Source: CHEM 212 LT6 Worksheet, Penn State
Tags: multi-step synthesis, organic synthesis, functional group transformation, alkyl halide, SN2, E2, alkynide, acetylide, C-C bond formation, pKa, retrosynthesis, CHEM 212
Difficulty: Intermediate | Prerequisites: SN1/SN2/E1/E2 mechanisms, alkene and alkyne reactions, acid-base chemistry (Chapters 7–11)
This is the point in CHEM 212 where individual reactions stop being isolated facts and start working together. Everything you have learnt about substitution, elimination, addition, and acid-base chemistry now feeds into a single skill: designing a sequence of reactions to build a target molecule from simpler starting materials. If you have fallen behind on SN1/SN2/E1/E2 or alkene reactions, revisit those before tackling this material. The payoff is large: synthesis problems appear on every organic chemistry exam, and the thinking here carries straight through to carbonyl chemistry and beyond.
Organic synthesis is about choosing the right sequence of reactions to convert a simple starting material into a more complex target compound. This set of notes covers how to make alcohols, alkenes, and thioethers from different starting points, how alkyl halides serve as a central hub for building many functional groups, and how alkynide anions let you form new carbon-carbon bonds. The key skill is learning to work both forwards and backwards through reaction steps to find the most efficient route.
Target compound
The final desired product of a synthesis. Think of it as the molecule you are trying to build.
Functional group transformation (FGT)
A reaction that converts one functional group into another on an existing carbon skeleton. In simple terms, you are swapping one chemical "attachment" for a different one without changing the backbone.
Alkyl halide
An organic compound in which a halogen atom (F, Cl, Br, I) is bonded to an sp3-hybridised carbon (R-X). Think of it as the Swiss Army knife of organic synthesis: it can be converted into dozens of other functional groups.
Nucleophilic substitution (SN1 / SN2)
A reaction in which a nucleophile replaces a leaving group on an electrophilic carbon. In simple terms, one group kicks out another and takes its place.
Elimination (E1 / E2)
A reaction that removes a leaving group and a beta-hydrogen to form a pi bond (typically an alkene). Think of it as stripping atoms away to create a double bond.
Alkynide (alkyne anion)
The conjugate base of a terminal alkyne, R-C≡C⁻. It is a strong base (pKa of the conjugate acid is approximately 25) and an excellent nucleophile for SN2 reactions that form new C-C bonds.
Acetylide
The specific alkynide where R = H, i.e. HC≡C⁻. Prepared by deprotonating acetylene with a sufficiently strong base.
pKa
A measure of acid strength: the lower the pKa, the stronger the acid. For bases, the relevant number is the pKa of the conjugate acid. A base whose conjugate acid has a high pKa is a strong base.
Retrosynthetic analysis
A problem-solving strategy where you start from the target compound and work backwards, disconnecting bonds to identify simpler precursors and the reactions needed. Think of it as reverse-engineering the recipe.
The worksheet opens by asking: given a target functional group, what reaction types could produce it? This is the heart of synthesis planning.
Making an alcohol (-OH)
Addition to an alkene: acid-catalysed hydration (H₃O⁺) gives Markovnikov addition. Hydroboration-oxidation (1. BH₃·THF, 2. H₂O₂, NaOH, H₂O) gives anti-Markovnikov addition. Hydroboration-oxidation is often the better choice when you need the OH on the less substituted carbon.
Substitution on an alkyl halide: SN1 or SN2 with water or hydroxide. SN1 (H₂O) works for tertiary and some secondary substrates. SN2 (NaOH) works for primary and methyl substrates.
You cannot make an alcohol from an elimination reaction. Elimination creates a double bond; it removes atoms rather than installing -OH.
Making an alkene (C=C)
Elimination from an alkyl halide: E2 with a strong base (NaOH, NaOtBu, or NaH). The base removes a beta-hydrogen while the leaving group departs.
Dehydration of an alcohol: H₂SO₄ with heat (acid-catalysed elimination).
You cannot make an alkene from an addition or substitution reaction. Those reactions consume double bonds or swap groups; they do not create pi bonds.
Making a thioether (R-S-R')
Substitution on an alkyl halide with a thiolate nucleophile (e.g. NaSCH₃). Thiolates are excellent SN2 nucleophiles because sulphur is large, polarisable, and a good nucleophile.
