Organic Chemistry · Alkynes: An Introduction to Organic Synthesis

Alkylation of Acetylide Anions

7 min read

On this page 9 sections

In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

of acetylide anions is the payoff reaction of alkyne chemistry: it builds a new carbon–carbon bond by letting the acetylide ion, R-C ≡ C^-, attack an alkyl halide. The general reaction is an S_N2 substitution:

$R - C \equiv C - Na + + R' - X \to R - C \equiv C - R' + NaX$

The acetylide acts as a nucleophile, attacking the carbon that carries the halogen and displacing the halide ion. The product is an internal alkyne with a longer carbon chain — the carbon skeleton has been extended by the alkyl group R'. This reaction works cleanly only with methyl and primary alkyl halides; secondary and tertiary halides mostly give elimination instead. Because the acetylide is both a strong base and a good nucleophile, that substitution–elimination balance is the central practical question.

Why this matters

Carbon–carbon bond formation is the heart of organic synthesis — it is how chemists assemble small building blocks into large, complex molecules. Acetylide alkylation is one of the first truly general C–C bond-forming reactions students learn, and it demonstrates the whole toolkit of synthesis logic: deprotonate to make a nucleophile, react it with an electrophile, and extend the chain. The reaction is used to make longer-chain alkynes that are precursors to natural products, pharmaceuticals, and materials. It also introduces two ideas that recur throughout organic chemistry: the S_N2 mechanism (backside attack, inversion, steric sensitivity) and the competition between substitution and elimination. Mastering it builds a template for nucleophilic-carbon thinking that carries into Grignard, enolate, and organolithium chemistry.

The college version

Core Concepts

The S_N2 mechanism

Alkylation of an acetylide proceeds by the bimolecular substitution (S_N2) mechanism. The acetylide carbon's lone pair attacks the electrophilic carbon of the alkyl halide from the back side, opposite the , in one concerted step:

$R - C \equiv C - + R' - X \to [R - C \equiv C \dots R' \dots X] ‡ \to R - C \equiv C - R' + X -$

Because the nucleophile must reach the back side of the carbon bearing the halogen, the reaction is very sensitive to steric hindrance. Methyl halides (CH₃X) and primary halides (R'CH₂X) react readily; secondary halides are slow and give mostly elimination; tertiary halides give essentially only elimination. The leaving group is typically bromide or iodide (better leaving groups), and the reaction is usually run in liquid ammonia or an ether solvent that keeps the sodium acetylide soluble.

Substitution versus elimination

The acetylide ion is a strong base as well as a good nucleophile, so it can also pull a proton from the alkyl halide's β-carbon, triggering :

$R - C \equiv C - + CH 3 CH 2 - X E2 ⟶ R - C \equiv CH + CH 2 = CH 2 + X -$

The competition between S_N2 and E2 depends mostly on the structure of the alkyl halide: methyl and primary halides favor substitution; secondary halides give mixtures; tertiary halides give pure elimination. This is why acetylide alkylation is effectively limited to methyl and primary alkyl halides. It is a constraint worth memorizing: "acetylide + primary halide = internal alkyne; acetylide + tertiary halide = alkene plus recovered terminal alkyne."

Synthesis strategy: building bigger alkynes

Alkylation lets you construct any internal alkyne from acetylene or a simple terminal alkyne in a stepwise fashion:

Deprotonate a terminal alkyne with NaNH₂ to make the acetylide (previous topic).
Add a or methyl halide to alkylate it.

Each alkylation adds one alkyl group. Starting from acetylene, HC ≡ CH, two successive alkylations with different halides give an unsymmetrical internal alkyne — for example, acetylene + ethyl bromide gives 1-butyne, and deprotonation followed by methyl iodide gives 2-pentyne. This "build in pieces" logic is the essence of retrosynthetic thinking: to make a target alkyne, ask which two fragments (an acetylide and an alkyl halide) could be joined.

What cannot be alkylated

Aryl halides and vinyl halides do not undergo S_N2 reactions — the back-side attack needed is blocked by the π system, and the C–X bond has partial double-bond character. So acetylides cannot be alkylated with bromobenzene or bromoethene-type halides. If the target has an aromatic ring, the ring must be introduced by a different route (e.g., coupling reactions covered in later chapters). Recognizing which halides are "alkylatable" is a common exam discriminator.

Common Confusions

Do Not Confuse	With	The Difference
Alkylation (S_N2)	Elimination (E2)	S_N2 attacks the halogen-bearing carbon (works for 1° halides); E2 removes a β-H (dominates for 3° halides)
Acetylide as nucleophile	Acetylide as base	Same ion, two behaviors; the alkyl halide's structure decides which one wins
Primary halide reactivity	Tertiary halide reactivity	S_N2 loves 1° (open back side); 3° halides only eliminate
Internal alkyne product	Terminal alkyne starting material	After alkylation the acidic H is gone; the product is an internal alkyne that can no longer be deprotonated the same way
Alkyl halides	Aryl/vinyl halides	Alkyl halides undergo S_N2; aryl/vinyl halides cannot (blocked back side, partial double-bond character)

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine the acetylide ion is a person holding a strong magnet (the negative charge). An alkyl halide is a toy with a handle (the halogen). The magnet-person grabs the toy by the handle from behind, and the handle pops off — now the toy is stuck to the magnet. The molecule just got longer by the size of the toy. But if the toy is too big and clunky (a tertiary halide), the magnet-person can't get behind it to grab the handle, so instead they just knock the toy over (elimination) and nothing gets attached.

