Organic Chemistry 1 · Synthesis Foundation

Changing the Carbon Skeleton

7 min read

On this page 7 sections

In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Study tools

In 30 seconds

To make a molecule bigger you must form a carbon–carbon bond, and the workhorse of the introductory toolkit is the : a is deprotonated to give a strong carbon nucleophile that attacks a methyl or primary in an SN2 reaction, forming a longer alkyne that can later be reduced. Retrosynthetically, disconnect the target at the new C–C bond so one fragment is a terminal alkyne (nucleophile) and the other a primary alkyl halide (electrophile).

Why this matters

Controlled builds drug-like backbones one bond at a time, appending alkyl groups to a scaffold without touching other functional groups; choosing an unhindered electrophile so a strong nucleophile adds rather than eliminates is a recurring medicinal-chemistry decision. This section is conceptual planning; acetylide and alkyne work must follow approved institutional safety and waste-handling documentation.

The college version

1. Carbon–Carbon Bond Formation

Nearly all organic synthesis assembles a carbon skeleton, and only a few introductory reactions reliably make C–C bonds: acetylide alkylation, Grignard/organolithium addition to carbonyls, and (later) aldol reactions. Recognizing which a target calls for is the first planning decision. The acetylide route is special because it adds carbons to an alkyne, which can then be functionalized or reduced.

2. Acetylide Ions as Carbon Nucleophiles

Terminal alkynes have an acidic sp C–H (pKa ≈ 25), so strong bases (NaNH₂; BuLi conceptually) deprotonate them to acetylide ions, \( \mathrm{RC\equiv C^-} \). The negative charge sits in an sp orbital on carbon, making the acetylide a potent carbon nucleophile and a strong base. By SN2 it attacks the electrophilic carbon of an alkyl halide, displacing the halide and forging a new C–C bond.

3. Methyl/Primary Preference and Steric Limitations

SN2 needs backside approach, so the attacked carbon must be open. Only methyl and primary alkyl halides react cleanly; secondary halides are slow and give substantial elimination (the acetylide pulls a β-hydrogen instead), and tertiary halides give almost pure elimination. Vinylic and aryl halides fail because their C–X bond is strong and the backside is blocked. Practical rule: chain extension works best with R–CH₂–X or CH₃–X.

How it works

Locate the C–C bond(s) that must be created to grow the skeleton.
Disconnect that bond; label one fragment the nucleophile precursor (terminal alkyne) and the other the electrophile (alkyl halide).
Confirm the electrophile is methyl or primary; otherwise redesign.
Deprotonate the terminal alkyne with a strong base to form the acetylide.
Add the primary halide; the acetylide attacks by SN2, forming the new C–C bond.
Functionalize the alkyne as needed (hydrogenate to alkane, partial reduction, hydration).
Forward-check for stereochemistry, regiochemistry, and group tolerance.

Common confusions

Do not confuse	With	Difference
Acetylide nucleophilicity	Acetylide basicity	The same lone pair can attack carbon (SN2) or pull a proton (E2); sterics decide.
Methyl/primary preference	A rule about rate only	It is not just speed—crowded halides take a different path (elimination).
Terminal alkyne	Internal alkyne	Only terminal alkynes have an acidic C–H to deprotonate.
SN2 alkylation	SN2 hydrolysis	The nucleophile here is carbon, so the product is a new C–C bond, not an alcohol.
Disconnection	Actual cleavage	The disconnection is imaginary; the real step is the forward SN2.
Chain extension	Functional-group change	Extension adds carbons; FGI changes groups without changing carbon count.

Memory aids

"Hook, Socket, No Crowd." Make a carbon hook (deprotonate a terminal alkyne to an acetylide), plug it into an uncrowded socket (a methyl or primary alkyl halide), and keep the socket clear (no secondary/tertiary, no vinylic/aryl) so SN2 forms the bond instead of elimination.

Quick review

Topic Recap

Growing a carbon skeleton means forming a C–C bond, and the acetylide SN2 is the reliable introductory method: deprotonate a terminal alkyne to an acetylide ion, then alkylate it with a methyl or primary alkyl halide. Steric and basicity limits restrict the electrophile to unhindered substrates, and the at the new bond reveals the two fragments. The resulting alkyne is a versatile intermediate for further transformation.

Knowledge Check

Why must a terminal (not internal) alkyne be used to make an acetylide?
Which class of alkyl halide reacts cleanly with an acetylide, and why?
What competing reaction occurs if a tertiary alkyl halide is used?
In retrosynthesis of \( \mathrm{CH_3C\equiv CCH_2CH_3} \), which bond is disconnected, and what are the two fragments?
After forming 2-pentyne by alkylation, how would you convert it to pentane?

Answers and Rationales

Only a terminal alkyne has the acidic sp C–H proton; an internal alkyne has no such hydrogen, so no acetylide forms.
Methyl and primary halides: backside SN2 attack is unhindered, so the acetylide adds rather than acting as a base.
Elimination (E2): the acetylide acts as a base, removing a β-hydrogen to give an alkene, because substitution is sterically blocked.
The bond between the internal alkyne carbon and the CH₂ of the ethyl group; fragments are propyne (nucleophile precursor) and an ethyl halide such as ethyl bromide (electrophile).
Catalytic hydrogenation (H₂ with a metal catalyst) reduces the alkyne fully to pentane.

