantly more acidic than alkenes (pKa ~44) or alkanes
(pKa ~50). This acidity is due to the sp-hybridized carbon of the triple bond, which has 50% s-character.
Greater s-character means electrons are held closer to the nucleus, stabilizing the conjugate base (acetylide
anion).
, 2. Which proton on a terminal alkyne is acidic?
A. The H attached to the sp² carbon
B. The H attached to the sp³ carbon adjacent to the triple bond
C. The H directly attached to the sp carbon of the triple bond
D. All hydrogens on a terminal alkyne are equally acidic
CORRECT ANSWER C — The H directly attached to the sp carbon of the triple bond
RATIONALE Only the hydrogen bonded to the sp-hybridized carbon (C≡C–H) is acidic. The sp carbon has 50% s-character,
which stabilizes the conjugate base (acetylide anion, RC≡C:⁻) by holding the lone pair closer to the nucleus.
This is why NaNH₂ (a strong base) can deprotonate terminal alkynes but not alkenes or alkanes.
3. Which base is commonly used to deprotonate a terminal alkyne?
A. NaOH
B. NaNH₂
C. NaHCO₃
D. Pyridine
CORRECT ANSWER B — NaNH₂
RATIONALE Sodium amide (NaNH₂) is a strong base (conjugate acid NH₃, pKa ~38) that quantitatively deprotonates
terminal alkynes (pKa ~25). NaOH (pKa of H₂O ~15.7) is not strong enough to deprotonate an alkyne. The
resulting acetylide anion (RC≡C:⁻ Na⁺) is a powerful nucleophile used in alkylation and addition reactions.
4. When a terminal alkyne is deprotonated, what is formed?
A. A carbocation
B. A radical
C. An acetylide anion
D. An alkene
CORRECT ANSWER C — An acetylide anion
RATIONALE Deprotonation of a terminal alkyne (RC≡CH) with NaNH₂ produces an acetylide anion (RC≡C:⁻). This anion is
stabilized by the sp hybridization (50% s-character) and serves as a strong nucleophile. Acetylide anions
undergo SN2 reactions with primary alkyl halides to form new C–C bonds (alkylation), a key reaction in
organic synthesis.
5. Why are terminal alkynes more acidic than alkenes or alkanes?
A. Their conjugate base is sp³ hybridized, which is more stable
B. Their conjugate base is sp hybridized with 50% s character, which stabilizes negative charge
C. Alkynes have more hydrogen atoms than alkenes
D. Alkynes are actually less acidic than alkanes
CORRECT ANSWER B — Their conjugate base is sp hybridized with 50% s character, which stabilizes negative charge
RATIONALE Acidity trends: alkyne (pKa ~25) > alkene (pKa ~44) > alkane (pKa ~50). The key factor is hybridization of the
conjugate base's carbanion: sp (50% s-character) > sp² (33%) > sp³ (25%). Higher s-character means electrons
are held closer to the nucleus — the lone pair in the acetylide anion is more stable (lower energy), making
deprotonation more favorable.
012 • 3 MAXE
CH College of Natural Sciences & Mathematics
VERITAS ET SCIENTIA
EST. 1869
CHEM 210 — Exam 3
A L KY N E S , O R G A N O M E TA L L I C R E A G E N TS & R E A C T I O N M E C H A N I S M S
INSTITUTION Department of Chemistry EXAM CODE CHEM-210-EX3-2026
PROGRAM CHEM 210 — Organic Chemistry ACADEMIC YEAR
EXAM TITLE Exam 3 — Alkynes, Organometallics & TOTAL QUESTIONS 30 Questions
Oxidation/Reduction
COURSE TITLE Organic Chemistry I FORMAT Multiple Choice — Select the Single Best
Answer
EXAMINATION INSTRUCTIONS
▸ Select the single best answer for each question.
