Difficulty: Advanced | Prerequisites: Alkene addition reactions, nomenclature, carbocation stability, SN2 mechanism basics
This document covers the alkyne-specific reactions, multi-step synthesis problems, mechanism drawing, and the spectral analysis question from Dr. Salmon's Exam 4 practice. Alkynes add a layer of complexity over alkenes because they can undergo one or two rounds of addition, and they introduce acetylide chemistry and dissolving metal reductions. Synthesis questions tie everything together, and spectral analysis requires you to work backwards from data to structure.
Alkynes undergo many of the same additions as alkenes but can react with one or two equivalents of reagent, and they unlock acetylide chemistry (strong base + terminal alkyne gives a carbon nucleophile for SN2). Synthesis questions on this exam ask you to chain multiple reactions together. The spectral analysis question requires combining NMR, IR, and reaction clues to deduce a structure.
Acetylide ion
The conjugate base of a terminal alkyne, formed by deprotonating the sp C-H bond with a strong base such as NaNH2 or NaH. The resulting carbanion (RC≡C⁻) is a strong nucleophile and a strong base. In simple terms, you are turning a terminal alkyne into a carbon nucleophile that can attack electrophilic carbons in SN2 reactions.
Dissolving metal reduction (Na/NH3 or Li/NH3)
A method for reducing an internal alkyne to a trans alkene. Sodium (or lithium) metal in liquid ammonia donates electrons one at a time to the triple bond through a radical anion intermediate, and the geometry of the resulting alkene is exclusively trans (E). Think of it as the anti-Markovnikov, anti-addition cousin of catalytic hydrogenation.
Lindlar's catalyst
A "poisoned" palladium catalyst (Pd/CaCO3, quinoline) that reduces an alkyne to a cis alkene only, stopping at one equivalent of H2. The catalyst is deactivated enough that it cannot reduce the resulting alkene further. This gives the Z (cis) alkene.
Tautomerisation (keto-enol tautomerism)
The interconversion between a ketone (keto form) and an enol (vinyl alcohol). When an alkyne undergoes Markovnikov hydration (H2O, H2SO4, HgSO4), the initial product is an enol, which immediately tautomerises to the more stable ketone.
SN2 reaction
A bimolecular nucleophilic substitution: the nucleophile attacks the electrophilic carbon in one concerted step, displacing the leaving group with inversion of configuration. Acetylide ions attack methyl and primary alkyl halides by SN2. This is how you extend a carbon chain.
Infrared (IR) spectroscopy
Measures the absorption of infrared light by molecular bonds. Each functional group absorbs at a characteristic wavenumber (cm⁻¹). Key absorptions for this exam: O-H stretch (broad, 3200-3600 cm⁻¹), C=O stretch (strong, around 1700 cm⁻¹), C≡C stretch (2100-2260 cm⁻¹), C-H stretch for sp carbons (around 3300 cm⁻¹, sharp).
¹H NMR spectroscopy
Measures the chemical environment of hydrogen atoms. Chemical shift (ppm) tells you what type of hydrogen it is. Splitting pattern tells you how many neighbouring hydrogens there are. Integration tells you the relative number of each type of hydrogen. Key shifts: alkyl C-H (0-2 ppm), allylic/next to C=C (around 4-5 ppm), vinyl =C-H (4.5-6.5 ppm), aldehyde (around 9-10 ppm).
Hydration of alkynes (H2O, H2SO4, HgSO4)
Markovnikov addition of water across the triple bond.
The initial product is an enol, which tautomerises to a ketone.
For terminal alkynes, this gives a methyl ketone (the oxygen ends up on C2).
Exam context: Question 3a (right arrow).
Addition of 2 equivalents of HX to alkynes
The first HX adds Markovnikov across the triple bond, giving a vinyl halide.
The second HX adds Markovnikov across the resulting alkene.
Both halogens end up on the same carbon: a geminal dihalide.
Exam context: Question 3a (left arrow, 2 equiv. HCl), Question 5a (mechanism for 2 equiv. HBr).
Halogenation of alkynes (2 equiv. Br2)
Two equivalents of Br2 add across the triple bond to give a tetrahalide (four bromine atoms, two on each carbon).
The first equivalent adds anti across the triple bond. The second adds anti across the resulting alkene.
Exam context: Question 3d (left arrow).
Dissolving metal reduction (Na or Li in NH3)
Reduces an internal alkyne to a trans (E) alkene.
The mechanism proceeds through radical anion intermediates, and the trans geometry is favoured because the bulky groups prefer to be on opposite sides of the developing double bond.
This is the complement to Lindlar hydrogenation, which gives the cis (Z) alkene.
Exam context: Question 3d (right arrow).
Acetylide SN2 reactions
Deprotonate a terminal alkyne with a strong base (NaNH2) to form the acetylide ion.
The acetylide ion attacks a methyl or primary alkyl halide in an SN2 reaction, forming a new C-C bond.