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of growing a chain like adding a car to a train: you do not weld two finished halves at random; you need a coupling that works only when the two ends fit. A carbon chain is extended the same way—one fragment carries a reactive "hook" (a lone pair on carbon, the acetylide) and the other a "socket" (a carbon bonded to a leaving group such as bromine). The hook snaps in, one bond forms, and the chain is a carbon longer. The everyday image is plugging an extension cord into an outlet: prongs must line up with slots, and a bulky plug will not fit a crowded socket.

Where it stops being exact: chemical "plugs" are not reversible like cords. If the socket carbon is crowded (secondary/tertiary) or the leaving group sits on a ring or double bond, the reaction stalls from steric hindrance or takes a different path (elimination). So the plug-and-socket picture holds only for unhindered, methyl/primary electrophiles.

Simple Example

Extend acetylene to 1-pentyne. Deprotonate \( \mathrm{HC\equiv CH} \) with NaNH₂ to the acetylide \( \mathrm{HC\equiv C^-} \), then add 1-bromopropane (\( \mathrm{CH_3CH_2CH_2Br} \)), a primary halide. The acetylide's lone pair attacks the carbon bearing Br, Br⁻ leaves, and the new bond gives \( \mathrm{HC\equiv CCH_2CH_2CH_3} \)—three carbons longer.

Worked example

Synthesize 2-pentyne, \( \mathrm{CH_3C\equiv CCH_2CH_3} \), from propyne and a suitable alkyl halide.

Disconnect. Cut the C–C bond joining the internal alkyne carbon to the CH₂ of the ethyl group; backward gives propyne and an ethyl electrophile.
Select the electrophile. The ethyl fragment must be a primary alkyl halide, ethyl bromide (\( \mathrm{CH_3CH_2Br} \)), so SN2 proceeds cleanly.
State the nucleophile. Propyne is deprotonated by NaNH₂ to \( \mathrm{CH_3C\equiv C^- Na^+} \).
Trace electrons before products. The acetylide lone pair attacks the CH₂ carbon from the back; the C–Br σ bond breaks as bromide departs with both electrons—one concerted SN2 step, no carbocation.
Check accounting. Two reactants give one product plus Na⁺ and Br⁻ (balanced); backside attack inverts the (achiral, primary) carbon, unobservable here. Product: 2-pentyne.
Forward-check. Propyne's acidic proton is consumed first by NaNH₂, and the primary halide avoids elimination—a textbook chain extension.

Key takeaways

High yield: C–C bond formation is the core of skeleton-building; acetylide SN2 is the simplest reliable tool.
High yield: Only methyl and primary halides alkylate acetylides cleanly—secondary/tertiary eliminate instead.
High yield: Terminal alkynes are acidic (pKa ≈ 25); NaNH₂ is needed to form the acetylide.
High yield: The acetylide is both strong nucleophile and strong base—steric and elimination issues follow from that dual role.
High yield: Disconnect the target at the new C–C bond so one side is an alkyne, the other a primary halide.
The bond forms by backside SN2; no carbocation, so no rearrangements.
The alkyne product can be reduced to an alkane or a cis/trans alkene, extending the route's reach.
Vinylic and aryl halides do not undergo this SN2—strong C–X bond and blocked backside.

Keep learning

Ready to build on this? Continue to the next lesson.

Practice Organic Chemistry 1

This lesson has no separate scored set. Practice draws from the subject’s question bank.

Study toolsYou’ll learn to · Key vocabulary

You’ll learn to

Explain why carbon–carbon bond formation is the central operation in building molecular skeletons.
Use acetylide ions from terminal alkynes as nucleophiles to extend a chain by SN2 alkylation with alkyl halides.
Justify the methyl/primary alkyl halide preference and predict failure of secondary/tertiary substrates on steric and elimination grounds.
Plan a chain extension by retrosynthetic disconnection at the new bond, selecting the correct electrophile and nucleophile.

Key vocabulary

Carbon–carbon bond formation: Making a new bond between two carbon atoms.
Acetylide ion: The conjugate base of a terminal alkyne, \( \mathrm{RC\equiv C^-} \).
Terminal alkyne: An alkyne with the triple bond at the chain end.
Alkyl halide: An alkane with a halogen substituent; here the electrophile.
SN2 bond formation: A concerted backside displacement at an sp³ carbon.
Methyl/primary preference: Only methyl and primary halides react cleanly.
Steric limitations: Crowded electrophilic carbons block backside attack.
Chain extension: Adding carbons to lengthen the skeleton.
Retrosynthetic disconnection: Cutting the new bond backward to reveal the fragments.
Electrophile/nucleophile selection: Picking the halide and the acetylide that match.

Educational content only. It is not medical, legal or professional advice. Found an error? Tell us.