The SN2 mechanism proceeds via backside attack on the electrophilic carbon, with inversion of configuration.
Alkyl halides are the most versatile intermediates in introductory organic synthesis. From R-X, you can reach nearly every functional group covered in CHEM 212.
Starting from an alkyl halide (R-X), the following nucleophiles or bases give these products:
NaOH → alcohol (R-OH)
NaOCH₃ → methyl ether (R-OCH₃)
NaSH → thiol (R-SH)
NaSCH₃ → thioether (R-SCH₃)
NaC≡N → nitrile (R-C≡N), with a new C-C bond
NaC≡CH → terminal alkyne (R-C≡CH), with a new C-C bond
NaN₃ → azide (R-N₃)
NH₃ → amine (R-NH₂), though over-alkylation is a concern
NaI → alkyl iodide (R-I), useful for Finkelstein-type halide swaps
KOtBu → alkene (elimination product)
The key idea: alkyl halides sit at the centre of a web. Master the web and you can plan routes to almost any target from almost any starting material.
Forming new carbon-carbon bonds is one of the most important tasks in synthesis because it builds up the molecular skeleton. The alkynide anion (R-C≡C⁻) is the first C-C bond-forming nucleophile introduced in this course.
Structure and basicity
The alkynide is written as R-C≡C-Na, which dissociates to R-C≡C⁻ + Na⁺. The pKa of the conjugate acid (R-C≡CH) is approximately 25. Because this is well above 11, the alkynide is classified as a strong base.
Base strength thresholds to remember:
Strong base: pKa of conjugate acid > 11
Weak base: pKa of conjugate acid between 0 and 11
Very weak base: pKa of conjugate acid < 0
Reactions with alkyl halides
Because the alkynide is both a strong base and a strong nucleophile, both substitution and elimination can occur. The outcome depends on substrate class:
Primary alkyl halide (no branching at the alpha carbon): SN2 is the major product. E2 is minor. This is the substrate you want for C-C bond formation.
Secondary alkyl halide: E2 dominates. SN2 is minor. Not suitable for C-C bond formation with alkynides.
Tertiary alkyl halide: E2 only (steric hindrance blocks backside attack entirely). No substitution at all.
Critical rule: only primary alkyl halides and primary tosylates can be used for C-C bond forming via SN2 with alkynide nucleophiles.
To make an alkynide, you deprotonate a terminal alkyne (R-C≡C-H, pKa ≈ 25) with a base. The base's conjugate acid must have a pKa roughly 5 or more units higher than 25 to drive the equilibrium forward.
Base candidates from the worksheet:
KOtBu (pKa of conjugate acid ≈ 16): too weak, cannot deprotonate the alkyne
NaNH₂ (pKa ≈ 38): strong enough, this works ✓
NaOH (pKa ≈ 14): too weak
KCN (pKa ≈ 9): far too weak
CH₃OH (pKa ≈ -2.4 for conjugate acid CH₃OH₂⁺): far too weak, this is an acid
NaH (pKa ≈ 35): strong enough, this works ✓
The correct bases are NaNH₂ and NaH. Both have conjugate acids with pKa values well above 25.
Base strength classification by pKa of conjugate acid:
Strong base: pKa(BH) > 11
Weak base: pKa(BH) = 0 to 11
Very weak base: pKa(BH) < 0
Rule for deprotonation equilibrium:
To drive a deprotonation forward, the base's conjugate acid should have a pKa roughly 5 units higher than the acid being deprotonated. For terminal alkynes (pKa ≈ 25), the base needs a conjugate acid pKa of approximately 30 or above.
SN2 vs E2 outcome with strong base/nucleophile (alkynide):
Methyl or 1° R-X → SN2 dominates (C-C bond formed)
2° R-X → E2 dominates (alkene formed, no C-C bond)
3° R-X → E2 only (complete steric blockade of SN2)
Students often think you can make an alcohol through an elimination reaction. You cannot. Elimination creates a double bond; it removes atoms rather than installing -OH.
Students frequently try to use a secondary or tertiary alkyl halide for C-C bond formation with an alkynide. This fails because E2 dominates with bulky, strong bases on hindered substrates. Only primary substrates work for SN2-based C-C bond formation.