Worked example

Worked example 1 (synthesis plan). How would you make 2-pentyne, CH₃C ≡ CCH₂CH₃, from acetylene and simple halides? Step 1: deprotonate acetylene with NaNH₂ and alkylate with ethyl bromide to give 1-butyne:

$HC \equiv CH 1) NaNH 2 2) CH 3 CH 2 Br ⟶ CH 3 CH 2 C \equiv CH$

Step 2: deprotonate the 1-butyne with NaNH₂ and alkylate with methyl iodide:

$CH 3 CH 2 C \equiv CH 1) NaNH 2 2) CH 3 I ⟶ CH 3 CH 2 C \equiv CCH 3$

The two alkylations add the ethyl group first and the methyl group second, giving the unsymmetrical internal alkyne 2-pentyne.

Worked example 2 (predicting failure). Would acetylide alkylation with 2-bromo-2-methylpropane (tertiary butyl bromide) work? No. The tertiary halide cannot undergo S_N2 (the back side is blocked), and the acetylide's basicity dominates: it abstracts a β-proton and forms 2-methylpropene (isobutylene) by E2. The acetylide is simply converted back to the terminal alkyne.

Worked example 3 (stoichiometry with dimensional analysis). How many grams of 2-heptyne, CH₃C ≡ C(CH₂)₃CH₃, can be made in theory from 10.0 g of 1-hexyne, CH₃(CH₂)₃C ≡ CH, by deprotonation and alkylation with methyl iodide? The reaction is 1:1 in moles:

$n(1 - hexyne) = 10.0 g 82.15 g/mol = 0.122 mol$

One mole of 2-heptyne forms per mole of 1-hexyne:

$m(2 - heptyne) = 0.122 mol \times 96.17 g/mol = 11.7 g$

The theoretical yield is 11.7 g of 2-heptyne (the chain grew by one carbon — from C₆ to C₇ — and the molar mass increased accordingly).

Key takeaways

Reaction: R-C ≡ C^- + R'-X → R-C ≡ C-R' + X^- — an S_N2 that makes a C–C bond.
Scope: works with methyl and primary alkyl halides (bromides/iodides best); fails with secondary/tertiary (elimination) and aryl/vinyl halides (no S_N2).
Mechanism vocabulary: backside attack, concerted, inversion of configuration at the reacting carbon.
Competition: acetylide is a strong base, so E2 elimination competes; more hindered halides favor elimination.
Synthesis pattern: deprotonate (NaNH₂) → alkylate; repeat to add more carbons; acetylene can be alkylated twice to build unsymmetrical internal alkynes.
Product class: internal alkynes — valuable building blocks for later hydrogenation, hydration, and cleavage chemistry.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

What general reaction type is the alkylation of an with an alkyl halide?
Show answer
An S_N2 (bimolecular nucleophilic substitution): the acetylide attacks the carbon bearing the halogen from the back side, displacing the halide in one step.
Why do tertiary alkyl halides fail to give alkylation products with acetylides?
Show answer
The back side of a tertiary carbon is completely blocked by the three alkyl groups, so the S_N2 attack cannot happen; the acetylide instead acts as a base and removes a β-proton (E2), giving an alkene.
What two steps convert 1-butyne into 2-hexyne? (Name the reagents.)
Show answer
(1) Deprotonate with NaNH₂ to form the acetylide; (2) alkylate with ethyl bromide (CH₃CH₂Br) to add the two-carbon group, giving 2-hexyne.
What are the two products when an acetylide reacts with a secondary alkyl halide under forcing conditions?
Show answer
A mixture of the elimination product (an alkene) and the regenerated terminal alkyne — substitution is slow for secondary halides and elimination competes heavily.
Why can acetylene be alkylated twice, while an internal alkyne cannot be alkylated at all?
Show answer
Acetylene has two terminal hydrogens, so after one alkylation it becomes a terminal alkyne that can be deprotonated and alkylated again. An internal alkyne has no acidic hydrogens left, so there is nothing to deprotonate.
What is the molar ratio of alkyne to alkyl halide in the alkylation step?
Show answer
1:1 — one mole of acetylide reacts with one mole of alkyl halide; the sodium halide (NaX) is the by-product.

Keep learning

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

Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Alkylation: Adding an alkyl group to a molecule
Acetylide anion: R-C ≡ C^-, a carbon nucleophile from a terminal alkyne
S_N2: Bimolecular substitution: one-step backside attack with inversion
E2 elimination: Base pulls a β-proton while the halide leaves
Primary alkyl halide: RCH₂X — halogen on a carbon bonded to one carbon
Leaving group: The group (usually X^-) displaced in substitution
Retrosynthesis: Planning a molecule backward into simpler pieces

Sources & references

openstax.org — Organic Chemistry

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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