Synthesis questions ask you to chain individual reactions into a multi-step route. Spectral analysis asks you to work backwards from IR and NMR data to identify an unknown structure. Both require you to hold the full toolkit of reactions in your head and apply them strategically.
Difficulty: Advanced | Prerequisites: All material from the two companion study notes (nomenclature, properties, and all individual reactions/mechanisms). Familiarity with IR absorption frequencies and basic 1H NMR interpretation (chemical shift, integration, splitting patterns).
Synthesis and spectral analysis sit at the top of the difficulty ladder in Organic Chemistry I. Synthesis requires you to think forwards (what reagents achieve each transformation?) and backwards (retrosynthetic analysis: what starting material could give me this target?). Spectral analysis, meanwhile, tests whether you understand the physical properties that arise from molecular structure, such as which functional groups absorb at which IR frequencies and how different hydrogen environments appear in an NMR spectrum. These are the questions that separate students who have memorised individual reactions from those who can apply them flexibly.
Acetylide anion
The conjugate base of a terminal alkyne, formed by deprotonation with a strong base such as NaNH₂. The resulting RC≡C⁻ is a powerful carbon nucleophile. Think of it as a negatively charged carbon that can form new C-C bonds via SN2 reactions with primary alkyl halides.
SN2 alkylation of acetylides
The reaction of an acetylide anion (RC≡C⁻) with a primary alkyl halide (R'-CH₂-X) to form a new C-C bond and extend the carbon chain. Only primary (and methyl) halides work, because SN2 requires back-side attack at an unhindered carbon.
Retrosynthetic analysis
Working backwards from the target molecule to identify what starting materials and reactions could produce it. You mentally "disconnect" bonds in the target to reveal simpler precursors. Symbolised by a double-barred arrow (⇒).
Markovnikov hydration of alkynes (HgSO₄/H₂SO₄/H₂O)
Acid-catalysed addition of water to an alkyne in the presence of mercury(II) sulfate. For a terminal alkyne, this gives a methyl ketone (Markovnikov regiochemistry: the OH ends up on the more substituted carbon, which then tautomerises from an enol to a ketone).
IR spectroscopy (key absorptions for alkynes)
Infrared spectroscopy measures which frequencies of light a molecule absorbs, revealing functional groups. For alkynes:
sp C-H stretch: a sharp, medium peak near 3300 cm⁻¹ (diagnostic of a terminal alkyne).
C≡C stretch: appears near 2100-2200 cm⁻¹. Internal alkynes may show a weak or absent C≡C stretch (symmetric stretches are IR-inactive).
¹H NMR spectroscopy (basics)
Nuclear magnetic resonance reveals the number and types of hydrogen environments in a molecule. Three things to read from a spectrum:
Chemical shift (δ, in ppm): tells you what kind of environment the hydrogen is in. Alkyl H near 0.9-1.5 ppm; propargylic/allylic H near 1.5-2.5 ppm; terminal alkyne H near 1.8-2.0 ppm.
Integration: the area under each peak, proportional to the number of hydrogens in that environment.
Splitting pattern: caused by neighbouring non-equivalent hydrogens (n+1 rule).
Enol-keto tautomerism
The equilibrium between an enol (C=C-OH) and a ketone or aldehyde (C-C=O). In most cases the keto form is thermodynamically favoured. Hydration of an alkyne initially forms an enol, which tautomerises to the carbonyl compound.
The exam asks you to synthesise a target alcohol starting only from acetylene (HC≡CH) and bromoethane (CH₃CH₂Br). Every carbon in the product must come from these two building blocks.
Step-by-step route:
Step 1: Deprotonate acetylene with NaNH₂ to form the sodium acetylide (HC≡C⁻ Na⁺).
Step 2: React the acetylide with bromoethane (SN2). This gives 1-butyne (CH₃CH₂C≡CH). One equivalent of bromoethane has been added.
Step 3: Deprotonate 1-butyne again with NaNH₂ to form CH₃CH₂C≡C⁻ Na⁺.