▸ Questions cover terminal alkyne chemistry (acidity, alkylation), hydration and hydroboration-oxidation of alkynes, reduction of
alkynes (Lindlar, dissolving metal, catalytic hydrogenation), Grignard and Gilman reagents, SN2/E2 mechanisms,
stereochemistry, and radical reactions.
▸ Distinguish carefully between related reagents: HgSO₄/H₂SO₄/H₂O vs. bulky borane/H₂O₂/OH⁻; Lindlar catalyst vs. Li/Na in liquid
NH₃; Grignard vs. Gilman reagents.
▸ Correct answers and detailed rationales appear below each question for comprehensive review.
▸ All content is derived from CHEM 210 Exam 3 curriculum.
SECTION I — ALKYNES, ORGANOMETALLICS & REACTION MECHANISMS Questions 1 – 30
1. The pKa of a terminal alkyne is approximately:
A. 10
B. 25
C. 44
D. 50
CORRECT ANSWER B — 25
RATIONALE Terminal alkynes have a pKa of ~25, making them significantly more acidic than alkenes (pKa ~44) or alkanes
(pKa ~50). This acidity is due to the sp-hybridized carbon of the triple bond, which has 50% s-character.
Greater s-character means electrons are held closer to the nucleus, stabilizing the conjugate base (acetylide
anion).
, 2. Which proton on a terminal alkyne is acidic?
A. The H attached to the sp² carbon
B. The H attached to the sp³ carbon adjacent to the triple bond
C. The H directly attached to the sp carbon of the triple bond
D. All hydrogens on a terminal alkyne are equally acidic
CORRECT ANSWER C — The H directly attached to the sp carbon of the triple bond
RATIONALE Only the hydrogen bonded to the sp-hybridized carbon (C≡C–H) is acidic. The sp carbon has 50% s-character,
which stabilizes the conjugate base (acetylide anion, RC≡C:⁻) by holding the lone pair closer to the nucleus.
This is why NaNH₂ (a strong base) can deprotonate terminal alkynes but not alkenes or alkanes.
3. Which base is commonly used to deprotonate a terminal alkyne?
A. NaOH
B. NaNH₂
C. NaHCO₃
D. Pyridine
CORRECT ANSWER B — NaNH₂
RATIONALE Sodium amide (NaNH₂) is a strong base (conjugate acid NH₃, pKa ~38) that quantitatively deprotonates
terminal alkynes (pKa ~25). NaOH (pKa of H₂O ~15.7) is not strong enough to deprotonate an alkyne. The
resulting acetylide anion (RC≡C:⁻ Na⁺) is a powerful nucleophile used in alkylation and addition reactions.
4. When a terminal alkyne is deprotonated, what is formed?
A. A carbocation
B. A radical
C. An acetylide anion
D. An alkene
CORRECT ANSWER C — An acetylide anion
RATIONALE Deprotonation of a terminal alkyne (RC≡CH) with NaNH₂ produces an acetylide anion (RC≡C:⁻). This anion is
stabilized by the sp hybridization (50% s-character) and serves as a strong nucleophile. Acetylide anions
undergo SN2 reactions with primary alkyl halides to form new C–C bonds (alkylation), a key reaction in
organic synthesis.
5. Why are terminal alkynes more acidic than alkenes or alkanes?
A. Their conjugate base is sp³ hybridized, which is more stable
B. Their conjugate base is sp hybridized with 50% s character, which stabilizes negative charge
C. Alkynes have more hydrogen atoms than alkenes
D. Alkynes are actually less acidic than alkanes
CORRECT ANSWER B — Their conjugate base is sp hybridized with 50% s character, which stabilizes negative charge
RATIONALE Acidity trends: alkyne (pKa ~25) > alkene (pKa ~44) > alkane (pKa ~50). The key factor is hybridization of the
conjugate base's carbanion: sp (50% s-character) > sp² (33%) > sp³ (25%). Higher s-character means electrons
are held closer to the nucleus — the lone pair in the acetylide anion is more stable (lower energy), making
deprotonation more favorable.