This is the primary method for building longer carbon chains from alkynes.
The mechanism is SN2 because the acetylide is a strong nucleophile and a good base, and it works best with unhindered substrates.
Exam context: Question 2a (acetylide + methyl bromide is SN2).
Synthesis questions (Question 4, worth 21 points) ask you to convert one compound into another using any reagents. The approach:
Work backwards from the product. What bonds need to form? What functional groups need to change?
Identify the key transformation: do you need to add carbons (use acetylide SN2), change stereochemistry (use a stereospecific reaction), convert a functional group (use the right reagent)?
Write the forward synthesis with all intermediates shown.
Exam example 4a: Cyclopentene to a cyclopentane bearing -CH3 and -OH (with specific stereochemistry, + enantiomer)
The product has both an -OH and a -CH3 on a cyclopentane ring with anti stereochemistry (implied by "+ enantiomer").
One approach: acid-catalysed hydration or oxymercuration-demercuration would add -OH with Markovnikov selectivity. But getting both -CH3 and -OH with the right regiochemistry and stereochemistry requires careful choice of reactions.
Consider: the starting material is an alkene. You might need to convert it to an epoxide (mCPBA), then open the epoxide with a methyl nucleophile (e.g. MeMgBr or MeLi), which would give anti addition of -OH and -CH3.
Exam example 4b: Cyclopentene to an epoxycyclopentane
The product is a cyclopentane with an epoxide fused to it.
Direct epoxidation with mCPBA converts the cyclopentene double bond to an epoxide in one step, with syn stereochemistry.
The spectral analysis question gives you clues from multiple sources and asks you to deduce structures.
How to approach it:
Start with the molecular formula or degree of unsaturation if available.
Use the NMR data: a signal near 4.7 ppm (singlet, 2H) suggests a =CH2 group (vinylic or allylic hydrogens). The singlet means no neighbouring H atoms, and 2H means two equivalent hydrogens.
Use the reaction clues: if compound X reacts with HCl to give a specific chloride, work backwards from that product. If ozonolysis of X gives compound Y and formaldehyde, then X has a =CH2 group that becomes the formaldehyde.
Use the IR spectrum: a broad O-H stretch at 3200-3600 cm⁻¹ tells you compound Y is an alcohol or carboxylic acid. A C=O stretch near 1700 cm⁻¹ indicates a ketone or aldehyde.
For this specific exam problem:
Compound X has a singlet at 4.7 ppm (2H), suggesting a =CH2 group.
Ozonolysis of X gives Y + formaldehyde, confirming the =CH2 (terminal alkene where the =CH2 becomes CH2O).
Reaction of X with HCl gives a chloride product (shown on the exam). Working backwards from the chloride product and the ozonolysis products allows you to reconstruct X.
Y's IR spectrum shows what functional groups Y contains, completing the puzzle.
Mechanism questions require you to show every electron-movement step with curved arrows.
Key rules for full marks:
Show every curved arrow starting from the electron source (lone pair or bond) and ending at the electron sink (electrophilic atom or bond being broken).
Show all intermediates. If a carbocation forms, draw it explicitly.
Show resonance structures of intermediates where relevant (e.g. allylic carbocation resonance).
For proton transfers, show the proton source and the base explicitly.
Exam example 5a: 2 equiv. HBr adding to a terminal alkyne
Step 1: Pi electrons of the triple bond attack H of HBr. Arrow from the triple bond to the H.
Intermediate: a vinyl carbocation (Markovnikov, more stable) with Br⁻ as counterion.
Step 2: Br⁻ attacks the carbocation. Arrow from Br⁻ lone pair to the positive carbon.
Product of first addition: a vinyl bromide (Markovnikov, on the more substituted carbon).
Steps 3-4: Repeat with the second equivalent of HBr on the resulting alkene.
Exam example 5b: Oxidation of an allylic alcohol to a ketone using base (-OH)
This involves deprotonation of the alcohol, then tautomerisation or elimination to reach the ketone product.
Show the base removing the O-H proton, forming an alkoxide, then the mechanism by which the alkoxide converts to the carbonyl product.
Students often confuse dissolving metal reduction (Na/NH3, gives trans alkene) with Lindlar hydrogenation (H2/Lindlar catalyst, gives cis alkene). A useful mnemonic: Na/NH3 has an "a" for anti, giving the trans product; Lindlar has an "L" for... well, there is no neat mnemonic, so just remember Lindlar = cis, dissolving metal = trans.
Assuming acetylide SN2 works on secondary and tertiary substrates. It does not. Acetylide ions are strong bases as well as strong nucleophiles, so with hindered substrates you get elimination (E2) instead of substitution. Stick to methyl and primary halides.
Forgetting keto-enol tautomerism after alkyne hydration. The Markovnikov hydration of a terminal alkyne does not stop at the enol. The enol spontaneously tautomerises to the ketone under the acidic conditions. The product you draw should be the ketone, not the enol.