Step 4: React this acetylide with a second equivalent of bromoethane (SN2). This gives 3-hexyne (CH₃CH₂C≡CCH₂CH₃). Now all six carbons are in place.
Step 5: Convert the internal alkyne to the target. If the target is an alcohol (3-hexanol), you need to reduce the triple bond to a double bond or single bond and then hydrate. The exact final steps depend on the target structure. For 3-hexyn-1-ol, you would use hydroboration-oxidation on the alkyne.
Key strategic points:
Acetylide alkylation only works with primary (or methyl) alkyl halides. Never use a secondary or tertiary halide (you would get elimination instead of substitution).
You can add one equivalent of alkyl halide at a time, deprotonating between additions.
Count carbons in the target and match them to acetylene (2 C) plus however many equivalents of bromoethane (2 C each) are needed.
The exam gives you the following clues about compound A:
Compound A, when treated with H₂O/HgSO₄/H₂SO₄, gives 2-butanone (methyl ethyl ketone).
IR spectrum shows a medium peak at ~3300 cm⁻¹.
¹H NMR shows three signals: 2H, 1H, and 3H.
Working through the problem:
The conversion to 2-butanone via mercury-catalysed hydration tells you compound A is a terminal alkyne. Markovnikov hydration of a terminal alkyne gives a methyl ketone. For the product to be 2-butanone (CH₃COCH₂CH₃, four carbons), the starting alkyne must also be four carbons: 1-butyne (CH₃CH₂C≡CH).
The IR peak at ~3300 cm⁻¹ confirms a terminal alkyne (sp C-H stretch).
The ¹H NMR is consistent with 1-butyne:
3H signal (~1.0 ppm): the CH₃ group (triplet, coupled to adjacent CH₂).
2H signal (~2.2 ppm): the CH₂ group (a multiplet, coupled to both CH₃ and ≡CH).
1H signal (~1.8 ppm): the terminal ≡C-H (a triplet, coupled to adjacent CH₂).
Answers:
(a) Compound A is 1-butyne (CH₃CH₂C≡CH).
(b) The fragment responsible for the ~3300 cm⁻¹ peak is the sp C-H bond (≡C-H).
(c) The second characteristic IR peak would be the C≡C stretch, at approximately 2100-2200 cm⁻¹.
Acetylide alkylation is the basis of many pharmaceutical and materials chemistry syntheses where precise C-C bond formation is needed. Spectroscopic identification (IR + NMR) is how every organic chemist in a lab confirms that a reaction has produced the intended product. These are not just exam skills; they are the daily tools of the discipline.
Students often try to alkylate an acetylide with a secondary halide. This fails because SN2 does not work at secondary or tertiary carbons; you get elimination (E2) instead. Only primary and methyl halides are suitable.
Many students confuse the hydration of alkynes with the hydration of alkenes. Mercury-catalysed hydration of a terminal alkyne gives a ketone (via enol tautomerism), not an alcohol directly. Hydroboration-oxidation of an alkyne gives an aldehyde (anti-Markovnikov).
In spectral analysis, students sometimes mistake the O-H stretch (broad, 3200-3600 cm⁻¹) for the sp C-H stretch (sharp, ~3300 cm⁻¹). The shape of the peak matters: the alkyne C-H is sharp and narrow, while O-H is broad.
Students forget that the C≡C stretch (~2100-2200 cm⁻¹) is the second diagnostic peak for an alkyne in IR. When asked for a second characteristic absorption, this is the answer.
⚠️ The synthesis problem is worth 10 points. You must show every reagent and every isolable intermediate. Missing one intermediate or one set of reagents costs points even if the final product is correct.
⚠️ Acetylide alkylation is the go-to method for building longer carbon chains from short fragments. Expect to see it on every exam that covers alkynes.
⚠️ The spectral analysis problem (8 points) requires you to work from the chemical transformation backwards: if the product is a methyl ketone and the reaction is mercury-catalysed hydration, the starting material must be a terminal alkyne.