A common error is assuming any strong base can deprotonate a terminal alkyne. Bases like NaOH (pKa ≈ 14) and KOtBu (pKa ≈ 16) are strong bases in general, but they are far too weak to deprotonate an alkyne (pKa ≈ 25). You need a base whose conjugate acid has a pKa well above 25, such as NaNH₂ (38) or NaH (35).
Students sometimes confuse "strong base" in the general sense with "strong enough to deprotonate this specific acid." These are different questions. Whether a base can deprotonate something depends on the relative pKa values, not just whether the base is classified as "strong."
⚠️ Expect questions asking you to identify which reaction type (addition, substitution, elimination) can produce a given functional group. Know the "cannot" rules: no alcohol from elimination, no alkene from addition or substitution.
⚠️ The alkyl halide reaction web is almost certainly testable. You should be able to predict the product of R-X with any of the nucleophiles listed (NaOH, NaOCH₃, NaSH, NaSCH₃, NaCN, NaC≡CH, NaN₃, NH₃, NaI, KOtBu).
⚠️ Predicting SN2 vs E2 outcomes with alkynide nucleophiles on primary, secondary, and tertiary substrates is a high-probability exam question.
⚠️ Base selection for alkynide preparation (which bases can deprotonate a terminal alkyne, and why) tests your understanding of pKa-based equilibrium reasoning.
Real-world connection: Alkynide-based C-C bond formation is used in pharmaceutical synthesis. The worksheet references vildagliptin, a type 2 diabetes drug whose synthesis involves installing a nitrile group via SN2 with cyanide, a reaction that follows the same logic as alkynide substitution.
True or false: An elimination reaction can produce an alcohol. → False.
Fill in the blank: The pKa of a terminal alkyne (R-C≡CH) is approximately ___. → 25.
True or false: NaOH is a strong enough base to deprotonate a terminal alkyne. → False.
Fill in the blank: Only ___ alkyl halides or tosylates should be used for C-C bond formation with alkynide nucleophiles via SN2. → Primary.
True or false: An alkynide anion reacting with a tertiary alkyl halide will give an E2 elimination product, not a substitution product. → True.
Q: Name two ways to synthesise an alcohol from an alkene. Which gives Markovnikov regiochemistry and which gives anti-Markovnikov?
A: Acid-catalysed hydration (H₃O⁺) gives Markovnikov addition (OH on the more substituted carbon). Hydroboration-oxidation (1. BH₃·THF, 2. H₂O₂/NaOH/H₂O) gives anti-Markovnikov addition (OH on the less substituted carbon).
Q: You react NaC≡CH with a secondary alkyl bromide. What is the major product, and why?
A: The major product is an alkene (E2 elimination), not the desired alkyne (SN2). The alkynide is a strong base, and on a secondary substrate, E2 dominates over SN2 because of steric hindrance at the electrophilic carbon.
Q: Which of the following bases can deprotonate a terminal alkyne to form an alkynide: NaNH₂, KOtBu, NaH, NaOH?
A: NaNH₂ (pKa conjugate acid ≈ 38) and NaH (pKa ≈ 35). Both have conjugate acids with pKa values well above the alkyne's pKa of 25. KOtBu (pKa ≈ 16) and NaOH (pKa ≈ 14) are too weak.
Q: An alkyl halide is treated with NaSCH₃. What functional group is formed, and by what mechanism?
A: A thioether (R-SCH₃) is formed via SN2. The thiolate (⁻SCH₃) is an excellent nucleophile due to the large, polarisable sulphur atom.
Q: Why can you not form a C-C bond by treating a tertiary alkyl halide with sodium acetylide?
A: The tertiary carbon is too sterically hindered for backside attack (SN2). The acetylide, being a strong base, reacts exclusively via E2 elimination, producing an alkene instead.
This material builds directly on the SN1/SN2/E1/E2 decision framework from earlier in the course. If you are comfortable predicting which mechanism dominates based on substrate class, nucleophile/base strength, and solvent, the synthesis problems here become a matter of applying those rules in sequence.
Alkynide chemistry connects forward to later topics on alkyne reactions (partial reduction with Lindlar catalyst or dissolving metal to give cis or trans alkenes), ozonolysis, and ultimately to carbonyl chemistry where C-C bond formation becomes even more central.
The retrosynthetic analysis approach introduced here (working backwards from the target) is the same strategy used in more advanced organic chemistry courses and in real pharmaceutical synthesis.
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