In spectral analysis, confusing the =CH2 singlet in NMR with other types of hydrogens near 4.7 ppm. Vinylic hydrogens on a terminal =CH2 appear in that region, but so do some benzylic or allylic protons. Use the ozonolysis and reaction data together with the NMR to confirm.
⚠️ Synthesis (Question 4) is worth 21 points. You must show all isolable intermediates, not just the reagents. Skipping an intermediate loses marks even if the final product is correct.
⚠️ Mechanism (Question 5) is worth 27 points. Curved arrows are required for every electron-movement step. Missing arrows, skipped intermediates, or missing resonance structures all cost points.
⚠️ Spectral analysis (Question 6) is worth 6 points but is often the question students leave until last and rush. Practise working backwards from ozonolysis products and NMR signals before the exam so you can do it quickly under time pressure.
⚠️ Acetylide chemistry is tested in Question 2a (multiple choice: the mechanism is SN2). It is also a likely tool in synthesis problems where you need to form a new C-C bond.
⚠️ The exam gives you 50 minutes for 100 points. Budget roughly 30 seconds per point: 4 minutes for nomenclature, 4 for multiple choice, 15 for contrasting reactions, 10.5 for synthesis, 13.5 for mechanisms, 3 for spectral analysis.
True or false: Dissolving metal reduction (Na/NH3) of an internal alkyne gives a cis alkene.
False. It gives a trans (E) alkene. Lindlar hydrogenation gives the cis (Z) alkene.
Fill in the blank: An acetylide ion reacting with a primary alkyl halide proceeds by the ______ mechanism.
SN2.
True or false: Markovnikov hydration of a terminal alkyne gives an aldehyde.
False. It gives a methyl ketone (the enol tautomerises to the ketone with the carbonyl on the internal carbon).
Fill in the blank: In IR spectroscopy, a broad absorption between 3200 and 3600 cm⁻¹ indicates an ______ group.
O-H (hydroxyl) group.
True or false: Adding 2 equivalents of HBr to a terminal alkyne gives a vicinal dibromide (bromines on adjacent carbons).
False. It gives a geminal dibromide (both bromines on the same carbon), because both additions follow Markovnikov's rule.
Q: Show how you would convert a terminal alkyne (e.g. 1-pentyne) into a longer internal alkyne (e.g. 3-heptyne). What reagents do you need?
A: Step 1, deprotonate 1-pentyne with NaNH2 to form the sodium acetylide (pentynide ion). Step 2, react the acetylide with 1-bromoethane (a primary alkyl halide) in an SN2 reaction. The acetylide attacks the electrophilic carbon, displacing bromide and forming 3-heptyne.
Q: What is the product of treating 2-butyne with Na in liquid NH3?
A: Trans-2-butene (the E isomer). Dissolving metal reduction of an internal alkyne gives the trans alkene selectively.
Q: Compound X shows a singlet at 4.7 ppm (2H) in its ¹H NMR. Ozonolysis of X gives formaldehyde and another compound Y. What can you deduce about the structure of X?
A: The singlet at 4.7 ppm (2H) indicates a terminal =CH2 group (two equivalent vinylic hydrogens with no adjacent H neighbours to cause splitting). The ozonolysis producing formaldehyde confirms the =CH2: when ozone cleaves a terminal C=CH2, the CH2 fragment becomes formaldehyde (CH2O). Compound X therefore has the partial structure R-C(=CH2)-... where R is whatever remains as compound Y after the double bond is cleaved.
Q: Write the mechanism for the first equivalent of HBr adding to propyne. Include curved arrows.
A: The pi electrons of the C≡C triple bond act as a nucleophile and attack the electrophilic H of HBr (curved arrow from the triple bond to H). This generates a vinyl carbocation at C2 (the more substituted position, Markovnikov) and bromide ion. Then Br⁻ attacks the positive carbon (curved arrow from Br⁻ lone pair to C+). Product: 2-bromopropene.
Q: You need to convert cyclohexene into an epoxide. What single reagent accomplishes this?
A: mCPBA (meta-chloroperoxybenzoic acid). It delivers an oxygen atom to one face of the alkene in a concerted mechanism, forming 1,2-epoxycyclohexane with syn stereochemistry.
alkyne reactions organic chemistry, dissolving metal reduction Na NH3 trans alkene, Lindlar catalyst cis alkene, acetylide ion SN2, terminal alkyne deprotonation NaNH2, keto-enol tautomerism alkyne hydration, geminal dihalide 2 equiv HBr alkyne, multi-step synthesis organic chemistry, retrosynthetic analysis, epoxide opening mechanism, curved arrow mechanism organic chemistry, IR spectroscopy functional groups, ¹H NMR chemical shift alkene, ozonolysis structure determination, spectral analysis practice, CHEM 2301 Exam 4, Dr. Salmon, University of Minnesota organic chemistry, exam 4 synthesis strategy, how to draw mechanisms with curved arrows