⚠️ Know the two diagnostic IR peaks for a terminal alkyne: sp C-H stretch (~3300 cm⁻¹) and C≡C stretch (~2100-2200 cm⁻¹). Both are fair game.
⚠️ For NMR, be able to match integration ratios (1H:2H:3H) and chemical shift ranges to the proposed structure.
Fill in the blank: To deprotonate a terminal alkyne, use the base ___. (NaNH₂, sodium amide)
True or false: Acetylide anions can be alkylated with secondary alkyl halides via SN2. (False. Only primary and methyl halides.)
Fill in the blank: Mercury-catalysed hydration of 1-butyne gives ___ via enol-keto tautomerism. (2-butanone)
True or false: A sharp, medium IR absorption at ~3300 cm⁻¹ is characteristic of an O-H stretch. (False. A sharp peak at 3300 cm⁻¹ is the sp C-H stretch of a terminal alkyne. The O-H stretch is broad.)
Fill in the blank: The C≡C stretch appears in the IR at approximately - cm⁻¹. (2100-2200)
Q: Starting from acetylene and bromoethane only, synthesise 3-hexyne. Show all reagents and intermediates.
A: (1) HC≡CH + NaNH₂ → HC≡C⁻ Na⁺. (2) HC≡C⁻ + CH₃CH₂Br → CH₃CH₂C≡CH (1-butyne). (3) CH₃CH₂C≡CH + NaNH₂ → CH₃CH₂C≡C⁻ Na⁺. (4) CH₃CH₂C≡C⁻ + CH₃CH₂Br → CH₃CH₂C≡CCH₂CH₃ (3-hexyne).
Q: An unknown compound shows a sharp medium peak at 3300 cm⁻¹ and a weak absorption at 2150 cm⁻¹ in its IR spectrum. What functional group is present?
A: A terminal alkyne. The 3300 cm⁻¹ peak is the sp C-H stretch and the 2150 cm⁻¹ peak is the C≡C stretch.
Q: Compound B gives 2-pentanone when treated with H₂O/HgSO₄/H₂SO₄. What is compound B?
A: 1-pentyne (CH₃CH₂CH₂C≡CH). Mercury-catalysed Markovnikov hydration of a terminal alkyne places the OH on the internal carbon (C1 of the triple bond becomes the carbonyl carbon), giving a methyl ketone. For 2-pentanone (five carbons, C=O at C2), the starting material is a five-carbon terminal alkyne: 1-pentyne.
Q: Why can you not use 2-bromopropane instead of bromoethane in an acetylide alkylation?
A: 2-Bromopropane is a secondary alkyl halide. Acetylide anions are strong bases as well as nucleophiles, and with a secondary halide the E2 elimination pathway dominates over SN2 substitution. You would get propene (an alkene) as the major product instead of the desired alkylation product.
Q: In the ¹H NMR of 1-butyne, how many distinct hydrogen environments are there, and what are their approximate integrations?
A: Three environments. The terminal ≡C-H (1H), the CH₂ adjacent to the triple bond (2H), and the CH₃ group (3H). Total integration ratio is 1:2:3.
Synthesis tightly connects to every individual reaction covered in the companion reactions study notes. If you cannot recall the product of hydroboration-oxidation or the regiochemistry of HBr/ROOR, you will struggle with multi-step routes. Spectral analysis connects forward to the more advanced NMR and mass spectrometry interpretation in Organic Chemistry II, and backward to hybridisation and functional group identification from earlier in the course.
acetylide anion, sodium amide NaNH2, SN2 alkylation alkyne, retrosynthetic analysis, multi-step synthesis organic chemistry, mercury catalysed hydration alkyne, HgSO4 H2SO4 H2O, enol keto tautomerism, methyl ketone from terminal alkyne, IR spectroscopy sp C-H stretch 3300, C triple bond C stretch 2100, 1H NMR chemical shift integration, 1-butyne NMR, spectral analysis organic chemistry, CHEM 2301 exam 4, Novikov, University of Minnesota